System Dynamics 2nd Edition Solution manual
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Transcript of System Dynamics 2nd Edition Solution manual
Solutions Manual c©
to accompany
System Dynamics, Second Edition
by
William J. Palm III
University of Rhode Island
Solutions to Problems in Chapter One
c©Solutions Manual Copyright 2010 The McGraw-Hill Companies. All rightsreserved. No part of this manual may be displayed, reproduced, or distributedin any form or by any means without the written permission of the publisheror used beyond the limited distribution to teachers or educators permitted byMcGraw-Hill for their individual course preparation. Any other reproductionor translation of this work is unlawful.
1.1 W = mg = 3(32.2) = 96.6 lb.
1.2 m = W/g = 100/9.81 = 10.19 kg. W = 100(0.2248) = 22.48 lb. m = 10.19(0.06852) =0.698 slug.
1.3 d = (50 + 5/12)(0.3048) = 15.37 m.
1.4 n = 1/[60(1.341× 10−3)] = 12.43, or approximately 12 bulbs.
1.5 5(70− 32)/9 = 21.1 C
1.6 ω = 3000(2π)/60 = 314.16 rad/sec. Period P = 2π/ω = 60/3000− 1/50 sec.
1.7 ω = 5 rad/sec. Period P = 2π/ω = 2π/5 = 1.257 sec. Frequency f = 1/P = 5/2π =0.796 Hz.
c©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
1.8 Physical considerations require the model to pass through the origin, so we seek a modelof the form f = kx. A plot of the data shows that a good line drawn by eye is given byf = 0.2x. So we estimate k to be 0.2 lb/in.
Skipping ahead to Section 1.5, we can solve this problem using the least squares method,based on equation (1.5.3). The script file is
x = [4.7,7.2,10.6,12.9]-4.7;f = [0,0.47,1.15,1.64];num = sum(x.*f);den = sum(x.^2);k = num/den
The result is k = 0.1977.
c©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
1.9 The script file is
x = [0:0.01:1];subplot(2,2,1)plot(x,sin(x),x,x),xlabel(′x (radians)′),ylabel(′x and sin(x)′),...gtext(′x′),gtext(′sin(x)′)subplot(2,2,2)plot(x,sin(x)-x),xlabel(′x (radians)′),ylabel(′Error: sin(x) - x′)subplot(2,2,3)plot(x,100*(sin(x)-x)./sin(x)),xlabel(′x (radians)′),...ylabel(′Percent Error′),grid
The plots are shown in the figure.
0 0.5 10
0.2
0.4
0.6
0.8
1
x (radians)
x an
d si
n(x)
x
sin(x)
0 0.5 1−0.2
−0.15
−0.1
−0.05
0
x (radians)
Err
or: s
in(x
) − x
0 0.5 1−20
−15
−10
−5
0
x (radians)
Per
cent
Err
or
Figure : for Problem 1.9.
From the third plot we can see that the approximation sin x ≈ x is accurate to within5% if |x| ≤ 0.5 radians.
c©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
1.10 For θ near π/4,
f(θ) ≈ sinπ
4+
(cos
π
4
) (θ − π
4
)
For θ near 3π/4,
f(θ) ≈ sin3π
4+
(cos
3π
4
) (θ − 3π
4
)
c©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
1.11 For θ near π/3,
f(θ) ≈ cosπ
3−
(sin
π
3
) (θ − π
3
)
For θ near 2π/3,
f(θ) ≈ cos2π
3−
(sin
2π
3
) (θ − 2π
3
)
c©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
1.12 For h near 25,
f(h) ≈√
25 +1
2√
25(h − 25) = 5 +
110
(h − 25)
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
1.13 For r near 5,f(r) ≈ 52 + 2(5)(r− 5) = 25 + 10(r − 5)
For r near 10,f(r) ≈ 102 + 2(10)(r− 10) = 100 + 20(r − 10)
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
1.14 For h near 16,
f(h) ≈√
16 +1
2√
16(h − 16) = 4 +
18(h − 16)
f(h) ≥ 0 if h > −16.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
1.15 Construct a straight line the passes through the two endpoints at p = 0 and p = 900.At p = 0, f(0) = 0. At p = 900, f(900) = 0.002
√900 = 0.06. This straight line is
f(p) =0.06900
p =1
15, 000p
c©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
1.16 (a) The data is described approximately by the linear function y = 54x − 1360. Theprecise values given by the least squares method are y = 53.5x−1354.5 (see Problem 1.34a).
(b) Only the loglog plot of the data gives something close to a straight line, so the datais best described by a power function y = bxm where the approximate values are m = −0.98and b = 3600. The precise values given by the least squares method are y = 3582.1x−0.9764
(see Problem 1.34b).(c) Both the loglog and semilog plot (with the y axis logarithmic) give something close
to a straight line, but the semilog plot gives the straightest line, so the data is best describedby a exponential function y = b(10)mx where the approximate values are m = −0.007 andb = 2.1 × 105. The precise values given by the least squares method are y = 2.0622 ×105(10)−0.0067x (see Problem 1.34c).
c©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
1.17 With this problem, it is best to scale the data by letting x = year − 1990, to avoidraising large numbers like 1990 to a power. Both the loglog and semilog plot (with they axis logarithmic) give something close to a straight line, but the semilog plot gives thestraightest line, so the data is best described by a exponential function y = b(10)mx. Theapproximate values are m = 0.035 and b = 9.98.
Set y = 20 to determine how long it will take for the population to increase from 10 to20 million. This gives 20 = 9.98(10)0.03x. Solve it for x: x = (log(20) − log(9.98))/0.035.The answer is 8.63 years, which corresponds to 8.63 years after 1990.
More precise values are given by the least squares method (see Problem 1.35).
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
1.18 (a) If C(t)/C(0) = 0.5 when t = 500 years, then 0.5 = e−5500b, which gives b =− ln(0.5)/5500 = 1.2603× 10−4.
(b) Solve for t to obtain t = − ln[C(t)/C(0)]/b using C(t)/C(0) = 0.9 and b = 1.2603×10−4. The answer is 836 years. Thus the organism died 836 years ago.
(c) Using b = 1.1(1.2603 × 10−4) in t = − ln(0.9)/b gives 760 years. Using b =0.9(1.2603× 10−4) in t = − ln(0.9)/b gives 928 years.
c©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
1.19 Only the semilog plot of the data gives something close to a straight line, so thedata is best described by an exponential function y = b(10)mx where y is the temperaturein degrees C and x is the time in seconds. The approximate values are m = −3.67 andb = 356. The alternate exponential form is y = be(m ln 10)x = 356e−8.451x. The timeconstant is 1/8.451 = 0.1183 s.
The precise values given by the least squares method are y = 356.0199(10)−3.6709x (seeProblem 1.37).
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
1.20 Only the semilog plot of the data gives something close to a straight line, so the data isbest described by an exponential function y = b(10)mx where y is the bearing life thousandsof hours and x is the temperature in degrees F. The approximate values are m = −0.007and b = 142. The bearing life at 150 F is estimated to be y = 142(10)−0.007(150) = 12.66,or 12,600 hours. The alternate exponential form is y = be(m ln 10)x = 142e−0.0161x. Thetime constant is 1/0.0161 = 62.1 or 6.21× 104 hr.
The precise values given by the least squares method are y = 141.8603(10)−0.0070x (seeProblem 1.38).
c©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
1.21 Only the semilog plot of the data gives something close to a straight line, so thedata is best described by an exponential function y = b(10)mx where y is the voltage andx is the time in seconds. The first data point does not lie close to the straight line onthe semilog plot, but a measurement error of ±1 volt would account for the discrepancy.The approximate values are m = −0.43 and b = 96. The alternate exponential form isy = be(m ln 10)x = 96e−0.99x. The time constant is 1/0.99 = 1.01 s.
The precise values given by the least squares method are y = 95.8063(10)−0.4333x (seeProblem 1.39).
c©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
1.22 A semilog plot generated by the following script file shows that the exponential functionT − 70 = bemt fits the data well.
t = [0:300:3000];temp = [207,182,167,155,143,135,128,123,118,114,109];DT = temp-70;semilogy(t,DT,t,DT,’o’)
Fitting a line by eye gives the approximate values m = −4 × 10−4 and b = 125. Thecorresponding function is T (t) = 70 + 125e−4×10−4t.
The precise values given by the least squares method are m = −4.0317 × 10−4 andb = 125.1276 (see Problem 1.40).
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
1.23 Plots of the data on a log-log plot and rectilinear scales both give something close toa straight line, so we try both functions. (Note that the flow should be 0 when the heightis 0, so we do not consider the exponential function and we must force the linear functionto pass through the origin by setting b = 0.) The three lowest heights give the same time,so we discard the heights of 1 and 2 cm.
The power function fitted by eye in terms of the height h is approximately f = 4h0.9.Note that the exponent is not close to 0.5, as it is for orifice flow. This is because theflow through the outlet is pipe flow. For the linear function f = mh, the best fit by eye isapproximately f = 3.2h.
Using the least squares method gives more precise results: f = 4.1595h0.8745 and f =3.2028h (see Problem 1.41)
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
1.24 Plots of the data on a log-log plot and rectilinear scales both give something close toa straight line, so we try both functions. (Note that the flow should be 0 when the heightis 0, so we do not consider the exponential function and we must force the linear functionto pass through the origin by setting b = 0.) The variable x is the height and the variabley is the flow rate. The three lowest heights give the same time, so we discard the heightsof 1 and 2 cm.
The power function fitted by eye in terms of the height h is approximately f = 4h0.9.Note that the exponent is not close to 0.5, as it is for orifice flow. This is because theflow through the outlet is pipe flow. For the linear function f = mh, the best fit by eye isapproximately f = 3.7h.
Using the least squares method gives more precise results: f = 4.1796h0.9381 and f =3.6735h (see Problem 1.42).
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
1.25 Fitting a straight line by eye gives the approximate values m = 15 and b = 7. Theprecise values given by the least squares method are m = 15.0750, b = 7.1500, J = 43.5750,S = 4.5451× 103, r2 = 0.9904 (see Problem 1.43).
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
1.26 Plot the data on a loglog plot. We must delete the first data point to avoid takingthe logarithm of 0. The power function fitted by eye is approximately y = 7x3. The precisevalues given by the least squares method are m = 2.9448, b = 7.4053, J = 2.7494 × 103,S = 1.1218× 105, and r2 = 0.9755 (see Problem 1.44).
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
1.27 Plot the data on a semilog plot. The exponential function fitted by eye is approximatelyy = 6e3x. The precise values given by the least squares method are y = 6.4224e2.8984x,J = 77.4488, S = 2.5496× 104, and r2 = 0.9970 (see Problem 1.45).
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
1.28 Fitting a straight line by eye through the origin gives the approximate values m = 17and b = 0. The precise values given by the least squares method are m = 16.6071, J =116.6071, S = 5.5420× 103, and r2 = 0.9790 (see Problem 1.46).
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
1.29 Plot the data on a loglog plot. We must delete the first data point to avoid takingthe logarithm of 0. The power function fitted by eye is approximately y = 7x3. The precisevalues given by the least squares method are b = 7.4793, J = 799.4139, S = 1.6176× 105,and r2 = 0.9951 (see Problem 1.47).
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
1.30 a) The equation for b is obtained as follows.
J =n∑
i=1
(bemxi − yi)2
∂J
∂b= 2b
n∑
i=1
e2mxi − 2n∑
i=1
yiemxi = 0
b =∑
yemx
∑e2mx
b) Plot the data on a semilog plot. The exponential function fitted by eye is approxi-mately y = 6e3x. The precise values given by the least squares method are y = 5.8449e3x
(see Problem 1.48).
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
1.31 The integral form of the sum of squares to fit the function y = 5x2 is
J =∫ 4
0
(5x2 − mx − b
)2dx =
∫ 4
0
[25x4 − 10mx3 + (m2 − 10b)x2 + 2mbx + b2
]dx
This evaluates to
J = 5120− 640m + (m2 − 10b)643
+ 16mb + 4b2
Thus∂J
∂m= −640 +
1283
m + 16b = 0
∂J
∂b= −640
3+ 16m + 8b = 0
These give m = 20 and b = −13.333. Thus the line is y = 20x− 13.333.
c©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
1.32 a) The integral form of the sum of squares to fit the function y = Ax2 + Bx is
J =∫ L
0
(Ax2 + Bx − mx − b
)2dx
or
J =∫ L
0
[A2x4 + (2AB − 2mA)x3 + (−2bA + B2 + m2 − 2mB)x2 + (2mb− 2bB)x + b2
]dx
This evaluates to
J = A2 L5
5+ 2A(B − m)
L4
4+ (−2bA + B2 + m2 − 2mB)
L3
3+ 2b(m− B)
L2
2+ b2L
Thus∂J
∂m= −AL4
2+
23mL3 − 2
3BL3 + bL2 = 0
∂J
∂b= −2
3AL3 + (m − B)L2 + 2bL = 0
These give4Lm + 6b = 4BL + 3AL2
3Lm + 6b = 2AL2 + 3BL
which can be solved for m and b, given values of A, B, and L.b) With L = 2, A = 3, and B = 5, we obtain m = 11 and b = −2. The fitted straight
line is y = 11x − 2.
c©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
1.33 a) The integral form of the sum of squares to fit the function y = BeMx is
J =∫ L
0
(BeMx − mx − b
)2dx
or
J =∫ L
0
[B2e2Mx − 2mBxeMx − 2bBeMx + m2x2 + 2bmx + b2
]dx =
This evaluates to
J =B2
2M
(e2ML − 1
)− 2mBL
MeML +
2mB
M2
(eML − 1
)
+2bB
M
(1 − eML
)− m2 L3
3+ bmL2 + b2L
Thus∂J
∂m= −2BL
MeML +
2B
M2
(eML − 1
)+
2mL3
3+ bL2 = 0
∂J
∂b= −2B
MeML +
2B
M+ mL2 + 2bL = 0
These give2L3
3m + L2b =
2BL
MeML − 2B
M2
(eML − 1
)
L2m + 2Lb =2B
MeML − 2B
M
which can be solved for m and b, given values of M , B, and L.b) With L = 1, M = −5, and B = 15, we obtain m = −10.973 and b = 8.466. The
fitted straight line is y = −10.973x + 8.466.
c©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
1.34 (a) The script file is
x = [25:5:45];y = [5, 260, 480, 745, 1100];p = polyfit(x,y,1)p =
1.0e+003 *0.0535 -1.3545
The function is y = 53.5x− 1354.5.
c©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
1.34 (b) Only the loglog plot of the data gives something close to a straight line, so thedata is best described by a power function y = bxm. The script file to find the coefficientsm and b is
x = [2.5:0.5:6,7:10];y = [1500,1220,1050,915,810,745,690,620,520,480,410,390];p = polyfit(log10(x),log10(y),1);m = p(1)b = 10^p(2)
The results are m = −0.9764 and b = 3582.1. Thus the power function is y = 3582.1x−0.9764.
c©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
1.34 (c) Both the loglog and semilog plot (with the y axis logarithmic) give somethingclose to a straight line, but the semilog plot gives the straightest line, so the data is bestdescribed by a exponential function y = b(10)mx. The script file to find the coefficients mand b is
x = [550:50:750];y =[41.2,18.62,8.62,3.92,1.86];p = polyfit(x,log10(y),1);m = p(1)b = 10^p(2)m =
-0.0067b =
2.0622e+005
This gives the results m = −0.0067 and b = 2.0622× 105. Thus the exponential function isy = 2.0622× 105(10)−0.0067x.
The results for the power function are obtained from
p = polyfit(log10(x),log10(y),1);m = p(1)b = 10^p(2)
This gives the results m = −9.9949 and b = 1.0601 × 1029. Thus the power function isy = 1.0601× 1029x−9.9949.
A plot of the data and the two functions shows that both functions describe the datawell, but the exponential curve passes closer to the first data point than the power curve.In addition, we should be careful about using this power function because its coefficient bhas a very large value. This large coefficient means that any predictions of y values madewith this function will be very prone to error unless we use very precise values of x. Thusthe exponential function is the best choice to describe this data.
c©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
1.35 With this problem, it is best to scale the data by letting x = year − 1990, to avoidraising large numbers like 1990 to a power. Both the loglog and semilog plot (with they axis logarithmic) give something close to a straight line, but the semilog plot gives thestraightest line, so the data is best described by a exponential function y = b(10)mx. Thescript file to find the coefficients m and b is
year = [1990:1995];x = x = year-1990;pop=[10,10.8,11.7,12.7,13.8,14.9];p=polyfit(x,log10(pop),1)p =
0.0349 0.9992m=p(1);b=10^p(2)
This gives the results m = 0.0349 and b = 9.9817. Thus the exponential function isy = 9.9817(10)0.0349x. Set y = 20 to determine how long it will take for the populationto increase from 10 to 20 million. This gives 20 = 9.9817(10)0.0349x. Solve it for x: x =(log(20)− log(9.9817))/0.0349 = 8.6483 years, which corresponds to 8.6483 years after 1990.
c©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
1.36 (a) If C(t)/C(0) = 0.5 when t = 500 years, then 0.5 = e−5500b, which gives b =− ln(0.5)/5500. In Matlab this calculation is
b = -log(0.5)/5500
The answer is b = 1.2603× 10−4.(b) Solve for t to obtain t = − ln[C(t)/C(0)]/b using C(t)/C(0) = 0.9 and b = 1.2603×
10−4. In MATLAB this calculation is
t = -log(0.9)/b
The answer is 836.0170 years. Thus the organism died 836 years ago.(c) Using b = 1.1(1.2603 × 10−4) in t = − ln(0.9)/b gives 760 years. Using b =
0.9(1.2603× 10−4) in t = − ln(0.9)/b gives 928 years.
c©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
1.37 Only the semilog plot of the data gives something close to a straight line, so thedata is best described by an exponential function y = b(10)mx. The script file to find thecoefficients m and b is
time = [0:0.1:0.6];temp = [300,150,75,35,12,5,2];p=polyfit(time,log10(temp),1)m=p(1)b=10^p(2)
This gives the results: m = −3.6709 and b = 356.0199. Thus the exponential function isy = 356.0199(10)−3.6709x, where y is the temperature in degrees C and x is the time inseconds. The alternate exponential form is y = be(m ln 10)x = 356.0199e−8.4526x. The timeconstant is 1/8.4526 = 0.1183 s.
c©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
1.38 Only the semilog plot of the data gives something close to a straight line, so thedata is best described by an exponential function y = b(10)mx. The script file to find thecoefficients m and b is
temp = [100:20:220];life = [28,21,15,11,8,6,4];p = polyfit(temp,log10(life),1);m = p(1)b = 10^p(2)
This gives the results: m = −0.0070 and b = 141.8603. Thus the exponential function is y =141.8603(10)−0.0070x, where y is the bearing life thousands of hours and x is the temperaturein degrees F. The bearing life at 150 F is estimated to be y = 141.8603(10)−0.0070(150) =12.6433, or 12,643 hours.
The alternate exponential form is y = be(m ln 10)x = 141.8603e−0.0161x. The time con-stant is 1/0.0161 = 62.0421 or 6.20421× 104 hr.
c©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
1.39 Only the semilog plot of the data gives something close to a straight line, so the datais best described by an exponential function y = b(10)mx. The first data point does not lieclose to the straight line on the semilog plot, but a measurement error of ±1 volt wouldaccount for the discrepancy. The script file to find the coefficients m and b is
time = [0:0.5:4];voltage = [100,62,38,21,13,7,4,2,3];p = polyfit(time,log10(voltage),1);m=p(1)b = 10^p(2)
This gives the results: m = −0.4333 and b = 95.8063. Thus the exponential functionis y = 95.8063(10)−0.4333x, where y is the voltage and x is the time in seconds. Thealternate exponential form is y = be(m ln 10)x = 95.8063e−0.9977x. The time constant is1/0.9977 = 1.002 s.
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1.40 The semilog plot generated by the following script file shows that the exponentialfunction T − 70 = bemt fits the data well.
t = [0:300:3000];temp = [207,182,167,155,143,135,128,123,118,114,109];DT = temp-70;semilogy(t,DT,t,DT,’o’)p = polyfit(t,log(DT),1)m = p(1)b = exp(p(2))
The results are m = −4.0317 × 10−4 and b = 125.1276. The time constant is 1/4.0317 ×10−4 = 2.4803× 103 s, or 0.689 hr.
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1.41 The calculations are shown in the following script file. Plots of the data on a log-logplot and rectilinear scales both give something close to a straight line, so we try all bothfunctions. (Note that the flow should be 0 when the height is 0, so we do not consider theexponential function and we must force the linear function to pass through the origin bysetting b = 0.) The variable x is the height and the variable y is the flow rate. The threelowest heights give the same time, so we discard the heights of 1 and 2 cm.
For the power function, we use (1.5.1) and (1.5.2) in terms of the variables x andY = log y.
t = [7,8,9,10,11,13,15,17,23]; x = [11:-1:3];y = 250./t; X = log10(x); Y = log10(y);a1 = sum(X.^2); a2 = sum(X);a3 = sum(Y.*X); a4 = sum(Y); n = length(x);A = [a1, a2; a2, n]; C = [a3; a4];solution = A\C;m = solution(1)b = 10^solution(2)J = sum((b*x.^m-y).^2)S = sum((b*x.^m-mean(y)).^2)r2 = 1 - J/S
The results are m = 0.8745, b = 4.1595, J = 5.3644, S = 493.5634, and r2 = 0.9891. So thefitted function in terms of the height h is f = 4.1595h0.8745. Note that the exponent is notclose to 0.5, as it is for orifice flow. This is because the flow through the outlet is pipe flow.
For the linear function f = mx, we use (1.5.3).
m = sum(x.*y)/sum(x.^2)J = sum((m*x-y).^2)S = sum((m*x-mean(y)).^2)r2 = 1 - J/S
The results are m = 3.2028, J = 8.0247, S = 615.9936, and r2 = 0.9870. The fitted functionin terms of the height h is f = 3.2028h. Using only r2 as the criterion, it is impossible todecide whether the linear or the power function is the best model.
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1.42 The calculations are shown in the following script file. Plots of the data on a log-logplot and rectilinear scales both give something close to a straight line, so we try all bothfunctions. (Note that the flow should be 0 when the height is 0, so we do not consider theexponential function and we must force the linear function to pass through the origin bysetting b = 0.) The variable x is the height and the variable y is the flow rate. The threelowest heights give the same time, so we discard the heights of 1 and 2 cm.
For the power function, we use (1.5.1) and (1.5.2)in terms of the variables x and Y =log y.
t = [6, 7, 8, 9, 9, 11, 13, 17, 21]; x = [11:-1:3];y = 250./t; X = log10(x); Y = log10(y);a1 = sum(X.^2); a2 = sum(X);a3 = sum(Y.*X); a4 = sum(Y); n = length(x);A = [a1, a2; a2, n]; C = [a3; a4];solution = A\C;m = solution(1)b = 10^solution(2)J = sum((b*x.^m-y).^2)S = sum((b*x.^m-mean(y)).^2)r2 = 1 - J/S
The results are m = 0.9381, b = 4.1796, J = 13.5531, S = 729.3505, and r2 = 0.9814. Sothe fitted function in terms of the height h is f = 4.1796h0.9381.
For the linear function f = mx, we use (1.5.3).
t = [6, 7, 8, 9, 9, 11, 13, 17, 21]; x = [11:-1:3];y = 250./t;m = sum(x.*y)/sum(x.^2)J = sum((m*x-y).^2)S = sum((m*x-mean(y)).^2)r2 = 1 - J/S
The results are m = 3.6735, J = 14.7546, S = 809.8919, and r2 = 0.9818. The fittedfunction in terms of the height h is f = 3.6735h. Using only r2 as the criterion, it isimpossible to decide whether the linear or the power function is the best model.
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1.43 The script file is
x = [0:2:6];y = [4.5, 39, 72, 94];a1 = sum(x.^2);a2 = sum(x);a3 = sum(y.*x);a4 = sum(y);n = length(x);A = [a1, a2; a2, n];B = [a3; a4];solution = A\B;m = solution(1)b = solution(2)J = sum((m*x+b-y).^2)S = sum((m*x+b-mean(y)).^2)r2 = 1 - J/S
The results are m = 15.0750, b = 7.1500, J = 43.5750, S = 4.5451× 103, r2 = 0.9904.
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1.44 Following the procedure shown in Example 1.5.1, we fit X = log x and Y = log y toa linear function. We must delete the first data point to avoid taking the logarithm of 0.The script file is
x = [1:4];y = [8, 50, 178, 490];X = log10(x);Y = log10(y);a1 = sum(X.^2);a2 = sum(X);a3 = sum(Y.*X);a4 = sum(Y);n = length(x);A = [a1, a2; a2, n];C = [a3; a4];solution = A\C;m = solution(1)b = 10^solution(2)J = sum((b*x.^m-y).^2)S = sum((b*x.^m-mean(y)).^2)r2 = 1 - J/S
The results are m = 2.9448, b = 7.4053, J = 2.7494×103, S = 1.1218×105, and r2 = 0.9755.
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1.45 We fit x and Y = log y to the linear function Y = mx + B, where B = log b. Thescript file is
x = [0:0.4:1.2];y = [6.3, 22, 60, 215];Y = log10(y);a1 = sum(x.^2);a2 = sum(x);a3 = sum(Y.*x);a4 = sum(Y);n = length(x);A = [a1, a2; a2, n];C = [a3; a4];solution = A\C;m = solution(1)/log10(exp(1))b = 10^solution(2)J = sum((b*exp(m*x)-y).^2)S = sum((b*exp(m*x)-mean(y)).^2)r2 = 1 - J/S
The results are m = 2.8984, b = 6.4244, J = 77.4488, S = 2.5496× 104, and r2 = 0.9970.The fitted function is y = 6.4244e2.8984x.
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1.46 The equation for m, obtained from (1.5.3), is
m =∑
xiyi∑x2
i
The script file is
x = [0:2:6];y = [4.5, 39, 72, 94];m = sum(x.*y)/sum(x.^2)J = sum((m*x-y).^2)S = sum((m*x-mean(y)).^2)r2 = 1 - J/S
The results are m = 16.6071, J = 116.6071, S = 5.5420× 103, and r2 = 0.9790.
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1.47 The equation for b is obtained from equation (1) in Example 1.5.3 with m = 3.
b =∑
x3y∑x6
x = [0:4];y = [1, 8, 50, 178, 490];b = sum((x.^3).*y)/sum(x.^6)J = sum((b*x.^3-y).^2)S = sum((b*x.^3-mean(y)).^2)r2 = 1 - J/S
The results are b = 7.4793, J = 799.4139, S = 1.6176× 105, and r2 = 0.9951.
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1.48 a) The equation for b is obtained as follows.
J =n∑
i=1
(bemxi − yi)2
∂J
∂b= 2b
n∑
i=1
e2mxi − 2n∑
i=1
yiemxi = 0
b =∑
yemx
∑e2mx
b) The script file is
x = [0:0.4:1.2];y = [6.3, 22, 60, 215];b = sum(y.*exp(3*x))/sum(exp(6*x))J = sum((b*exp(3*x)-y).^2)S = sum((b*exp(3*x)-mean(y)).^2)r2 = 1 - J/S
The results are b = 5.8449, J = 27.7380, S = 2.7279× 104, and r2 = 0.9990.
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Solutions Manual c©
to accompany
System Dynamics, Second Edition
by
William J. Palm III
University of Rhode Island
Solutions to Problems in Chapter Two
c©Solutions Manual Copyright 2010 The McGraw-Hill Companies. All rightsreserved. No part of this manual may be displayed, reproduced, or distributedin any form or by any means without the written permission of the publisheror used beyond the limited distribution to teachers or educators permitted byMcGraw-Hill for their individual course preparation. Any other reproductionor translation of this work is unlawful.
2.1 mv = mg. Thus
v(t) = gt = 32.2t x(t) =12gt2 = 16.1t2 h(t) = 20 − x(t) = 20 − 16.1t2
Thus
t =
√20− h(t)
16.1
For h = 10, t = 0.788 sec. For h = 0, t = 1.115 sec.
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2.2
t =60/5280
903600 = 0.455 sec
x =12gt2 = 16.1(0.455)2 = 3.326 ft
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2.3 Summing forces in the direction parallel to the plane gives
mv = f1 − mg sin φ − µmg cos φ
Substituting the given values,
10v = f1 − 98.1 sin 25 − 0.3(98.1) cos 25 = f1 − 68.132
Thus v > 0 if f1 > 68.132. If f1 = 100 the block will continue to move up the plane. Iff1 = 50, v = −1.813 and the speed is given by
v(t) = −1.813t + v(0) = −1.813t + 2
Thus v(0) = 0 at t = 1.103 s.
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2.4 Let d be the vertical distance dropped by the time the mass leaves the surface. See thefigure on the following page. Then
d =L
tan θ
The speed v0 of the mass when it leaves the surface is found from conservation of energy:
KE =12mv2
0 = PE =mgL
tan θ
Thus
v0 =
√2gL
tan θ(1)
The horizontal and vertical velocity components are
vx = v0x sin θ
v0y = v0 cos θ (2)
Establish a coordinate system at the point where the mass leaves the surface, with xpositive to the right and y positive down. In the vertical direction : y = g and
y =12gt2 + v0yt
The mass hits the ground when y = H , and time-to-hit tH is found from
12gt2H + 2v0ytH = H
or
tH =−2v0y ±
√4v2
0y + 8gH
2g(3)
There will be one positive solution and one negative solution. Take the positive solution.(Continued on the next page)
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Problem 2.4 continued:In the horizontal direction: mx = 0 and
x = (v0 sin θ) t
andD = (v0 sin θ) tH (4)
The solution is given by (1) through (4).
Figure : for Problem 4
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2.5 Let the origin of the launch point be at x = y = 0. Assuming the projectile is launchedwith a speed v0 at an angle θ from the horizontal, Newton’s law in the x and y directionsgives
x = 0 y = −g
vx = v0 cos θ vy = v0 sin θ − gt
x = (v0 cos θ)t y = −g
2t2 + (v0 sin θ)t
These can be manipulated to show that
v2y = (v0 − sin θ)2 − 2gy
which givesv0 sin θ =
√v2y − 2gy
Using this relation, the above can also be manipulated to show that
vy = −gx
vx+√
v2y − 2gy
Solve for x.x =
vx
g
√v2y − 2gy − vxvy
g(1)
where y is computed fromy = R sin φ (2)
The desired distance D can be computed from
D = x + R cos φ
where x is computed from (1).
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2.6 IG = 2mr2/5. Apply the parallel axis theorem.
IO = IG + mR2 =25mr2 + mR2
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2.7 a) Let point O be the pivot point and G be the center of mass. Let L be the distancefrom O to G. Treat the pendulum as being composed of three masses:
1) m1, the rod mass above point O, whose center of mass is 1 ft above point O;2) m2, the rod mass below point O, whose center of mass is 1.5 ft below point O, and3) m3, the mass of the 10 lb block.
Then, summing moments about G gives
m1g(L + 1) − m2g(1.5− L)− m3g(3.5− L) = 0 (1)
wherem1g =
253 =
65
lb
m2g =353 =
95
lb
m3g = 10 lb
From equation (1):
65(L + 1) − 9
5(1.5− L) − 10(3.5− L) = 0
which gives L = 2.808 ft.b) Summing moments about the pivot point O gives
IOθ = −mgL sin θ
where m is the total mass. From the parallel-axis theorem, treating the rod as a slenderrod, we obtain
IO =112
(3g
)(5)2 +
(3g
)(0.5)2 +
(10g
)(3.5)2 = 4.022 slug − ft2
and mgL = 13(2.808) = 36.504 ft-lb. Thus the equation of motion is
4.022θ = −36.504 sin θ
orθ + 9.076 sin θ = 0
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2.8 a)IA = mL2
2 + mCL21
IAθ = −mCL1g sin θ + mgL2 cos(β − θ)
b) Set θ = 0 and solve for mg.
mg =mCL1g sin θ
L2 cos(β − θ)
c) Substitute the given values to obtain
mg =5(0.2)g sin 20
0.15 cos 10= 2.315g = 22.713 N
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2.9 The easiest way is to use the energy-equivalence method. Let ωB the speed of pulley B(positive counterclockwise) and ωC the speed of pulley C (positive clockwise). The kineticenergy of the entire system is
KE =12IBω2
B +12ICω2
C +12mLv2
C +12mCv2
C =12mev
2A
where me is the mass of the equivalent translational system. The potential energy is
PE = (mL + mC)g(−sC) = FesA
where Fe is the equivalent gravity force. Thus
Fe = −mL + mC
2g
Using the kinematic relations:
sC = −12sA vC = −1
2vA ωB =
1RB
vA ωC =1
RCvC
the kinetic energy expression becomes
KE =12
(IB
R2B
+IC
4R2C
+mL + mC
4
)v2A =
12mev
2A
Thusme =
IB
R2B
+IC
4R2C
+mL + mC
4
The model ismevA = FA − mL + mC
2g
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2.10 a) Let T be the tension in the cable attached to mass m2. See the figure on thefollowing page. Then the cable force pulling up on m1 is T/2 because of the pulleys. Notealso that because of the pulleys, x = 2y. Summing forces acting on m2 parallel to the plane,we obtain
m2y = T − m2g sin θ (1)
Summing the vertical forces acting on m1, we obtain
m1x = m1g − 12T (2)
Since x = 2y, this becomes
2m1y = m1g − 12T (3)
Solve for T :T = 2m1g − 4m1y (4)
Substitute this into (1) and collect the y terms to obtain
(4m1 + m2)y = 2m1g − m2g sin θ (5)
The mass m1 will lift m2 if y > 0; that is, if 2m1 − m2 sin θ > 0.
Figure : for Problem 10
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Problem 2.10 continued:b) Follow the same procedure as in part (a) but include the friction force. Equation (1)
becomesm2y = T − m2g sin θ − µdm2g cos θ (6)
Equations (2) through (4) remain the same, but (5) becomes
(4m1 + m2)y = 2m1g − m2g(µd cos θ + sin θ) (7)
The mass m1 will lift m2 if y > 0; that is, if
2m1 − m2(µd cos θ + sin θ) > 0
For the case m1 = m2/2, this becomes
1 − (µd cos θ + sin θ) > 0
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2.11 The assumption in Example 2.3.2 is that the only inertia on each shaft is the concen-trated inertia I1 and I2 at each end. This implies that the inertias of the shafts and gearsare negligible. If, on the other hand, the inertias of the shafts are not negligible, then thekinetic energy expression becomes
KE =12
(I1 + Is1)ω21 +
12
(I2 + Is2)ω22
or
KE =12
(I1 + Is1)ω21 +
12
(I2 + Is2)(
ω1
N
)2
Therefore the equivalent inertia felt on the input shaft is
Ie = I1 + Is1 +I2 + Is2
N2
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2.12 Let F be the contact force between the two gears. For gear 1,
IG1ω1 = T1 − r1F
For gear 2,IG2ω1 = T2 − r2F
If ω1 = ˙ ω2 = 0, or if IG1 = IG2 = 0,
T1 = r1F T2 = r2F
which give
T1 =r1
r2T2 =
1N
T2
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2.13 The equivalent inertia felt on shaft 1 is
Ie = I1 +1
N2I2 +
1N2
m2R2 +
1N2
m3R2
With N = 2,
Ie = I1 +14
(I2 + m2R
2 + m3R2)
The equation of motion is
Ieω1 = T1 −m2gR
N+
m3gR
N= T1 −
gR
2(m3 − m2)
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2.14 The total kinetic energy is
KE =12
(Is + I) θ2 +12mx2
Substituting x = Rθ we obtain
KE =12
(Is + I + mR2
)θ2
Thus the equivalent inertia isIe = Is + I + mR2
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2.15 The total kinetic energy is
KE =12
(I1 + IS1)ω21 +
12
(I2 + IS2)ω22 +
12mv2
Substituting ω2 = r1ω1/r2 and v = r1ω1 we obtain
KE =12
(I1 + IS1) ω21 +
12
(I2 + IS2)(
r1ω1
r2
)2
+12m (r1ω1)
2
or
KE =12
[I1 + IS1 + (I2 + IS2)
(r1
r2
)2
+ mr21
]ω2
1
Thus the equivalent inertia is
Ie = I1 + IS1 + (I2 + IS2)(
r1
r2
)2
+ mr21
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2.16 a) With everything reflected to the load shaft (shaft 2), Newton’s law gives
ω2 =T2 + NT1
I2 + N2I1
b) To maximize ω2, differentiate the above expression with respect to N , set the deriva-tive equal to 0. This gives
∂ω2
∂N=
(I2 + N2I1)T1 − 2I1N(T2 + NT1)(I2 + N2I1)2
= 0
This is true if the numerator is 0. Thus
I1T1N2 + 2I1T2N − I2T1 = 0
The positive solution for N is
N = −T2
T1+
√(T2
T1
)2
+I2
I1
This is the ratio that maximizes ω2. (This can be confirmed to give a maximum ratherthan a minimum by showing that ∂2ω2/∂N2 < 0).
If the load torque T2 is 0, the optimal ratio for this case is denoted No and is: No =√I2/I1, or I1 = I2/N
2o . This says that the ratio that maximizes the load acceleration is the
ratio that makes the reflected load inertia (felt by the motor) equal to the motor’s inertia.This is the principle of inertia matching.
If T2 = 0 and the actual ratio N differs from the optimal value No such that N = γNo,the efficiency E is
E =actual ω2
max ω2=
2γ
1 + γ2
Because E(γ) = E(1/γ), the efficiency when “overgearing” is the same as when “under-gearing”. For example, if γ = 2 or γ = 1/2, E = 0.8, so the acceleration is 80% of themaximum possible.
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2.17 With I1 = I2 = I3 = 0, the total kinetic energy is
KE =12I4ω
21 +
12I5ω
23
Substituting ω2 = 1.4ω3 and ω1 = 1.4ω2 = (1.4)2ω3 = 1.96ω3, and I4 = 0.02, I5 = 0.1, weobtain
KE =12(0.177)ω2
3
and the equivalent inertia is Ie = 0.177 kg·m2.The equation of motion is Ieω3 = (1.4)2T , or 0.177ω3 = 1.96T
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2.18 The total kinetic energy is
KE =12
(I4 + I1)ω21 +
12I2ω
22 +
12
(I3 + I5)ω23
Substituting ω2 = 1.4ω3 and ω1 = 1.4ω2 = (1.4)2ω3 = 1.96ω3 and the given values of theinertias, we obtain
KE =12(0.203)ω2
3
and the equivalent inertia is Ie = 0.203 kg·m2.The equation of motion is Ieω3 = (1.4)2T , or 0.203ω3 = 1.96T
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2.19 a)ω4
ω1=
ω4
ω3
ω3
ω2
ω2
ω1=
76
353 = 2.1
b) The torque T1 felt on shaft 4 is T1/2.1 and the equation of motion is
Iω4 =T1
2.1
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2.20 Using kinetic energy equivalence,
KE =12mv2 +
12Isω
2 =12Ieω
2
The mass translates a distance x when the screw rotates by θ radians. When θ = 2π, x = L.Thus x = Lθ/2π and x = v = Lθ/2π. Because ω = θ, we have
KE =12m
(Lω
2π
)2
+12Isω
2 =12Ieω
2
Solve for Ie to obtain
Ie =mL2
4π2+ Is
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2.21 The expression for the kinetic energy is
KE = KE of 2 rear wheels + KE of front wheel + KE of body
= 2(
12Irω
2r
)+
12Ifω2
f +12mbv
2
But for the rear wheels, v = Rrωr = 4ωr, and for the front wheel, v = Rfωf = 2ωf . Theinertias are calculated as follows. For the rear wheels:
Ir =12mrR
2r =
12
500g
42 =4000
g
For the front wheel:If =
12mfR2
f =12
800g
22 =1600
g
Also, mb = 9000/g. Thus
KE =4950
gv2
and the equivalent mass is me = 2(4950)/g = 9000/g.The equation of motion is mev = mg sin 10, where
m = 2mr + mf + mb = 2500g
+800g
+9000
g=
10, 800g
Thus9900
gv = 10, 800 sin 10
or v = 0.1894. Thus v = 0.1894t ft/sec.
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2.22 Using the equivalent mass approach, the equivalent mass referenced to the coordinatex is
me = m1 + m2 +I
R2
where I is the inertia of the cylinder about its center. The force acting on me due to theweight of the cylinder is m1g sin β. The force acting on me due to the weight of m2 ism2g sin φ. See the following figure. The equation of motion of the equivalent system is
mex = m1g sin β − m2g sin φ
Figure : for Problem 22
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2.23 Let y be the translational displacement of the cylinder to the right. Using the equiv-alent mass approach, the kinetic energy of the system is
KE =12m2y
2 +12Iω2 +
12m1x
2
where I = m2R2/2 is the inertia of the cylinder about its center. Because y = 2x and
ω = ˙ y/R, we have
KE =12m2
(4x2
)+
12
(m2R
2
2
)(2x
R
)2
+12m1x
2 =12
(6m2 + m1) x2
Thus the effective mass referenced to the coordinate x is me = 6m2 + m1.The equation of motion is
mex = m1g
or(6m2 + m1)x = m1g
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2.24 Summing moments about the pivot gives
IOθ = T − mgL sin θ
where the effect of the motor torque is T = [2(1.5)]Tm = 3Tm and the inertia is
IO = mL2 + Im[2(1.5)]2 + IG1[2(1.5)]2 + IG2(1.5)2 + IG3(1.5)2 + IG4 = 1.936 kg · m2
Thus1.936θ = 3Tm − 29.43 sin θ
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2.25 a) The speed ratio of the sprocket drive at the driving shaft is the ratio of the sprocketdiameters, which is 0.05/0.15 = 1/3. The equivalent inertia felt at the motor shaft is
Ie = I1 + I2 +[IS1 + mc1(0.05)2
] ( 110
)2
+[IS2 + Id + 4Iw + 2mc2r
2d + mLr2
d
](13
110
)2
where IS1, IS2, Id, and Iw are the inertias of the shafts 1 and 2, the drive shaft, and thedrive wheels. The masses mc1, mc2, and mL are the masses of the sprocket chain, the drivechain, and the load. This expression evaluates to Ie = 0.0143 kg·m2. The magnitude of thefriction torque felt at the motor shaft is 54/[3(10)] = 1.8 N·m. The equation of motion is
0.0143ω1 = T1 − 1.8sgn(ω1) (1)
b) If T1 = 10, ω1 = 573.427 and ω1 = 573.427t rad/s.c) Solve (1) for T1 assuming ω1 > 0:
T1 = 0.0143ω1 + 1.8
From the trapezoidal profile,
ω1 =
300 0 ≤ t ≤ 0.50 0.5 < t < 2.5−300 2.5 ≤ t ≤ 3
Thus
T1 =
300Ie + 1.8 0 ≤ t ≤ 0.51.8 0.5 < t < 2.51.8− 300Ie 2.5 ≤ t ≤ 3
=
6.09 0 ≤ t ≤ 0.51.8 0.5 < t < 2.5−249 2.5 ≤ t ≤ 3
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2.26 a) The equation of motion for rotation is
Iω = Rft
where I = mR2/2 = 1.631 kg·m2 is the inertia of the cylinder about its center, and ft
is the tangential force between the cylinder and the ground. The equation of motion fortranslation is
mx = f cos φ − ft = f cos φ − Iω
R
Substituting x = Rω we obtain(mR2 + I
)ω = Rf cos φ
Noting that m = 800/9.81 = 81.549 kg, and substituting the given values, we obtain
16.636ω = f cos φ
b) Using the equivalent mass approach, the equivalent mass referenced to the coordinatex is
me = m +I
R2
where I = mR2/2 = 1.631 kg·m2 is the inertia of the cylinder about its center. The equationof motion is
mex = f cos φ
Substituting the given values, we obtain
83.18x = f cos φ
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2.27 See the following figure for the coordinate definitions and the definition of the reactionforce R. Let P be the point on the axle. Note that yP = 0. The coordinates of the masscenter of the rod are
xG = xP − L
2sin θ yG = −L
2cos θ
ThusxG = xP − L
2θ cos θ +
L
2θ2 sin θ
yG =L
2θ sin θ +
L
2θ2 cos θ
Let m be the mass of the rod. Summing forces in the x direction:
mxG = f or m
(xP − L
2θ cos θ +
L
2θ2 sin θ
)= f (1)
Summing forces in the y direction:
myG = R − mg or m
(L
2θ sin θ +
L
2θ2 cos θ
)= R − mg (2)
Figure : for Problem 27
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Problem 2.27 continued:Summing moments about the mass center:
IGθ = fL
2cos θ − R
L
2sin θ
Substituting for R from (2) and using the fact that IG = mL2/12, we obtain
112
mL2θ =fL
2cos θ − mgL
2sin θ − mL2
4sin2 θ θ − mL2
4θ2 sin θ cos θ (3)
The model consists of (1) and (3) with m = 20 kg and L = 1.4 m.
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2.28 See the following figure for the coordinate definitions and the definition of the reactionforces Rx, Ry, N , and ft. Let P be the point on the axle. Note that yP = 0. The coordinatesof the mass center of the rod are
xG = xP − L
2sin θ yG = −L
2cos θ
ThusxG = xP − L
2θ cos θ +
L
2θ2 sin θ
yG =L
2θ sin θ +
L
2θ2 cos θ
Figure : for Problem 28
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Problem 2.28 continued:Let m1 be the mass of the rod, m2 the mass of the wheel, and I the moment of inertia
of the wheel about its center. Summing forces on the wheel in the x direction:
m2xP = f − Rx − ft (1)
Summing forces on the wheel in the y direction:
m2yP = N − Ry − m2g (2)
Summing moments about the mass center of the wheel gives :Iω = Rft. But xP = Rω andxP = Rω. Thus
I
R2xP = ft (3)
Substitute this into (1) to obtain(m2 +
I
R2
)xP = f − Rx (4)
Summing forces on the rod in the x direction:
m1xG = Rx or m1
(xP − L
2θ cos θ +
L
2θ2 sin θ
)= Rx (5)
Summing forces on the rod in the y direction:
m1yG = Ry − m1g or m1
(L
2θ sin θ +
L
2θ2 cos θ
)= Ry − m1g (6)
(Continued on the next page)
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Problem 2.28 continued:Summing moments about the mass center of the rod:
IGθ = RxL
2cos θ − Ry
L
2sin θ
Substituting for Rx and Ry from (5) and (6), canceling terms using the identity sin2 θ +cos2 θ = 1, and using the fact that IG = m1L
2/12, we obtain
112
m1L2θ =
m1L
2xP cos θ − m1gL
2sin θ − m1L
2
4θ
which can be rearranged as follows:
13m1L
2θ =m1L
2xP cos θ − m1gL
2sin θ (7)
Substituting for Rx from (4) into (5) gives(
m1 + m2 +I
R2
)xP = F − m1L
2
(θ2 sin θ − θ cos θ
)(8)
The model consists of (7) and (8) with m1 = 20 kg, m2 = 3 kg, L = 1.4 m, R = 0.05 m,and I = m2R
2/2.
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2.29 Summing forces in the x direction:
maGx = mg sin θ − F (1)
Summing forces in the y direction:
maGy = N − mg cos θ − F (2)
For no bounce, aGy = 0 andN = mg cos θ (3)
Summing moments about G:IGα = Fr (4)
From (1) and (4):
maGx = mg sin θ − IGα
r(5)
With slipping, aGx 6= rα, butF = µdN (6)
From (4) and (6):IGα = rµdN = rµdmg cos θ (7)
Thus,
α =rµdmg cos θ
IG
From (1),aGx = g sin θ − µdg cos θ
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2.30 Follow the procedure of Example 2.4.2. Summing forces in the x direction:
maGx = mg sin θ − F (1)
Summing forces in the y direction:
maGy = N − mg cos θ − F (2)
For no bounce, aGy = 0 andN = mg cos θ (3)
a) Summing moments about G:
IGα = Fr (4)
From (1) and (4):
maGx = mg sin θ − IGα
r(5)
For no slip,aGx = rα (6)
From (5) and (6):
aGx =mgr2 sin θ
mr2 + IG(7)
Since IG = mr2,
aGx =g sin θ
2(8)
andα =
aGx
r=
g sin θ
2r
b) Summing moments about P :
MP = (mg sin θ)r = IGα + maGxr = IGaGx
r+ maGxr
Thus
aGx =mgr2 sin θ
mr2 + IG=
g sin θ
2
andα =
aGx
r=
g sin θ
2r
(Continued on the next page)
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Problem 2.30 continued:c) The maximum friction force is Fmax = µsN = µsmg cos θ. From (4), (6), and (7),
F =IGaGx
r2=
IGmg sin θ
mr2 + IG
If Fmax > F there will be no slip; that is, if
µs cos θ >IG sin θ
mr2 + IG=
sin θ
2
there will be no slip.
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2.31 See the following figure. Summing forces in the x and y directions gives:
maGx = µsNB − mg sin θ (1)
NA + NB = mg cos θ (2)
Summing moments about G:
NALA − NBLB + µsNBH = 0 (3)
Solve (2) and (3):
NA =mg cos θ (µsH − LB)
µsH − LA − LB=
16, 677 cos θ (µs − 2.1)µs − 4.6
NB = − mgLA cos θ
µsH − LA − LB=
40, 025µs cos θ
4.6− µs
The maximum acceleration is, from (1),
aGx =23.544µs cos θ
4.6− µs− 9.81 sin θ
Figure : for Problem 31
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2.32 Use (2.4.6) with MP = −mgL sin θ, IP = mL2, aPx = −a(t), and aPy = 0. Also,referring to Figure 2.4.1, rx = L sin θ and ry = −L cos θ. Thus
mL2θ − mLa(t) cos θ = −mgL sin θ
Rearrange to cancel mL and to put the input a(t) on the right-hand side.
Lθ + g sin θ = a(t) cos θ
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2.33 Use (2.4.6) with MP = mgL sin θ, IP = mL2, aPx = 0, and aPy = a(t). Also, referringto Figure 2.4.1, rx = −L sin θ and ry = L cos θ. Thus
mL2θ − mLa(t) sin θ = mgL sin θ
Rearrange to cancel mL and to put the input a(t) on the right-hand side.
Lθ − g sin θ = a(t) sin θ
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Solutions Manual c©
to accompany
System Dynamics, Second Edition
by
William J. Palm III
University of Rhode Island
Solutions to Problems in Chapter Three
c©Solutions Manual Copyright 2010 The McGraw-Hill Companies. All rightsreserved. No part of this manual may be displayed, reproduced, or distributedin any form or by any means without the written permission of the publisheror used beyond the limited distribution to teachers or educators permitted byMcGraw-Hill for their individual course preparation. Any other reproductionor translation of this work is unlawful.
3.1 a) Nonlinear because of the yy term. b) Nonlinear because of the sin y term. c)Nonlinear because of the
√y term.
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3.2 a)
4∫ x
2dx = 3
∫ t
0t dt
x(t) = 2 +38t2
b)
5∫ x
3dx = 2
∫ t
0e−4t dt
x(t) = 3.1− 0.1e−4t
c) Let v = ˙ x.
3∫ v
7dv = 5
∫ t
0t dt
v(t) =dx
dt= 7 +
56t2
∫ x
2dx =
∫ t
0
(7 +
56t2)
dt
x(t) = 2 + 7t +518
t3
d) Let v = x.
4∫ v
2dv = 7
∫ t
0e−2t dt
v(t) =238
− 78e−2t
∫ x
4dx =
∫ t
0
(238
− 78e−2t
)dt
x(t) =5716
+238
t +716
e−2t
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3.3 a) ∫ x
3
dx
25 − 5x2=∫ t
0dt = t
110
√5
ln
(√5 + x√5 − x
)∣∣∣∣∣
x
3
Let
A = ln√
5 + 3√5− 3
Thus
ln√
5 + x√5 − x
= 10√
5t + A
x(t) =√
5eAe10
√5t − 1
eAe10√
5t + 1b) ∫ x
10
dx
36 + 4x2=∫ t
0dt = y
112
tan−1 x
3
∣∣∣∣x
10= t
x(t) = 3 tan(12t + C) C = tan−1 103
c) ∫ x
4
dx
5x + 25=∫ t
0dt
15
ln(5x + 25)∣∣∣∣x
4= t
ln(5x + 25) = 5t + ln 45
x(t) = 9e5t − 5
(continued on the next page)
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Problem 3.3 continued:
d) ∫ x
5
dx
x= 2
∫ t
0e−4t dt
ln x|x5 = − 12e−4t
∣∣∣∣t
0
ln x = ln 5− 12
(e−4t − 1
)
x(t) = 5e−(e−4t−1)/2
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3.4 a) The roots are −3 and −5. The form of the free response is
x(t) = A1e−3t + A2e
−5t
Evaluating this with the given initial conditions gives
x(t) = 27e−3t − 17e−5t
The steady-state solution is xss = 30/15 = 2. Thus the form of the forced response is
x(t) = 2 + B1e−3t + B2e
−5t
Evaluating this with zero initial conditions gives
x(t) = 2 − 5e−3t + 3e−5t
The total response is the sum of the free and the forced response. It is
x(t) = 2 + 22e−3t − 14e−5t
The transient response consists of the two exponential terms.
(continued on the next page)
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Problem 3.4 continued:
b) The roots are −5 and −5. The form of the free response is
x(t) = A1e−5t + A2te
−5t
Evaluating this with the given initial conditions gives
x(t) = e−5t + 9te−5t
The steady-state solution is xss = 75/25 = 3. Thus the form of the forced response is
x(t) = 3 + B1e−5t + B2te
−5t
Evaluating this with zero initial conditions gives
x(t) = 3 − 3e−5t − 15te−5t
The total response is the sum of the free and the forced response. It is
x(t) = 3 − 2e−5t − 6te−5t
The transient response consists of the two exponential terms.
(continued on the next page)
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Problem 3.4 continued:
c) The roots are ±5j. The form of the free response is
x(t) = A1 sin 5t + A2 cos 5t
Evaluating this with the given initial conditions gives
x(t) =45
sin 5t + 10 cos 5t
The form of the forced response is
x(t) = B1 + B2 sin 5t + B3 cos 5t
Thus the entire forced response is the steady-state forced response. There is no transientforced response. Evaluating this function with zero initial conditions shows that B2 = 0and B3 = −B1. Thus
x(t) = B1 − B1 cos 5t
Substituting this into the differential equation shows that B1 = 4 and the forced responseis
x(t) = 4− 4 cos 5t
The total response is the sum of the free and the forced response. It is
x(t) = 4 + 6 cos 5t +45
sin 5t
The entire response is the steady-state response. There is no transient response.
(continued on the next page)
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Problem 3.4 continued:
d) The roots are −4 ± 7j. The form of the free response is
x(t) = A1e−4t sin 7t + A2e
−4t cos 5t
Evaluating this with the given initial conditions gives
x(t) =447
e−4t sin 7t + 10e−4t cos 7t
The form of the forced response is
x(t) = B1 + B2e−4t sin 7t + B3e
−4t cos 7t
The steady-state solution is xss = 130/65 = 2. Thus B1 = 2. Evaluating this function withzero initial conditions shows that B2 = −8/7 and B3 = −2. Thus the forced response is
x(t) = 2 − 87e−4t sin 7t − 2e−4t cos 7t
The total response is the sum of the free and the forced response. It is
x(t) = 2 +367
e−4t sin 7t + 8e−4t cos 7t
The transient response consists of the two exponential terms.
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3.5 a) The root is s = 5/3, which is positive. So the model is unstable.b) The roots are s = 5 and −2, one of which is positive. So the model is unstable.c) The roots are s = 3 ± 5j, whose real part is positive. So the model is unstable.d) The root is s = 0, so the model is neutrally stable.e) The roots are s = ±2j, whose real part is zero. So the model is neutrally stable.f) The roots are s = 0 and −5, one of which is zero and the other is negative. So the
model is neutrally stable.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
3.6 a) The system is stable if both of its roots are real and negative or if the roots arecomplex with negative real parts. Assuming that m 6= 0, we can divide the characteristicequation by m to obtain
s2 +c
ms +
k
m= s2 + as + b = 0
where a = c/m and b = k/m. The roots are given by the quadratic formula:
s =−a ±
√a2 − 4b
2
Thus the condition that m, c, and k have the same sign is equivalent to a > 0 and b > 0.There are three cases to be considered:
1. Complex roots (a2 − 4b < 0). In this case the real part of both roots is −a/2 and isnegative if a > 0.
2. Repeated, real roots (a2 − 4b = 0). In this case both roots are −a/2 and are negativeif a > 0.
3. Distinct, real roots (a2 − 4b > 0). Let the two roots be denoted r1 and r2. We canfactor the characteristic equation as s2 + as + b = (s − r1)(s − r2) = 0. Expandingthis gives
(s − r1)(s− r2) = s2 − (r1 + r2)s + r1r2 = 0
Comparing the two forms shows that
r1r2 = b (1) and r1 + r1 = −a (2)
If b > 0, condition (1) shows that both roots have the same sign. If a < 0, condition(2) shows that the roots must be negative. Therefore, if the roots are distinct andreal, the roots will be negative if a > 0 and b > 0.
b) Neutral stability occurs if either 1) both roots are imaginary or 2) one root is zerowhile the other root is negative. Imaginary roots occur when a = 0 (the roots are s = ±
√b)
In this case the free response is a constant-amplitude oscillation. Case 2 occurs when b = 0and a > 0 (the roots are s = 0 and s = −a). In this case the free response decays to anon-zero constant.
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3.7 From the transform definition, we have
L[mt] = limT→∞
[∫ T
0mte−stdt
]= m lim
T→∞
[∫ T
0te−stdt
]
The method of integration by parts states that∫ T
0u dv = uv|T0 −
∫ T
0v du
Choosing u = t and dv = e−stdt, we have du = dt, v = −e−st/s, and
L[mt] = m limT→∞
[∫ T
0te−stdt
]= m lim
T→∞
t
e−st
−s
∣∣∣∣∣
T
0
−∫ T
0
e−st
−sdt
= m limT→∞
t
e−st
−s
∣∣∣∣∣
T
0
− e−st
(−s)2
∣∣∣∣∣
T
0
= m lim
T→∞
[Te−sT
−s− 0 − e−sT
(−s)2+
e0
(−s)2
]
=m
s2
because, if we choose the real part of s to be positive, then from Table 3.3-2,
limT→∞
Te−sT = 0
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
3.8 From the transform definition, we have
L[t2] = limT→∞
[∫ T
0t2e−stdt
]
The method of integration by parts states that∫ T
0u dv = uv|T0 −
∫ T
0v du
Choosing u = t2 and dv = e−stdt, we have du = 2t dt, v = −e−st/s, and
L[t2] = limT→∞
[∫ T
0t2e−stdt
]= lim
T→∞
t2
e−st
−s
∣∣∣∣∣
T
0
−∫ T
0
e−st
−s2t dt
= limT→∞
[−T 2e−st
s+
2s
∫ T
0te−stdt
]= lim
T→∞
[−T 2 e−st
s
]+
2s
(1s2
)
=2s3
because, if we choose the real part of s to be positive, then,
limT→∞
T 2e−sT = 0
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3.9 a)
X(s) =10s
+2s3
b)
X(s) =6
(s + 5)2+
1s + 3
c) From Property 8,
X(s) = −dY (s)ds
where y(t) = e−3t sin 5t. Thus
Y (s) =5
(s + 3)2 + 52=
5s2 + 6s + 34
dY (s)ds
= − 10s + 30(s2 + 6s + 34)2
ThusX(s) =
10s + 30(s2 + 6s + 34)2
d) X(s) = e−5sG(s), where g(t) = t. Thus G(s) = 1/s2 and
X(s) =e−5s
s2
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3.10f(t) = 5us(t) − 7us(t − 6) + 2us(t − 14)
Thus
F (s) =5s− 7
e−6s
s+ 2
e−14s
s
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3.11 a)
x(0+) = lims→∞
s5
3s + 7=
53
x(∞) = lims→0
s5
3s + 7= 0
b)
x(0+) = lims→∞
s10
3s2 + 7s + 4= 0
x(∞) = lims→0
s10
3s2 + 7s + 4= 0
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3.12 a)
X(s) =32
(1s− 1
s + 4
)
x(t) =32
(1− e−4t
)
b)
X(s) =53
1s
+313
1s + 3
x(t) =53
+313
e−3t
c)
X(s) = −13
1s + 2
+133
1s + 5
x(t) = −13e−2t +
133
e−5t
d)
X(s) =5/2
s2(s + 4)=
58
1s2
− 532
1s
+532
1s + 4
x(t) =58t − 5
32+
532
e−4t
e)
X(s) =25
1s2
+1325
1s− 13
251
s + 5
x(t) =25t +
1325
− 1325
e−5t
f)
X(s) = −314
1(s + 3)2
+7916
1s + 3
− 7916
1s + 7
x(t) = −314
te−3t +7916
e−3t − 7916
e−7t
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3.13 a)
X(s) =7s + 2
(s + 3)2 + 52= C1
5(s + 3)2 + 52
+ C2s + 3
(s + 3)2 + 52
orX(s) = −19
55
(s + 3)2 + 52+ 7
s + 3(s + 3)2 + 52
x(t) = −195
e−3t sin 5t + 7e−3t cos 5t
b)
X(s) =4s + 3
s[(s + 3)2 + 52]=
C1
s+ C2
5(s + 3)2 + 52
+ C3s + 3
(s + 3)2 + 52
orX(s) =
334
1s
+127170
5(s + 3)2 + 52
− 334
s + 3(s + 3)2 + 52
x(t) =334
+127170
e−3t sin 5t − 334
e−3t cos 5t
c)
X(s) =4s + 9
[(s + 3)2 + 52][(s + 2)2 + 42]
= C15
(s + 3)2 + 52+ C2
s + 3(s + 3)2 + 52
+ C34
(s + 2)2 + 42+ C4
s + 2(s + 2)2 + 42
or
X(s) = − 44205
5(s + 3)2 + 52
− 1982
s + 3(s + 3)2 + 52
+69328
4(s + 2)2 + 42
+1982
s + 2(s + 2)2 + 42
x(t) = − 44205
e−3t sin 5t − 1982
e−3t cos 5t +69328
e−2t sin 4t +1982
e−2t cos 4t
d)
X(s) = 2.6251
s + 2− 18.75
1s + 4
+ 21.1251
s + 6
x(t) = 2.625e−2t − 18.75e−4t + 21.125e−6t
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3.14 a) x = 7t/5 ∫ x
3dx =
75
∫ t
0t dt
x(t) =710
t2 + 3
b) x = 3e−5t/4 ∫ x
0dx =
34
∫ t
0e−5t dt
x(t) =320
(1 − e−5t
)
c) x = 4t/7
x(t) − x(0) =47
∫ t
0t dt
x(t) =414
t2 + 5
∫ x
3dx =
∫ t
0
(414
t2 + 5)
dt
x(t) =442
t3 + 5t + 3
d) x = 8e−4t/3
x(t)− x(0) =83
∫ t
0e−4t dt
x(t) =173
− 812
e−4t
∫ x
3dx =
∫ t
0
(173
− 812
e−4t)
dt
x(t) =173
t +16e−4t +
176
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3.15 a) The root is −7/5 and the form is x(t) = Ce−7t/5. With x(0) = 4, C = 4 andx(t) = 4e−7t/5
b) The root is −7/5 and the form is x(t) = C1e−7t/5 + C2. At steady state, x = 15/7 =
C2. With x(0) = 0, C1 = −15/7. Thus
x(t) =157
(1 − e−7t/5
)
c) The root is −7/5 and the form is x(t) = C1e−7t/5 + C2. At steady state, x = 15/7 =
C2. With x(0) = 4, C1 = 13/7. Thus
x(t) =137
(1 + e−7t/5
)
d)
sX(s)− x(0) + 7X(s) =4s2
X(s) =5s2 + 4
s2(s + 7)=
47s2
− 449
+24949
e−7t
x(t) =47t − 4
49+
24949
e−7t
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3.16 a) The roots are −7 and −3. The form is
x(t) = C1e−7t + C2e
−3t
Evaluating C1 and C2 for the initial conditions gives
x(t) = −94e−7t +
254
e−3t
b) The roots are −7 and −7. The form is
x(t) = C1e−7t + C2te
−7t
Evaluating C1 and C2 for the initial conditions gives
x(t) = e−7t + 10te−7t
c) The roots are −7 ± 3j. The form is
x(t) = C1e−7t sin 3t + C2e
−7t cos 3t
Evaluating C1 and C2 for the initial conditions gives
x(t) =203
e−7t sin 3t + 4e−7t cos 3t
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3.17 a) The roots are −3 and −7. The form is
x(t) = C1e−3t + C2e
−7t + C3
At steady state, x = 5/63 so C3 = 5/63. Evaluating C1 and C2 for the initial conditionsgives
x(t) = − 35252
e−3t +15252
e−7t +563
b) The roots are −7 and −7. The form is
x(t) = C1e−7t + C2te
−7t + C3
At steady state, x = 98/49 = 2 so C3 = 2. Evaluating C1 and C2 for the initial conditionsgives
x(t) = −2e−7t − 16te−7t + 2
c) The roots are −7 ± 3j. The form is
x(t) = C1e−7t sin 3t + C2e
−7t cos 3t + C3
At steady state, x = 174/58 = 3 so C3 = 3. Evaluating C1 and C2 for the initial conditionsgives
x(t) = −7e−7t sin 3t − 3e−7t cos 3t + 3
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3.18 a)X(s)F (s)
=15
5s + 7
The root is s = −7/5.b)
X(s)F (s)
=5
3s2 + 30s + 63
The roots are s = −7 and s = −3.c)
X(s)F (s)
=4
s2 + 10s + 21
The roots are s = −7 and s = −3.d)
X(s)F (s)
=7
s2 + 14s + 49
The roots are s = −7 and s = −7.e)
X(s)F (s)
=6s + 4
s2 + 14s + 58
The roots are s = −7 ± 3j.f)
X(s)F (s)
=4s + 155s + 7
The root is s = −7/5.
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3.19 Transform each equation using zero initial conditions.
3sX(s) = Y (s)
sY (s) = F (s) − 3Y (s) − 15X(s)
Solve for X(s)/F (s) and Y (s)/F (s).
X(s)F (s)
=1
3s2 + 9s + 15
Y (s)F (s)
=3s
3s2 + 9s + 15
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3.20 Transform each equation using zero initial conditions.
sX(s) = −2X(s) + 5Y (s)
sY (s) = F (s) − 6Y (s)− 4X(s)
Solve for X(s)/F (s) and Y (s)/F (s).
X(s)F (s)
=5
s2 + 8s + 32
Y (s)F (s)
=s + 2
s2 + 8s + 32
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3.21 a)6
s(s + 5)=
65s
− 65
1s + 5
x(t) =65
(1− e−5t
)
b)4
s + 3)(s + 8)=
45
1s + 3
− 45
1s + 8
x(t) =45
(e−3t − e−8t
)
c)8s + 5
2s2 + 20s + 48=
12
8s + 5(s + 4)(s + 6)
= −274
1s + 4
+434
1s + 6
x(t) = −274
e−4t +434
e−6t
d) The roots are s = −4 ± 10j.
4s + 13s2 + 8s + 116
+4s + 13
(s + 4)2 + 102= C1
10(s + 4)2 + 102
+ C2s + 4
(s + 4)2 + 102
= − 310
10(s + 4)2 + 102
+ 4s + 4
(s + 4)2 + 102
x(t) = − 310
e−4t sin 10t + 4e−4t cos 10t
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3.22 a)3s + 2
s2(s + 10)=
15
1s2
− 150
1s
+150
1s + 10
x(t) =15t − 1
50
(1− e−10t
)
b)5
(s + 4)2(s + 1)= 1.25
1(s + 4)2
− 0.31251
s + 4+ 0.3125
1s + 1
x(t) = 1.25te−4t − 0.3125e−4t + 0.3125e−t
c)s2 + 3s + 5s3(s + 2)
=54
1s3
+716
1s2
+964
1s− 9
641
s + 4
x(t) =58t2 +
716
t +964
− 964
e−4t
d)s3 + s + 6s4(s + 2)
= 31s4
− 1s3
+12
1s2
+14
1s− 1
41
s + 2
x(t) =12t3 − 1
2t2 +
12t +
14− 1
4e−2t
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3.23 a)
5[sX(s)− 2] + 3X(s) =10s
+2s3
X(s) =10s3 + 10s2 + 2
5s3(s + 3)=
2s3 + 2s2 + 2/5s3(s + 3/5)
=23
1s3
− 109
1s2
+1409
1s− 86
271
s + 3/5
x(t) =13t2 − 10
9t +
14027
− 8627
e−3t/5
b)
4[sX(s)− 5] + 7X(s) =6
(s + 5)2+
1s + 3
X(s) =14
20s3 + 261s2 + 1116s + 1525(s + 5)2(s + 7/4)(s + 3)
=14
[6013
1(s + 5)2
− 474169
1s + 5
+16904845
1s + 7/4
+145
1s + 3
]
x(t) = 1.1538te−5t − 0.7012e−5t + 5.0012e−7t/4 + 0.7e−3t
c) This simple-looking problem actually requires quite a lot of algebra to find the solu-tion, and thus it serves as a good motivating example of the convenience of using MATLAB.The algebraic complexity is due to a pair of repeated complex roots.
First obtain the transform of the forcing function. Let f(t) = te−3t sin 5t. From Prop-erty 8,
F (s) = −dY (s)ds
where y(t) = e−3t sin 5t. Thus
Y (s) =5
(s + 3)2 + 52=
5s2 + 6s + 34
dY (s)ds
= − 10s + 30(s2 + 6s + 34)2
ThusF (s) =
10s + 30(s2 + 6s + 34)2
(1)
(continued on the next page)
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Problem 3.23 continued:
Using the same technique, we find that the transform of te−3t cos 5t is
2s2 + 12s + 18(s2 + 6s + 34)2
− 1s2 + 6s + 34
(2)
This fact will be useful in finding the forced response.From the differential equation,
4[s2X(s)− 10s + 2] + 3X(s) = F (s) =10s + 30
(s2 + 6s + 34)2
Solve for X(s).
X(s) =40s − 84s2 + 3
+10s + 30
[(s + 3)2 + 25]2(4s2 + 3)
The free response is given by the first fraction, and is
xfree(t) = − 4√3
sin√
32
t + 10 cos√
32
t (3)
The forced response is given by the second fraction, which can be expressed as
2.5s + 7.5[(s + 3)2 + 25]2(s2 + 3/4)
(4)
The roots of this are s = ±j√
3/2 and the repeated pair s = −3 ± 5j. Thus, referring to(1), (2), and (3), we see that the form of the forced response will be
xforced(t) = C1te−3t sin 5t + C2te
−3t cos 5t
+ C3e−3t sin 5t + C4e
−3t cos 5t
+ C5 sin√
32
t + C6 cos√
32
t (5)
(continued on the next page)
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
Problem 3.23 continued:
The forced response can be obtained several ways. 1) You can substitute the form (5)into the differential equation and use the initial conditions to obtain equations for the Ci
coefficients. 2) You can use (1) and (2) to create a partial fraction expansion of (4) in termsof the complex factors. 3) You can perform an expansion in terms of the six roots, of theform
A1
(s + 3 + 5j)2+
A2
s + 3 + 5j+
A3
(s + 3 − 5j)2+
A4
s + 3 − 5j
+√
3A5/2s2 + 3/4
+A6s
s2 + 3/4
4) You can use the MATLAB residue function.The solution for the forced response is
xforced(t) = −0.0034te−3t sin 5t + 0.0066te−3t cos 5t
− 0.0026e−3t sin 5t + 2.308× 10−4e−3t cos 5t
+ 0.00796 sin 0.866t− 2.308× 10−4 cos 0.866t
The initial condition x(0) = 0 is not exactly satisfied by this expression because of thelimited number of digits used to display it.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
3.24 The denominator roots are s = −3 and s = −5, which are distinct. Factor thedenominator so that the highest coefficients of s in each factor are unity:
X(s) =7s + 4
2s2 + 16s + 30=
12
[7s + 4
(s + 3)(s + 5)
]
The partial-fraction expansion has the form
X(s) =12
[7s + 4
(s + 3)(s + 5)
]=
C1
s + 3+
C2
s + 5
Using the coefficient formula (3.5-4), we obtain
C1 = lims→−3
[(s + 3)
7s + 42(s + 3)(s + 5)
]= lim
s→−3
[7s + 4
2(s + 5)
]= −17
4
C2 = lims→−5
[(s + 5)
7s + 42(s + 3)(s + 5)
]= lim
s→−5
[7s + 4
2(s + 3)
]=
314
Using the LCD method we have
12
7s + 4(s + 3)(s + 5)
=C1
s + 3+
C2
s + 5=
C1(s + 5) + C2(s + 3)(s + 3)(s + 5)
=(C1 + C2)s + 5C1 + 3C2
(s + 3)(s + 5)
Comparing numerators, we see that C1 + C2 = 7/2 and 5C1 + 3C2 = 4/2 = 2, which giveC1 = −17/4 and C2 = 31/4.
The inverse transform is
x(t) = C1e−3t + C2e
−5t = −174
e−3t +314
e−5t
In this example the LCD method requires more algebra, including the solution of twoequations for the two unknowns C1 and C2.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
3.25 a)7[sX(s)− 3] + 5X(s) = 4
X(s) =25
7s + 5=
25/7s + 5/7
x(t) =257
e−5t/7
Note that this gives x(0+) = 25/7. From the initial value theorem
x(0+) = lims→∞
s25/7
s + 5/7)=
257
which is not the same as x(0).b)
(3s2 + 30s + 63)X(s) = 5
X(s) =5
3s2 + 30s + 63=
5/3s2 + 10s + 21
=512
1s + 3
− 512
1s + 7
x(t) =512
(e−3t − e−7t
)
From the initial value theorem
x(0+) = lims→∞
s5/3
s2 + 10s + 21= 0
which is the same as x(0). Also
x(0+) = lims→∞
s2 5/3s2 + 10s + 21
=53
which is not the same as x(0).
(continued on the next page)
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
Problem 3.25 continued:
c)s2X(s)− 2s − 3 + 14[sX(s)− 2] + 49X(s) = 3
X(s) =2s + 34
s2 + 14s + 49= 20
1(s + 7)2
+ 21
s + 7
x(t) = 20te−7t + 2e−7t
From the initial value theorem
x(0+) = lims→∞
s2s + 35
s2 + 14s + 49= 2
which is the same as x(0). However, the initial value theorem is invalid for computingx(0+) and gives an undefined result because the orders of the numerator and denominatorof sX(s) are equal.
d)s2X(s)− 4s − 7 + 14[sX(s)− 4] + 58X(s) = 4
X(s) =4s + 67
s2 + 14s + 58=
4s + 67(s + 7)2 + 32
= 133
(s + 7)2 + 32+ 4
s + 7(s + 7)2 + 32
x(t) = 13e−7t sin 3t + 4e−7t cos 3t
From the initial value theorem
x(0+) = lims→∞
s4s + 67
s2 + 14s + 58= 4
which is the same as x(0). However, the initial value theorem is invalid for computing x(0+)and gives an undefined result because the order of the numerator of sX(s) is greater thanthe denominator.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
3.26 a)
7[sX(s)− 3] + 5X(s) = 4s1s
= 4
X(s) =25
7s + 5=
25/7s + 5/7
x(t) =257
e−5t/7
From the initial value theorem
x(0+) = lims→∞
s25/7
s + 5/7=
257
b)
7[sX(s)− 3] + 5X(s) = 4s1s
+6s
X(s) =25s + 6
s(7s + 5)=
17
25s + 6s(s + 5/7)
=65
1s
+8335
1s + 5/7
x(t) =65
+8335
e−5t/7
which gives x(0+) = 25/7. However, the initial value theorem is invalid for computingx(0+) and gives an undefined result because the orders of the numerator and denominatorof X(s) are equal (see the bottom of page 146).
c)
3[s2X(s)− 2s − 3] + 30[sX(s)− 2] + 63X(s) = 4s1s
= 4
X(s) =13
6s + 73(s + 3)(s + 7)
=5512
1s + 3
− 3112
1s + 7
x(t) =5512
e−3t − 3112
e−7t
From the initial value theorem
x(0+) = lims→∞
s13
6s + 73(s + 3)(s + 7)
= 2
which is the same as x(0). However, the initial value theorem is invalid for computing x(0+)and gives an undefined result because the order of the numerator of sX(s) is greater thanthe denominator (see the bottom of page 146).
(continued on the next page)
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
Problem 3.26 continued:
d)
3[s2X(s)− 4s − 7] + 30[sX(s)− 4] + 63X(s) = 4s1s
+6s
X(s) =13
12s2 + 145s + 6s(s2 + 10s + 21)
= 0.09521s
+ 8.91671
s + 3− 5.0119
1s + 7
x(t) = 0.0952 + 8.9167e−3t − 5.0119e−7t
The initial value theorem gives x(0+) = 4 but is invalid for computing x(0+) because theorders of the numerator and denominator of sX(s) are equal (see the bottom of page 146).
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
3.27 Transform each equation.
3[sX(s)− 5] = Y (s)
sY (s) − 10 = 4− 3Y (s) − 15X(s)
Solve for X(s) and Y (s).
X(s) =15s + 59
3s2 + 9s + 15=
13
15s + 59s2 + 3s + 5
Y (s) =42s − 225
3s2 + 9s + 15=
14s − 75s2 + 3s + 5
The denominator roots are s = −1.5± 1.658j. Thus
X(s) =13
[C1
1.658(s + 1.5)2 + 2.75
+ C2s + 1.5
(s + 1.5)2 + 2.75
]
= 7.33681.658
(s + 1.5)2 + 2.75+ 5
s + 1.5(s + 1.5)2 + 2.75
andx(t) = 7.3368e−1.5t sin 1.658t + 5e−1.5t cos 1.658t
Also,
Y (s) = C11.658
(s + 1.5)2 + 2.75+ C2
s + 1.5(s + 1.5)2 + 2.75
= −57.9011.658
(s + 1.5)2 + 2.75+ 14
s + 1.5(s + 1.5)2 + 2.75
andy(t) = −57.901e−1.5t sin 1.658t + 14e−1.5t cos 1.658t
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3.28 Transform each equation.
sX(s) − 5 = −2X(s) + 5Y (s)
sY (s) − 2 = −6Y (s) − 4X(s) +10s
Solve for X(s) and Y (s).
X(s) =5s2 + 40s + 50s3 + 8s2 + 32s
Y (s) =2s2 − 6s + 20s3 + 8s2 + 32s
The denominator roots are s = 0 and s = −4 ± 4j. Thus
X(s) =C1
s+ C2
4(s + 4)2 + 42
+ C3s + 4
(s + 4)2 + 42
=2516s
+5516
4(s + 4)2 + 42
+5516
s + 4(s + 4)2 + 42
x(t) =2516
+5516
e−4t sin 4t +5516
e−4t cos 4t
Also,
Y (s) =C1
s+ C2
4(s + 4)2 + 42
+ C3s + 4
(s + 4)2 + 42
=58s
− 338
4(s + 4)2 + 42
+118
s + 4(s + 4)2 + 42
y(t) =58− 33
8e−4t sin 4t +
118
e−4t cos 4t
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3.29 Transforming both sides of the equation we obtain
s2Y (s) − sy(0)− y(0) + Y (s) =1
s + 1
which gives
Y (s) =(s + 1) [sy(0) + y(0)] + 1
(s + 1)(s2 + 1)=
s2y(0) + [y(0) + y(0)] + y(0) + 1(s + 1)(s2 + 1)
This can be expanded as follows.
Y (s) = C11
s + 1+ C2
1s2 + 1
+ C3s
s2 + 1
We find the coefficients following the usual procedure and obtain C1 = 1/2, C2 = y(0)+1/2,and C3 = y(0)− 1/2. Thus the solution is
y(t) =12e−t +
[y(0) +
12
]sin t +
[y(0)− 1
2
]cos t
Because the initial values can be arbitrary, the general form of the solution is
y(t) =12e−t + A1 sin t + A2 cos t (1)
This form can be used to obtain a solution for cases where y(t) or y(t) are specified at pointsother than t = 0. For example, suppose we are given that y(0) = 5/2 and y(π/2) = 3. Thenevaluation of equation (1) at t = 0 and at t = π/2 gives
y(0) =12
+ A2 =52
y
(π
2
)=
12e−π/2 + A1 = 3
The solution of these two equations is A1 = 3 − e−π/2/2 = 2.896 and A2 = 2, and thesolution of the differential equation is
y(t) =12e−t + 2.896 sin t + 2 cos t
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3.30 For zero initial conditions, the transform gives
(s2 + 4)X(s) =3s2
orX(s) =
3s2(s2 + 4)
=C1
s2+
C2
s+ C3
2s2 + 4
+ C4s
s2 + 4
The solution form is thus
x(t) = C1t + C2 + C3 sin 2t + C4 cos 2t
This form satisfies the differential equation if C1 = 3/4 and C2 = 0. From x(0) = 10, weobtain C4 = 10. From x(5) = 30, we obtain C3 = −63.675. Thus
x(t) =34t − 63.675 sin 2t + 10 cos 2t
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3.31 The denominator roots are s = −3 ± 5j and s = ±6j. Thus we can express X(s) asfollows.
X(s) =30
[(s + 3)2 + 52] (s2 + 62)
which can be expressed as the sum of terms that are proportional to entries 8 through 11in Table 3.3-1.
X(s) = C15
(s + 3)2 + 52+ C2
s + 3(s + 3)2 + 52
+ C36
s2 + 62+ C4
s
s2 + 62(1)
We can obtain the coefficients by noting that X(s) can be written as
X(s) =5C1(s2 + 62) + C2(s + 3)(s2 + 62) + 6C3
[(s + 3)2 + 52
]+ C4s
[(s + 3)2 + 52
]
[(s + 3)2 + 52] (s2 + 62)(2)
Comparing the numerators of equations (1) and (2), and collecting powers of s, we see that
(C2 + C4)s3 + (5C1 + 3C2 + 6C3 + 6C4)s2 + (36C2 + 36C3 + 34C4)s
+180C1 + 108C2 + 204C3 = 30
orC2 + C4 = 0 5C1 + 3C2 + 6C3 + 6C4 = 0
36C2 + 36C3 + 34C4 = 0 180C1 + 108C2 + 204C3 = 30
These are four equations in four unknowns. Note that the first equation gives C4 = −C2.Thus we can easily eliminate C4 from the equations and obtain a set of three equations inthree unknowns. The solution is C1 = 6/65, C2 = 9/65, and C3 = −1/130, and C4 = −9/65.
The inverse transform is
x(t) = C1e−3t sin 5t + C2e
−3t cos 5t + C3 sin 6t + C2 cos 6t
=665
e−3t sin 5t +965
e−3t cos 5t − 1130
sin 6t − 965
cos 6t
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3.32 Transform the equation.
(s2 + 12s + 40)X(s) = 35
s2 + 25
The characteristic roots are s = −6 ± 2j. Thus
X(s) =15
(s2 + 25)(s2 + 12s + 40)
= C15
s2 + 25+ C2
s
s2 + 25+ C3
2(s + 6)2 + 4
+ C4s + 6
(s + 6)2 + 4
orX(s) =
185
5s2 + 25
− 485
s
s2 + 25+
19170
2(s + 6)2 + 4
+485
s + 6(s + 6)2 + 4
Thusx(t) =
185
sin 5t − 485
cos 5t +19170
e−6t sin 2t +485
e−6t cos 2t
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3.33 From Example 3.7.1, the form A sin(ωt + φ) has the transform
As sin φ + ω cos φ
s2 + ω2
For this problem, ω = 5. Comparing numerators gives
s sin φ + 5 cos φ = 4s + 9
ThusA sin φ = 4 5A cosφ = 9
With A > 0, φ is seen to be in the first quadrant.
φ = tan−1 sin φ
cos φ= tan−1 4/A
9/5A= tan−1 20
9= 1.148 rad
Because sin2 φ + cos2 φ = 1, (4A
)2
+(
95A
)2
= 1
which gives A = 4.386. Thus
x(t) = 4.386 sin(5t + 1.148)
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3.34 Taking the transform of both sides of the equation and noting that both initial con-ditions are zero, we obtain
s2X(s) + 6sX(s) + 34X(s) = 56
s2 + 62
Solve for X(s).
X(s) =30
(s2 + 6s + 34)(s2 + 62)
The inverse transform was obtained in Problem 3.31. It is
x(t) =665
e−3t sin 5t +965
e−3t cos 5t − 1130
sin 6t − 965
cos 6t
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3.35 Transform the equation.
(s2 + 12s + 40)X(s) =10s
or, since the characteristic roots are s = −6 ± 2j,
X(s) =10
s[(s + 6)2 + 22](1)
From Example 3.7.1(b), the form Ae−at sin(ωt + φ) has the transform
As sin φ + a sin φ + ω cos φ
(s + a)2 + ω2
For this problem, a = 6 and ω = 2. Thus
X(s) =10
s[(s + 6)2 + 22]=
C1
s+ C2
s sin φ + 6 sin φ + 2 cos φ
(s + 6)2 + 22
or
X(s) =C1(s2 + 12s + 40) + C2s
2 sin φ + 6C2s sin φ + 2C2s cos φ
s[(s + 6)2 + 22](2)
Collecting terms and comparing the numerators of equations (1) and (2), we have
(C1 + C2 sinφ)s2 + (12C1 + 6C2 sin φ + 2C2 cos φ)s + 40C1 = 10
Thus comparing terms, we see that C1 = 1/4 and
14
+ C2 sin φ = 0
3 + 6C2 sin φ + 2C2 cos φ = 0
SoC2 sinφ = −1
4C2 cos φ = −3
4
(continued on the next page)
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Problem 3.35 continued:
Thus φ is in the third quadrant and
φ = tan−1 −1/4−3/4
= 0.322 + π = 3.463 rad
Because sin2 φ + cos2 φ = 1, (1
4C2
)2
+(
34C2
)2
= 1
which gives C2 = 0.791. Thus
x(t) =14
+ 0.791e−6t sin(2t + 3.463)
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3.36 Transform the equation.
X(s) =F (s)
s2 + 8s + 1Thus
F (s) − X(s) = F (s) − F (s)s2 + 8s + 1
=s2 + 8s
s2 + 8s + 1F (s)
Because F (s) = 6/s2,
F (s) − X(s) =s2 + 8s
s2 + 8s + 16s2
=s + 8
s2 + 8s + 16s
From the final value theorem,
fss − xss = lims→0
s[F (s) − X(s)] = lims→0
ss + 8
s2 + 8s + 16s
= 8
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3.37 The roots are s = −2 and −4. Thus
X(s) =1 − e−3s
(s + 2)(s + 4)
LetF (s) =
1(s + 2)(s + 4)
=12
(1
s + 2− 1
s + 4
)
sof(t) =
12
(e−2t − e−4t
)
From Property 6 of the Laplace transform,
x(t) =12
(e−2t − e−4t
)− 1
2
[e−2(t−3) − e−4(t−3)
]us(t − 3)
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3.38f(t) =
C
Dtus(t) −
2C
D(t − D)us(t − D) +
C
D(t − 2D)us(t − 2D)
From Property 6 of the Laplace transform,
F (s) =C
Ds2− 2C
Ds2e−Ds +
C
Ds2e−2Ds =
C
Ds2
(1− 2e−Ds + e−2Ds
)
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3.39f(t) =
C
Dtus(t) −
C
D(t − D)us(t − D)− Cus(t − D)
From Property 6 of the Laplace transform,
F (s) =C
Ds2− C
Ds2e−Ds − C
se−Ds
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3.40f(t) = Mus(t) − 2Mus(t − T ) + Mus(t − 2T )
From Property 6,
F (s) =M
s− 2M
se−Ts +
M
se−2Ts
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3.41P (t) = 3us(t) − 3us(t − 5)
From Property 6,
P (s) =3s− 3
se−5s
X(s) =P (s)4s + 1
=3(1 − e−5s
)
s(4s + 1)=
34
1 − e−5s
s(s + 1/4)
LetF (s) =
34
1s(s + 1/4)
= 3(
1s− 1
s + 1/4
)
Thenf(t) = 3
(1 − e−t/4
)
SinceX(s) = F (s)
(1 − e−5s
)
we have
x(t) = f(t) − f(t − 5)us(t − 5) = 3(1 − e−t/4
)− 3
[1 − e−(t−5)/4
]us(t − 5)
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3.42 Let
f(t) = t +t3
3+
2t5
15Then
F (s) =1s2
+2s4
+16s6
=s4 + 2s2 + 16
s6
From the differential equation,
X(s) =F (s)s + 1
=s4 + 2s2 + 16
s6(s + 1)
=16s6
− 16s5
+18s4
− 18s3
+19s2
− 19s
+19
s + 1
Thusx(t) =
215
t5 − 23t4 + 3t3 − 9t2 + 19t − 19 + 19e−t
On a plot of this and the solution obtained from the lower-order approximation, the twosolutions are practically indistinguishable.
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3.43 From the derivative property of the Laplace transform, we know that
L[x(t)] =∫ ∞
0x(t)e−st dt = sX(s)− x(0)
Thereforelims→∞
[sX(s)] = lims→∞
[x(0) +
∫ ∞
0x(t)e−st dt
]
= lims→∞
x(0) + lims→∞
lim
ε→0+
[∫ ε
0x(t)e−st dt
]+ lim
ε→0+
∫ ε
0lim
s→∞
[x(t)e−st dt
]
The limits on ε and s can be interchanged because s is independent of t. Within the interval[0, 0+], e−st = 1, and so
lims→∞
[sX(s)] = x(0) + lims→∞
lim
ε→0+
[∫ ε
0x(t) dt
]+ lim
ε→0+
∫ ε
0lims→∞
[x(t)e−st dt
]
= x(0) + x(t)|t=0+t=0 + 0 = x(0+)
This proves the theorem.
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3.44 From the derivative property of the Laplace transform, we know that
L[x(t)] =∫ ∞
0x(t)e−st dt = sX(s)− x(0)
Therefore,
lims→0
[sX(s)] = lims→0
x(0) + lims→0
[∫ ∞
0x(t)e−st dt
]
= x(0) +∫ ∞
0lims→0
[x(t)e−st dt
]= x(0) +
∫ ∞
0x(t) dt
because s is independent of t and lims→0 e−st = 1. Thus
lims→0
[sX(s)] = x(0) + limT→∞
[∫ T
0x(t) dt
]= x(0) + lim
T→∞
[x(t)|t=T
t=0
]
= x(0) + limT→∞
x(T )− x(0) = limT→∞
x(T ) = limt→∞
x(t)
This proves the theorem.
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3.45 Letg(t) =
∫ t
0x(t) dt
ThenL[∫ t
0x(t) dt
]= L[g(t)] =
∫ t
0g(t)e−st dt
To use integration by parts we define u = g and dv = e−stdt, which give du = dg = x(t) dtand v = −e−st/s. Thus
∫ t
0g(t)e−st dt =
g(t)e−st
−s
∣∣∣∣∣
t=∞
t=0
−∫ ∞
0
e−st
−sx(t) dt
= 0 +g(0)s
+1s
∫ ∞
0x(t)e−st dt =
g(0)s
+X(s)
s
=1s
∫x(t) dt
∣∣∣∣t=0
+X(s)
s
This proves the property.If there is an impulse in x(t) at t = 0, then g(0) equals the strength of the impulse. If
there is no impulse at t = 0, then g(0) = 0.
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3.46 a)
[r,p,k] = residue([8,5],[2,20,48])
The result is r = [10.7500, -6.7500], p = [-6.0000, -4.0000], and k = [ ]. Thesolution is
x(t) = 10.75e−6t − 6.75e−4t
b)
[r,p,k] = residue([4,13],[2,8,116])
The result is r = [1.0000 - 0.1701i, 1.0000 + 0.1701i], p = [-2.0000 + 7.3485i,-2.0000 - 7.3485i], and k = [ ]. The solution is
x(t) = (1 − 0.1701j)e(−2+7.3485j)t + (1 + 0.1701j)e(−2−7.3485j)t
From (3.8.4) and (3.8.5), the solution is
x(t) = 2e−2t (cos 7.3485t + 0.1701 sin 7.3485t)
c)
[r,p,k] = residue([3,2],[1,10,0,0])
The result is r = [ -0.2800, 0.2800, 0.2000], p = [-10, 0, 0], and k = [ ]. Thesolution is
x(t) = −0.28e−10t + 0.28 + 0.2t
d)
[r,p,k] = residue([1,0,1,6],[1,2,0,0,0,0])
The result is r = [-0.2500, 0.2500, 0.5000, -1.0000, 3.0000], p =[ -2, 0, 0, 0,0], and k = [ ]. The solution is
x(t) = −0.25e−2t + 0.25 + 0.5t − 12t2 +
12t3
(continued on the next page)
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Problem 3.46 continued:
e)
[r,p,k] = residue([4,3],[1,6,34,0])
The result is r = [-0.0441 - 0.3735i, -0.0441 + 0.3735i, 0.0882], p = [-3.0000+ 5.0000i, -3.0000 - 5.0000i, 0], and k = [ ].The solution is
x(t) = (−0.0441− 0.3735j)e(−3+5j)t + (−0.0441 + 0.3735j)e(−3−5j)t + 0.0882
From (3.8.4) and (3.8.5), the solution is
x(t) = 2e−3t (−0.0441 cos 5t + 0.3735 sin 5t) + 0.0882
f)
[r,p,k] = residue([5,3,7],[1,12,144,48])
The result is r = [2.4759 + 1.3655i, 2.4759 - 1.3655i, 0.0482], p = -5.8286 +10.2971i,-5.8286 -10.2971i, -0.3428], and k = [ ]. The solution is
x(t) = (2.4759+1.3655j)e(−5.8286+10.2971j)t+(2.4759−1.3655j)e(−5.8286−10.2971j)t+0.0482e−0.3428t
From (3.8.4) and (3.8.5), the solution is
x(t) = 2e−5.8286t (2.4759 cos 10.29t− 1.3655 sin 10.29t) + 0.0482e−0.3428t
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3.47 a)
[r,p,k] = residue(5,conv([1,8,16],[1,1]))
The result is r = [-0.5556, -1.6667, 0.5556], p = [-4.0000, -4.0000, -1.0000], k= [ ]. The solution is
x(t) = −0.5556e−4t − 1.6667te−4t + 0.5556e−t
b)
[r,p,k] = residue([4,9],conv([1,6,34],[1,4,20]))
The result is r = [-0.1159 + 0.1073i, -0.1159 - 0.1073i, 0.1159 - 0.1052i, 0.1159+ 0.1052i], p = -3.0000 + 5.0000i, -3.0000 - 5.0000i, -2.0000 + 4.0000i, -2.0000- 4.0000i], and k = [ ]. The solution is
x(t) = (−0.1159 + 0.1073j)e(−3+5j)t + (−0.1159− 0.1073j)e(−3−5j)t
+ (0.1159− 0.1052j)e(−2+4j)t + (0.1159 + 0.1052j)e(−2−4j)t
From (3.8.4) and (3.8.5), the solution is
x(t) = 2e−3t (−0.1159 cos 5t − 0.1073 sin 5t) + 2e−2t (0.1159 cos 4t + 0.1052 sin 4t)
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3.48 a)
sys = tf(1,[3,21,30]);step(sys)
b)
sys = tf(1,[5,20, 65]);step(sys)
c)
sys = tf([3,2],[4,32,60]);step(sys)
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3.49 a)
sys = tf(1,[3,21,30]);impulse(sys)
b)
sys = tf(1,[5,20, 65]);impulse(sys)
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3.50
sys = tf(5,[3,21,30]);impulse(sys)
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3.51
sys = tf(5,[3,21,30]);step(sys)
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3.52 a)
sys = tf(1,[3,21,30]);t = [0:0.001:1.5];f = 5*t;[x,t] = lsim(sys,f,t);plot(t,x)
b)
sys = tf(1,[5,20,65]);t = [0:0.001:1.5];f = 5*t;[x,t] = lsim(sys,f,t);plot(t,x)
c)
sys = tf([3,2],[4,32,60]);t = [0:0.001:1.5];f = 5*t;[x,t] = lsim(sys,f,t);plot(t,x)
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3.53 a)
sys = tf(1,[3,21,30]);t = [0:0.001:6];f = 6*cos(3*t);[x,t] = lsim(sys,f,t);plot(t,x)
b)
sys = tf(1,[5,20,65]);t = [0:0.001:6];f = 6*cos(3*t);[x,t] = lsim(sys,f,t);plot(t,x)
c)
sys = tf([3,2],[4,32,60]);t = [0:0.001:6];f = 6*cos(3*t);[x,t] = lsim(sys,f,t);plot(t,x)
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Solutions Manual c©
to accompany
System Dynamics, Second Edition
by
William J. Palm III
University of Rhode Island
Solutions to Problems in Chapter Four
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4.1 For a helical coil spring
k =Gd4
64nR3= 1.7 × 109 (1/24)4
64(6)(2/12)3= 2.88× 103 lb/ft
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4.2 Let D be the length of the dead space to the left of spring k2. The plot follows.
Figure : for Problem 4.2
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4.3 a) Series.b) Let kb be the stiffness of the beam and kc be the stiffness of the cable.
ke =kbkc
kb + kc
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4.4 Sum moments about the pivot.
0 = fL1 − k1(L2θ)L2 − k2(L3θ)L3
Thus
f =k1L
22 + k2L
23
L1θ
But x = L1θ, so
f =k1L
22 + k2L
23
L21
x = kex
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4.5 Series combination.ke =
k1k2
k1 + k2
where
ki =EAi
Li=
E
Liπ
(Di
2
)2
=EπD2
i
4Li
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4.6
ke = 4(
EA
L
)=
4Eπ(d/2)2
L=
4(2× 1011)π(0.03)2
4(1)= 1.8π × 108 N/m
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4.7 The beam stiffness is
kb =4Ewh3
L3=
4(3× 107)(144)(1)(1/12)3
(3)3= 3.7037× 105 lb/ft
a) The springs are in parallel, so ke = kb + k. We want ke = 2kb, so we must requirethat k = kb.
b)
ωn =
√ke
m=
√2(3.7037× 105)
40= 136.1 rad/sec
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4.8 First reduce the system to the equivalent one shown in part (a) of the figure, where
1k1
=12k
+1k
=32k
Thus k1 = 2k/3.From part (b) of the figure,
1ke
=1k
+1
k1 + k=
k1 + 2k
k(k1 + k)
Thus, solving for ke and substituting for k1, we obtain
ke =k(k1 + k)k1 + 2k
=5k
8
Figure : for Problem 4.8
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4.9 The stiffnesses are in series. Thus
kTe =kT1kT2
kT1 + kT2
where
kT1 =πG(D4 − d4)
32L=
π(8× 1010)(0.44 − 0.34)32(2)
= 6.874× 107
kT2 =πG(D4 − d4)
32L=
π(8× 1010)(0.354 − 0.254)32(3)
= 2.906× 107
ThuskTe = 2.043× 107 N ·m/rad
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4.10 The plot is the following.
Figure : for Problem 4.10
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4.11 Since x1 and x2 are measured from the equilibrium positions, the gravity forces arecanceled by the static spring forces in parts (a) and (b). Thus the equations of motion arethe same for all three cases. They are
m1x1 = −k1x1 + k2(x2 − x1)
m2x2 = f − k2(x2 − x1)
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4.12 Sum moments about the pivot to obtain
mL23θ = −(k1L1θ)L1 + k2(x − L2θ)L2 − mgL3θ
Collecting terms we obtain
mL23θ + (k1L
21 + k2L
22 + mgL3)θ = k2L2x
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4.13 Let T be the tension in the cable. See the figure. Since the pulley is consideredmassless,
2T =kx
2For the mass m,
mx = f − T + mg = f − kx
4+ mg
Thusmx +
kx
4= f + mg
Note that the static spring force does not cancel the weight mg because x is not measuredfrom the equilibrium position.
Figure : for Problem 4.13
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4.14 a) The buoyancy force is
B = (ρV )g = ρgπ
(D
2
)2
x
or
B =πρgD2
4x
b) Summing forces in the vertical direction,
mx = −B
or
mx = −πρgD2
4x
Thus, the natural frequency is
ωn =
√πρgD2
4m=
D
2
√πρg
m
c) The period is
P =2π
ωn=
4π
D
√m
πρg=
4π
gD
√W
πρ
or
P =4π
32.2(2)
√10001.94π
= 2.5 sec
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4.15 a) Sum moments about the mass center G to obtain
Iθ = −(B sin θ)h
Because B = mg = W , the ship’s weight,
Iθ + Wh sin θ = 0
where I is the ship’s moment of inertia about G.b) For small angles, sin θ ≈ θ, and the equation becomes
Iθ + Whθ = 0
The characteristic roots are s = ±j√
Wh/I , and the roll period is given by 2π√
I/Wh.
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4.16 Sum moments about the rotation axis.
Iθ = k1(φ − θ) − k2θ
Collecting terms we obtainIθ + (k1 + k2)θ = k1φ
The transfer function isΘ(s)Φ(s)
=k1
Is2 + k1 + k2
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4.17 a) Refer to the following figure. Summing forces in the x direction gives
mx = k(y − x)− ft (1)
Summing moments about the mass center of the wheel gives Iθ = Rft. But x = Rθ, andthus θ = x/R. Therefore
ft =I
Rθ =
I
R2x (2)
Combine (1) and (2):
mx + kx = ky − I
R2x
or (m +
I
R2
)x + kx = ky
where I = mR2/2. Thus1.5mx + kx = ky (3)
Figure : for Problem 4.17
(continued on the next page)
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Problem 4.17 continued:
b) Substituting the values into (3) gives
15x + 1000x = 1000y
or3x + 200x = 200y
The roots are s = ±j10√
2/3. The response to a unit-step input is
x(t) = 1 − cos 10√
23t
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4.18 Summing forces in the horizontal direction on each mass gives
m1x1 = −k1x1 + k2(x2 − x1)
m2x2 = −k3x2 − k2(x2 − x1)
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4.19 Summing moments about the axis of rotation of each inertia gives
I1θ1 = −k1θ1 − k2(θ1 − θ2)
I2θ2 = T2 + k2(θ1 − θ2) − k3θ2
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4.20 Summing moments about each pivot point and assuming small angles, we obtain
m1L22θ1 = −m1gL2θ1 − kL1(θ1 − θ2)L1
m2L22θ2 = −m2gL2θ2 + kL1(θ1 − θ2)L1
Collecting terms givesm1L
22θ1 + (m1gL2 + kL2
1)θ1 = kL21θ2
m2L22θ2 + (m2gL2 + kL2
1)θ2 = kL21θ1
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4.21 The torsional stiffness of the torsion bar is
kT =πGD4
32L=
1.7 × 109π(1.5/12)4
32(4)= 104 lb − ft/rad
The characteristic equation isIs2 + kT = 0
where I = mL2 = (40/32.2)(2)2 = 4.9689. Thus the natural frequency is
ωn =
√kT
I= 44.86 rad/sec
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4.22 The kinetic energy of the system is
KE =12mx2 +
12Iω2 =
12mx2 +
12I
(x
R
)2
=12(1.5m)x2
since x = Rω and I = mR2/2. The potential energy is
PE =12(k1 + k2)x2
From conservation of mechanical energy,
KE + PE =12(1.5m)x2 +
12(k1 + k2)x2 = constant
Differentiating with respect to time gives
1.5mxx + (k1 + k2)xx = 0
Canceling x gives the equation of motion.
1.5mx + (k1 + k2)x = 0
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4.23 Because x = 0 at equilibrium, the static spring force cancels the constant weight m2g.Taking the potential energy to be zero at x = 0, we obtain
PE =12kx2
The kinetic energy of the system is
KE =12m1x
2 +12m2y
2 =12(m1 + 4m2)x2
since y = 2x.From conservation of energy,
KE + PE = constant
or12(m1 + 4m2)x2 +
12kx2 = constant
Differentiating with respect to time and setting the derivative to 0, we obtain
(m1 + 4m2)xx + kxx = 0
Canceling x gives the equation of motion.
(m1 + 4m2)x + kx = 0
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4.24 Because x = 0 at equilibrium, the static spring force cancels the constant weight m1g.Taking the potential energy to be zero at x = 0, and noting that the spring extension y
from its equilibrium position is y = 2x, we obtain
PE =12k(2x)2 = 2kx2
Note that y = Rω = 2x and the inertia of the cylinder is I = m2R2/2. The kinetic energy
of the system is
KE =12m2y
2 +12Iω2 +
12m1x
2
=12
(4m2 + 4
I
R2+ m1
)x2
=12
(4m2 + 2m2 + m1) x2
=12
(6m2 + m1) x2
From conservation of energy,
KE + PE = constant
or12
(6m2 + m1) x2 + 2kx2 = constant
Differentiating with respect to time and setting the derivative to 0, we obtain
(6m2 + m1)xx + 4kxx = 0
Canceling x gives the equation of motion.
(6m2 + m1)x + 4kx = 0
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4.25 Because x = 0 at equilibrium, the static spring force cancels the constant weight mg.Taking the potential energy to be zero at x = 0, and noting that the spring extension y isy = x/2, we obtain
PE =12ky2 =
18kx2
Note that x = Rω, where ω is the angular velocity of the pulley, and that the inertia of thecylinder is I = mR2/2. The kinetic energy of the system is
KE =12mx2 +
12Iω2
=12mx2 +
12
mR2
2x2
R2
=12
(m +
m
2
)x2
=12(1.5m)x2
Let A = |xmax|. Then |x| = ωnA for simple harmonic motion, and Rayleigh’s principlegives
12(1.5m)ω2
nA2 =18kA2
This gives
ωn =
√k
6m
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4.26 The kinetic energy of the system is
T =12mx2 +
12Iω2 =
12mx2 +
12I
(x
R
)2
=12(1.5m)x2
since x = Rω and I = mR2/2. The potential energy is
V =12kx2
From the Rayleigh principle, Tmax = Vmax − Vmin. For simple harmonic motion, x = ωnA,and this implies that
12(1.5m)ω2
nA2 =12kA2
Thus
ωn =
√k
1.5m
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4.27 The kinetic energy of the system is
T =12mv2 =
12m
L21
L22
x2
since v = L1θ and x = L2θ.The potential energy is
V =12kx2
From the Rayleigh principle, Tmax = Vmax − Vmin. For simple harmonic motion, x = ωnA,and this implies that
12m
L21
L22
ω2nA2 =
12kA2
Thus
ωn =L2
L1
√k
m
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4.28 See the figure. Assuming that θ is small, summing moments about the pivot gives
IOθ = fcL1 − k2L2x = fcL1 − k2L2(L2θ)
The equivalent mass at x due to the valve and spring is
me = mv +13ms
Since IO = Ir + meL22, we have
(Ir + meL22)θ = fcL1 − k2L
22θ
Thus the force on the cam is given by
fc =(Ir + meL
22)θ + k2L
22θ
L1
Knowing the cam profile and the cam rotation speed, we can compute θ(t) and θ(t) to usein the expression for fc.
Figure : for Problem 4.28
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4.29 The equivalent mass is me = 12 + 0.23(3) = 12.69 kg. From statics,
k(0.02) = 12(9.81)
or k = 5886 N/m.
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4.30 The equivalent mass is me = 3 + 0.38(1) = 3.38 slug. From statics,
k(0.01) = 3(32.2)
or k = 9660 lb/ft.
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4.31 The equation given in the problem statement is
xy =P
6EIAy2(3L− y) (1)
The deflection xL at the end of the beam, where y = L, is found from (1) to be:
xL =PL3
3EIA(2)
Comparing (1) and (2) shows that
xy =y2(3L− y)
2L3xL (3)
Differentiate this equation with respect to time for a fixed value of y to obtain
xy =y2(3L− y)
2L3xL (4)
Because (1) describes the static deflection, it does not account for inertia effects, and (4) isnot exactly true. However, lacking another reasonable expression for xy , we will use (4).
The kinetic energy of a beam mass element dm at position y is x2ydm/2. Let ν be the
beam’s mass per unit length. Then dm = ν dy, and the total kinetic energy KE in thebeam is
KE =12
∫ L
0x2
y dm
=ν
8L6x2
L
∫ L
0y4(3L − y)2 dy
=ν
8L6x2
L
33L7
35=
33νL
2(140)x2
L
Because the beam mass mb = νL,
KE =33mb
2(140)x2
L
If a mass me is located at the end of the beam, and moves with a velocity xL, its kineticenergy is mex
2L/2. Comparing this with the beam’s energy, we see that me = (33/140)mb ≈
0.23mb. Thus the equivalent mass of the cantilever spring is 23% of the beam mass.
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4.32 a) See part (a) of the figure. The equivalent mass of the beam and winch is me =0.23mb + mw. Assuming that x1 and x2 are measured from the equilibrium positions, theequation of motion for me is
mex1 = T − kx1 (1)
where T is the tension in the rope and the beam stiffness is k = Ewh3/4L3.For the hoisted mass,
mhx2 = −T (2)
Solve this for T and substitute into (1) to obtain
mex1 + mhx2 = −kx1 (3)
If the rope does not stretch, then x1 = x2 and (3) becomes
(me + mh)x1 = −kx1
Figure : for Problem 4.32
(continued on the next page)
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Problem 4.32 continued:
b) If the rope acts like a spring, then T = kr(x2 − x1), and we obtain the free-bodydiagrams shown in part (b) of the figure. For me,
mex1 = kr(x2 − x1) − kx1
For mh,mhx2 = −kr(x2 − x1)
These are the equations of motion.
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4.33 a) Summing forces parallel to the plane, we obtain
mv = mg sin 30 − cv
or6v + v = 29.43
b)v(t) = 29.43− 25.43e−t/6
c) The steady-state speed is 29.43 m/s, and it takes approximately 4τ = 4(6) = 24 s toreach that speed.
d) Because the steady-state speed can be found directly from the equation of motion,by setting v = 0, it is independent of the initial speed.
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4.34 a) The characteristic equation is 40s2 + 680s + 1200 = 0, which has the roots s = −2and s = −15. The form of the solution is
x(t) = A1e−2t + A2e
−15t + 6000/1200
For zero initial conditions, A1 = −75/13 and A2 = 10/13, so
x(t) = −7513
e−2t +1013
e−15t + 5
b) The characteristic equation is 40s2 + 400s + 1200 = 0, which has the roots s =−5 ± j
√5. The form of the solution is
x(t) = A1e−5t sin
√5t + A2e
−5t cos√
5t + 6000/1200
For zero initial conditions, A1 = −5√
5 and A2 = −5, so
x(t) = −5√
5e−5t sin√
5t − 5e−5t cos√
5t + 5
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4.35 a) Summing the horizontal forces on each mass gives
m1x1 = −cx1 + k(x2 − x1)
m2x2 = f − k(x2 − x1)
b) Summing the horizontal forces on each mass gives
m1x1 = −kx1 + c(x2 − x1)
m2x2 = f − c(x2 − x1)
c) Summing the horizontal forces on each mass gives
m1x1 = −cx1 − k1x1 + k2(x2 − x1)
m2x2 = f − k2(x2 − x1)
d) Summing the horizontal forces on each mass gives
m1x1 = −cx1 − k1x1 + k2(x2 − x1) + c2(x2 − x1)
m2x1 = f − k2(x2 − x1) − c2(x2 − x1)
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4.36 a) See the diagram. Let F be the tension in the cable. Summing moments about thedrum center gives
Iω = T − cTω − FR (1)
Summing vertical forces on the mass m gives
mv = F − mg (2)
Solve (2) for F , substitute for F in (1), and use the fact that v = Rω to obtain
(I + mR2)v + cTv = TR − mgR2
Figure : for Problem 4.36
(continued on the next page)
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Problem 4.36 continued:
b) Using the given values, we obtain
24v + v = 243.3
The solution has the formv(t) = 243.3 + Ae−t/24
For v(0) = 0, A = −243.3, and
v(t) = 243.3(1 − e−t/24
)
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4.37 Summing moments about the pivot, and assuming small, angles, we obtain
Iθ = fL1 − c(L2θ
)L2 − k (L3θ) L3
orIθ + cL2
2θ + kL23θ = fL1
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4.38 Summing horizontal forces gives
mx = −k2x + k1(y − x) + c(y − x)
Collect terms to obtainmx + cx + (k1 + k2)x = k1y + cy
The transfer function isX(s)Y (s)
=cs + k1
ms2 + cs + k1 + k2
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4.39 Summing moments about the rotation axis of the pulley gives
Ipθp = kT (φ − θp) − cT
(θp − θd
)
Summing moments about the center of mass of the damper gives
Idθd = cT
(θp − θd
)
These two equations form the system model.
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4.40 Summing moments about the pivot, and assuming small, angles, we obtain
IOθ = −mgL3θ − c(L2θ
)L2 + k (z − L1θ) L1
where IO = mL23 + I . Collecting terms gives
(mL23 + I)θ + cL2
2θ + (kL21 + mgL3)θ = kL1z
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4.41 Summing horizontal forces on m gives
mx = f + k(xA − x)
From statics, at point A,0 = −k(xA − x) − cxA
These can be combined into a single equation by using the Laplace transform.
ms2X(s) = F (s) + k [XA(s) − X(s)]
0 = −k [XA(s) − X(s)]) − csXA(s)
These can be solved for the transfer function relating the given output xA to the given inputf .
XA(s)F (s)
=k
s(mcs2 + mks + ck)
This is the required model.
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4.42 Summing horizontal forces on m gives
mx = c(xA − x)− k2x
From statics, at point A,0 = k1(y − xA) − c(xA − x)
These can be combined into a single equation by using the Laplace transform.
ms2X(s) = csXA(s) − csX(s)− k2X(s)
0 = k1 [Y (s) − XA(s)]) − csXA(s) + csX(s)
These can be solved for the transfer function relating the given output x to the given inputy.
X(s)Y (s)
=ck1s
mcs3 + mk1s2 + c(k1 + k2)s + k1k2
This is the required model.
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4.43 See the diagram. Let F be the force on the pinion due to the rack. Summing momentson the pinion gives
(Im + Ip)θ = T − RF (1)
Summing horizontal forces on the rack gives
mrx = F − cx− kx (2)
Solving (2) for F , substituting in (1), and using the fact that x = Rθ, gives
(Im + Ip + mrR2)θ + cR2θ + kR2θ = T
Figure : for Problem 4.43
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4.44 We can represent the system as shown in the diagram, where
I1e = N2I1 T1e = NT1
Summing moments on I1e gives
I1eθ2 = T1e + kT (θ3 − θ2) (1)
Summing moments on I2 gives
I2θ3 = −kT (θ3 − θ2) − cT θ3 (2)
To obtain the model in terms of θ1, substitute θ2 = θ1/N into (1) and (2) to obtain
I1e
Nθ1 = T1e + kT
(θ3 −
θ1
N
)(3)
andI2θ3 = −kT
(θ3 −
θ1
N
)− cT θ3 (4)
The system model consists of (3) and (4).
Figure : for Problem 4.44
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4.45 a) Assume that the equilibrium position corresponds to θ = 0. Let φ = 45 + θ andδ be the static spring deflection at equilibrium. Summing moments about the pivot andassuming small angles, we obtain
mL2θ = (mg sin φ)L − 2k(δ + L1θ)L1
But
sin φ = sin(45 + θ) =√
22
cos θ +√
22
sin θ ≈√
22
+√
22
θ
for small angles. Thus
mL2θ = mgL
(√2
2+
√2
2θ
)− 2kL1δ − 2k1L
21θ
At equilibrium,
mgL
√2
2− 2kL1δ = 0
and thus the equation of motion becomes
mL2θ +
(mgL
√2
2− 2k1L
21
)θ = 0
The characteristic equation is
mL2s2 + mgL
√2
2− 2k1L
21 = 0
This will be neutrally stable if
mgL
√2
2− 2k1L
21 ≥ 0
and unstable if
mgL
√2
2− 2k1L
21 < 0
(continued on the next page)
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Problem 4.44 continued:
b) Assume that the equilibrium position corresponds to θ = 0. Let φ = 135 + θ andδ be the static spring deflection at equilibrium. Summing moments about the pivot andassuming small angles, we obtain
mL2θ = (mg sin φ)L − 2k(δ + L1θ)L1
But
sin φ = sin(135 + θ) =√
22
cos θ −√
22
sin θ ≈√
22
−√
22
θ
for small angles. Thus
mL2θ = mgL
(√2
2−
√2
2θ
)− 2kL1δ − 2k1L
21θ
At equilibrium,
mgL
√2
2− 2kL1δ = 0
and thus the equation of motion becomes
mL2θ +
(mgL
√2
2+ 2k1L
21
)θ = 0
The characteristic equation is
mL2s2 + mgL
√2
2+ 2k1L
21 = 0
This will always be neutrally stable since
mgL
√2
2+ 2k1L
21 > 0
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4.46 Assume that the equilibrium position corresponds to θ = 0. Let φ = 135 + θ and δbe the static spring deflection at equilibrium. Let L be the distance from the pivot to themass m. Let L1 be the distance from the pivot to the connection point of the spring anddamper.
Summing moments about the pivot and assuming small angles, we obtain
mL2θ = (mg sin φ)L− k(δ + L1θ)L1 − c(L1θ)L1
But
sin φ = sin(135 + θ) =√
22
cos θ −√
22
sin θ ≈√
22
−√
22
θ
for small angles. Thus
mL2θ = mgL
(√2
2−
√2
2θ
)− kL1δ − k1L
21θ − cL2
1θ
At equilibrium,
mgL
√2
2− kL1δ = 0
and thus the equation of motion becomes
mL2θ + cL21θ +
(mgL
√2
2+ k1L
21
)θ = 0
The characteristic equation is
mL2s2 + cL21s + mgL
√2
2+ k1L
21 = 0
This will always be stable sincemL2 > 0
cL21 > 0
and
mgL
√2
2+ k1L
21 > 0
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4.47 Let I be the inertia of the pendulum about the pivot point. Then summing momentsabout the pivot point, and assuming small angles, we give
Iθ = −k1(L1θ)L1 + k2(y − L2θ)L2 + c(y − L2θ)L2
Collecting terms we obtain
Iθ + cL22θ + (k1L
21 + k2L
22)θ = k2L2y + cL2y
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4.48 Assuming that x1, x2, and x3 are measured from equilibrium, summing vertical forceson m1 gives
m1x1 = k1(x2 − x1) + c1(x2 − x1)− k3(x1 − x3) − c3(x1 − x3)
Summing vertical forces on m2 gives
m2x2 = k2(y − x2) − k1(x2 − x1) − c1(x2 − x1)
Summing vertical forces on m3 gives
m3x3 = k3(x1 − x3) + c3(x1 − x3)
The system model consists of these three equations.
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4.49 From impulse-momentum,
(m + 5m)v(0+)− [mv1 + 5m(0)] = 0
Thus v(0+) = v1/6.The equation of motion is 6mx + kx = 0, and its solution with x(0+) = 0, x(0+) =
v(0+) = v1/6 is
x(t) =v(0+)
ωnsin
√k
6mt =
v1
6
√6m
ksin
√k
6mt
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4.50 With m1 = m and m2 = 5m, (4.6.7) becomes
v3 =m − 5m
m + 5mv1 = −2
3v1
The change in momentum is
m
(−2
3v1
)− mv1 =
∫ 0+
0f(t) dt
Thus ∫ 0+
0f(t) dt = −5
3mv1
The linear impulse applied to m2 is (5/3)mv1, so
5mx + kx =53mv1δ(t)
orx + ω2
nx =13v1δ(t)
where ωn =√
k/5m. The response is given by
x(t) =v1
3ωnsin ωnt =
v1
3
√5m
ksin
√k
5mt
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4.51 a) Summing horizontal forces on each mass gives
m1x1 = f − k1x1 − k2(x1 − x2)
m2x2 = k2(x1 − x2)
Substituting the parameter values and collecting terms, we obtain
20x1 = f − 9 × 104x1 + 6 × 104x2
60x2 = 6 × 104(x1 − x2)
b) Apply the Laplace transform to both sides of each equation, using zero initial condi-tions: (
20s2 + 9× 104)
X1(s) − 6 × 104X2(s) = F (s)
−6 × 104X1(s) +(60s2 + 6 × 104
)X2(s) = 0
The solutions areX1(s)F (s)
=s2 + 103
20s4 + 1.1× 105s2 + 3 × 107
X2(s)F (s)
=103
20s4 + 1.1× 105s2 + 3 × 107
c) The MATLAB session is
sys = tf([1, 0, 1e+3],[20, 0, 1.1e+5, 0, 3e+7]);step(sys)
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4.52 a) Summing moments about the axis of rotation of each inertia gives
I1θ1 = −k1θ1 − k2(θ1 − θ2)
I2θ2 = T2 + k2(θ1 − θ2) − k3θ2
b) Substitute the given parameter values and collect terms to obtain
Iθ1 = −2kθ1 + kθ2
2Iθ2 = T2 − 2kθ2 + kθ1
Apply the Laplace transform to both sides of each equation, using zero initial conditions:
(Is2 + 2k)Θ1(s) − kΘ2(s) = 0
−kΘ1(s) + (2Is2 + 2k)Θ2(s) = T2(s)
The solutions areΘ1(s)T2(s)
=k
2I2s4 + 6kIs2 + 3k2
Θ2(s)T2(s)
=Is2 + 2k
2I2s4 + 6kIs2 + 3k2
c) With I = 10 and k = 60,
Θ1(s)T2(s)
=60
200s4 + 3600s2 + 10, 800
The MATLAB session is
sys = tf(60,[200, 0, 3600, 0, 10800]);impulse(sys)
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4.53 Summing horizontal forces on each mass gives
m1x1 = f − k1x1 − k2(x1 − x2)
m2x2 = k2(x1 − x2)
Substituting the parameter values and collecting terms, we obtain
20x1 = f − 9 × 104x1 + 6 × 104x2
60x2 = 6 × 104(x1 − x2)
Apply the Laplace transform to both sides of each equation, using zero initial conditions:
(20s2 + 9 × 104)X1(s) − 6 × 104X2(s) = F (s)
−6 × 104X1(s) + (60s2 + 6 × 104)X2(s) = 0
The transfer functions can be found by using Cramer’s method. The session is
D = conv([20, 0, 9e+4], [60, 0, 6e+4])-[0, 0, 0, 0, 3.6e+9];D1 = [60, 0, 6e+4];D2 = 6e+4;sys1 = tf(D1, D)sys2 = tf(D2, D)
The displayed transfer functions agree with the answers found by hand in Problem 4.51.
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4.54 Summing moments about the axis of rotation of each inertia gives
I1θ1 = −k1θ1 − k2(θ1 − θ2)
I2θ2 = T2 + k2(θ1 − θ2) − k3θ2
Substitute the given parameter values and collect terms to obtain
10θ1 = −120θ1 + 60θ2
20θ2 = T2 − 120θ2 + 60θ1
Apply the Laplace transform to both sides of each equation, using zero initial conditions:
(10s2 + 120)Θ1(s) − 60Θ2(s) = 0
−60Θ1(s) + (20s2 + 120)Θ2(s) = T2(s)
The transfer functions can be found by using Cramer’s method. The session is
D = conv([10, 0, 120], [20, 0, 120])-[0, 0, 0, 0, 60*60];D1 = 60;D2 = [10, 0, 120];sys1 = tf(D1, D)sys2 = tf(D2, D)
The displayed transfer functions agree with the answers found by hand in Problem 4.52.
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4.55 a) Summing moments about each pivot point and assuming small angles, we obtain
m1L22θ1 = −m1gL2θ1 − kL1(θ1 − θ2)L1
m2L22θ2 = −m2gL2θ2 + kL1(θ1 − θ2)L1
Collecting terms givesm1L
22θ1 + (m1gL2 + kL2
1)θ1 = kL21θ2
m2L22θ2 + (m2gL2 + kL2
1)θ2 = kL21θ1
b) Substitute the given values: m1 = 1, m2 = 4, L1 = 2, L2 = 5, and k = 10, we obtain
25θ1 + 89.05θ1 − 40θ2 = 0
400θ2 + 118.48θ2 − 40θ1 = 0
Apply the Laplace transform to both sides of each equation, using the given initial condi-tions: θ1(0) = 0.1, θ2(0) = 0, θ1(0) = 0, and θ2(0) = 0.
(25s2 + 89.05)Θ1(s) − 40Θ2(s) = 2.5s
−40Θ1(s) + (400s2 + 118.48)Θ2(s) = 0
The transfer functions can be found by using Cramer’s method. Note that we must appenda 0 to the numerator polynomial D1, so that we can use the step function to obtain thefree response. See Example 4.7.1. The session is
D = conv([25, 0, 89.05], [400, 0, 118.48]) - [0, 0, 0, 0, 40*40];D1 = conv([2.5, 0, 0], [400, 0, 118.48]);sys1 = tf(D1, D);step(sys1)
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4.56 a) Summing horizontal forces on each mass gives
m1x1 = −k1x1 − k2(x1 − x2)
m2x2 = −k3x2 − k2(x1 − x2)
b) Substitute the parameter values and collect terms to obtain
x1 = −3.2 × 104x1 + 1.6× 104x2
2x2 = −3.2x2 + 1.6× 104x2
Apply the Laplace transform to both sides of each equation, using the initial conditions:x1(0) = 0.1, x2(0) = 0, x1(0) = 0, and x2(0) = 0.
(s2 + 3.2× 104)X1(s)− 1.6× 104X2(s) = 0.1s
−1.6× 104X1(s) + (2s2 + 3.2 × 104)X2(s) = 0
The transfer functions can be found by using Cramer’s method. Note that we must appenda 0 to the numerator polynomial D1, so that we can use the step function to obtain thefree response. See Example 4.7.1. The session is
D = conv([1, 0, 3.2e+4], [2, 0, 3.2e+4]) - [0, 0, 0, 0, 2.56e+8];D1 = conv([0.1, 0, 0], [2, 0, 3.2e+4]);sys1 = tf(D1, D);step(sys1)
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Solutions Manual c©
to accompany
System Dynamics, Second Edition
by
William J. Palm III
University of Rhode Island
Solutions to Problems in Chapter Five
c©Solutions Manual Copyright 2010 The McGraw-Hill Companies. All rightsreserved. No part of this manual may be displayed, reproduced, or distributedin any form or by any means without the written permission of the publisheror used beyond the limited distribution to teachers or educators permitted byMcGraw-Hill for their individual course preparation. Any other reproductionor translation of this work is unlawful.
5.1 Define the following state variables: x1 = x and x2 = ˙ x. Then the state equations are
x1 = x2 x2 =15[−4x1 − 7x2 + f(t)]
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5.2 Define the following state variables: x1 = y, x2 = y, and x3 = y. Then the stateequations are
x1 = x2 x2 = x3 x3 =12[f(t) − 7x1 − 4x2 − 5x3]
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5.3 Define the following state variables: x1 = x and x2 = ˙ x. Then the state equations are
x1 = x2 x2 =12(4y − 4x1 − 5x2)
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5.4 From the transfer function we obtain:
3y + 6y + 10y = 6f(t)
Define the following state variables: x1 = x and x2 = ˙ x. Then the state equations are
x1 = x2 x2 =13[6f(t) − 10x1 − 6x2]
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5.5 Define the following state variables: z1 = x1, z2 = ˙ x1, z3 = x2, and z4 = x2. Then thestate equations are
z1 = z2 z2 =1
m1[f(t)− k1z1 + k1z3]
z3 = z4 z4 =1
m2[k1z1 − (k1 + k2)z3]
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
5.6 Define the following state variables: z1 = x1, z2 = ˙ x1, z3 = x2, and z4 = x2. Then thestate equations are
z1 = z2 z2 =110
(−40z1 − 8z2 + 25z3 + 5z4)
z3 = z4 z4 =15[f(t) + 25z1 + 5z2 − 25z3 − 5z4]
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
5.7 Define the following state variables: x1 = x and x2 = ˙ x. Then the state equations are
x1 = x2 x2 =12[4y(t) − 4x1 − 5x2]
Then
A =
[0 1−2 −5/2
]B =
[02
]
C =[
1 0]
D = [0]
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
5.8
A =
[−5 30 −4
]B =
[2 00 6
]
C =
[1 30 1
]D =
[2 00 0
]
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
5.9 a)
A =
[−5 31 −4
]B =
[05
]
C =
[1 00 1
]D =
[00
]
b)
A =
[−5 30 −4
]B =
[4 00 5
]
C =
[1 00 0
]D =
[0 00 0
]
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
5.10 Define the following state variables: z1 = x1, z2 = x1, z3 = x2, and z4 = ˙ x2. Then thestate equations are
z1 = z2 z2 =110
(−40z1 − 8z2 + 25z3 + 5z4)
z3 = z4 z4 =15[f(t) + 25z1 + 5z2 − 25z3 − 5z4]
The matrices are
A =
0 1 0 0−4 −4/5 5/2 1/20 0 0 15 1 −5 −1
B =
000
1/5
C =
[1 00 1
]D =
[00
]
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
5.11 First way: Isolate a 1 in the denominator:
Y (s)F (s)
=6 + 7/s
1 + 3/s
ThusY (s) = −3
sY (s) + 6F (s) +
7sF (s) =
1s
[7F (s) − 3Y (s)] + 6F (s) (1)
LetX(s) =
1s
[7F (s) − 3Y (s)] (2)
Thus, from (1),Y (s) = X(s) + 6F (s) (3)
Substitute this into (2):
X(s) =1s7F (s) − 3 [X(s) + 6F (s)] =
1s
[−3X(s)− 11F (s)]
Thusx = −3x − 11f(t)
wherey = x + 6f
Thus x(0) = y(0)− 6f(0) = y(0) if we take f(0) = 0.(Continued on the next page)
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Problem 5.11 continued:
Second way: Write the equation as
Y (s) = (6s + 7)F (s)s + 3
(1)
LetX(s) =
F (s)s + 3
(2)
Then(s + 3)X(s) = F (s)
which givessX(s) = −3X(s) + F (s) (3)
andx = −3x + f(t)
From (1), (2), and (3),
Y (s) = 6sX(s) + 7X(s) = 6[−3X(s) + F (s)] + 7X(s) = −11X(s) + 6F (s)
Thusy(t) = −11x(t) + 6f(t)
Sox(0) = − 1
11y(0) +
611
f(0) = − 111
y(0)
if we take f(0) = 0.Thus the state model is
x = −3x + f(t)
wherex(0) = − 1
11y(0)
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5.12 First way: Divide by s2 to obtain a 1 in the denominator.
Y (s)F (s)
=2s−2 + s−1
1 + 4s−1 + 3s−2
Use the 1 in the denominator to solve for Y (s).
Y (s) =(2s−2 + s−1
)F (s) −
(4s−1 + 3s−2
)Y (s)
=1s
−4Y (s) + F (s) +
1s
[2F (s) − 3Y (s)]
(2)
This equation shows that Y (s) is the output of an integration. Thus y can be chosen as astate-variable.
X1(s) = Y (s)
The term within square brackets in (2) is the input to an integration, and can be chosen asthe second state-variable:
X2(s) =1s
[2F (s) − 3Y (s)] =1s
[2F (s) − 3X1(s)]
Then from equation (2)
X1(s) =1s
[−4X1(s) + F (s) + X2(s)]
The state equations arex1 = −4x1 + x2 + f (3)
x2 = 2f − 3x1 (4)
and the output equation is y = x1.(Continued on the next page)
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
Problem 5.12 continued.
The matrices of the standard form are
A =
−4 1
−3 0
B =
[12
]
C =[
1 0]
D = [0]
Because x1 = y, we have x1(0) = y(0). From (3), x2 = x1 + 4x1 − f = y + 4y − f . Thus,taking f(0) = 0, we obtain x2(0) = y(0) + 4y(0).
Second way: Express the model as
Y (s) =s + 2s + 1
F (s)s + 3
Let the first state variable beX1(s) =
F (s)s + 3
(1)
Thenx1 = f − 3x1 (2)
ThenY (s) =
s + 2s + 1
X1(s) (3)
(Continued on the next page)
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
Problem 5.12 continued:
Let the second state variable be
X2(s) =X1(s)s + 1
(4)
This givesx2 = x1 − x2 (5)
From (3)Y (s) = (s + 2)X2(s)
andy = ˙ x2 + 2x2
From (5),y = x1 − x2 + 2x2 = x1 + x2 (6)
Thus the model isx1 = f − 3x1
x2 = x1 − x2
y = x1 + x2
For the initial conditions, note that (1) and (3) give
y + y = ˙ x1 + 2x1 = f − x1
orx1 = f − y − y
(continued on the next page)
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Problem 5.12 continued:
Taking f(0) = 0, this gives
x1(0) = −y(0)− y(0) (7)
From (6),x2 = y − x1
which givesx2(0) = y(0)− x1(0) = 2y(0) + y(0) (8)
This initial conditions are given by (7) and (8).The state model matrices are
A =
−3 0
1 −1
B =
[10
]
C =[
1 1]
D = [0]
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
5.13 a) The matrices are
A =
[−5 30 −4
]B =
[05
]
C =
[1 00 1
]D =
[00
]
The MATLAB session is
A = [-5, 3; 1, -4]; B = [0; 5];C = [1, 0; 0, 1]; D = [0;0];sys = ss(A,B,C,D);systf = tf(sys)
The resulting transfer functions displayed on the screen are for x1 and x2 in that order.
X1(s)U(s)
=15
s2 + 9s + 17
X2(s)U(s)
=5s + 25
s2 + 9s + 17
(Continued on the next page)
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
Problem 5.13 continued:
b) The matrices are
A =
[−5 31 −4
]B =
[4 00 5
]
C =[
1 0]
D =[
0 0]
The MATLAB session is
A = [-5, 3; 1, -4]; B = [4, 0;0, 5];C = [1, 0]; D = [0, 0];sys = ss(A,B,C,D);systf = tf(sys)
The resulting transfer functions displayed on the screen are for u1 and u2 in that order.They are
X1(s)U1(s)
=4s + 16
s2 + 9s + 17
X1(s)U2(s)
=15
s2 + 9s + 17
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5.14 a) The session is
sys1 = tf(1, [2, 5, 4, 7]);sys2 = ss(sys1)
The matrices displayed on the screen are
A =
−2.5 −0.5 −0.21884 0 00 4 0
B =
0.12500
C =[
0 0 0.25]
D = [0]
b) The session is
sys1 = tf(6, [3, 6, 10]);sys2 = ss(sys1)
The matrices displayed on the screen are
A =
[−2 −0.41678 0
]B =
[0.50
]
C =[
0 0.5]
D = [0]
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5.15 a) The session is
sys1 = tf([6, 7], [1, 3]);sys2 = ss(sys1)
The matrices, which are scalars in this case, displayed on the screen are
A = −3 B = 4
C = −2.75 D = 6
So the model isx = −3x + 4f
y = −2.75x + 6f
b) The session is
sys1 = tf([1, 2], [1, 4, 3]);sys2 = ss(sys1)
The matrices displayed on the screen are
A =
[−4 −0.3758 0
]B =
[10
]
C =[
1 0.25]
D = [0]
So the model isx1 = −4x1 − 0.375x2 + f x2 = 8x1
y = x1 + 0.25x2
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5.16 The matrices are
A =
[−5 31 −4
]B =
[05
]
C =[
1 0]
D = [0]
The session is
A = [-5 ,3; 1, -4]; B = [0; 5]; C = [1, 0]; D = 0;sys = ss(A,B,C,D);initial(sys, [3, 5])
b) The session is
A = [-5 ,3; 1, -4]; B = [0; 5]; C = [1, 0]; D = 0;sys = ss(A,B,C,D);step(sys)
c) The session is
A = [-5 ,3; 1, -4]; B = [0; 5]; C = [1, 0]; D = 0;sys = ss(A,B,C,D);t = [0:0.01:2];f = 3*sin(10*pi*t);lsim(sys,f,t)
This plots both the input f(t) and the output x1(t) on the same plot. To plot only theoutput, replace the last line with
y = lsim(sys, f, t); plot(t, y)
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5.17 To obtain the roots, all we need is the A matrix, which is
A =
[−5 30 −4
]
The session is
A = [-5, 3; 0, -4];poly(A)ans =
1 9 20eig(A)ans =-5 -4
Thus the characteristic equation is s2 + 9s + 20 = 0 and the roots are s = −5 and s = −4.
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5.18 a) Let z1 = x1, z2 = ˙ x1, z3 = x2, and z4 = ˙ x2. Then the state equations are
z1 = z2
z2 =1
m1[−k1z1 − c1z2 + k1z3 + c1z4]
z3 = z4
z4 =1
m2[k1z1 + c1z2 − k1z3 − c1z4 + k2y − k2z3]
The following script file creates the state model.
m1 = 36;m2 = 240;k1 = 1.6e+5;k2 = 1.6e+4;c1 = 98;A = [0, 1, 0, 0; -k1/m1, -c1/m1, k1/m1, c1/m1; ....
0, 0, 0, 1; k1/m2, c1/m2, -(k1+k2)/m2, -c1/m2];B = [0; 0; 0; k2/m2];C = [1, 0, 0, 0; 0, 0, 1, 0]; D = [0; 0];sys = ss(A,B,C,D);
(continued of the next page)
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Problem 5.18 continued:
b), c) To compute the impulse response, add the following lines to the script file.
impulse(sys)format longpoly(A)eig(A)
We must use the format long command to see the results adequately. The characteristicpolynomial given by the poly(A) function is
s4 + 3.13106s3 + 517777s2 + 181.481s + 2.96296× 105 = 0
The roots given by the eig(A) function are
s = −1.56527± 71.5364j s = −3.07931× 10−5 ± 7.60733j
The period of the first root pair is 2π/71.5364 = 0.0878 s, whereas the dominant timeconstant of the system is 1/(3.07931×10−5) = 3.247×104 s. So there will be approximately4(3.247×104)/0.0878 = 1.479×105 oscillations before the impulse response disappears. Thusthe impulse(sys) will produce a plot on which the individual oscillations are impossibleto discern. To remedy this, you can use the syntax [y, t] = impulse(sys); plot(t,y),axis([0 1000 -10 10]). This will show several oscillations.
d) Add the following line to the script file.
tfsys = tf(sys)
The following transfer functions are displayed on the screen.
X1(s)Y (s)
=181.5s + 2.963× 105
s4 + 3.131s3 + 5178s2 + 181.5s + 2.963× 105
X2(s)Y (s)
=66.67s2 + 181.5s + 2.963× 105
s4 + 3.131s3 + 5178s2 + 181.5s + 2.963× 105
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5.19 (a) The small displacement assumption implies that the suspension forces are nearlyperpendicular to the centerline of the mass m, and thus are nearly vertical. To obtain theequations of motion, assume arbitrarily that
y1 > x − L1θ
y1 > x − L1θ
y2 > x + L2θ
y2 > x + L2θ
We obtain the following moment equation about the mass center G.
IGθ = − c1(y1 − x + L1θ)L1 − k1(y1 − x + L1θ)L1
+ c2(y2 − x − L2θ)L2 + k2(y2 − x − L2θ)L2
Rearranging gives
IGθ + (c2L22 + c1L
21)θ + (k1L
21 + k2L
22)θ
− (c1L1 − c2L2)x − (k1L1 − k2L2)x= −c1L1y1 + c2L2y2 − k1L1y1 + k2L2y2 (1)
Summing forces in the vertical direction gives
mx = c1(y1 − x + L1θ) + k1(y1 − x + L1θ) + c2(y2 − x − L2θ) + k2(y2 − x − L2θ)
Rearranging gives
mx + (c1 + c2)x + (k1 + k2)x− (c1L1 − c2L2)θ − (k1L1 − k2L2)θ= c1y1 + c2y2 + k1y1 + k2y2 (2)
Equations (1) and (2) are the desired model.(Continued on the next page)
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Problem 5.19 continued:
b) The transformed equations of motion have the following forms:
AΘ(s) + BX(s) = CY1(s) + DY2(s)
EΘ(s) + FX(s) = GY1(s) + HY2(s)
whereA = IGs2 + (c2L
22 + c1L
21)s + k1L
21 + k2L
22
B = (c2L2 − c1L1)s + k2L2 − k1L1
C = −c1L1s + k1L1
D = c2L2s + k2L2
E = −B
F = ms2 + (c1 + c2)s + k1 + k2
G = c1s + k1
H = c2s + k2
Because these models have input derivatives, we must be careful in deriving a statevariable model. First obtain the transfer functions using Cramer’s method, and then usethe MATLAB functions tf and ss to obtain the state model.(continued on the next page)
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Problem 5.19 continued:
For Cramer’s method, let
M =
∣∣∣∣∣A B
E F
∣∣∣∣∣ = AF − BE
N1 =
∣∣∣∣∣C BG F
∣∣∣∣∣ = CF − BG N2 =
∣∣∣∣∣D BH F
∣∣∣∣∣ = DF − BH
N3 =
∣∣∣∣∣A C
E G
∣∣∣∣∣ = AG − CE N4 =
∣∣∣∣∣A D
E H
∣∣∣∣∣ = AH − DE
Then the transfer functions are found as follows
Θ(s)Y1(s)
=N1
M
Θ(s)Y2(s)
=N2
M
X(s)Y1(s)
=N3
M
X(s)Y2(s)
=N4
M
We can implement Cramer’s method in MATLAB by using the conv function. To dothis, note that A and F are second-order polynomials, and that B, C, D, E, G, and Hare first-order polynomials. This distinction is important because will need to add leadingzeros to subtract polynomials of different orders. Note that with a multi-input, multi-output(MIMO) model whose transfer functions all have the same denominator, we must store thenumerators in a cell array. The MATLAB script file is shown on the following page.(continued on the next page)
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Problem 5.19 continued:
% Store the parameter values.k1 = 1100; k2 = 1525; c1 = 4; c2 = c1;L1 = 4.8; L2 = 3.6; m = 50; IG = 1000;% Create the coefficients of the transformed equations.A = [IG, c2*L2^2 + c1*L1^2, k1*L1^2 + k2*L2^2];B = [c2*L2 - c1*L1, k2*L2 - k1*L1];C = -[c1*L1, k1*L1];D = [c2*L2, k2*L2];E = -B;F = [m, c1 + c2, k1 + k2];G = [c1, k1];H = [c2, k2];% Create the Cramer determinants.M = conv(A, F) - [0, 0, conv(E, B)];N1 = conv(C, F) - [0, conv(G, B)];N2 = conv(D, F) - [0, conv(B, H)];N3 = conv(A, G) - [0, conv(C, E)];N4 = conv(A, H) - [0, conv(D, E)];% Create the transfer functions.% Store the numerators in a cell array.numerators = N1, N2; N3, N4 ;tfsys = tf(numerators, M);% Create the state variable models.statesys = ss(tfsys)
(continued on the next page)
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Problem 5.19 continued:
The results are (note that MATLAB returns the state model matrices as lowercasesymbols, a, b, c, d. We use these symbols to avoid confusion with the equation coefficientsA, B,C, and D defined above.)
a =
[a1 a2
a3 a4
]
where
a1 =
−0.304 −6.102 −0.1151 −2.31416 0 0 00 8 0 00 0 8 0
a2 =
0 0 0 00 0 0 00 0 0 00 0 0 0
a3 = a2 a4 = a1
b =
2 00 00 00 00 20 00 00 0
c =
[−0.0096 −0.1651 −0.006891 −0.1376 0.0072 0.1716 0.006891 0.1376
0.04 0.6879 0.02814 0.4737 0.04 0.9534 0.02943 0.683
]
d =
[0 00 0
]
(continued on the next page)
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Problem 5.19 continued:
c) With y1 = 0, we must redefine b as
b =
0 00 00 00 00 20 00 00 0
To do this, and to compute and plot the impulse response, add the following lines to thescript file.
[a, b, c, d] = ssdata(statesys)% Obtain unit-impulse response for y_1 = 0.b(:,1) = zeros(size(b(:,1)));statesys2 = ss(a,b,c,d);impulse(statesys2)
Note that we do not need the state model to compute the impulse response, which canalso be computed from the transfer functions N2/M and N4/M . The problem statement,however, specifically asks for the state space model.
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5.20 a) Create the following function file.
function ydot = problem20(t,y)ydot = cos(t);
Then type
[ta, ya] = ode45(′problem20′,[0, 12], 6);
b) The closed-form solution is found as follows.∫ y
6dy =
∫ t
0cos t dt
Thusy(t) − 6 = sin t
To compare the two solutions, continue the session as follows:
tb = [0:0.01:12];yb = 6 + sin(tb);plot(ta,ya,tb,yb)
The plots are identical.
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5.21 a) Create the following function file.
function ydot = problem21(t,y)ydot = 5*exp(-4*t);
Then type
[ta, ya] = ode45(′problem21′,[0, 1], 2);
b) The closed-form solution is found as follows.∫ y
2dy = 5
∫ t
0e−4t dt = −5
4e−4t
∣∣∣t
0
Thusy(t) =
134
− 54e−4t
To compare the two solutions, continue the session as follows:
tb = [0:0.01:1];yb = 5*exp(-4*tb);plot(ta,ya,tb,yb)
The plots are identical.
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5.22 a) Create the following function file.
function ydot = problem22(t,y)ydot = 5*exp(4*t)-3*y;
Then type
[ta, ya] = ode45(′problem22′,[0, 1], 10);
b) The closed-form solution is found with the Laplace transform.
sY (s) − y(0) + 3Y (s) =5
s − 4
With y(0) = 10, this gives
Y (s) =10s − 35
(s + 3)(s− 4)=
65/7s + 3
+5/7
s − 4
andy(t) =
657
e−3t +57e4t
To compare the two solutions, continue the session as follows:
tb = [0:0.01:1];yb = (65*exp(-3*tb)+5*exp(4*tb))/7;plot(ta,ya,tb,yb)
The plots are identical.
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5.23 a) Create the following function file.
function ydot = problem23(t,y)ydot = -sin(y);
Then type
[ta, ya] = ode45(′problem23′,[0, 4], 0.1);
b) The closed-form solution to the approximate equation y = −y is found as follows.The root is s = −1, so the solution form is
y(t) = Ae−t
Because y(0) = 0.1, we obtain A = 0.1. Thus the solution is
y(t) = 0.1e−t
To compare the two solutions, continue the session as follows:
tb = [0:0.01:4];yb = 0.1*exp(-tb);plot(ta,ya,tb,yb)
The plots are identical.
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5.24 a) Create the following function file.
function ydot = problem24(t,y)if t <= 2
f = 3*t;elseif t <= 5
f = 6;else
f = -3*(t - 5) + 6;endydot = f - 2*y;
Then type
[ta, ya] = ode45(′problem24′,[0, 7], 2);
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5.25 A MATLAB solution is as follows. First create the following function file.
function vdot = problem25(t,v)vdot = (8000-20*v-0.05*v.^2)/50;
Then type
[t,v] = ode45(′problem25′,[0, 5], 0);plot(t,v),xlabel(′t (sec)′),ylabel(′v (ft/sec)′)
The final time used (5 sec) was an estimate. The solution actually reaches steady-state inabout 4.6 sec.
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5.26 Put the model into state variable form. Let x1 = y and x2 = y. Then x1 = x2 andx2 = 9.81 − 180x1 − 340x3
1. A MATLAB solution is as follows. First create the followingfunction file.
function xdot = problem26(t,x)xdot = [x(2);9.81-180*x(1)-340*x(1).^3;
Then type
[ta,xa] = ode45(′problem26′,[0, 2], [0.06, 0]);[tb,xb] = ode45(′eqn′,[0, 2], [0.1, 0];plot(ta,xa(:,1),tb,xb(:,1)),xlabel(′t (s)′),ylabel(′y(m)′)
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5.27 The state variable form, with x1 = y and x2 = ˙ y, is
x1 = x2
x2 = −x1 + b(1− x21)x2
Create the following function file:
function xdot = vander(t,x)global bxdot(1) = x(2);xdot(2) = b*(1-x(1)^2)*x(2) - x(1);xdot = [xdot(1);xdot(2)];
Then use the following script file to solve the equation and plot x1 = y. For Case 1:
global bb = 0.1;[t, x] = ode45(′vander′, [0, 25], [1, 1]);plot(t,x(:,1)),xlabel(′t′),ylabel(′y(t)′)
For Case 2:
global bb = 0.1;[t, x] = ode45(′vander′, [0, 25], [3, 3]);plot(t,x(:,1)),xlabel(′t′),ylabel(′y(t)′)
For Case 3:
global bb = 3;[t, x] = ode45(′vander′, [0, 25], [1, 1]);plot(t,x(:,1)),xlabel(′t′),ylabel(′y(t)′)
(continued on the next page)
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Problem 5.27 continued:
Note on Passing Additional Parameters to an ODE Function. Rather than using theglobal statement, the extended syntax of the ODE solvers using the options argument letsyou pass any input parameters that follow the options argument to the ODE function andany function you specify in options. This syntax uses function handles. For example, youcan pass the parameter b directly to the van der Pol function, as follows:
function xdot = vander(t,x,b)xdot(1) = x(2);xdot(2) = b*(1-x(1)^2)*x(2) - x(1);xdot = [xdot(1);xdot(2)];
Then pass the parameter b to the function vander by specifying it after the options argumentin the call to the solver, using options = [ ] as a placeholder, as follows:
[t,x] = ode45(@vander, [0, 25], [1, 1], [ ],b)
The function handle @vander calls vander(t,x,b).
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5.28 Put the model in state variable form as follows. Let x1 = y and x2 = ˙ y.
x1 = x2
x2 = −x1 + b(1− x21)x2
Create the following function file.
function xdot = vander(t,x);b = 1000;xdot = [x(2); -x(1) + b*(1-x(1)^2)*x(2)];
The following script file uses the ode23s function.
[t, x] = ode23s(′vander′, [0, 3000], [2, 0]);plot(t,x(:,1)),xlabel(′t′),ylabel(′y′)
The ode23 function successfully solves the equation, but the functions ode45 and ode23fail to converge to a solution for this problem. They can, however, obtain a solution for“nonstiff” values of b, such as b = 1.
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5.29 The state variable form, with x1 = θ and x2 = θ, is
x1 = x2
x2 =a(t) cos θ − g sin θ
L
Express the acceleration a(t) as the linear function a(t) = mt + b. Thus for cases (a)and (b), m = 0 and b = 5. For case (c), m = 0.5 and b = 0.
Create the following function file:
function xdot = accbase(t,x)global m bL = 1; g = 9.81;xdot(1) = x(2);xdot(2) = ((m*t + b)*cos(x(1)) -g*sin(x(1)))/L;xdot = [xdot(1);xdot(2)];
To solve the equation and plot the solution, create the following script file.
global m bm = 0; b = 5;[t, x] = ode45(′accbase′, [0, 10], [0.5, 0]);plot(t,x(:,1)),xlabel(′t (seconds)′),ylabel(′theta(t) (radians)′)
For case (b), change the third line to
[t, x] = ode45(′accbase′, [0, 10], [3, 0]);
For case (c), change the second line to
m = 0.5; b = 0;
(continued on the next page)
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Problem 5.29 continued:
Note on Passing Additional Parameters to an ODE Function. Rather than using theglobal statement, the extended syntax of the ODE solvers using the options argument letsyou pass any input parameters that follow the options argument to the ODE function andany function you specify in options. This syntax uses function handles. For example, youcan pass the parameters L, g, m, and b directly to the accbase function, as follows:
function xdot = accbase(t,x,L,g,m,b)xdot(1) = x(2);xdot(2) = ((m*t + b)*cos(x(1)) -g*sin(x(1)))/L;xdot = [xdot(1);xdot(2)];
Then pass the parameters L, g, m, and b to the function accbase by specifying them afterthe options argument in the call to the solver, using options = [ ] as a placeholder, asfollows:
[t,x] = ode45(@accbase, [0, 10], [3, 0], [ ], L, g, m, b)
The function handle @accbase calls accbase(t,x,L,g,m,b).
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5.30 The cosine function is obtained with the Sine Wave Function block by using a phaseshift of π/4 (pi/2). Make sure the Sine Wave blocks are set to use simulation time. Oth-erwise a Clock block will be required. Set the initial condition of the Integrator block to 1and the initial condition of the Integrator1 block to 4.
Figure : for Problem 5.35.
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5.31 The time variable tout is automatically placed in the workspace by selecting DataImport/Export under the Configuration Parameters menu. Thus a Clock block is notneeded. You can enter the initial condition x(0) as 100*cos(30*pi/180), with a similarexpression for y(0).
To plot the trajectory y versus x, type
plot(simout(:,1),simout(:,2)),xlabel(′x′),ylabel(′y′)
Figure : for Problem 5.36.
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5.32 The model is shown in the following figure. To plot the percent relative error in theseries solution, type
t = tout;x = (t.^3)/3-t.^2+3*t-3+3*exp(-t);plot(t,100*(simout-x)/simout),xlabel(′t′),ylabel(′Percent Error′),grid
The plot shows that the percent relative error in the series solution is less than 1% fort < 0.75. It is about 6% at t = 1.
Figure : for Problem 5.32.
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5.33 The model is shown in the following figure. Set the Step time and Final value for theStep block to 0 and 4, respectively. Set the Step time and Final value for the Step1 blockto 2 and 4, respectively. Set the Initial condition of the Integrator to 2. You can plot theresults by typing
plot(tout,simout),xlabel(′t′),ylabel(′x′)
Figure : for Problem 5.33.
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5.34 The model is shown in the following figure. Set the Step time and Final value for theStep block to 0 and 5, respectively. Set the Step time and Final value for the Step1 blockto 2 and 5, respectively. Set the numerator of the Transfer Function block to [1] and thedenominator to [2, 12, 10]. You can plot the results by typing
plot(tout,simout),xlabel(′t′),ylabel(′x′)
Figure : for Problem 5.34.
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5.35 The Simulink model is shown in the figure. Set the Amplitude of the Sine Wave blockto 5 and the Frequency to 0.8. Make sure the Initial conditions of the Integrator blocks areset to 0. The Save format of the To Workspace block has been selected as Array, and thesimulation time tout has been saved to the workspace (Use the Workspace I/O tab on theSimulation Parameters menu under the Simulation menu to do this). Then the output canbe plotted in MATLAB by typing
plot(tout,simout)
Figure : for Problem 5.35.
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5.36 The model is shown in the following figure. Set the Sine Wave block to use simulationtime. Set the Amplitude to 10 and the Frequency to 3. In the Saturation block set thelimits to ±8. Set the Initial condition of the Integrator to 2. You can plot the results bytyping
plot(tout,simout),xlabel(′t′),ylabel(′x′)
Figure : for Problem 5.36.
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5.37 The model is shown in the following figure. Set the Sine Wave block to use simulationtime. Set the Amplitude to 10 and the Frequency to 4. In the Saturation block set thelimits to ±5. Set the Initial condition of the Integrator to 0. You can plot the results bytyping
plot(tout,simout),xlabel(′t′),ylabel(′x′)
Figure : for Problem 5.37.
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5.38 The model is shown in the following figure. Set the Sine Wave block to use simulationtime. Set the Amplitude to 2 and the Frequency to 4. In the Math Function block selectthe square function. Set the Initial condition of the Integrator to 1. You can plot the resultsby typing
plot(tout,simout),xlabel(′t′),ylabel(′x′)
Using a stop time of 3 results in an error message that indicates that Simulink is havingtrouble finding a small enough step size to handle the rapidly decreasing solution. Byexperimenting with the stop time, we find that, to two decimal places, a stop time of 1.38will not generate an error.
This model is unstable, and the output x → −∞ if the time span is long enough. Wecan see this by writing the equation as
x = 2 sin 4t − 10x2
Once x drops below√
2/10, x remains negative and thus x continues to decrease throughnegative values.
Figure : for Problem 5.38.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
5.39 The model is shown in the following figure. Set the Start and End of the Dead Zoneto −0.5 and 0.5 respectively. Set the Sine Wave block to use simulation time. Set theAmplitude to 2 and the Frequency to 4. In the Math Function block select the squarefunction. Set the Initial condition of the Integrator to 1. You can plot the results by typing
plot(tout,simout),xlabel(′t′),ylabel(′x′)
Using a stop time of 3 results in an error message that indicates that Simulink is havingtrouble finding a small enough step size to handle the rapidly decreasing solution. Byexperimenting with the stop time, we find that, to two decimal places, a stop time of 1.49will not generate an error.
This model is unstable, and the output x → −∞ if the time span is long enough. Wecan see this by writing the equation as (not including the effect of the dead zone)
x = 2 sin 4t − 10x2
Once x drops below√
2/10, x remains negative and thus x continues to decrease throughnegative values.
Figure : for Problem 5.39.
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5.40 The model is shown in the following figure. Set the Initial condition of Integrator to 0.5and the Initial condition of Integrator1 to 0. In the Fcn block type 900*u(1)+1700*u(1)^3for the expression. You can plot the results by typing
plot(tout,simout),xlabel(′t′),ylabel(′x′)
Figure : for Problem 5.40.
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5.41 The model is shown in the following figure. Set the Initial condition of Integra-tor to 30*pi/180 and the Initial condition of Integrator1 to 0. In the Fcn block type-17500*cos(u(1))+626000*sin(1.33+u(1))/sqrt(2020+1650*cos(1.33+u(1)) for the ex-pression. You can plot the results by typing
plot(tout,simout),xlabel(′t′),ylabel(′x′)
Figure : for Problem 5.41.
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5.42 The model is shown in the following figure. Set the Step time and Final value for theStep block to 0 and f1 − mg cos(30π/180) = 5 − 3(9.8) cos(30π/180), respectively. Set thegain in the top gain block to µmg sin(30π/180) = 0.5(2)(9.8) sin(30π/180). Set the Initialcondition of the Integrator to 3. The block comes to rest in about 0.35 s. You can plot theresults by typing
plot(tout,simout),xlabel(′t′),ylabel(′x′)
Figure : for Problem 5.42.
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5.43 a) The model is shown in the following figure. Set the Step time and Final value forthe Step block to 0 and 4, respectively. Set the Step time and Final value for the Step1block to 2 and 4, respectively. Set the Initial condition of the Integrator to 2. You can plotthe results by typing
plot(tout,simout),xlabel(′t′),ylabel(′x′)
b) Delete the Step block and replace it with a Sine Wave block. Set the Amplitude to10 and the frequency to 2.5.
Figure : for Problem 5.43.
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Solutions Manual c©
to accompany
System Dynamics, Second Edition
by
William J. Palm III
University of Rhode Island
Solutions to Problems in Chapter Six
c©Solutions Manual Copyright 2010 The McGraw-Hill Companies. All rightsreserved. No part of this manual may be displayed, reproduced, or distributedin any form or by any means without the written permission of the publisheror used beyond the limited distribution to teachers or educators permitted byMcGraw-Hill for their individual course preparation. Any other reproductionor translation of this work is unlawful.
6.1 See the following figure. From part (a) of the figure,
R1 = R +1
1R + 1
R
=3R
2
From part (b) of the figure,1
R2=
1R
+1
R1
which gives R2 = 3R/5.From part (c) of the figure,
Re = R + R2 =8R
5Thus
vs =8R
5i
Figure : for Problem 6.1
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
6.2 From the figure on the following page,
1R1
=15
+12
=710
So R1 = 10/7. Also, R2 = 8 + 2 = 10. From conservation of charge,
i1 + i2 + i3 = i (1)
Kirchhoff’s voltage law applied to the three loops gives
vs = 1i + 4i3 (2)
4i3 = R1(i1 + i2) + 3i2 (3)
3i2 = R2i1 (4)
Also,v1 = 2i1 (5)
Use (1) and (2) to eliminate i:
i1 + i2 + 5i3 = vs (6)
Collect terms, substitute the values of R1 and R2, and rearrange (3) and (4):
107
i1 +317
i2 − 4i3 = 0 (7)
10i1 − 3i2 + 0i3 = 0 (8)
Equations (6), (7), and (8) have the solution i1 = 0.1193vs. (These three equations can besolved with MATLAB by setting vs = 1.) From (5) we have the answer: v1 = 0.2386vs.
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Figure : for Problem 6.2
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
6.3 Let i1 be the current through R1 and R4, and i2 be the current through R2 and R3.Then vs = (R1 + R4)i1 = (R2 + R3)i2 and
v1 = R4i1 − R3i2 =(
R4
R1 + R4− R3
R2 + R3
)vs
This gives
v1 =R2R4 − R1R3
(R1 + R4)(R2 + R3)vs
If the changes in R3 and R4 are small enough so that (R1 + R4)(R2 + R3) ≈ constant = K,then
v1 =R2vs
KR4 −
R1vs
KR3 = αR4 − βR3
where α and β are approximately constant.
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6.4 The two resistors in series are equivalent to one 70 kΩ resistor, which is in parallel withthe 80 kΩ resistor. Thus the total equivalent resistance R is found from
1R
=180
+170
which gives R = 112/3 kΩ.Thus
i =9
112/3× 103= 2.411× 10−4 A
The power P is computed from
P = iv = (2.411× 10−4)9 = 2.17× 10−3 W
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6.5 The model isv1 =
1C
∫ (is −
1R
v1
)dt
Differentiate both sides with respect to t to obtain a differential equation.
RCdv1
dt+ v1 = Ris
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
6.6 a) Kirchhoff’s voltage law gives
vs = R1i + vo + R2i = (R1 + R2)i + vo (1)
For the capacitor:
vo =1C
∫i dt
which gives
Cdvo
dt= i (2)
Solve equation (1) for i and substitute into equation (2) to obtain the answer:
Cdvo
dt=
vs − vo
R1 + R2
which, in standard form, is
(R1 + R2)Cdvo
dt+ vo = vs
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6.7 a) From Kirchhoff’s voltage law,
vs = R2i2 = R1i1 + vo
which gives
i1 =vs − vo
R1(1)
For the capacitor,
vo =1C
∫i1 dt (2)
Differentiate (2) with respect to time, and substitute for i1 from (1):
Cdvo
dt= i1 =
vs − vo
R1
Thus the model isR1C
dvo
dt+ vo = vs (3)
Note the we did not need to find i2 or i3, and that the model is independent of R2
because there is no element between vs and R2, so the voltage input to the right-most loopis vs.
b) From (3) with vs = 0, we obtain the free response:
vo(t) = vo(0)e−t/R1C
If vs(t) = V us(t) and if vo(0) = 0, the response is
vo(t) = V(1− e−t/R1C
)
This is the forced response. So the total response is
vo(t) = vo(0)e−t/R1C + V(1 − e−t/R1C
)
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6.8 a) Kirchhoff’s voltage law gives
vs = R1i1 + vo (1)
For the capacitor:
vo = R2i2 =1C
∫i3 dt (2)
From conservation of charge,i1 = i2 + i3
Substitute for the currents to obtain:
vs − vo
R1=
vo
R2+ C
dvo
dt
This gives
R1R2Cdvo
dt+ (R1 + R2)vo = R2vs
b) The free response isvo(t) = vo(0)e−t/τ
whereτ =
R1R2C
R1 + R2
The forced response is
vo(t) =R2V
R1 + R2
(1 − e−t/τ
)
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
6.9 a) The model of the circuit is
Ri + Ldi
dt= vs
b) If vs(t) = V us(t) and i(0) = 0, then
I(s) =Vs(s)
Ls + R=
V
s(Ls + R)=
V
R− V
R
1s + R/L
Thusi(t) =
V
R
(1 − e−Rt/L
)
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6.10 For the circuit,
Ldi
dt+ Ri = vi(t)
or5di
dt+ 10i = vi(t)
Since the applied voltage is 12 volts, the impulse strength is 12(0.3) = 3.6. So we modelvi(t) as 3.6δ(t), and thus
(5s + 10)I(s) = 3.6
which gives
I(s) =3.6
5s + 10=
0.72s + 2
andi(t) = 0.72e−2t
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6.11 Define i to be the current passing through R and L. From Kirchhoff’s voltage law,
vs = Ri + v0
Solve for i:i =
vs − v0
R
For the inductor:v0 = L
di
dt
Substitute for i to obtain
v0 = Ld
dt
(vs − v0
R
)=
L
R
(dvs
dt− dv0
dt
)
Move the output terms (the v0 terms) to the left side, putting the highest-order derivativefirst, and move the input terms to the right side, to obtain the answer in a somewhatstandard form:
L
R
dv0
dt+ v0 =
L
R
dvs
dt
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6.12 Define i to be the current passing through R1, C, and R2. Then from Kirchhoff’svoltage law:
vs = R1i + v0 (1)
From the definition of v0,
v0 =1C
∫i dt + R2i (2)
Differentiate this equation to obtain:
dv0
dt=
1C
i + R2di
dt(3)
Solve equation (1) for i:
i =vs − v0
R1(4)
and differentiate to obtain:
di
dt=
1R1
(dvs
dt− dv0
dt
)(5)
Substitute equations (4) and (5) into equation (3) to obtain
dv0
dt=
1R1C
(vs − v0) +R2
R1
(dvs
dt− dv0
dt
)
Putting this into standard form gives the answer:
(R1 + R2)Cdv0
dt+ v0 = R2C
dvs
dt+ vs
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6.13 From the voltage law,
vs = Ri + Ldi
dt+
1C
∫i dt
Differentiate this with respect to time:
LCd2i
dt2+ RC
di
dt+ i = C
dvs
dt
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6.14is = i1 + i2
vo = Ri1
Thusv1 =
1C
∫i2 dt =
1C
∫(is − i1) dt =
1C
∫(is −
vo
R) dt
Also,
Ldi1dt
= v1 − vo =1C
∫(is −
vo
R) dt − vo
orL
R
dvo
dt=
1C
∫(is −
vo
R) dt− vo
ThusL
R
d2vo
dt2=
1C
(is −
vo
R
)− dvo
dt
Rearranging gives
LCd2vo
dt2+ RC
dvo
dt+ vo = Ris
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6.15v1 = L
di1dt
+ Ri3 (1)
Ri3 =1C
∫i2 dt + v2
ThusRC
di3dt
= i2 + Cdv2
dt(2)
i1 = i2 + i3 (3)
Transform (1) through (3) to obtain
LsI1(s) + RI3(s) = V1(s)
I2(s)− RCsI3(s) = −CsV2(s)
I1(s) − I2(s) − I3(s) = 0
These have the following solution
I1(s) =(RCs + 1)V1(s) − RCsV2(s)
D(s)
I2(s) =RCsV1(s) − (RCs + LCs2)V2(s)
D(s)
I3(s) =V1(s) + LCs2V2(s)
D(s)
whereD(s) = LRCs2 + Ls + R
The differential equation model is
LRCd2i1dt2
+ Ldi1dt
+ Ri1 = RCdv1
dt+ v1 − RC
dv2
dt
LRCd2i2dt2
+ Ldi2dt
+ Ri2 = RCdv1
dt− RC
dv2
dt− LC
d2v2
dt2
LRCd2i3dt2
+ Ldi3dt
+ Ri3 = v1 + LCd2v2
dt2
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6.16v1 = R1i1 + L1
di1dt
+ v3 (1)
v3 =1C
∫i3 dt
which gives
i3 = Cdv3
dt
v3 = L2di2dt
+ v2 (2)
i1 = i2 + i3 = i2 + Cdv3
dt(3)
Transform (1) through (3) to obtain
(R + L1s)I1(s) + V3(s) = V1(s)
L2sI2(s) − V3(s) = −V2(s)
I1(s) − I2(s) − CsV3(s) = 0
These have the following solution
I1(s) =(L2Cs2 + 1)V1(s)− V2(s)
D(s)
I2(s) =V1(s) − (L1Cs2 + RCs + 1)V2(s)
D(s)
V3(s) =L2sV1(s) + (L1s + R)V2(s)
D(s)
whereD(s) = L1L2Cs3 + RL2Cs2 + (L1 + L2)s + R
(continued on the next page)
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Problem 6.16 continued:
The differential equation model is
L1L2Cd3i1dt3
+ RL2Cd2i1dt2
+ (L1 + L2)di1dt
+ Ri1 = L2Cd2v1
dt2+ v1 − v2
L1L2Cd3i2dt3
+ RL2Cd2i2dt2
+ (L1 + L2)di2dt
+ Ri2 = v1 − L1Cd2v2
dt2− RC
dv2
dt− v2
L1L2Cd3v3
dt3+ RL2C
d2v3
dt2+ (L1 + L2)
dv3
dt+ Rv3 = L2
dv1
dt+ L1
dv2
dt+ Rv2
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
6.17 a) Choose v1 and i1 as the state variables because they describe the energy storage inthe circuit. The basic circuit equations are
v1 =1C
∫(is − i1) dt
v1 = Ldi1dt
+ vo vo = Ri1
These give the state equations
Cdv1
dt= is − i1 L
di1dt
= v1 − Ri1
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
6.18 Choose i1 and vc as the state variables because they describe the energy storage inthe circuit (vc is the voltage drop across the capacitor). The basic circuit equations are
v1 = L1di1dt
+ Ri3 Ri3 = vc + v2
vc =1C
∫i2 dt i1 = i2 + i3
These give the state equations
Ldi1dt
= v1 − v2 − vc Cdvc
dt= i1 −
1R
vc −1R
v2
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
6.19 a) Choose i1, i2, and v3 as the state variables because they describe the energy storagein the circuit. The basic circuit equations are
v1 = R1i1 + L1di1dt
+ v3 v3 =1C
∫i3 dt
v3 = L2di2dt
+ v2 i1 = i2 + i3
These give the state equations
L1di1dt
= v1 − v3 − Ri1 L2di2dt
= v3 − v2
Cdv3
dt= i1 − i2
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
6.20 It can be seen that R2 and C are in parallel. Thus, their equivalent impedance Z(s)is found from
1Z(s)
=1
1/Cs+
1R2
orZ(s) =
R2
R2Cs + 1
It can be seen that Z and R1 are in series. Thus,
Vs(s)I(s)
= Z(s) + R1
andVo(s) = R1I(s)
Eliminating I(s) from the last two relations yields the desired transfer function.
Vs(s) =Vo(s)R1
[Z(s) + R1]
or
T (s) =Vo(s)Vs(s)
=R1
Z(s) + R1
=R1R2Cs + R1
R1R2Cs + R2 + R1
The network is seen to be a first-order system with numerator dynamics.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
6.21 The impedance of R1 and C is their parallel combination:
11
R1+ 1
1/Cs
=R1
R1Cs + 1
The series combination of this impedance and R2 is:
R2 +R1
R1Cs + 1=
R1R2Cs + R1 + R2
R1Cs + 1=
Vs(s)I1(s)
and the required transfer function is
I1(s)Vs(s)
=R1Cs + 1
R1R2Cs + R1 + R2
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
6.22 Kirchhoff’s voltage law givesvs = Ri + vo
Solve this for i:i =
vs − vo
R(1)
For the capacitor:
vo =1C
∫i2 dt
which gives
Cdvo
dt= i2 (2)
For the inductor:vo = L
di1dt
Solve this for i1:
i1 =1L
∫vo dt (3)
From conservation of charge:i = i1 + i2 (4)
Substitute i, i1, and i2 from equations (1), (2), and (3) into equation (4) to obtain:
vs − vo
R=
1L
∫vo dt + C
dvo
dt
Differentiate both sides to get
1R
(vs − vo) =vo
L+ Cvo
Rearrange in standard form to get the answer:
RLCvo + Lvo + Rvo = Lvs
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
6.23 From conservation of charge:
is = i1 + i2 + i3 (1)
R, L, and C have the same voltage drop vo across them, so
vo = Ldi2dt
= Ri1 =1C
∫i3 dt
Integrate the first equation for i2 to obtain:
i2 =1L
∫vo dt (2)
Solve the second equation for i1:i1 =
vo
R(3)
Solve the third equation for i3:
i3 = Cdvo
dt(4)
Substitute equations (2), (3), and (4) into (1) to obtain:
is = Cdvo
dt+
vo
R+
1L
∫vo dt
Differentiate both sides to obtain
disdt
= Cd2vo
dt2+
1R
dvo
dt+
1L
vo
If desired, this can be “cleaned up” somewhat by multiplying both sides by RL to obtain
RLCd2vo
dt2+ L
dvo
dt+ Rvo = RL
disdt
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
6.24 Let Z1(s) = Vs(s)/I(s) be the series impedance of L, C1, C2, and R.
Z1(s) = Ls +1
C1s+
1C2s
+ R =LC1C2s
2 + RC1C2s + C1 + C2
C1C2s
Also,
Vo(s) =1
C2sI(s) =
1C2s
Vs(s)Z1(s)
=C1s
LC1C2s3 + RC1C2s2 + (C1 + C2)sVs(s)
and
LC1C2d3vo
dt3+ RC1C2
d2vo
dt2+ (C1 + C2)
dvo
dt= C1
dvs
dt
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6.25 Let i1 be the current through R1; i2 be the current through the left-side C; i3 be thecurrent through R2; i4 be the current through the right-side C; and i5 be the current to theoutput vo. Then from Kirchhoff’s voltage law,
Vs(s) =1
CsI2(s) + R2I2(s) R2I3(s) =
1Cs
I4(s) + Vo(s)
R1I1(s) =1
CsI2(s) +
1Cs
I4(s)
From conservation of charge,
I2(s) = I3(s) + I4(s) I5(s) = I1(s) + I4(s) = 0
Thus I4(s) = −I1(s). Use these equations to eliminate the current variables, and find theratio Vo(s)/Vs(s). The answer is
Vo(s)Vs(s)
=R1R2C
2s2 + 2R2Cs + 1R1R2C2s2 + (R1 + 2R2)Cs + 1
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6.26 The impedance of the R1C1 combination is
Z1(s) =R1
R1C1s + 1
For the R2C2 combination:
Z2(s) = R2 +1
C2s=
R2C2s + 1C2s
ThusVs(s) = Z1(s)I(s) + Z2(s)I(s) = [Z1(s) + Z2(s)] I(s)
andVo(s) = Z2(s)I(s) =
Z2(s)Z1(s) + Z2(s)
Vs(s)
ThusVo(s)Vs(s)
=Z2(s)
Z1(s) + Z2(s)=
(R1C1s + 1)(R2C2s + 1)R1C2s + (R1C1s + 1)(R2C2s + 1)
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6.27 The transfer function isVo(s)Vi(s)
= −Tf(s)Ti(s)
where Ti(s) = R1 and
Tf (s) = R2 +1
Cs=
R2Cs + 1Cs
ThusVo(s)Vi(s)
= −R2Cs + 1R1Cs
andR1C
dvo
dt= −R2C
dvi
dt− vi
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6.28 The transfer function isVo(s)Vi(s)
= −Tf(s)Ti(s)
where Tf (s) = R3 andTi(s) = Z(s) + R2
1Z(s)
=1
R1+
11
Cs
ThusZ(s) =
R1
R1Cs + 1
andTi(s) =
R1
R1Cs + 1+ R2 =
R1R2Cs + R1 + R2
R1Cs + 1Thus
Vo(s)Vi(s)
= − R3(R1Cs + 1)R1R2Cs + R1 + R2
andR1R2C
dvo
dt+ (R1 + R2)vo = −R1R3C
dvi
dt− vi
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6.29 The transfer function isVo(s)Vi(s)
= −Tf(s)Ti(s)
whereTf(s) = R2 +
1C2s
=R2C2s + 1
C2s
andTi(s) = Z(s) + R3
1Z(s)
=1R1
+11
C1s
ThusZ(s) =
R1
R1C1s + 1
andTi(s) =
R1
R1C1s + 1+ R3 =
R3R1C1s + R3 + R1
R1C1s + 1Thus
Vo(s)Vi(s)
= − (R2C2s + 1)(R1C1s + 1)C2s(R3R1C1s + R3 + R1)
and
R3R1C1C2d2vo
dt2+ (R3 + R1)C2
dvo
dt= −R2R1C1C2
d2vi
dt2− (R1C1 + R2C2)
dvi
dt− vi
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6.30 Let Z1(s) be the forward impedance, and Z2(s) be the feedback impedance. Then
1Z1(s)
=1
R1+
11
C1s
Z1(s) =R1
R1C1s + 1
Similarly,
Z2(s) =R2
R2C2s + 1and
Vo(s)Vi(s)
= −Z2(s)Z1(s)
= −R2C2s + 1R1C1s + 1
R1
R2
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6.31 Let Z1(s) be the impedance of the series R1C1 connection, Z2(s) the forward impedance,Z3(s) the impedance of the series R3C2 connection, and Z4(s) the feedback impedance.Then
Z1(s) = R1 +1
C1s=
R1C1s + 1C1s
1Z2(s)
=1
R2+
1Z1(s)
=(R1C1 + R2C1)s + 1
R1C1s + 1
Z3(s) =R3C2s + 1
C2s
Z4(s) =R3C2s + 1
(R3C2 + R4C2)s + 1
ThusVo(s)Vi(s)
= −Z4(s)Z2(s)
= − R3C2s + 1(R3C2 + R4C2)s + 1
(R1C1 + R2C1)s + 1R1C1s + 1
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6.32 a) From equations (3) and (4) of Example 6.5.1,
(Is2 + cs + KT )Θ(s) = nBLrI(s)
(Ls + R)I(s) + nBLrsΘ(s) = Vi(s)
Let A = nBLr. The solution for Θ(s) gives
Θ(s)Vi(s)
=A
(Is2 + cs + KT )(Ls + R) + A2=
A
LIs3 + (RI + Lc)s2 + (LKT + Rc)s + RKT + A2
b) Assuming all the parameters are positive, the system is stable if and only if
(RI + Lc)(LKT + Rc) > LI(RKT + A2)
If the system is stable, the final value theorem can be applied. It gives
θss = lims→0
sA
LIs3 + (RI + Lc)s2 + (LKT + Rc)s + RKT + A2
Vi
s=
A
RKT + A2Vi
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6.33 a) With vf = 0, if = 0, and
IsΩ(s) = −cΩ(s)− TL(s)
ThusΩ(s)TL(s)
= − 1Is + c
b)
Id2θ
dt2= KT if − c
dθ
dt− TL
Thus (Is2 + cs
)Θ(s) = KT If (s) − TL(s)
Also,Vf(s) = (Lfs + Rf)If (s)
and therefore (Is2 + cs
)Θ(s) = KT
Vf(s)(Lfs + Rf)
− TL(s)
This gives
Θ(s)Vf(s)
=KT
(Lfs + Rf)(Is2 + cs)=
KT
s[ILfs2 + (RfI + Lfc)s + Rfc]
andΘ(s)TL(s)
= − 1s(Is + c)
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6.34 The equivalent inertia and damping at the motor shaft are
Ie = Im +IL
N2ce =
cL
N2+ cm
The transfer functions are
Ωf(s)Vf(s)
=KT /N
(Lfs + Rf)(Ies + ce)
Ωf(s)TL(s)
=1/N2
Ies + ce
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6.35 Summing moments on I1 and I2 gives
I1d2θ1
dt2= T − kT (θ1 − θ2) = KT if − kT (θ1 − θ2)
I2d2θ2
dt2= kT (θ1 − θ2)− cθ2
For the field circuit,
vf = Lfdifdt
+ Rif
Transforming these equations and solving for Θ2(s) gives
Θ2(s) =kTKT
s2(Lfs + Rf)[I1I2s2 + cI1s + kT (I1 + I2)]
The differential equation is
I1I2Lfd5θ2
dt5+(cI1I2Lf+Rf )
d4θ2
dt4+[LfkT (I1+I2)+RfcI1]
d3θ2
dt3+RfkT (I1+I2)
d2θ2
dt2= kT KTvf
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6.36 There are two circuits and one inertia in this system, and we must write an equationfor each. For the generator circuit, Kirchhoff’s voltage law gives
vf = Rf if + Lfdifdt
For the motor circuit, Kirchhoff’s voltage law gives
va = Raia + Ladiadt
+ Kbω
where va = Kf if . For the inertia I , Newton’s law gives:
Idω
dt= T − cω − TL
where T = KT ia. Substitute for va and T , and rearrange to obtain:
Lfdifdt
+ Rf if = vf
Idω
dt+ cω = KT ia − TL
Ladiadt
+ Raia = Kf if − Kbω
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6.37 Equations (6.6.5) and (6.6.6) give
ia =5 × 10−4Va + 0.2TL
(0.8)5× 10−4 + (0.2)2=
7.5 + 200TL
40.4(1)
ω =0.2Va − 0.8TL
4.04× 10−2=
300− 80TL
4.04(2)
Setting TL = 0 in (2) gives the no-load speed ω = 300/4.04 = 74.26 rad/s or 709 rpm. From(1) we obtain the no-load current: ia = 7.5/40.4 = 0.186 A.
Setting ω = 0 in (2) gives the stall torque TL = 0.2(15)/0.8 = 3.75 N·m. From (1) weobtain the stall current: ia = [7.5 + 200(15/4)]/4.04 = 187.5 A.
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6.38 (a) The characteristic equation is
LaIs2 + (Lac + RaI)s + Rac + KbKT = 0
The roots are
s =−(Lac + RaI)±
√R2
aI2 − 2RaILac + L2
ac2 − 4LaIKbKT
2LaI
b) For c = 0 the roots are −219 and −47.5. The speed and the current will not oscillate.It will take about 4(1/47.5) = 0.08 second for the speed and the current to become constant.
For c = 0.01 the roots are −196 ± 73.5i. The speed and current will oscillate with afrequency of 73.5 rad/s and a period of 0.085 s. It will take about 4(1/196) = 0.02 secondfor the speed and current to become constant. Because this time is less than the period, wewill not see any oscillations in the plots.
For c = 0.1 the roots are −1240 and −277. The speed and current will not oscillate. Itwill take about 4(1/277) = 0.014 second for them to become constant.
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6.39 Use the transfer functions given by (6.6.1) through (6.6.4). All four transfer functionshave the same denominator, which is
D(s) = 20× 10−7s2 + 4.02× 10−4s + 4× 10−4
Current Response to Command Input Assuming va is a unit step input, we have from(6.6.1):
Ia(s) =5 × 10−4(s + 1)
sD(s)=
250(s + 1)s(s2 + 201s + 2020)
(1)
The characteristic roots are s = −100.5± 100.5j. Thus the denominator can be expressedas (s + 100.5)2 + (100.5)2. Expanding Ia(s) in a partial fraction expansion gives
Ia(s) =C1
s+
100.5C2
(s + 100.5)2 + (100.5)2+
(s + 100.5)C3
(s + 100.5)2 + (100.5)2(2)
This gives a solution of the form:
ia(t) = C1 + C2e−100.5t sin 100.5t + C3e
−100.5t cos 100.5t
Reducing (2) to a single fraction using the least common denominator s[(s + 100.5)2 +(100.5)2], we obtain
Ia(s) =C1[(s + 100.5)2 + (100.5)2] + [100.5C2 + (s + 100.5)C3]s
s[(s + 100.5)2 + (100.5)2](3)
Comparing the numerators of (1) and (3), we obtain
C1[(s + 100.5)2 + (100.5)2] + [100.5C2 + (s + 100.5)C3]s = 250(s + 1)
or(C1 + C3)s2 + (201C1 + 100.5C2 + 100.5C3)s + 20, 200C1 = 250s + 250
Comparing like powers of s gives
C1 = −C3 = 0.01238 C2 = 2.47518
Thus, after multiplying by 10, the magnitude of the step input, we have
ia(t) = 0.1238 + 24.7518e−100.5t sin 100.5t− 0.1238e−100.5t cos 100.5t
(continued on the next page)
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Problem 6.39 continued:
Speed Response to Command Input Assuming va is a unit step input, we have from(6.6.3):
Ω(s) =0.2
sD(s)=
105
s(s2 + 201s + 2020)(4)
Expanding Ω(s) in a partial fraction expansion gives
Ω(s) =C1
s+
100.5C2
(s + 100.5)2 + (100.5)2+
(s + 100.5)C3
(s + 100.5)2 + (100.5)2(5)
This gives a solution of the form:
ω(t) = C1 + C2e−100.5t sin 100.5t + C3e
−100.5t cos 100.5t
Reducing (5) to a single fraction using the least common denominator s[(s + 100.5)2 +(100.5)2], we obtain
Ω(s) =C1[(s + 100.5)2 + (100.5)2] + [100.5C2 + (s + 100.5)C3]s
s[(s + 100.5)2 + (100.5)2](6)
Comparing the numerators of (5) and (6), we obtain
C1[(s + 100.5)2 + (100.5)2] + [100.5C2 + (s + 100.5)C3]s = 105
or(C1 + C3)s2 + (201C1 + 100.5C2 + 100.5C3)s + 20, 200C1 = 105
Comparing like powers of s gives
C1 = −C3 = 4.9505 C2 = −C1 = −4.9505 (7)
Thus, after multiplying by 10, the magnitude of the step input, we have
ω(t) = 49.505(1 − e−100.5t sin 100.5t− e−100.5t cos 100.5t
)
(continued on the next page)
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Problem 6.39 continued:
Current Response to Disturbance Input Comparing (6.6.2) with (6.6.3), and notingthat Kb = KT , we see that
Ia(s)TL(s)
=Kb
KT
Ω(s)Va(s)
=Ω(s)Va(s)
Thus the response of ia to a unit-step disturbance TL will be the same as the response of ωto a unit-step voltage va. Thus we can use the coefficients given in (7) and multiply themby 0.2, the magnitude of TL, to obtain
ıa(t) = 0.9901(1 − e−100.5t sin 100.5t− e−100.5t cos 100.5t
)
(continued on the next page)
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Problem 6.39 continued:
Speed Response to Disturbance Input To obtain the response of ω to TL, use (6.6.4).Assuming TL is a unit step input, we have from (6.6.4):
Ω(s) =4 × 10−3s + 0.8
sD(s)=
5 × 10−4s + 0.1s(s2 + 201s + 2020)
(8)
Expanding Ω(s) in a partial fraction expansion gives
Ω(s) =C1
s+
100.5C2
(s + 100.5)2 + (100.5)2+
(s + 100.5)C3
(s + 100.5)2 + (100.5)2(9)
This gives a solution of the form:
ω(t) = C1 + C2e−100.5t sin 100.5t + C3e
−100.5t cos 100.5t
Reducing (9) to a single fraction using the least common denominator s[(s + 100.5)2 +(100.5)2], we obtain
Ω(s) =C1[(s + 100.5)2 + (100.5)2] + [100.5C2 + (s + 100.5)C3]s
s[(s + 100.5)2 + (100.5)2](10)
Comparing the numerators of (9) and (10), we obtain
C1[(s + 100.5)2 + (100.5)2] + [100.5C2 + (s + 100.5)C3]s = 5× 10−4s + 0.1
or(C1 + C3)s2 + (201C1 + 100.5C2 + 100.5C3)s + 20, 200C1 = 5 × 10−4s + 0.1
Comparing like powers of s gives
C1 = −C3 = 4.9505× 10−6 C2 = −4.187× 10−7
Thus, after multiplying by 0.2, the magnitude of the step input, we have
ω(t) = 9.901× 10−7 − 8.37× 10−8e−100.5t sin 100.5t− 9.901× 10−7e−100.5t cos 100.5t
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6.40 Given Va = 20 and
Istall = 25 ino load = 0.6 ωno load =240060
2π = 80π rad/s
ThusRa =
Va
istall=
2025
= 0.8 Ω
From (6.6.6) and (6.6.6) with TL = 0 and Kb = KT ,
ino load = 0.6 =20c
0.8c + K2T
ωno load = 80π =20KT
0.8c + K2T
These two equations have the solution KT = Kb = 0.078 N·m/A and c = 1.85 × 10−4
N·m·s/rad.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
6.41 Refer to Table 6.6.1. θf = 3π/4 rad. Also,
I = 0.215 Td = 4.2 tf = 2
t1 = 0.3 t2 = 1.7 R = 4
KT = 0.3 = Kb
Note that L is not needed.The calculated quantities are
E = 35.7 J per cycle
ωmax = 1.386 rad/s = 13.2 rpm
Tmax = 5.19 N · m
Trms = 4.24 N ·m
imax = 17.3 A
irms = 14.1 A
vmax = 69.6 V
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6.42 Refer to Table 6.6.1. θf = 11(2π) = 22π rad. Also,
I = 0.05 Td = 3.6 tf = 3
t1 = 0.5 t2 = 2.5 R = 3
KT = 0.4 = Kb
Note that L is not needed.The calculated quantities are
E = 872 J per cycle
ωmax = 27.65 rad/s = 264 rpm
Tmax = 6.36 N · m
Trms = 3.94 N ·m
imax = 15.9 A
irms = 9.85 A
vmax = 58.8 V
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6.43 a) The transfer function is
Y (s)Z(s)
= T (s) =s2
s2 + 18s + 100
With s = 120j we obtain
M = |T (120j)| =∣∣∣∣∣
−(120)2
100− (120)2 + 18(120)j
∣∣∣∣∣ = 0.9957
φ = 6 T (120j) = − tan−1 18(120)100− (120)2
= −2.992
Thus the steady state response is
y(t) = 9.957 sin(120t− 2.992)
b) The transfer function is
Y (s)Z(s)
= T (s) =s2
s2 + 1800s + 106
With s = 120j we obtain
M = |T (120j)| =∣∣∣∣∣
−(120)2
106 − (120)2 + 1800(120)j
∣∣∣∣∣ = 0.0144
φ = 6 T (120j) = − tan−1 1800(120)106 − (120)2
= −0.2157
Thus the steady state response is
y(t) = 0.144 sin(120t− 0.2157)
For case (a), the amplitude of the response (9.957) is almost identical to the amplitudeof the displacement input (10). Thus the instrument functions as a vibrometer.
For case (b), the amplitude of the response (0.144) is equal to 10−6 times the amplitudeof the acceleration of the input, which is 10(120)2 = 0.144 × 106. Thus the instrumentfunctions as an accelerometer with a gain of 10−6.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
6.44 Summing forces on the mass m gives
md2x
dt2+ c
dx
dt+ kx = fs + Kf i
or(ms2 + cs + k)X(s) = Fs(s) + KfI(s) (1)
Since the applied voltage is zero, the circuit equation is
Ldi
dt+ Ri + Kb
dx
dt= 0
or(Ls + R)I(s) + KbsX(s) = 0 (2)
Solving (1) for X(s) and substituting into (2), and rearranging, gives the transfer func-tion.
I(s)Fs(s)
=−Kbs
mLs3 + (cL + mR)s2 + (kL + cR + KbKf )s + kR
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6.45 a) The roots of the second-order model are s = −4333± 13 462j.b) The roots of the third-order model are s = −1643 ± 16 513j, which is the dominant
pair, and s = −8715.
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6.46 The script file is
KT = 0.2; Kb = 0.2; c = 5e-4; Ra = 0.8;La = 4e-3; I = 5e-4;den = [La*I, Ra*I + c*La, c*Ra + Kb*KT];% Speed transfer functionsys1 = tf(KT, den);% Current transfer functionsys2 = tf([I, c], den);[om, t1] = step(sys1);[ia, t2] = step(sys2);subplot(2,1,1)plot(t1, 10*om),xlabel(′t (s)′),ylabel(′\omega (rad/s)′)subplot(2,1,2)plot(t1, 10*ia),xlabel(′t (s)′),ylabel(′i_a (A)′)
The plot is shown in the following figure. The peak current is approximately 8 A. The moreaccurate value of 8.06 A can be found with the step(sys2) function by right-clicking onthe plot, selecting characteristics, then peak response, and multiplying the answer by 10.
0 0.01 0.02 0.03 0.04 0.05 0.060
10
20
30
40
50
60
t (s)
ω (ra
d/s)
0 0.01 0.02 0.03 0.04 0.05 0.06−2
0
2
4
6
8
10
t (s)
i a (A)
Figure : for Problem 6.46
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
6.47 The script file is
KT = 0.2; Kb = 0.2; c = 5e-4; Ra = 0.8; La = 4e-3; I = 5e-4;den = [La*I, Ra*I + c*La, c*Ra + Kb*KT];% Speed transfer function: sys1 = tf(KT, den);% Current transfer function: sys2 = tf([I, c], den);% Applied voltaget = [0:0.001:0.05];va = 10*ones(size(t));om = lsim(sys1, va, t);ia = lsim(sys2, va, t);subplot(2,1,1)plot(t, om),xlabel(′t (s)′),ylabel(′\omega (rad/s)′)subplot(2,1,2)plot(t, ia),xlabel(′t (s)′),ylabel(′i_a (A)′)
The plot is shown in the following figure. The peak current is approximately 8 A. Themore accurate value of 8.06 A can be found with the lsim(sys2, va, t) function byright-clicking on the plot, selecting characteristics, then peak response.
0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.050
10
20
30
40
50
60
t (s)
ω (ra
d/s)
0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05−2
0
2
4
6
8
10
t (s)
i a (A)
Figure : for Problem 6.47
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
6.48 See Example 6.3.3 for the transfer functions. The script file is
R = 1e+3; C = 2e-6; L = 2e-3;den = [L*R*C, L, R];% Transfer function for v1:sys1 = tf(1, den);% Transfer function for v2:sys2 = tf([R*C, 0], den);% Response to unit-step voltage v1:t = [0:0.00001:0.025];i31 = step(sys1, t);% Voltage v2:v2 = 4*sin(2*pi*60*t);% Response to voltage v2:i32 = lsim(sys2, v2, t);% Total response:i3 = 5*i31 + i32;plot(t, i3),xlabel(′t (s)′),ylabel(′i_3 (A)′)
The plot is shown in the figure on the following page. The steady-state response is a sinewave.
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0 0.005 0.01 0.015 0.02 0.0250
0.002
0.004
0.006
0.008
0.01
0.012
0.014
0.016
t (s)
i 3 (A)
Figure : for Problem 6.48
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
6.49 See equations (6.6.1) through (6.6.4) for the transfer functions, where I and c mustbe replaced with the equivalent inertia and equivalent damping: Ie = I + 4× 10−4/N2 andce = c + 1.8× 10−3/N2. The script file is
KT = 0.2; Kb = 0.2; c = 3e-4; Ra = 0.8;La = 4e-3; I = 4e-4; N = 3;% Equivalent inertia:Ie = I + 1e-3/N^2;% Equivalent damping:ce = c + 1.8e-3/N^2;den = [La*Ie, Ra*Ie + ce*La, ce*Ra + Kb*KT];% Speed command transfer functionsys1c = tf(KT, den);% Current command transfer functionsys2c = tf([I, c], den);% Speed disturbance transfer functionsys1d = tf([La, Ra], den);% Current disturbance transfer functionsys2d = tf(Kb, den);% Unit Command response:[omc, t1c] = step(sys1c);[iac, t2c] = step(sys2c);% Unit Disturbance response:omd = step(sys1d, t1c);iad = step(sys2d, t2c);% Total response:om = 20*omc + 0.04*omd;ia = 20*iac + 0.04*iad;subplot(2,1,1)plot(t1c, om),xlabel(′t (s)′),ylabel(′\omega (rad/s)′)subplot(2,1,2)plot(t2c, ia),xlabel(′t (s)′),ylabel(′i_a (A)′)
The plot is shown in the figure on the following page. The peak current is 12.73 A, whichcan be found by displaying the values in the array ia.
(continued on the next page)
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Problem 6.49 continued:
0 0.01 0.02 0.03 0.04 0.05 0.060
20
40
60
80
100
120
t (s)
ω (ra
d/s)
0 0.01 0.02 0.03 0.04 0.05 0.06−5
0
5
10
15
t (s)
i a (A)
Figure : for Problem 6.49
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
6.50 The main script file is Problem6p50.m.
% Program Problem6p50.mKT = 0.05; Kb = KT; c = 0;Ra = 0.8; La = 3e-3; I = 8e-5;trapezoidmotortfcurrent = lsim(currenttf, v, t);speed = lsim(speedtf, v, t);subplot(2,1,1)plot(t,current),xlabel(′t (s)′),ylabel(′Current (A)′)subplot(2,1,2)plot(t,speed),xlabel(′t (s)′),ylabel(′Speed (rad/s)′)performance
This program calls the following files.
% trapezoid.m Trapezoidal voltage profilet1 = 0.5; t2 = 2; tfinal = 2.5; t3 = 4;max_v = 30;dt = t3/1000;t = [0:dt:t3];for k = 1:1001
if t(k) <= t1v(k) = (max_v/t1)*t(k);
elseif t(k) <= t2v(k) = max_v;
elseif t(k) <= tfinalv(k) = (max_v/t1)*(tfinal - t(k));
elsev(k) = 0;
endend
(continued on the next page)
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Problem 6.50 continued:
% motortf.m Motor transfer functions for voltage input.% current:currenttf = tf([I,c],[La*I,Ra*I+c*La,c*Ra+Kb*KT]);% speed:speedtf = tf(KT,[La*I,Ra*I+c*La,c*Ra+Kb*KT]);
% performance.m Computes motor performance measures.ia = current;dt = t(2) - t(1);E = trapz(t,Ra*ia.^2) + trapz(t,c*speed.^2)i_max = max(ia)i_rms = sqrt(trapz(t,ia.^2)/t3)T_max = KT*i_maxT_rms = KT*i_rmsspeed_max = max(speed)v_max = Ra*i_max+Kb*speed_max
The plots are shown in the figure on the following page. The performance results are
E = 2.8203 J/cycle imax = 1.9200 A irms = 0.9388 A
Tmax = 0.0960 N · m Trms = 0.0469 N · m
speedmax = 600 rad/s vmax = 31.536 V
(continued on the next page)
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Problem 6.50 continued:
0 0.5 1 1.5 2 2.5 3 3.5 4−2
−1
0
1
2
t (s)
Cur
rent
(A)
0 0.5 1 1.5 2 2.5 3 3.5 40
100
200
300
400
500
600
700
t (s)
Spe
ed (r
ad/s
)
Figure : for Problem 6.50
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
6.51 The equations are
Ladiadt
= va − Raia − Kbω (1)
dθ
dt= ω (2)
Iedω
dt= KT ia − ceω − 4.2 sin θ (3)
Choose the three state variables to be x1 = ia, x2 = θ, and x3 = ω.The slew speed required for a rotation angle of 3π/4 rad is ωslew = 3π/4/t2 = 1.386
rad/s. Thus, letting A represent ωslew, the specified trapezoidal speed profile is
ω(t) =
Att1
0 ≤ t ≤ t1A t1 ≤ t ≤ t2−A
t1(t − t2) + A t2 ≤ t ≤ tfinal
The specified angle θ(t) resulting from the trapezoidal speed profile can be found by inte-grating that profile. The result is
θ(t) =
At2
2t10 ≤ t ≤ t1
A(t − t1) + At12 t1 ≤ t ≤ t2
−A(t−t2)2
2t1+ A(t − t2) + At2 − At1
2 t2 ≤ t ≤ tfinal
(continued on the next page)
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Problem 6.51 continued:
The voltage required to move the arm at the desired angle θ(t) is found from (1):
va = Ladiadt
+ Raia − Kbω (4)
We obtain ia from (3):
ia(t) =1
KT
(Ie
dω
dt+ ceω + 4.2 sin θ
)(5)
Note that dω/dt = α, the angular acceleration, where
α(t) =
At1
0 ≤ t ≤ t10 t1 ≤ t ≤ t2−A
t1t2 ≤ t ≤ tfinal
Thus (5) can be expressed as
ia(t) =1
KT(Ieα + ceω + 4.2 sin θ) (6)
Differentiate this to obtain dia/dt:
diadt
=1
KT
(Ie
dα
dt+ ceα + 4.2θ cos θ
)(7)
Note that θ = ω and that dα/dt = 0. Also, for this problem, ce = 0. Thus (7) becomes
diadt
=1
KT4.2ω cos θ (8)
Thus, with ce = 0, (4) becomes
va =4.2La
KTω cos θ +
Ra
KT(Ieα + 4.2 sin θ) − Kbω (9)
In summary, va is computed from (9) using the expressions for α(t), ω(t) and θ(t). Thecomputed values of va are then used with (2) and (3) to compute the actual values of ω(t)and θ(t). The m-files required for this are shown on the next page.
(continued on the next page)
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Problem 6.51 continued:
% Program problem6p51.mglobal KT Kb La Ra Ieglobal t1 t2 tfinal AKT = 0.3; Kb = 0.3; Ra = 4;La = 3e-3; Ie = 0.215;t1 = 0.3;t2 = 1.7; tfinal = 2;% A is the slew speedA = (3*pi/4)/t2;[t, x] = ode45(′problem6p52a′, [0, tfinal], [0, 0, 0]);subplot(3,1,1)plot(t, x(:,1)),ylabel(′i_a (A)′)subplot(3,1,2)plot(t, x(:,2)),ylabel(′\theta (rad)′)subplot(3,1,3)plot(t, x(:,3)),xlabel(′t (s)′),ylabel(′\omega (rad/s)′)
(continued on the next page)
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Problem 6.51 continued:
function xdot = problem6p51a(t,x)% Differential equation file for program problem6p52.mglobal KT Kb La Ra Ieglobal t1 t2 tfinal A% Evaluate trapezoidal speed profile% om is the speedif t <= t1
om = (A/t1)*t;elseif t <= t2
om = A;else om = -(A/t1)*(t-t2) + A;end% Evaluate angle profile% theta is the angleif t <= t1
theta = (A/(2*t1))*t.^2;elseif t <= t2
theta = A*(t-t1)+A*t1/2;else theta = -(A/(2*t1))*(t-t2).^2 + A*(t-t2) + A*t2-A*t1/2;end% Evaluate angular acceleration profile% alpha is the accelerationif t <= t1
alpha = A/t1;elseif t <= t2
alpha = 0;else alpha = -A/t1;end% Evaluate the voltage profile% va is the voltageva = 4.2*(La/KT)*om*cos(theta) + (Ra/KT)*(Ie*alpha + 4.2*sin(theta)) - Kb*om;% x(1) is the current; x(2) is the angle; x(3) is the speedxdot = [(-Ra*x(1)-Kb*x(3)+va)/La;x(3);(KT*x(1)-4.2*sin(x(2)))/Ie];
(continued on the next page)
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Problem 6.51 continued:
The simulation results are shown in the following figure. The angle θ does not quitereach the desired value of 3π/4. From the velocity plot we see that the velocity starts tofall below that specified by the trapezoidal profile. This is due to the approximation madeby neglecting dα/dt at the corners of the velocity profile, at the time t1 and t2. Because α,the slope of the profile, changes suddenly at these times, dα/dt is undefined, and thus ourapproximation does not represent a true solution of the differential equations.
This problem is an example of open-loop control in which the voltage is precalculatedfrom the differential equations. Thus the effectiveness of this method depends on the ac-curacy of the differential equation model and on any approximation made to solve for thevoltage. The method is also tedious, as this problem illustrates. A better way to controlthe arm angle is to use feedback control, which is discussed in Chapter 10.
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20
5
10
15
i a (A)
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20
1
2
3
θ (r
ad)
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2−2
−1
0
1
2
t (s)
ω (r
ad/s
)
Figure : for Problem 6.51
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
6.52 From Example 6.3.1,
I3(s) =V1(s) + RCsV2(s)LRCs2 + Ls + R
To avoid using a Transfer Function block containing the illegal transfer function RCs (whichis a differentiator), we can write the equation as follows:
I3(s) =RCsV1(s)
RCs(LRCs2 + Ls + R)+
RCsV2(s)LRCs2 + Ls + R
The corresponding Simulink model is shown in the following figure. In the Sine block, setthe Amplitude to 4 and the Frequency to 120*pi. In the Pulse Generator block, set theAmplitude to 5, the Period to 1, and the Pulse Width to 5. Set the Stop Time to whatevervalue you want, greater than 0.05 to see the full pulse but less than 1 so that the pulse willnot repeat; say 0.075. In the To Workspace block, set the Save Format to array. Beforerunning the model, type the following in the MATLAB Command window.
R = 1e4; C = 2e-6; L = 2e-3;
You can plot the response by typing plot(tout, simout(:,1)). Note that if you use alarger Stop time, the buffer for tout may overflow. The number of points to be retained intout can be increased if necessary. This is done under the Configuration Parameters menu.
Figure : for Problem 6.52
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6.53 From Figure 6.3.2b we can create the corresponding Simulink model, which is shownin the following figure. In the Sine block, set the Amplitude to 3 and the Frequency to120*pi. In the Step block, set the Step time to 0, the Initial value to 0, and the Final valueto 12. Set the Stop Time to an appropriate value, say 0.5. In the To Workspace block, setthe Save Format to Array. Before running the model, type the following in the MATLABCommand window.
R = 2e4; C = 3e-6;
You can plot the response v1(t) by typing plot(tout, simout(:,1)). You can plot theresponse v2(t) by typing plot(tout, simout(:,2)).
Figure : for Problem 6.53
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6.54 The Simulink model is like that shown in Figure 6.9.5, with the disturbance torque Tdset to 0. In the Step block, set the Step time to 0, the Initial value to 0, and the Final valueto 10. Set the Stop Time to an appropriate value, say 0.1. In the To Workspace block, setthe Save Format to Array. Before running the model, in the MATLAB Command window,type
KT = 0.2; Kb = KT; c = 5e-4;Ra = 0.8; La = 4e-3; I = 5e-4;
For part (a), set the limits on the saturation block very high, say ±100. You can plot thetorque response by typing plot(tout, simout(:,1)). You can plot the speed response bytyping plot(tout, simout(:,2)). The following figure shows the two plots created withthe subplot function. The maximum torque is 1.6 N·m.
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1−0.5
0
0.5
1
1.5
2
t (s)
Tor
que
(N m
)
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10
10
20
30
40
50
60
t (s)
Spe
ed (
rad/
s)
Figure : for Problem 6.54a
(continued on the next page)
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Problem 6.54 continued:
For part (b), set the limits on the saturation block to ±0.8. The resulting responses areshown in the following figure.
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1−0.2
0
0.2
0.4
0.6
0.8
t (s)
Tor
que
(N m
)
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10
10
20
30
40
50
60
t (s)
Spe
ed (
rad/
s)
Figure : for Problem 6.54b
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
6.55 The model is shown in the following figure. Set the Initial output to 0 in all fourRamp blocks. In the topmost Ramp block, set the Slope to 60 and the Start time to 0. Inthe Ramp1 block, set the Slope to −60 and the Start time to 0.5. In the Ramp2 block, setthe Slope to −60 and the Start time to 2. In the Ramp3 block, set the Slope to 60 and theStart time to 2.5. Set the Stop Time to 4. In the To Workspace block, set the Save Formatto array. Before running the model, type the following in the MATLAB Command window.
KT = 0.05;Kb = 0.05;c = 0;Ra = 0.8;La = 3e-3;I = 8e-5;
You can plot the current response by typing plot(tout, simout(:,1)) and the speedresponse by typing plot(tout, simout(:,2)).
Figure : for Problem 6.55
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Solutions Manual c©
to accompany
System Dynamics, Second Edition
by
William J. Palm III
University of Rhode Island
Solutions to Problems in Chapter Eight
c©Solutions Manual Copyright 2010 The McGraw-Hill Companies. All rightsreserved. No part of this manual may be displayed, reproduced, or distributedin any form or by any means without the written permission of the publisheror used beyond the limited distribution to teachers or educators permitted byMcGraw-Hill for their individual course preparation. Any other reproductionor translation of this work is unlawful.
8.1 a)
yss =25√
(14ω)2 + 18215 =
25√[14(1.5)]2 + 182
15 = 13.5582
b)
yss =15ω√
[3ω]2 + 425 =
15(2)√(3(2))2 + 42
5 = 20.8013
c)
yss =√
(ω)2 + 502
√(ω)2 + 1502
3 =√
104 + 502
√104 + 1502
3 = 1.8605
d)
yss =33200
√ω2 + 1002
√ω2 + 332
8 =33200
√502 + 1002
√502 + 332
8 = 2.46
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
8.2 a) In standard form of (8.1.2), the transfer function is
T (s) =15
6s + 2=
152
13s + 1
Here τ = 3 and the multiplicative factor 15/2 shifts the m curve up by 20 log 152 = 17.501
dB. The m plot looks like that in Figure 8.1.6, but shifted up by 17.501 dB. Because6 (15/2) = 0, the phase curve is identical to that shown in Figure 8.1.6.
b) In standard form of τs/(τs + 1), the transfer function is
T (s) =9s
8s + 4=
92
2s
2s + 1
Here τ = 2 and the multiplicative factor 9/2 shifts the m curve up by 20 log 92 = 13.064 dB.
The m plot looks like that in Figure 8.1.9, but shifted up by 13.064 dB. Because 6 (9/2) = 0,the phase curve is identical to that shown in Figure 8.1.9.
c) In standard form of (8.1.13), the transfer function is
T (s) = 614s + 710s + 2
=72
2s + 15s + 1
Here τ1 = 2 and τ2 = 5. The multiplicative factor 7/2 shifts the m curve up by 20 log 72 =
10.881 dB. The m plot looks like that in part (a) of Figure 8.1.12 (because τ1 < τ2), butshifted up by 10.881 dB. Note that Figure 8.1.12 applies to the case where K = 1 in (8.1.13).
Because 6 (7/2) = 0, the phase curve is given by
φ(ω) = 6 (1 + 2ωj)− 6 (1 + 5ωj)
At low frequencies, φ ≈ 0 but negative. At ω = 1/τ2 = 1/5 and at ω = 1/τ1 = 1/2,φ = −23 (this identical value is a coincidence because here τ1 = 1/τ2). As ω → ∞, φ → 0
through negative values.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
8.3 The input is
hi(t) = 10 + 3 sin2π
12t
We are given that the steady-state response is
h(t) = 10 + 2 sin(
2π
12t + φ
)
Thus the amplitude ratio is M = 2/3.The model is
Ah = − g
R(h − hi)
so that the time constant is τ = RA/g. Thus the model can be written as
τ h + h = hi
The amplitude ratio for this model is
M =1√
1 + ω2τ2=
23
where ω = 2π/12 = π/6. Solving for τ gives
τ =2π
√5 = 1.423 hr
The phase shift isφ = − tan−1(ωτ) = −0.641 rad
and the time shift is |φ|/ω = 1.22 hr.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
8.4 The total wall surface area is A = 4(5×3) = 60 m2, and the air volume is V = 3(5×5) =75 m3. The thermal capacitance of the room air is
C = mcp = ρV cp = 1.289(75)(1004) = 9.704× 104
The outside temperature is To = 15 + 5 sin ωt where
ω =2π
24(3600)= 7.272× 10−5 rad/sec
The model isC
dT
dt=
1RT
(To − T )
orRTC
dT
dt+ T = To
where RT is the total wall resistance, which is
RT = AR = 60(4.5× 10−3) = 0.27
The time constant isτ = RTC = 2.62× 104 s
The amplitude ratio is
M =1√
1 + ω2τ2= 0.4647
Thus the amplitude of oscillation of the indoor temperature is 5(0.4647) = 2.32C. Thephase shift is
φ = − tan−1 ωτ = −1.087 rad
ThusT (t) = 15 + 2.32 sin(7.272× 10−5t − 1.087)
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
8.5 The model is Iω + cω = T , or 2ω + 4ω = T . The transfer function is
Ω(s)T (s)
=1
2s + 4=
0.250.5s + 1
and the amplitude ratio is
M(ω) =0.25√
1 + 0.25ω2
The phase shift is φ(ω) = − tan−1(ωτ) = − tan−1(0.5ω).Because there are three terms in the forcing function, there are three corresponding
terms in the steady-state response. These are
ωss(t) = A1M1 + A2M2 sin(3t + φ2) + A3M3 cos(5t + φ3)
From the M(ω) and φ(ω) equations we obtain
M1 = M(0) = 0.25 M2 = M(3) = 0.1387 M3 = M(5) = 0.0928
φ2 = φ(3) = − tan−1(1.5) = −0.983 φ3 = φ(5) = − tan−1(2.5) = −1.19
Thus, noting from the input that A1 = 30, A2 = 5, and A3 = 2, we have
ωss(t) = 7.5 + 0.6935 sin(3t − 0.983) + 0.1856 cos(5t − 1.19)
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
8.6 The model isAh = − g
Rh + qvi
orRA
gh + h =
R
gqvi
With the given values,585.39h + h = 46.58qvi
The time constant is τ = 585.39 sec.The amplitude ratio is
M(ω) =46.58√
1 + ω2τ2
where ω = 0.002.The phase shift is
φ(ω) = − tan−1(ωτ)
Since the input consists of the sum of two terms, the steady-state response has the form
hss(t) = A1M1 + A2M2 sin(0.002t + φ2)
where A1 = 0.2, A2 = 0.1, and
M1 = M(0) = 46.58 M2 = M(0.002) = 30.25
φ2 = φ(0.002) = − tan−1[0.002(585.39)] = −0.8639
Thushss(t) = 9.316 + 3.025 sin(0.002t− 0.8639)
The lag is given by |φ2|/ω = 432 sec.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
8.7 The transfer function given in the text is
T (s) =R1R2Cs + R2
R1R2Cs + R1 + R2=
R2
R1 + R2
τ1s + 1τ2s + 1
whereτ1 = R1C τ2 =
R1R2C
R1 + R2
For the circuit to be a low-pass filter, the corner frequency of the numerator must begreater than the corner frequency of the denominator; that is,
1τ1
>1τ2
This implies that τ2 > τ1, which says
R1R2C
R1 + R2> R1C
Canceling R1C givesR2
R1 + R2> 1
which is impossible to satisfy. Thus the circuit cannot be a low-pass filter.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
8.8 a)
yss =10√
(10ω)2 + 1√
(4ω)2 + 110 =
10√[10(0.2)]2 + 1
√[4(0.2]2 + 1
10 = 34.92
b)
yss =1√
(200− 4ω2)2 + (20ω)216 =
1√(200− 100)2 + [20(5)]2
16 = 0.1131
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8.9 a)
T (jω) =8
jω(100− ω2 + 10ωj)
M(ω) =8
ω√
(100− ω2)2 + 100ω2
Thus M(9) = 0.0096.
φ(ω) = −6 jω − 6 [(100− ω2) + 10ωj]
So φ(9) = π/2− tan−1(90/19) = 0.2081. Thus
yss(t) = 6M(9) sin[9t + φ(9)] = 0.0576 sin(9t − 0.2081)
b)
T (jω) =10
−ω2(1 + ωj)
M(ω) =10
ω2√
1 + ω2
Thus M(2) = 1.118.
φ(ω) = −6 (−10/ω2) − 6 [(1 + ωj)
So φ(2) = π − tan−1(2) = 2.0344. Thus
yss(t) = 9M(2) sin[2t + φ(2)] = 10.062 sin(2t + 2.0344)
(continued on the next page)
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Problem 8.9 continued:
c)
T (jω) =ωj
(1 + 2ωj)(1 + 5ωj)
M(ω) =ω√
1 + 4ω2√
1 + 25ω2
Thus M(0.7) = 0.1118.
φ(ω) = 6 ωj − 6 (1 + 2ωj)− 6 (1 + 5ωj)
So φ(0.7) = π/2− tan−1(1.4)− tan−1(3.5) = −0.6722. Thus
yss(t) = 9M(0.7) sin[0.7t + φ(0.7)] = 1.0062 sin(0.7t− 0.6722)
d)
T (jω) =−ω2
(1 + 2ωj)(1 + 5ωj)
M(ω) =ω2
√1 + 4ω2
√1 + 25ω2
Thus M(0.7) = 0.0782.
φ(ω) = 6 (−ω2) − 6 (1 + 2ωj)− 6 (1 + 5ωj)
So φ(0.7) = −π/2 − tan−1(1.4)− tan−1(3.5) = −3.8138. Thus
yss(t) = 9M(0.7) sin[0.7t + φ(0.7)] = 0.7038 sin(0.7t− 3.8138)
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8.10 The resonant frequency and peak magnitude are given by
ωr = ωn
√1 − 2ζ2 Mr =
12ζ
√1 − ζ2
Note that ωn =√
2. Thusa) For ζ = 0.1, ωr = 1.4 and Mr = 5.025.b) For ζ = 0.3, ωr = 1.28 and Mr = 1.747.
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8.11 The maximum amplitude occurs at ω = ωr and is 22Mr, where
Mr =1
2ζ√
1 − ζ2
The damping ratio isζ =
c
2√
50=
c
10√
2Thus
22Mr =22
2ζ√
1 − ζ2= 3
Solve for ζ by squaring each side to obtain
4ζ4 − 4ζ2 − 13.44 = 0
which has the positive root ζ2 = 4.2, or ζ = 2.05. Thus c = 10√
2(2.05) = 28.99.
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8.12 The transfer function isT (s) =
113s2 + 2s + k
From (8.2.20)
ωr = ωn
√1− 2ζ2
where
ωn =
√k
13
ζ =2
2√
13k=
1√13k
We want ωr = 4. Thus we must solve the following for k.
4 =
√k
13
√1 − 2
13k=
113
√13k − 2
This gives k = 208 lb/ft. This gives ζ = 0.0192.From (8.2.21),
Mr =1
2ζ√
1− ζ2= 26.295
Thus the amplitude of the steady-state response is 26.295(10) = 262.95The phase shift of the response is (with ω = 4)
φ = − 6 [(k − 13ω2) + 2ωj] = −6 [−8 + 8j] = −2.356
Thus the steady-state response is
xss(t) = 262.95 sin(4t − 2.356)
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8.13 a)
T (jω) =7
ωj(58− ω2 + 6ωj)
M(ω) =7
ω√
(58− ω2)2 + 36ω2
Since M(0) = ∞, the resonant frequency is at ω = 0.b)
T (jω) =7
(174− 3ω2 + 18ωj)(58− 2ω2 + 8ωj)
M(ω) =7√
(174− 3ω2)2 + 324ω2√
(58− 2ω2)2 + 64ω2
A plot of M versus ω shows that M has a single peak near ω = 4.97, which is the resonantfrequency. We might have expected two peaks since there are two quadratic factors in thedenominator. These quadratic factors have resonant frequencies of 4.58 and 6.32. Becausethese frequencies are so close in value, their two peaks merge into a single peak at ω = 4.97.
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8.14 Applying the voltage law to each loop gives
v1 − Ri1 − Ldi3dt
= 0
v2 −1C
∫i2 dt − L
di3dt
= 0
From conservation of charge, i3 = i1 + i2.Take the Laplace transform of each equation using zero initial conditions. Solve the first
two for I1(s) and I2(s):
I1(s) =V1(s) − LsI3(s)
R
I2(s) = CsV2(s) − LCs2I3(s)
Substitute these into the third equation to obtain
(RLCs2 + Ls + R)I3(s) = V1(s) + RCsV2(s)
The two transfer functions are
I3(s)V1(s)
=1
RLCs2 + Ls + R=
110−5s2 + 0.1s + 100
I3(s)V2(s)
=RCs
RLCs2 + Ls + R=
10−4s
10−5s2 + 0.1s + 100
The roots are s = −1127 and s = −8873. Thus the corner frequencies are ω1 = 1127 rad/sand ω2 = 8873 rad/s.(continued on the next page)
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Problem 8.14 continued:
First consider the transfer function I3(s)/V1(s). At low frequencies the m curve ap-proaches 20 log(105/[(1127)(8873)] = −40 dB. The m curve breaks down at ω1 = 1127 andagain at ω2 = 8873 rad/s. So the system acts like a low pass filter for the input voltage v1,and it tends to filter out frequency components in v1 that are higher than ω1 = 1127 rad/s.The low frequency gain is M = 10m/20 = 10−40/20 = 0.01.
Now consider the transfer function I3(s)/V2(s). At low frequencies the m curve ap-proaches 20 log(0) = −∞. At ω = 10,
m = 20 log(
10(10)√11272 + 102
√88732 + 102
)= −100 dB
The m curve has a slope of 20 dB/decade until ω1 = 1127, when the slope becomes approx-imately zero. The curve breaks downward at ω2 = 8873 rad/s, and the slope becomes −20dB/decade for higher frequencies. So the system acts like a band pass filter for the inputvoltage v2, and it passes frequency components in v2 that are between ω1 = 1127 rad/s andω2 = 8873 rad/s.
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8.15 a) The equation of motion is
mx = k(y − x) − cx
orx + cx + 600x = 600y
The transfer function isT (s) =
X(s)Y (s)
=600
s2 + cs + 600
This has the same form as (8.2.14) with ωn =√
600 = 10√
6 and ζ = c/(20√
6). So,provided that c is such that ζ ≤ 0.707, we can use (8.2.20) and (8.2.21) to obtain
ωr = ωn
√1 − 2ζ2 = 10
√6√
1 − 2ζ2 rad/s
Mr =1
2ζ√
1 − ζ2
Since c = 20ζ√
6, ζ will be no greater than 0.707 if 0 ≤ c ≤ 34.64 N·s/m.For 0 ≤ c ≤ 34.64, a plot of ωr versus c shows that the resonant frequency varies from
3.6 (for c = 34.64) to 24.5 rad/s (for c = 0). For 0 ≤ c ≤ 34.64, a plot of Mr versus c showsthat Mr varies from ∞ (for c = 0) to 1 rad/s (for c = 34.64).
For values of ζ > 0.707 (for c > 34.64), there is no resonant peak and thus no resonantfrequency.
b) The transfer function is
T (s) =X(s)Y (s)
=k
ms2 + cs + k=
k/m
s2 + (c/m)s + k/m=
ω2n
s2 + 2ζωns + ω2n
This has the same form as (8.2.14), so provided ζ ≤ 0.707, we can use (8.2.20) and (8.2.21)to compute ωr and Mr. For values of ζ > 0.707, there is no resonant peak and thus noresonant frequency.
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8.16 Applying the voltage law gives
vs −1C
∫i dt − L
di
dt− vo = 0
where v0 = Ri. Use the latter equation to eliminate i:
LCd2vo
dt2+ RC
dvo
dt+ vo = RC
dvs
dt
For the given values,
5 × 10−8 d2vo
dt2+ 10−5R
dvo
dt+ vo = 10−5R
dvs
dt
The transfer function isVo(s)Vs(s)
=103Rs
5s2 + 103Rs + 108
For R = 10,Vo(s)Vs(s)
=104s
5s2 + 104s + 108
The roots are s = −1000± 4359j.For R = 100,
Vo(s)Vs(s)
=105s
5s2 + 105s + 108
The roots are s = −100.05 and s = −199, 900.The plots are shown on the next page. The peak is approximately the same for each. It
is −0.000185 dB, or M = 10−0.000185/20 = 1 at 5170 rad/s.(continued on the next page)
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100
101
102
103
104
105
106
107
−90
−80
−70
−60
−50
−40
−30
−20
−10
0
Mag
nitu
de (
dB)
Frequency (rad/sec)
R = 1000
R = 10
Figure : Log magnitude plots for Problem 8.16
Thus an increase in the value of R by a factor of 100 does not significantly change thepeak value or the peak frequency, but it does change the spread of the plot about the peak.
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8.17 The equations of motion are
I1ω1 = T1 − c1(ω1 − ω2)
I2ω2 = c1(ω1 − ω2) − c2ω2
Applying the Laplace transform for zero initial conditions, and eliminating Ω1(s), we obtainthe transfer function:
Ω2(s)T1(s)
=100
s2 + 5s + 2
From this we obtain the magnitude ratio and the phase angle.
M(ω) =100√
(2 − ω2)2 + (5ω)2
φ(ω) = − tan−1 5ω
2 − ω2
Evaluation of M and φ for ω = 0, 1.5, and 2 gives the following results.
M(0) = 50 M(1.5) = 13.3259 M(2) = 9.8058
φ(0) = 0 φ(1.5) = −1.604 rad φ(2) = −1.763 rad
Thus the steady state response is
ω2ss = 50(4) + 13.3259(2) sin(1.5t− 1.604) + 9.8058(0.9) sin(2t − 1.763)
orω2ss = 200 + 26.6518 sin(1.5t− 1.604) + 8.8252 sin(2t− 1.763)
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8.18 The forcing frequency is ω = 5.2 and the natural frequency is ωn =√
75/3 = 5. Thebeat period is
2π
|ω − ωn|=
2π
|5.2− 5|= 31.4159
The vibration period is4π
ω + ωn=
4π
10.2= 1.232
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8.19 From (8.3.4) with fo = 0.2, k = 64, and ωn = 8,
x(t) = 0.0125(
sin 8t
8− t cos 8t
)
The plot is shown below. It takes approximately 8.2 sec for |x(t)| to become greater than0.1 ft.
0 1 2 3 4 5 6 7 8 9 10−0.1
−0.08
−0.06
−0.04
−0.02
0
0.02
0.04
0.06
0.08
0.1
t (sec)
x(t)
(ft)
Figure : Plot for Problem 8.19
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8.20 From the given data
ω =3400(2π)
60= 356 rad/s
c = 2ζ√
mk = 548 N · s/m
AlsoFt(s)Fr(s)
=cs + k
ms2 + cs + k
Thus
|Ft(ω)| = |Fr||k + cωj|
|k − mω2 + cωj| = |Fr|√
k2 + c2ω2
√(k − mω2)2 + c2ω2
For rotating unbalance
|Fr| = muεω2 = 0.1(0.02)(356)2 = 253.5 N
Substituting the values of k, c, m, and ω, we obtain
|Ft| = 253.5(0.1027) = 26 N
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8.21 For base motion,X
Y=
√k2 + c2ω2
√(k − mω2)2 + c2ω2
Plotting this expression versus ω for the given values of m, c, and k, we find that the ratioX/Y is never less than 1. Therefore, X will never be greater than 2 mm as long as Y isnever greater than 2 mm.
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8.22 The vibration frequency of the base motion as a function the vehicle velocity v is
ω =(
528020
) (1
3600
)(2π)v = 0.4608v
From the results of Example 4.5.9,
D3 = (m1s2+c1s+k1)(m2s
2+c1s+k1+k2)−(c1s+k1)2 = m1m2s4+(m1k1+m1k2+m2k1)s2+k1k2
where we have taken c1 = 0. Using the given values, we obtain
114.472s4 + 4.0838× 105s2 + 8 × 107 = 0
The roots are s = ±14.4j and s = ±57.96j. Thus the resonant frequencies are 14.4 and57.96 rad/s, and the resonant speeds are
v =14.4
0.4608= 31.25 ft/sec and v =
57.960.4608
= 125.8 ft/sec
or v = 21.3 mph and 85.8 mph.Note that the amplitude of the road surface variation, although given in the problem
statement, was not needed.
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8.23 Using ρ = 2388 kg/m3 for the density of steel, the beam mass is calculated to be
mb = 2388(2)(0.3)(0.03) = 42.98 kg
The equivalent system mass is
me = 50 + 0.23mb = 59.886 kg
The stiffness of the beam is
k =Ewh3
4L3=
2 × 1011(0.3)(0.03)3
4(2)3= 1.6875× 106 N/m
The forcing frequency is
ω =3400(2π)
60= 356 rad/s
and the amplitude of the unbalance force is
|Fr| = muεω2 = 0.02(356)2 = 2535 N
Thus the displacement amplitude is, since c = 0,
X =|Fr|√
(k − meω2)2=
|Fr||k − meω2| = 4.29× 10−4 m
or 0.428 mm.
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8.24 Using ρ = 2388 kg/m3 for the density of steel, the beam mass is calculated to be
mb = 2388(2)wh = 4776wh
where w is the beam width and h is the beam thickness.The equivalent system mass is
me = 50 + 0.5mb = 50 + 2388wh
The stiffness of the beam is
k =16Ewh3
L3= 4 × 1011wh3
The forcing frequency is
ω =3400(2π)
60= 356 rad/s
and the amplitude of the unbalance force is
|Fr| = muεω2 = 0.02(356)2 = 2535 N
Thus the displacement amplitude is, since c = 0,
X = 0.01 =|Fr|√
(k − meω2)2=
|Fr||k − meω2| (1)
This expression is a complicated function of w and h since both k and me are functionsof w and h. So let us temporarily neglect the beam mass. Then me ≈ 50 kg and theprevious expression becomes
X = 0.01 =2535
|k − 6.3368× 106|
This gives k = 6.5903× 106 N/m. Thus
wh3 =6.5903× 106
4 × 1011= 1.6× 10−5
(continued on the next page)
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Problem 8.24 continued:
There is an infinite number of solutions to this equation, but we should choose one thatresults in an equivalent beam mass that is small compared to 50 kg, because we neglectedthe beam mass in obtaining this solution.
One solution is h = 0.04 m, which gives w = 0.25 m, both reasonable dimensions.These give an equivalent beam mass of 0.5(47.76) = 23.88 kg and a total equivalent massof me = 73.88 kg. Using this value in equation (1) we obtain
X =2535
|6.59× 106 − 73.88(356)2| = 9.14× 10−4 m
which is much less than the maximum allowable value of 0.01 m. So our solution is valid.
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8.25 We are given m = 1500 kg, k = 20 000 N/m, ζ = 0.04, and Y = 0.01 m. Thusc = 2ζ
√mk = 438.18. From (8.3.9), with s = jω,
|Ft| = |Y | mω2√
k2 + (cω)2√(k − mω2)2 + (cω)2
At resonance, from (8.2.20),
ω = ωr = ωn
√1− 2ζ2 = 0.998
√k
m= 3.6442
Thus|Ft| = 2500 N
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8.26 The static deflection is δ = mg/k. Thus k = mg/δ = 200/0.003 = 66 667 N/m. Also,
ω = 40 Hz = 251 rad/s
ωn =
√k
m=
√66 667
(200/9.81)= 57.2 rad/s
From (8.3.7) with c = 0,
|X ||Y | =
k√(k − mω2)2
= 0.055
Thus 5.5% of the airframe motion is transmitted to the module.
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8.27 a) Neglect damping in the isolator, and determine its required stiffness k. From (8.3.7)with c = 0,
|X ||Y | =
k√(k − mω2)2
=k/m√
(k/m− ω2)2=
ω2n√
(ω2n − ω2)2
= 0.1
which gives (ω/ωn)2 = 11. Thus
ω2n =
ω2
11=
[3000(2π)/60]2
11=
(314)2
11
and
k = mω2n =
232.2
(314)2
11= 556.7 lb/ft
b) Let r = ω/ωn. From part (a), ωn = 314/√
11, we have
r1 =2500(2π)/60
314/√
11= 2.77
andr2 =
3500(2π)/60314/
√11
= 3.87
From (8.3.7) with c = 0,|X ||Y | =
1|1− r2|
Thus the highest percentage of motion will be transmitted at the lowest r value, whichis r1, the value corresponding to 2500 rpm. For 2500 rpm,
|X ||Y | =
1|1 − r2
1|= 0.15
For 3500 rpm,|X ||Y | =
1|1 − r2
2|= 0.07
Thus at most, 15% of the crane motion will be transmitted to the module.
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8.28 The transmitted force is given by (8.3.14) and (8.3.17):
|Ft| = muεω2
√k2 + (cω)2√
(k − mω2)2 + (cω)2
If we letr = ω/ωn
then we can express the previous equation as
|Ft| = muεω2
√1 + 4ζ2r2
(1 − r2)2 + 4ζ2r2
We are given that mu = 0.05/32.2 = 0.00155 slug, m = 50/32.2 = 1.55 slug. ε =0.1.12 = 0.0083 ft, and ω = 1000(2π)/60 = 104.7 rad/sec. Thus
r =ω
ωn=
104.7√500/1.55
= 5.83
Thus
|Ft| = (0.00155)(0.0083)(104.7)2√
1 + 136ζ2
1088 + 136ζ2= 0.141
√1 + 136ζ2
1088 + 136ζ2
a) For ζ = 0.05, Ft = 0.0049 lb.b) For ζ = 0.7, Ft = 0.034 lb.
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8.29 The transmitted force is given by (8.3.14) and (8.3.17):
|Ft| = muεω2
√k2 + (cω)2√
(k − mω2)2 + (cω)2
If we letr = ω/ωn
then if c = 0, we can express the previous equation as
|Ft| = muεω2 1|1− r2|
Here Ft = 15 and ωR = 200(2π)/60 = 20.9 rad/s. Thus
r =ω
ωn=
20.9√k/m
=20.9√
2500/75= 3.62
Thus|Ft| = muε(20.9)2(0.0826) = 15
Solve for muε to obtain muε = 0.416 kg·m.
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8.30 For a vibrometer we require that the natural frequency ωn be much less than theforcing frequency ω. Here
ω = 2π(200) = 400π rad/s
and
ωn =
√k
0.1
So we require that √k
0.1 400π
ork 1.579× 105 N/m
Suppose that k = 1.6×104. Then the static deflection will be δst = mg/k = 6.1×10−5 m.The designer must verify that there will be enough space in the instrument to accommodatethis deflection.
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8.31 The equation of motion is
mx = k(y − x) − cx
or200x + 2000x + 2 × 104x = 2 × 104y
The transfer function is
T (s) =X(s)Y (s)
=2 × 104
200s2 + 2000s + 2 × 104=
100s2 + 10s + 100
This has the same form as (8.2.14) with ωn = 10 and ζ = 0.5. So we can use (8.2.20) and(8.2.21) to obtain
ωr = ωn
√1 − 2ζ2 = 7.07 rad/s
Mr =1
2ζ√
1 − ζ2= 1.155
For the bandwidth, equation (2) of Example 8.4.2 gives
r =
√1 − 2ζ2 ± 2ζ
√1 − ζ2 = 1.167, 0.605j
Because there is only one positive, real solution, the bandwidth extend from ω = 0 toω = 1.167ωn = 11.69 rad/s.
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8.32 The transfer function becomes
T (s) =10−6s
3 × 10−4s2 + 10−2s + 1
From the frequency response plot shown below we can tell that the bandwidth is between45 and 76 rad/s.
You can obtain this plot in MATLAB by typing:
sys = tf([1e-6,0],[3e-4,1e-2,1]);bodemag(sys)
100
101
102
103
−125
−120
−115
−110
−105
−100
−95
−90
−85
−80
Mag
nitu
de (
dB)
Bode Diagram
Frequency (rad/sec)
Figure : Plot for Problem 8.32
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
8.33 a) The transfer function for the circuit in part (a) of the figure was derived in Chapter6. It is
Vo(s)Vs(s)
=G
R2C2s2 + 2RCs + 1
Its natural frequency is
ωn =√
1R2C2
=1
RC
Its damping ratio is
ζ =2RC
2√
R2C2= 1
From Equation (1) in Example 8.4.2, with r = ω/ωn = RCω,
kM =1√
(1− r2)2 + 4r2
which has a maximum of 1 at r = 0. Setting kM = 1/√
2 and solving for r gives r = 0.644.Thus the bandwidth corresponds to 0 ≤ r ≤ 0.644 or
0 ≤ ω ≤ 0.644RC
(continued on the next page)
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Problem 8.33 continued:
b) The transfer function for the circuit in part (b) of the figure was derived in Chapter6. It is
Vo(s)Vs(s)
=G
R2C2s2 + 3RCs + 1
Its natural frequency is
ωn =√
1R2C2
=1
RC
Its damping ratio is
ζ =3RC
2√
R2C2= 1.5
From Equation (1) in Example 8.4.2, with r = ω/ωn = RCω,
kM =1√
(1− r2)2 + 9r2
which has a maximum of 1 at r = 0. Setting kM = 1/√
2 and solving for r gives r = 0.374.Thus the bandwidth corresponds to 0 ≤ r ≤ 0.374 or
0 ≤ ω ≤ 0.374RC
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8.34 The circuit equation isLCv1 + RCv1 + v1 = vs
or10−7v1 + 10−6Rv1 + v1 = vs
Its natural frequency and damping ratio are
ωn =√
107 = 103√
10
ζ =10R
2√
107=
5R
1000√
10The resonant frequency is
ωr = ωn
√1 − 2ζ2 if ζ ≤ 1/
√2
Thus there is a resonant peak and a resonant frequency only if
5R
1000√
10≤ 1√
2
which implies that
R ≤ 1000√
105√
2= 447 Ω
If there is a resonant peak, its value is
Mr =1
2ζ√
1 − ζ2
A plot of Mr versus R for 0 ≤ R ≤ 447 shows that Mr = 3.2026 for R = 100. As R isincreased to 447, Mr decreases to Mr = 1.
For R > 447, the maximum value of M occurs at ω = 0. The plots on the next pageshow the variation of M versus r = ω/ωn, for R = 100, 250, 400, 500, 750, and 1000 Ω.The plots show that as R is increased, the system changes from a bandpass filter to a lowpass filter. For R ≤ 447, increasing R decreases the bandwidth. For R > 447, increasing Rdecreases the bandwidth.(continued on the next page)
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0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20
0.5
1
1.5
2
2.5
3
3.5
r
M
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20
0.2
0.4
0.6
0.8
1
r
M
100
250
400
500 750
1000
Figure : Plot for Problem 8.34
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
8.35 The magnitude ratio is
M(ω) =1√
1 + R2C2ω2=
1√1 + 36 × 10−8ω2
and the phase angle is
φ(ω) = − tan−1 RCω = − tan−1 6 × 10−4ω
The bandwidth of this circuit is 0 ≤ ω ≤ 1/RC = 1667 rad/s. Thus the cos 720πt termand the higher terms lie outside the bandwidth. Evaluation of M and φ for ω = 0, 240π
and 480π gives the following results.
M(0) = 1 M(240π) = 0.9111 M(480π) = 0.7415
φ(0) = 0 φ(240π) = −0.425 rad φ(480π) = −0.735 rad
Thus the steady state response is
voss =20π
− 0.9111403π
cos(240πt− 0.425)− 0.74154015π
cos(480πt− 0.735)
orvoss = 6.3662− 3.8664 cos(240πt− 0.425)− 0.6294 cos(480πt− 0.735)
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8.36 The magnitude ratio is
M(ω) =1√
1 + R2C2ω2=
1√1 + 10−6ω2
and the phase angle is
φ(ω) = − tan−1 RCω = − tan−1 10−3ω
The bandwidth of this circuit is 0 ≤ ω ≤ 1/RC = 1000 rad/s. Thus the sin 360πt termand the higher terms lie outside the bandwidth. Evaluation of M and φ for ω = 0 and 120π
gives the following results.
M(0) = 1 M(120π) = 0.9357
φ(0) = 0 φ(120π) = −0.361 rad
Thus the steady state response is
voss = 5 + 0.935720π
sin(120πt− 0.361)
orvoss = 5 + 5.9569 sin(120πt− 0.361)
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8.37 The equation of motion is
mx + cx + kx = cy + ky
orx + 98x + 4900x = 98y + 4900y
The transfer function isX(s)Y (s)
=98s + 4900
s2 + 98s + 4900
The magnitude ratio is
M(ω) =√
49002 + 982ω2
√(4900− ω2)2 + 982ω2
and the phase angle is
φ(ω) = tan−1 98ω
4900− tan−1 98ω
4900− ω2
A plot of M versus ω shows that M has a peak of 1.2764 at ω = 55 rad/s. Noting that1.2764/
√2 = 0.9026 and that M(0) = 1, we see that the lower bandwidth frequency is 0.
The plot also shows that M(111) = 0.906 which is close to 0.9026. So the upper bandwidthfrequency is 111 rad/s. The circuit is a low pass filter with the bandwidth of 0 ≤ ω ≤ 111rad/s. So we see that only the terms up to and including the sin 30πt term lie within thebandwidth.
Evaluation of M and φ for ω = 0, 10π, 20π, and 30π gives the following results.
M(0) = 1 M(10π) = 1.1623 M(20π) = 1.263 M(30π) = 1.0395
φ(0) = 0 φ(10π) = −0.1057 rad φ(20π) = −0.5187 rad φ(30π) = 2.2467 rad
Thus the steady state response is
xss =120
−1.16231
10πsin(10πt−0.1057)−1.263
120π
sin(20πt−0.5187)−1.03951
30πsin(30πt+2.2467)
or
xss = 0.05− 0.037 sin(10πt− 0.1057)− 0.0201 sin(20πt− 0.5187)− 0.011 sin(30πt+2.2467)
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8.38 a) The transfer function is
T (s) =1
0.5s + 5=
0.20.1s + 1
The time constant is τ = 0.1 sec. and the bandwidth is 1/τ = 10 rad/sec.b) The magnitude ratio and phase angle are given by
M(ω) =0.2√
0.01ω2 + 1φ(ω) = − tan−1(1 + 0.1ωj)
The only components of the input that lie within the bandwidth are sin 4t and sin 8t. Thuswe evaluate M(ω) and φ(ω) at ω = 4 and ω = 8.
M(4) = 0.1856 φ(4) = −0.381 rad
M(8) = 0.1562 φ(8) = −0.675 rad
Thus,y(t) ≈ 0.1856 sin(4t − 0.381) + 4(0.1562) sin(8t − 0.675)
ory(t) ≈ 0.1856 sin(4t − 0.381) + 0.6248 sin(8t − 0.675)
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8.39 a) The transfer function is
T (s) =X(s)F (s)
=1
0.25s2 + 2s + 25=
0.040.01s2 + 0.08s + 1
From the frequency response plot we can determine that the bandwidth is from 0 to11.9 rad/sec.
b) Only the first two terms in the expansion have frequencies within the bandwidth.Thus
f(t) ≈ sin 3t +13
sin 9t
and at steady state:
x(t) ≈ M(3) sin[3t + φ(3)] + M(9) sin[9t + φ(9)]
whereM(3) = 0.0425 φ(3) = −0.2579 rad
M(9) = 0.0537 φ(9) = −1.3128 rad
Thusx(t) ≈ 0.0425 sin(3t − 0.2579) + 0.0179 sin(9t − 1.3128)
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8.40 First normalize the vo data by dividing by the amplitude of the input. Let
v =vo
20
Then compute m = 20 log v dB and log ω, and plot m versus log ω. The plot looks likea first order system plot (see Figure 8.1.6). The peak value of m is −11.25 dB, and m isapproximately 3 dB below that value at ω = 0.8. Thus we estimate the time constant tobe τ = 1/0/8 = 1.25 s. Thus we surmise that the form of the transfer function is
Vo(s)Vs(s)
=K
1.25s + 1
To estimate K, note that for low frequencies (ω << 0.8), |vo| = K|v|. So from the firstdata point, we estimate that K = |vo|/20 = 5.48/20 = 0.274. Thus we estimate the transferfunction to be
Vo(s)Vs(s)
=0.274
1.25s + 1
The MATLAB code to generate this analysis is shown below.
omega = [0.1:0.1:1,1.5,2:7];vo = [5.48,5.34,5.15,4.92,4.67,4.4,4.14,3.89,3.67,...
3.2,2.59,2.05,1.42,1.08,0.87,0.73,0.63];v = vo/20;m = 20*log10(v);logom = log10(om);plot(logom,m)
This result can be checked by comparing the m plot of the transfer function with theplot of the data, by continuing the above session as follows:
[M, ph, w]=bode(sys);M = M(:);plot(log10(w),20*log10(M),logom,m,’o’)
The plots agree very well.
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8.41 First convert the data to meters and then divide by 15, the amplitude of the input. Thisgives data for the magnitude ratio M(ω) = 10−3x/15. Then create the log magnitude ratioplot (see the following MATLAB program). The plot shows the presence of a denominatorterm s and a breakpoint frequency somewhere between 1 and 2 rad/s. This suggests atransfer function of the form
X(s)F (s)
=K
s(τs + 1)
This transfer function form agrees with the physical situation, because it can be due to amass attached to a damper, with no stiffness. This would have the following equation ofmotion: mx + cx = f(t) and a transfer function
X(s)F (s)
=1
s(ms + c)=
1/m
s(s + c/m)
The breakpoint frequency is at ω = c/m. From the data, we see that c/m is somewherebetween 1 and 2. The magnitude ratio is
M(ω) =1/m
ω√
ω2 + (c/m)2
To estimate the mass m, we can choose one of the data points. Choosing the first point,for which M = 0.0139 and ω = 0.1, we obtain
m =1
0.0139(0.1)√
ω2b + (0.1)2
where ωb is the estimate of the breakpoint frequency.The choice ωb = 2 gives a better fit to the data that ωb = 1. Thus the solution is
c/m = 2 and
m =1
0.0139(0.1)√
22 + (0.1)2= 359
from which we obtain c = 2m = 718. The estimated transfer function is
X(s)F (s)
=1
s(359s + 718)
(continued on the next page)
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Problem 8.41 continued:
The following program generates the plot, which shows good agreement with the data.
% File for problem 8.41x = (1e-3)*[209,52,28,19,7,2,1]/15; % Converted x dataw = [0.1,0.4,0.7,1,2,4,6]; % Frequency datam = 20*log10(x); % Decibel conversionwb = 2; % Estimated breakpoint frequencyM0 = x(1); % First converted data pointw0 = w(1); % First frequency data pointmass = 1/(M0*w0*sqrt(wb^2+w0^2)) % Estimated mass valuew1 = [0.1:0.01:6]; % Frequency vector for plottingm1 = 20*log10((1/mass)./(w1.*(sqrt(w1.^2+wb^2))));semilogx(w,m,’o’,w1,m1)
10−1
100
101
−85
−80
−75
−70
−65
−60
−55
−50
−45
−40
−35
ω (rad/s)
m (d
B)
Figure : Plot for Problem 8.41
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
8.42 The equation of motion is
mx + cx + kx = muεω2 sin ωt
The frequency transfer function for this system, with |x| normalized by m/muε is
m|x|muε
=r2
√(1− r2)2 + (2ζr)2
(1)
where
r = ω/ωn ωn =
√k
m
First plot the raw data after converting the frequency data to rad/sec and the displacementdata to meters. From this plot you can estimate that the resonant frequency is 19 rad/sec.Assuming that the damping is small enough so that the resonant frequency is close to ωn,we estimate that ωn = 19, and thus k = mω2
n = 100(192) = 36100 N/m.Next, normalize the frequency data to find r: r = ω/19. To plot the normalized
displacement m|x|/muε vs. r, we need to estimate muε. To do this we need to plot thetheoretical curve from (1), and this requires an estimate of ζ. After several attempts, usingdifferent values of muε and ζ, the estimates muε = 0.005 and ζ = 0.1 gave a good fit to thedata.
With ζ = 0.1 and k = 36100, we obtain c = 2√
mkζ = 2√
100(36100)(0.1) = 380 N·s/m.
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8.43 The equations of motion are
m1x1 = −k1(x1 − x2)
m2x2 = k1(x1 − x2) + k2(y − x2)
Convert these to state variable form by dividing by m1 and m2. The state vector is[x1, x1, x2, x2]. This gives the following state and input matrices for the given parametervalues.
A =
0 1 0 0−64 0 64 00 0 0 164 0 −128 0
B =
00064
Since the output is given to be x1, the output matrices are
C =[
1 0 0 0]
D = [0]
The MATLAB program is the following.
A = [0,1,0,0;-64,0,64,0;0,0,0,1;64,0,-128,0];B = [0;0;0;64];C = [1,0,0,0]; D = 0;sys1 = ss(A,B,C,D);sys2 = tf(sys1)eig(A)bode(sys1)
The transfer function is
X1(s)Y (s)
=4096
s4 + 192s2 + 4096
The roots are s = ±4.9443j and s = ±12.9443j. Thus the resonant frequencies in radianunits are 4.9443 and 12.9443. The Bode plots are shown on the next page. The resonantpeaks are very tight, so the bandwidths are very small. You can type bodemag(sys1) toobtain just the m plot, which can be expanded about the resonant peaks. After doing thisit is still difficult to identify the bandwidths, which appear to be about ±0.1 rad/unit timefor each peak.(continued on the next page)
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−100
−50
0
50
100
150
Mag
nitu
de (
dB)
10−1
100
101
102
−720
−675
−630
−585
−540
Pha
se (
deg)
Bode Diagram
Frequency (rad/sec)
Figure : Plot for Problem 8.43
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
8.44 The equations of motion are
I1θ1 = kT (φ − θ1) − cT1(θ1 − θ2)
I2θ2 = cT1(θ1 − θ2) − cT2θ2
Convert these to state variable form by dividing by I1 and I2. The state vector is [θ1, θ1, θ2, θ2].This gives the following state and input matrices for the given parameter values.
A =
0 1 0 0−1 −0.1 0 0.10 0 0 10 0.1 0 −0.2
B =
0100
Since the output is given to be x1, the output matrices are
C =[
1 0 0 0]
D = [0]
The MATLAB program is the following.
A = [0,1,0,0;-1,-0.1,0,0.1;0,0,0,1;0,0.1,0,-0.2];B = [0;1;0;0];C = [1,0,0,0]; D = 0;sys1 = ss(A,B,C,D);sys2 = tf(sys1)eig(A)bode(sys1)
The transfer function is
X1(s)Y (s)
=s + 0.2
s3 + 0.3s2 + 1.01s + 0.2
The roots are s = 0, s = −0.202, and s = −0.049± 0.9939j. Thus the resonant frequencyin radian units is 0.9939. The Bode plots are shown on the next page. You can typebodemag(sys1) to obtain just the m plot. The bandwidth can be computed by sliding thecursor along the plot and noting the frequencies corresponding to 3 dB below the peak.These are ω = 0.942 and ω = 1.04, so the bandwidth is 1.04− 0.942 = 0.098.(continued on the next page)
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−40
−30
−20
−10
0
10
20
30
40
Mag
nitu
de (
dB)
10−1
100
101
−180
−135
−90
−45
0
Pha
se (
deg)
Bode Diagram
Frequency (rad/sec)
Figure : Plot for Problem 8.44
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
8.45 The equation of motion is
mx + cx + kx = ky(t)
Thus the transfer function is
X(s)Y (s)
=k
ms2 + cs + k=
500050s2 + 200s + 5000
=100
s2 + 4s + 100
The MATLAB code to generate the m, plot is
[mag,phase,w]=bode(100,[1,4,100]);mag = mag(:);plot(w,20*log10(mag)),axis([6 15 0 10]),ylabel(’Magnitude (db)’),...xlabel(’Frequency (rad/sec)’),grid
From the plot we can determine that the resonant frequency is approximately 9.6 rad/s,and the peak in m is approximately 8 db. Thus
Mr = 10mr/20 = 2.51
The bandwidth frequencies are found from the plot to be from 7.2 rad/sec to 11.5 rad/sec.
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8.46 a) The MATLAB code is
KT = 0.04;Ke = 0.04;c = 7e-5;R = 0.6;L = 2e-3;Im = 2e-5;IL = 4e-5;I = Im + IL;den = [I*L,c*L+I*R,c*R+Ke*KT];num1 = [KT];sys1 = tf(num1,den);bode(sys1)[mag,phase,w] = bode(sys1);mag = mag(:);
To obtain the plots for the disturbance transfer function, type
num2 = [L,R];sys2 = tf(num2,den);bode(sys2)[mag,phase,w] = bode(sys2);mag = mag(:);
You can examine the data by typing [w,20*log10(mag)]. This enables you to determinethe resonant frequency and the bandwidth more precisely than from the plot.
The plots show that there is no resonant frequency. The bandwidth for the voltageinput is from 0 to 37 rad/s. The bandwidth for the disturbance torque input is from 0 to50 rad/s.
b) From the data generated by the above code, for the voltage input, m(0) = 27.733and φ(0) = 0. This m value corresponds to M = 1027.733/20 = 24.36. Thus the constantterm 10 in the input will produce a term 10(24.36) = 243.6 in the output. Also, from thedata, M(130) = 8.4853 and φ(130) = −94.7 = −1.653 rad. Thus the steady-state outputis
ωss(t) = 243.6 + 8.4853(2) sin(130t − 1.653) = 243.605 + 16.9706 sin(130t − 1.653)
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8.47 The program is given below, along with the resulting plot. The resonant frequencyis 2.51 × 104 rad/s. By moving the cursor along the plot we can determine the points atwhich the curve is 3 dB below the peak. These frequencies are 2.39 × 104 and 2.61 × 104,so the bandwidth is (2.61− 2.39)× 104 = 2.2× 103 rad/sec.
m = 0.002;k = 1e6;Kf = 20;Kb = 15;R = 10;L = 0.001;sys = tf(Kf,[m*L,m*R,k*L+Kf*Kb,k*R])roots([m*L,m*R,k*L+Kf*Kb,k*R])bode(sys)
−220
−200
−180
−160
−140
−120
−100
Mag
nitu
de (
dB)
102
103
104
105
106
−270
−180
−90
0
Pha
se (
deg)
Bode Diagram
Frequency (rad/sec)
Figure : Plot for Problem 8.47
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
8.48 Transform the equations and arrange them as follows.
(m1s2 + c1s + k1)X1(s) − (c1s + k1)X2(s) = 0
−(c1s + k1)X1(s) + (m2s2 + c1s + k1 + k2)X2(s) = k2Y (s)
Use Cramer’s method to find the transfer functions. You can use MATLAB to perform thealgebra after inserting the given numerical values for the parameters. The general resultsare
X1(s)Y (s)
=k2(c1s + k1)
D(s)
X2(s)Y (s)
=k2(m1s
2 + c1s + k1)D(s)
where
D(s) = m1m2s4 + c1(m1 + m2)s3 + (k1m2 + k1m1 + k2m1)s2 + c1k2s + k1k2
The MATLAB code is
m1 = 250;m2 = 40;k1 = 15000;k2 = 150000;c1 = 1000;D = [m1*m2,c1*(m1+m2),k1*m2+k1*m1+k2*m1,c1*k2,k1*k2];num1 = k2*[c1,k1];num2 = k2*[m1,c1,k1];bode(num1,D)[mag,phase,w] = bode(num1,D,[5:.05:9]);mag = mag(:);plot(w,mag),axis([0 20 0 3]),xlabel(’Frequency (rad/sec)’),ylabel(’M’),grid
The plots show that the major peak has a resonance frequency of 7.3 rad/sec and a band-width of 5.3 to 8.8 rad/sec. The secondary resonance is at 62 rad/sec.
b) The peak amplitude occurs at ω = 7.3 rad/sec. With a period of 10 m, this frequencycorresponds to a speed of
v =7.3(1)
2π= 11.62 m/s
or 41.8 kilometers per hour.
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Solutions Manual c©
to accompany
System Dynamics, Second Edition
by
William J. Palm III
University of Rhode Island
Solutions to Problems in Chapter Nine
c©Solutions Manual Copyright 2010 The McGraw-Hill Companies. All rightsreserved. No part of this manual may be displayed, reproduced, or distributedin any form or by any means without the written permission of the publisheror used beyond the limited distribution to teachers or educators permitted byMcGraw-Hill for their individual course preparation. Any other reproductionor translation of this work is unlawful.
9.1 From (9.1-5),
v(t) =270024
(1 − e−4t
)= 112.5
(1 − e−4t
)
Because x = v,
x(t) =∫ t
0v(t) dt = 112.5t + 28.125
(e−4t − 1
)
Set x(t) = 2000.2000 = 112.5t + 28.125
(e−4t − 1
)
Combine terms and solve the following equation for t.
112.5t + 28.125e−4t − 2028.125 = 0
This can be solved by plotting or by iteration, for example. The answer is t = 17.8 sec.
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9.2 τ = 0.25 sec. Thus the speed takes approximately 4(0.25) = 1 sec to reach constantspeed. Because 0.04 sec is small compared to 1 sec, the answer is yes.
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9.3a) τ = 8/7, x(t) = 6e−7t/8
b) τ = 12/5, x(t) = 3e−5t/12
c) τ = 13/6, x(t) = −2e−6t/13
d) No time constant is defined because the model is unstable. x(t) = 9e5t/7
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9.4 a) xss = 10, t = 4τ = 8b) xss = 10, t = 4τ = 8c) xss = 200, t = 4τ = 8
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9.5 Note that the initial condition does not affect the steady-state response.a) xss = 4, t = 4τ = 24/5b) xss = 4, t = 4τ = 4/5c) No steady-state response exists because the model is unstable.
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9.6 From (9.1-5),a) x(t) = 4
(1− e−5t
)
b) x(t) = e−5t + 4(1 − e−5t
)= 4− 3e−5t
c) x(t) = −2e6t/13 − 186
(1− e6t/13
)= −3 + e6t/13
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9.7 The time constant of the rotational system is τ1 = 100/5 = 20 s. The time constant ofthe field circuit is τ2 = 0.002/4 = 5 × 10−4 s. The steady-state current is ifss = 12/4 = 3A, and thus the steady-state torque is Tss = KT ifss = 15(3) = 45 N·m. It takes about4τ2 = 2 × 10−3 s for the torque to reach steady state, and since this time is much smallerthan τ1, we may treat the torque as a step input to the rotational system. The steady-statespeed is 45/5 = 9 rad/s, and it takes about 4τ1 = 80 s to reach the steady-state speed.
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9.8 RCv + v = vs, where RC = 3 s. From (9.1-5),
v(t) = 6e−t/3 + 12(1− e−t/3
)
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9.9 (AR/g)h + h = (R/g)qv. With the given values, this becomes
93.1677h + h = 1.8634qv
From (9.1-5),h(t) = 2e−t/93.1677 + 18.634
(1 − e−t/93.1677
)
Set h(t) = 15 ft and solve for e−t/93.1677:
e−t/93.1677 = 0.2184
which gives t = 142 sec.
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9.10 mcpRT + T = Tb, where mcpR = 100(500)0.09 = 4500 s. From (9.1-5),
T (t) = 20e−t/4500 + 80(1− e−t/4500
)
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9.11 Transforming the first model, 2v + v = ˙ g(t) + g(t), gives
2[sV (s) − v(0)] + V (s) = (s + 1)G(s) =10(s + 1)
s
With v(0) = 5, this becomes
V (s) = 10s + 0.5
s(s + 0.5)=
10s
Thus v(t) = 10us(t).Transforming the second model, 2v + v = g(t), gives
2[sV (s) − v(0)] + V (s) = G(s) =10s
With v(0) = 5, this becomes
V (s) = 5s + 1
s(s + 0.5)=
10s
− 5s + 0.5
Thus v(t) = 10us(t) − 5e−0.05t.
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9.12 Note that g = 0 for −∞ ≤ t ≤ ∞. Thus the model is equivalent to 5v + v = g. From(9.1-5),
v(t) = 10− 5e−0.2t
for both cases.
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9.13 a) Transforming the equation gives
(6s + 3)V (s) = (s + 1)G(s) =s + 1
s
orV (s) =
s + 1s(6s + 3)
=16
s + 1s(s + 0.5)
=13
1s− 1
61
s + 0.5
Thusv(t) =
13− 1
6e−0.5t
Note that v(0+) = 1/6.b) Transforming the equation gives
(6s + 3)V (s) = (s + 1)G(s) = (s + 1)[1s− 1
s + 5
]= 5
s + 1s(s + 5)
orV (s) =
56
s + 1s(s + 5)(s + 0.5)
=13
1s− 5
271
s + 0.5− 4
271
s + 5
Thusv(t) =
13− 5
27e−0.5t − 4
27e−5t
Note that v(0+) = 0.
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9.14 From (9.1.10),
v(t) =(
3 +52
)e−t/2 = 5.5e−t/2
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9.15 Transforming the equation gives
(2s + 1)V (s) =5s2
orV (s) =
52s2(s + 0.5)
=5s2
− 10s
+10
s + 0.5
Thusv(t) = 5t − 10 + 10e−0.5t
The transient response is 10e−0.5t. The steady-state response is 5t − 10.
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9.16 Transforming the equation gives
(9s + 3)V (s) =7s2
orV (s) =
79s2(s + 1/3)
=7
3s2− 7
s+
7s + 1/3
Thusv(t) =
73t − 7 + 7e−t/3
The steady-state response is 7t/3 − 7, which is not parallel to the input 7t.
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9.17 The equation reduces to x + 4x = 0, whose roots are s = ±2j. Thus the oscillationfrequency is 2 rad/sec for both cases, and
x(t) = B sin(2t + φ)
andx(0) = B sin φ
x(0) = B cos φ
a) B sin φ = 5 and B cos φ = 0. Thus, φ = π/2 rad and the amplitude is
B =5
sin π/2= 5
b) B sin φ = 0 and B cos φ = 5. Thus, φ = 0 and the amplitude is
B =5
cos 0= 5
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9.18 From Table 9.2.2,a) s = −2 ± 2j. x(t) = 0.5e−2t sin 2t.b) s = −6,−2. x(t) = 0.25e−2t − 0.25e−6t.c) s = −2, −2. x(t) = te−2t.
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9.19a) The roots are s = −2 ± 2j, so we use the trial solution
x(t) = Be−2t sin(2t + φ) + C
At steady state, xss = 2/8 = 0.25, so C = 0.25. We obtain B and φ from the initialconditions. The solution is
x(t) =14
[√2e−2t sin
(2t +
5π
4
)+ 1
]
b) The roots are s = −6, −2, so we use the trial solution
x(t) = C1e−6t + C2e
−2t + C3
At steady state, xss = 2/12 = 1/6, so C3 = 1/6. We obtain C1 and C2 from the initialconditions. The solution is
x(t) =16
(12e−6t − 3
2e−2t + 1
)
c) The roots are s = −2, −2, so we use the trial solution
x(t) = C1e−2t + C2te
−2t + C3
At steady state, xss = 2/4 = 0.5, so C3 = 0.5. We obtain C1 and C2 from the initialconditions. The solution is
x(t) =12
[1 − (2t + 1)e−2t
]
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9.20 Using Table 9.2.1,a) s = −2, −5.
x(t) =43e−2t − 1
3e−5t
b) s = −2, −2.x(t) = (1 + t)e−2t
c) s = −2 ± 5j.x(t) = 1.02e−2t sin (5t + 1.373)
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9.21a) s = −2, −5.
x(t) =130
(23e−5t − 5
3e−2t + 1
)
b) s = −2, −2.
x(t) =120
[1− (2t + 1)e−2t
]
c) s = −2 ± 5j.
x(t) =158
[15
√29e−2t sin (5t + 4.33) + 1
]
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9.22 a) The roots are s = −2, −5.
X(s) =4
3s2(s2 + 7s + 10)=
43
(0.1s2
− 0.07s
+0.0833s + 2
− 0.0133s + 5
)
Thusx(t) =
43
(0.1t − 0.07 + 0.0833e−2t − 0.0133e−5t
)
b) The roots are s = −2, −2.
X(s) =4
5s2(s2 + 4s + 4)=
45
(0.25s2
− 0.25s
+0.25
(s + 2)2+
0.25s + 2
)
Thusx(t) =
45
(0.25t− 0.25 + 0.25te−2t + 0.25e−2t
)
c) The roots are s = −2 ± 5j.
X(s) =5
2s2[(s + 2)2 + 25]=
52(292)
[29s2
− 4s− 21
(s + 2)2 + 25+ 4
s + 2(s + 2)2 + 25
]
Thusx(t) =
51682
(29t − 4 +
215
e−2t sin 5t + 4e−2t cos 5t
)
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9.23X(s) =
a
s2(ms2 + cs + k)
Let e(t) be the difference between the input and the response: e(t) = f(t) − x(t). Then
E(s) = F (s) − X(s) =a
s2
(1 − 1
ms2 + cs + k
)=
a
s2
(ms2 + cs + k − 1
ms2 + cs + k
)
From the final value theorem,
ess = lims→0
sa
s2
(ms2 + cs + k − 1
ms2 + cs + k
)=
∞ if k 6= 1ac if k = 1
The response will be parallel to the input only if k = 1. In that case, the difference betweenthe input and the response is ac.
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9.24 a) ζ = cos[tan−1(6/2)] = 0.316. τ = 1/2. ωd = 6. ωn =√
4 + 36 =√
40.b) Unstable so ζ and τ are not defined. ωd = 5. ωn =
√1 + 25 =
√26.
c) ζ = 1 because the roots are real and equal. τ = 1/10. ωd and ωn are not definedbecause the free response is not oscillatory.
d) ζ is not defined because the system is first order. τ = 1/10. ωd and ωn are notdefined because the free response is not oscillatory.
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9.25 a) τ = 0.5, but ζ, ωn, and ωd do not apply because the dominant root is real.b) τ = 0.5, ζ = 0.707, ωn = 2
√2, ωd = 2.
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9.26 The dominant root pair is s = −2 ± 4j. For this root, ζ = cos[tan−1(4/2)] = 0.447,τ = 1/2, and ωd = 4.
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9.27 a) The Routh-Hurwitz criterion implies that the system is stable if and only if −(µ +2) > 0 and 2µ + 5 > 0, which gives −2.5 < µ < −2 for stability. Neutral stability occurs ifeither the s term or the constant term is missing in the characteristic equation. This occursif µ + 2 = 0 (the roots are s = ±j) or if 2µ + 5 = 0 (the roots are s = 0, −0.5).
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9.28 We have 3x + cx + 27x = 0. Using Table 9.2.1,a) (c = 0)
x(t) = cos 3t
b) (c = 9)x(t) = e−1.5t cos 2.6t − 0.5774e−1.5t sin 2.6t
c) (c = 18)x(t) = −3te−3t + e−3t
d) (c = 22)x(t) = 1.3695e−5.774t − 0.3695e−1.558t
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9.29 a) Stable if and only if 6d > 0 and 25d2 > 0 (from the Routh-Hurwitz criterion).Thus, stable if and only if d > 0.
b) The damping ratio is
ζ =6d
2√
100d2=
310
< 1 for all d > 0
Thus there will be damped oscillations for any d > 0.
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9.30 a) Stable if and only if 6b > 0 and 5b − 10 > 0 (from the Routh-Hurwitz criterion).Thus, stable if and only if b > 2.
b) The roots ares = −3b ±
√9b2 − 5b + 10
There will be imaginary parts if f(b) = 9b2−5b+10 < 0. Solving df/db = 0 gives b = 5/18.At b = 5/18, df2/db2 = 18 > 0, and thus f(b) has a minimum value of f(5/18) = 9.305.Therefore there is no real value of b for which 9b2 − 5b + 10 < 0, and thus no decayingoscillations.
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9.31 Transforming each equation gives
Ω(s) =T (s)
50s + 10
T (s) = 25If(s)
If (s)V (s)
=1
0.0001s + 5
Thus
Ω(s)V (s)
=25
50s + 101
0.001s + 5=
500(s + 0.2)(s + 5000)
=500
s2 + 5000.2s + 1000
The damping ratio is
ζ =5000.22√
1000= 79.06
The time constants are τ1 = 5 and τ2 = 1/5000 s. The undamped natural frequency isωn =
√1000/1 = 31.62 rad/s.
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9.32 The characteristic equation becomes
24× 10−8s2 + (3.6× 10−5 + 4c× 10−3)s + 0.6c + 0.01 = 0
The damping ratio is
ζ =3.6× 10−5 + 4c × 10−3
2√
24× 10−8(0.6c + 0.010=
0.09 + 10c√6(0.6c + 0.01)
The undamped natural frequency is
ωn =
√0.6c + 0.0124× 10−8
= 5000√
0.6c + 0.016
If ζ < 1, the real part of the roots is
−3.6 × 10−5 + 4c × 10−3
24× 10−8= −900 + c × 105
6
Thus the time constant isτ =
6900 + c × 105
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9.33 Transforming the equation and using Y (s) = 1/s, we obtain
X(s) =s + 3
s2 + 5s + 41s
=C1
s+
C2
s + 1+
C3
s + 4
where C1 = 3/4, C2 = −1, and C3 = 1/4. Thus
x(t) =34− e−t +
14e−4t for t > 0
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9.34 Transforming the equation and using Y (s) = 1/s, we obtain
X(s) =s + 3
s2 + 4s + 851s
=C1
s+ C2
9(s + 2)2 + 81
+ C3s + 2
(s + 2)2 + 81
since the roots are s = −2 ± 9j. The coefficients are C1 = 3/85, C2 = 79/765, andC3 = −3/85. Thus
x(t) = C1 + C2e−2t sin 9t + C3e
−2t cos 9t =385
+79765
e−2t sin 9t − 385
e−2t cos 9t for t > 0
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9.35 Transforming the equation and using Y (s) = 5, we obtain
X(s) =3(5)
s2 + 5s + 4=
C1
s + 1+
C2
s + 4
where C1 = 5 and C2 = −5. Thus
x(t) = 5e−t − 5e−4t for t > 0
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9.36 ζ = 1/√
2 = 0.707 which implies from Figure 9.3.5a that the maximum percentovershoot is ≈ 5%. Because xss = 2/8 = 0.25, the overshoot is 0.05(0.25) ≈ 0.01.
For ζ = 0.707, Figure 9.3.5c shows that ωntr ≈ 3.2. Because ωn =√
8/1 = 2√
2,tr = 3.2/2
√2 = 1.1.
For ζ = 0.707, Figure 9.3.5a shows that ωntp ≈ 4.6. Because ωn = 2√
2, tp = 4.6/2√
2 =1.6.
For ζ = 0.707 and ωn = 2√
2, Table 9.3.2 gives td = 0.53.The roots are −2± 2j, so the time constant is τ = 0.5. The 2% settling time is 4τ = 2.
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9.37 ωn =√
4/1 = 2. For tr ≤ 3, ωntr ≤ 2(3) = 6. From Figure 9.3.5c, we see that thisrequires that ζ ≤ 0.9. Thus ζ must be no greater than 0.9 if tr is to be no greater than 3.
Percent overshoot ≤ 20% implies from Figure 9.3.5a that ζ must be ≥ 0.43. Thus ζ mustbe in the range 0.43 ≤ ζ ≤ 0.9 to satisfy both specifications. To minimize the overshoot,Figure 9.3.5a shows that we should choose ζ as large as possible (within the above range).Thus we choose ζ = 0.9. The damping c is found from the definition of ζ: ζ = c/2
√4 = c/4.
Thus c = 0.9(4) = 3.6.
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9.38 The approach is similar to that of Problem 9.37, but the two specifications cannot bothbe satisfied, so we give priority to the overshoot specification, as directed. The calculationsare as follows. ωn =
√4/9 = 2/3
For tr ≤ 3, ωntr ≤ 2(3)/3 = 2. From Figure 9.3.5c, we see that this requires thatζ ≤ 0.35. Thus ζ must be no greater than 0.35 if tr is to be no greater than 3.
Percent overshoot ≤ 20% implies from Figure 9.3.5a that ζ must be ≥ 0.43. Thus ζ
cannot satisfy both specifications. So we set ζ = 0.43, giving priority to the overshootspecification, as directed. (We could choose ζ > 0.43 to minimize the overshoot, but thiswould cause the rise time to become much larger than desired.)
Thus we choose ζ = 0.43. The damping c is found from the definition of ζ: ζ =c/2√
9(4) = c/12. Thus c = 0.43(12) = 5.16.
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9.39tp =
π
ωn
√1 − ζ2
But the imaginary part of the roots is ωd = ωn
√1 − ζ2. Thus tp = π/ωd, and thus tp
depends only on the imaginary part of the roots.
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9.40 For the system in part (a) of the figure,
mx + cx + kx = cy + ky
With the given values,3x + 18x + 10x = 18y + 10y
Note that Table 9.3.1 cannot be used because of the input derivative term y. Transformingthis equation with zero initial conditions gives
X(s) =18s + 10
3s2 + 18s + 10Y (s) =
6s + 10/3s(s + 0.6195)(s + 5.3805)
This expands to
X(s) =1s
+0.13
s + 0.6195− 1.13
s + 5.3805Thus
x(t) = 1 + 0.13e−0.6195t − 1.13e−5.3805t
For the system in part (b) of the figure,
mx + cx + kx = ky
With the given values,3x + 18x + 10x = 10y
Note that Table 9.3.1 can be used here. This result is
x(t) = 1 + 0.13e−0.5.3805 − 1.13e−0.6195t
The responses are shown in the plot on the following page. The two responses are verydifferent, due to the numerator dynamics of the model in part (a) of the figure.
(continued on the next page)
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Problem 9.40 continued:
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50
0.2
0.4
0.6
0.8
1
1.2
1.4
t
x(t)
(a)
(b)
Figure : for Problem 9.40
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9.41 For part (a) of the figure,
X(s)Y (s)
=18s + 10
3s2 + 18s + 10
In MATLAB, type
sysa = tf([18,10],[3,18,10]);
When the plot appears, right click on it and select “Characteristics”. From there you canselect the desired items to display on the plot. The results are: maximum per cent overshoot= 7%, peak time = 0.924, 2% settling time = 3.02, and rise time = 0.298. Note that youcannot use the formulas in Table 9.3.2 because this model has numerator dynamics.
For part (a) of the figure,
X(s)Y (s)
=10
3s2 + 18s + 10
In MATLAB, type
sysb = tf(10,[3,18,10]);
When the plot appears, right click on it and select “Characteristics”. From there youcan select the desired items to display on the plot. The results are: maximum per centovershoot = 0% (there is no overshoot), so the peak time does not apply. Also, 2% settlingtime = 6.51 and rise time = 3.6. Note how much larger the settling time at the rise timeare, as compared to the model having numerator dynamics. Thus numerator dynamics canreduce response time but can cause overshoot.
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9.42 The free response of the circuit is
vC(t) = vC(0)e−t/τ
where τ = RC = 3 × 106C. A plot of ln vC(t) versus t gives a straight line whose slope−1/τ is computed from
−1τ
=ln 12 − ln 6.2
0 − 20= −0.033
Thus τ = 1/0.033 = 30.286 and C = τ/R = 10−5 F.
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9.43 The free response is∆T (t) = ∆T (0)e−t/τ
A plot of ln ∆T (t) versus t gives a straight line whose slope −1/τ is computed from
−1τ
=ln(178− 68)− ln(82 − 68)
0 − 3000= −6.8714× 10−4
Thus τ = 1455 sec.Note that ∆T (0) = 178 − 68 = 110. The model is
T (t) = 68 + 110e−t/1455
For T = 135, we have135 = 68 + 110e−t/1455
ort = −1455 ln
135− 68110
= 721 sec
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9.44 From the graph we can easily identify the following three characteristics: the steady-state response, xss = 0.025 m, the peak time, tp = 0.13 s, and the maximum per centovershoot,
M% =0.0395− 0.025
0.025100 = 58%
From Table 9.3.2,
R = ln100M%
= 0.5447
Thusζ =
R√π2 + R2
= 0.171
Thus√
1− ζ2 = 0.985.Because the applied force is 1000 N, xss = 1000/k, and thus k = 1000/0.025 = 40 000
N/m.From
tp =π
ωn
√1 − ζ2
we obtainωn =
π
tp√
1− ζ2= 24.53 rad/s
so thatm =
k
ω2n
=40 000(24.53)2
= 66.476 kg
Becauseζ =
c
2√
mk
we havec = 2ζ
√mk = 2(0.171)
√66.476(40 000) = 557.7 N · s/m
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9.45 From the log decrement relations (9.4.12) and (9.4.14),
δ =1n
lnB1
Bn+1ζ =
δ√4π2 + δ2
Thusδ =
130
ln 5 = 0.0536 ζ = 0.0085
Also,ζ =
c
2√
mk=
c
2√
100k=
c
20√
k= 0.0085
Thus c = 0.17√
k.We are told the time to complete the 30 cycles is 60 s, so we can compute the period P
for one cycle from P = 60/30 = 2 s. Then,
k = mω2n =
mω2d
1 − ζ2=
m(2π/P )2
1 − ζ2=
100(2π/2)2
1− (0.0085)2= 987 N/m
and thus c = 0.17√
k = 5.34 N s/m.
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9.46 From the diagram:
X(s) =1s6 [F (s) − X(s)] + 4X(s)
Solve for the ratio:X(s)F (s)
=6
s + 2
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9.47 From the problem figure:
X(s) =1
s + 3G(s) + 10 [F (s) − X(s)]
Set G(s) = 0 and solve for the ratio:
X(s)F (s)
=10
s + 13
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9.48 From the problem figure:
X(s) =1s
−G(s) − 8X(s) +
4s
[F (s) − 6X(s)]
Set G(s) = 0 and solve for the ratio:
X(s)F (s)
=4
s2 + 8s + 7
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9.49 The following diagram is one of several possible ones:
Figure : for Problem 9.49
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9.50 The following diagram is one of several possible ones:
Figure : for Problem 9.50
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9.51 From the diagram,
C(s) =1
7s + 1
4s + 103s + 1
[R(s) − C(s)] − D(s)
Solve for C(s) to obtain
C(s) =4s + 10A(s)
R(s)− 3s + 1A(s)
D(s) (1)
whereA(s) = 21s2 + 14s + 11 (2)
From the diagram, and using (1),
E(s) = R(s) − C(s) = R(s) − 4s + 10A(s)
R(s) +3s + 1A(s)
D(s)
or
E(s) =21s2 + 10s + 1
A(s)R(s) +
3s + 1A(s)
D(s) (3)
From the diagram, and using (3),
M(s) =4s + 103s + 1
E(s) =(4s + 10)(7s + 1)
A(s)R(s) +
4s + 10A(s)
D(s)
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9.52 From the diagram,
C(s) =1
3s + 2
(1
4s + 1
6s
[R(s) − C(s)] − 10C(s)− D(s)
)
Solve for C(s) to obtain
C(s) =6
A(s)R(s) − 4s2 + s
A(s)D(s) (1)
whereA(s) = 12s3 + 11s2 + 12s + 6 (2)
From the diagram, and using (1),
E(s) = R(s) − C(s) = R(s) − 6A(s)
R(s) +4s2 + s
A(s)D(s)
or
E(s) =12s3 + 11s2 + 12s
A(s)R(s) +
4s2 + s
A(s)D(s) (3)
From the diagram,
C(s) =1
3s + 2[M(s) − D(s)]
Thus, using (1),
M(s) = (3s + 2)C(s) + D(s) =18s + 12
A(s)R(s) +
10s + 6A(s)
D(s)
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9.53 For one loop,
v1 = Ldi1dt
+ Ri3 (1)
For the other loop:
Ri3 =1C
∫i2 dt + v2 (2)
From conservation of charge:i1 = i2 + i3 (3)
Solve (1) and (2) for i1 and i2:
Ldi1dt
= v1 − Ri3 (4)
1C
∫i2 dt = Ri3 − v2 (5)
Transform (3), (4), and (5), and solve for I1(s) and I2(s):
I1(s) =V1(s)− RI3(s)
Ls(6)
I2(s) =RI3(s) − V2(s)
Cs(7)
I2(s) = I3(s)− I1(s) (8)
The left side of the diagram on the following page is drawn from equation (6); the right sidefrom (7), and the middle summer from (8).
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Figure : for Problem 9.53
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9.54 From the first loop:
v1 = R1i1 + L1di1dt
+ v3 (1)
From the second loop:
v3 = L2di2dt
+ v2 (2)
From conservation of charge:i1 = i2 + i3 (3)
Transform (1) through (3) to obtain
(R + L1s)I1(s) + V3(s) = V1(s) (4)
L2sI2(s) − V3(s) = −V2(s) (5)
I1(s) = I2(s) + I3(s) (6)
Solve (4) for I1(s):
I1(s) =1
L1s + R[V1(s) − V3(s)]
This gives the left side of the diagram on the following page.Solve (5) for I2(s):
I2(s) =1
L2s[V3(s) − V2(s)]
This gives the top of the diagram.Solve (6) for I3(s):
I3(s) = I1(s) − I2(s)
This gives the summer in the middle of the diagram.Finally, use the fact that
v3 =1C
∫i3 dt
to obtainV3(s) =
1Cs
I3(s)
This gives the right side of the diagram.
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Figure : for Problem 9.54
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9.55 See the following figure. The equivalent inertia and damping at the motor shaft are
Ie = Im +IL
N2ce =
cL
N2+ cm
The transfer functions are
Ωf(s)Vf(s)
=KT /N
(Lfs + Rf)(Ies + ce)
Ωf(s)TL(s)
=1/N2
Ies + ce
Figure : for Problem 9.55
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9.56 The Simulink digram is identical to that shown in Figure 9.8.6. Edit the Signal Builderand the Look-Up Table blocks to reflect the new data. Change the m-file damper.m to readas follows.
function f = damper(v)if v <= 0
f = -500*(abs(v)).^(1.2);else
f = 50*(abs(v)).^(1.2);end
A Stop Time of 15 seconds gives a suitable plot. The maximum overshoot is 0.522−0.3 =0.222 m at approximately t = 0.57 s. The maximum undershoot is 0− 0.1765 = −0.1765 mat approximately t = 4.73 s.
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9.57 a) The state variable representation is much easier to do because it does not requirethe algebra needed to find the transfer functions. The state space matrices are
A =
0 1 0 0−5 −4 4 10 0 0 14 1 −4 −1
B =
0001
C =
[1 0 0 00 0 1 0
]D =
[00
]
The transfer functions are
X1(s)F (s)
=s + 4
s4 + 5s3 + 12s2 + 13s + 4
X2(s)F (s)
=s2 + 4s + 5
s4 + 5s3 + 12s2 + 13s + 4
b) The figure on the following page shows the Simulink model using both representations.For each you must create the following MATLAB function.
function f = func(t)if t <= 1
f = t;elseif t < 2
f = 2 - t;else
f = 0;end
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Figure : for Problem 9.57
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Solutions Manual c©
to accompany
System Dynamics, Second Edition
by
William J. Palm III
University of Rhode Island
Solutions to Problems in Chapter Ten
c©Solutions Manual Copyright 2010 The McGraw-Hill Companies. All rightsreserved. No part of this manual may be displayed, reproduced, or distributedin any form or by any means without the written permission of the publisheror used beyond the limited distribution to teachers or educators permitted byMcGraw-Hill for their individual course preparation. Any other reproductionor translation of this work is unlawful.
10.1 a) A traffic light may be either closed or open loop, depending on whether or not ituses a sensor to detect the presence of vehicles in the roadway.
b) Most washing machines are open loop, but some now have sensors that detect theamount of dirt in the wash water and adjust their cycle accordingly. These would beclosed-loop devices.
c) If the toaster uses a timer, it is open loop. If it uses a sensor to tell when the desiredtemperature has been reached, it is closed loop.
d) Cruise control is closed loop because it uses a measurement of the vehicle speed tocontrol the engine.
e) A aircraft autopilot is closed loop because it uses measurements of a variety of vari-ables (speed, altitude, angle, etc.) to adjust the control surfaces (ailerons, rudder, elevators,etc.).
f) Closed loop because it reacts to changes in the environment to keep the temperaturenear 98.6.
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10.2 The diagram is like Figure 10.1.7 with
Gc(s) = KP
Ga(s) = 1
Gp(s) =1
4s2 + 6s + 3The command transfer function is found from the diagram as follows.
C(s) =1
4s2 + 6s + 3[Gf (s)R(s) + KP R(s) − KPC(s)]
This givesC(s)R(s)
=Gf (s) + KP
4s2 + 6s + 3 + KP
The system will give perfect response if C(s)/R(s) = 1. This requires that
Gf (s) = 4s2 + 6s + 3
The difficulty with implementing this design is that it requires the first and secondderivatives of the input, r and r, to be computed. This would be difficult to accomplish,especially for step inputs.
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10.3 The diagram is a combination of Figures 10.1.7 and 10.1.8 with
Gc(s) = KP
Ga(s) = 1
Gp(s) =10s
Gf(s) = Kf
Gd(s) = Kd
The command transfer function is found from the diagram as follows.
C(s) =10s
[−D(s) + KdD(s) + KfR(s) + KP R(s)− KP C(s)]
orC(s) =
10(Kf + KP )s + 10KP
R(s) +10(Kd − 1)s + 10KP
D(s)
The disturbance will have no effect if Kd = 1. The command transfer function is
C(s)R(s)
=10(Kf + KP )
s + 10KP
If R(s) = 1/s, the steady-state response is
css =10(Kf + KP )
10KP
which will be 1 if Kf +KP = KP , which is true if Kf = 0. Thus Kf is not needed to satisfythe given specifications.
The time constant expression is
τ =1
10KP
Setting KP = 0.05 will give the desired time constant τ = 2. So the design is KP = 0.05,Kd = 1, and Kf = 0.
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10.4 From the diagram,
C(s) =6
15s + 2[M(s) − D(s)] (1)
M(s) =4K
3s + 1E(s) (2)
E(s) = R(s) − C(s) (3)
To obtain the output C(s) in terms of R(s) and D(s), eliminate M(s) and E(s) fromequations (1), (2), and (3) to obtain
C(s)R(s)
=24K
45s2 + 21s + 2 + 24K
C(s)D(s)
= − 6(3s + 1)45s2 + 21s + 2 + 24K
To obtain the error E(s) in terms of R(s) and D(s), eliminate C(s) and M(s) fromequations (1), (2), and (3) to obtain
E(s)R(s)
=45s2 + 21s + 2
45s2 + 21s + 2 + 24K
E(s)D(s)
=6(3s + 1)
45s2 + 21s + 2 + 24K
To obtain M(s) in terms of R(s) and D(s), eliminate C(s) and E(s) from equations (1),(2), and (3) to obtain
M(s)R(s)
=4K(45s2 + 21s + 2)
(3s + 1)(45s2 + 21s + 2 + 24K)=
4K(15s + 2)45s2 + 21s + 2 + 24K
M(s)D(s)
=24K
45s2 + 21s + 2 + 24K
The characteristic polynomial is the denominator of the transfer functions: 45s2 +21s+2 + 24K.
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10.5 The command transfer function is
C(s)R(s)
=KP
5τas2 + (5 + τa)s + 1 + KP
If τa = 0.05, the characteristic roots are
s = −11.1 ±√
98.01− 4KP
The roots are real if KP ≤ 98.01/4 = 24.5. If KP > 24.5 the roots are complex and theclosed-loop time constant is 1/11.1 = 0.09.
If we neglect τa by setting it equal to zero, the characteristic equation becomes a first-order equation: 5s + 1 + KP = 0. Thus the root is always real and the closed-loop timeconstant is
τ =5
1 + KP
which goes to zero as we increase KP . So if we neglect the actuator time constant, we areled to believe that the step response will not oscillate and that we can make the closed-looptime constant as small as we like by increasing KP . However, in fact the time constant canbe made no smaller that 0.09 and the response will oscillate if KP > 24.5.
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10.6 a) Summing forces parallel to the plane gives
mv = f − F − Wx
b) The block diagram is shown below. The control algorithm is Gc(s), and the motorconstants are KT and Rf . The second diagram is a simplified form where
G1(s) = Gc(s)KaKT
Rf
G2(s) =1
ms
Figure : for Problem 10.6b
(continued on the next page)
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Problem 10.6 continued:
c) From the diagram on the following page,
V (s) = G2(s)[T (s)R
− D(s)]
(1)
T (s) = G1(s)E(s) (2)
E(s) = Vr(s) − V (s) (3)
To obtain the output V (s) in terms of Vr(s) and D(s), eliminate T (s) and E(s) fromequations (1), (2), and (3) to obtain
V (s)Vr(s)
=G1(s)G2(s)
R + G1(s)G2(s)=
KaKTGc(s)mRfRs + KaKTGc(s)
V (s)D(s)
= − RG2(s)R + G1(s)G2(s)
= − RRf
mRfRs + KaKTGc(s)
To obtain the error E(s) in terms of Vr(s) and D(s), eliminate T (s) and V (s) fromequations (1), (2), and (3) to obtain
E(s)Vr(s)
=mRfRs
mRfRs + KaKTGc(s)
E(s)D(s)
=RRf
mRfRs + KaKTGc(s)
To obtain the torque T (s) in terms of Vr(s) and D(s), eliminate E(s) and V (s) fromequations (1), (2), and (3) to obtain
T (s)Vr(s)
=mKaKTRsGc(s)
mRfRs + KaKTGc(s)
T (s)D(s)
=KaKT RGc(s)
mRfRs + KaKTGc(s)
(continued on the next page)
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Problem 10.6 continued:
Figure : for Problem 10.6c
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10.7 a) The diagram is shown in the following figure. Since no load torque is mentioned,we assume that TL = 0.
b)
Ie = I1 +122
I2 +1
[2(3)]2I3 = I1 +
14I2 +
136
I3
Ne = 2(3) = 6
Figure : for Problem 10.7
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10.8
KP +KI
s+ KDs =
KDs2 + KP s + KI
s
Thus KD = 15, KP = 6, and KI = 4, and
TI =KP
KI=
64
= 1.5
TD =KD
KP=
156
= 2.5
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10.9 From Figure 10.4.4,
KP =Rf
Ri= 4
KI =1
RiC= 0.08
Using C = 10−6, we obtain
Ri =1
0.08× 10−6= 1.25× 107 Ω
Rf = 4Ri = 5 × 107 Ω
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10.10 a) From Figure 10.4.6,
KP =R
R1 + R2= 2 (1)
TD = R2C = 2 (2)
α =R1
R1 + R2= 0.1 (3)
The denominator 1+αTDs causes the m curve to level off for ω > 1/αTD. So we choose
1αTD
= 5
which gives α = 0.2/TD = 0.2/2 = 0.1.Using C = 10−6, we obtain from (2): R2 = 2× 106 Ω. We then solve (3) for R1:
R1 =29× 106 Ω
and solve (1) for R:
R =409
× 106 Ω
b) The transfer function isVo(s)Vi(s)
= −2(1 + 2s)1 + 0.2s
The MATLAB code to obtain the frequency response plot is
sys = tf([4,2],[0.2,1]);bodemag(sys),grid
The plot is shown in the figure on the following page.
(continued on the next page)
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Problem 10.10 continued:
10−2
10−1
100
101
102
5
10
15
20
25
30
Mag
nitu
de (d
B)
Bode Diagram
Frequency (rad/sec)
Figure : for Problem 10.10
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10.11 From Figure 10.4.7,
KP = βRC + R2C1
R2C= 10 (1)
KI =β
R2C= 1.4 (2)
KD = βRC1 = 4 (3)
β =R2
R1 + R2(4)
From (2),
R2 =β
1.4C(5)
From (3),
R =4
βC1(6)
From (4),
R1 = R21 − β
β=
1 − β
1.4C(7)
Substituting (2), (5), and (6) into (1) gives
1.4(
4C
βC1+
βC1
1.4C
)= 10
This can be rearranged as follows:
(βC1)2 − 10C(βC1) + 5.6C2 = 0
Choosing C = 10−6 we obtain
(βC1)2 − 10−5(βC1) + 5.6× 10−12 = 0
which has the solutionsβC1 = 5.9546× 10−7, 9 × 10−6
(continued on the next page)
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Problem 10.11 continued:
Choosing the first solution we obtain
R = 6.7175× 106 Ω
R2 =0.4253
C1
R1 = 7.1429× 105 − 0.4253C1
(8)
From this, since we require that R1 > 0, we see that C1 must satisfy
C1 > 5.9546× 10−7 (9)
To limit the response above 100 rad/s, we require that
1βR1C1
≥ 100
or1
5.9546× 10−7R1≥ 100
This requires that R1 ≤ 1.679× 104. Combining this with (8) and (9) shows that C1 mustlie in the range
5.9546× 10−7 < C1 ≤ 6.3× 10−7
Choosing C1 = 6 × 10−7 F gives R1 = 5407 Ω, R2 = 7.089 × 105 Ω, and R = 4/βC1 =6.7175× 106 Ω.
The second solution, βC1 = 9 × 10−6, gives a much smaller allowable range for C1.
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10.12 a)
Y (s) =6
7s + 314s
=84
(7s + 3)s
Thusyss = lim
s→0s
84(7s + 3)s
=843
= 28
b)
Y (s) =7s − 3
10s2 + 6s + 95s
Thusyss = lim
s→0s
7s − 310s2 + 6s + 9
5s
= −159
= −53
c)
Y (s) =3s + 5s2 − 9
12s
ThussY (s) = 12
3s + 5s2 − 9
The roots of the denominator are s = ±3. Since one root is positive, the final value theoremcannot be applied.
d)
Y (s) =4s + 3
s2 + 2s − 78s
ThussY (s) = 8
4s + 3s2 + 2s − 7
One of the roots of the denominator is positive, so the final value theorem cannot be applied.
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10.13 For each case, the error transform is
E(s) = R(s) − C(s) = R(s) − T (s)R(s) = [1− T (s)]R(s)
whereT (s) =
C(s)R(s)
a)
E(s) =3s
3s + 16s2
=18
3s + 11s
ess = lims→0
s18s
3s + 11s2
= lims→0
183s + 1
= 18
b)
E(s) =3s − 43s + 1
6s2
ess = lims→0
s3s − 43s + 1
6s2
= ∞
c)
E(s) =3s2 + 5s
3s2 + 5s + 412s2
=3s + 5
3s2 + 5s + 412s2
ess = lims→0
s3s + 5
3s2 + 5s + 412s2
= lims→0
123s + 5
3s2 + 5s + 4= 15
d)
E(s) =2s2 + 4s − 52s2 + 4s + 5
8s2
ess = lims→0
s2s2 + 4s − 52s2 + 4s + 5
8s2
= lims→0
2s2 + 4s − 52s2 + 4s + 5
8s
= ∞
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10.14 Divide the equation by 3 and obtain the characteristic equation:
s2 − (b + 2)s + 2b + 5 = 0
The Routh-Hurwitz criterion implies that the system is stable if and only if −(b+2) > 0and 2b + 5 > 0, which gives −2.5 < b < −2 for stability. Neutral stability occurs if eitherthe s term or the constant term is missing in the characteristic equation. This occurs ifb + 2 = 0 (the roots are s = ±j) or if 2b + 5 = 0 (the roots are s = 0,−0.5.).
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10.15 From the Routh-Hurwitz criterion it is necessary and sufficient that K > 0 and9(26)− K > 0. Thus, the system is stable if and only if 0 < K < 234.
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10.16 The time constant requirement means that no root can lie to the right of s = −2.Translate the origin of the s plane to s = −2 by substituting s = p−2 into the characteristicequation. This gives
(p− 2)3 + 9(p − 2)2 + 26(p− 2) + K = 0
Expanding and collecting terms gives
p3 + 3p2 + 2p + K − 24 = 0
We can apply the Routh-Hurwitz criterion to this equation to determine when all roots phave negative real parts (and thus when all s roots lie to the left of s = −2). This occurswhen K − 24 > 0 and 3(2)− (K − 24) > 0. Thus K must be in the range 24 < K < 30.
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10.17 a) From the Routh-Hurwitz criterion,
a > 0 K > 0 b > 0
and 2aK > 2b, or aK > b.b) From the Routh-Hurwitz criterion,
a > 0 b > 0 K > 0
and 5aK > 25K, or ab > 5K.c) From the Routh-Hurwitz criterion,
4 + K > 0
and 12(12) > 4(4 + K). Thus−4 < K < 32
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10.18 Referring to Example 10.3.5, we obtain the following transfer function.
Ω(s)Ω(s)
=KP KT
D(s)
where D(s) is given by equation (3) in that example. The equivalent inertia is
Ie = Im +IL + It
N2= 17.75× 10−4
The equivalent damping isce = cm +
cL
N2= 10−3
The characteristic polynomial is
D(s) = 1.42× 10−5s2 + 2.848× 10−3s + 0.0816 + 0.2KP
The steady-state requirement is
ωLss
ωr=
0.2KP
0.0816 + 0.2KP= 0.9
This gives KP = 3.672. The resulting roots are
s = −100.282± 217.735j
So the time constant is 1/100.282 = 0.00997 s and the damping ratio is 0.418.The disturbance transfer function is
ΩL(s)TL
= −(Las + Ra)/ND(s)
The steady-state deviation caused by the disturbance torque is
ωLss = − Ra/N
NRace + NKTKb + KP KTTL = −0.8/2
0.8160.2 = −0.49(0.2) = 0.098
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10.19 The steady-state offset error is
offset error = ωr − ωss = 1 − KP
3 + KP= 0.2
if KP = 12. The time constant for this value of KP is
τ =I
c + KP=
23 + KP
=215
The steady-state response due to the disturbance is
disturbance response =−1
c + KP=
−115
= −0.067
If both inputs are applied, the actual steady-state speed is
ωss = command response + disturbance response = (1− 0.2) + (−0.067) = 0.733
Note that we can make the offset error, the disturbance response, and the time constantsmaller only by making KP larger than 12, but this increases the maximum required torqueand probably the cost of the system.
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10.20 Substitute I = 2, c = 3, and KP = 12 into equations (1) and (2) of Example 10.6.1to obtain:
Ω(s)Ωr(s)
=12
2s + 15
Ω(s)Td(s)
= − 12s + 15
Using Td(s) = 1/s2 with the final value theorem, we find the steady-state disturbanceresponse to be
ωss = lims→0
s−1
2s + 151s2
= −∞
Thus the controller cannot keep the output near its desired value if the ramp disturbancelasts too long.
To investigate the command response, use Ωr(s) = 1/s2 with the first transfer functionto obtain
Ω(s) =12
2s + 151s2
=4
5s2− 8
75s+
875(s + 15/2)
Thus the response is
ω(t) =45t − 8
75+
875
e−15t/2
Since the slope is 4/5 and is less than the slope of the input, we can see that the speed ω(t)never catches up with the command input ωr(t), and that the steady-state error is infinite.We could have also obtained this result from the final value theorem, using the followingexpression for the error when no disturbance is present.
E(s) = Ωr(s) − Ω(s) = Ωr(s)(1 − 12
2s + 15
)= Ωr(s)
(2s + 32s + 15
)
Thus, with Ωr(s) = 1/s2,
ess = lims→0
s1s2
(2s + 32s + 15
)= ∞
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10.21 The transfer functions are
Ω(s)Ωr(s)
=KI
20s2 + cs + KI
Ω(s)Td(s)
= − s
20s2 + cs + KI
The steady-state unit-step response for the command input is
ωss =KI
KI= 1
which is perfect. The steady-state deviation caused by a unit-step disturbance is
∆ωss = 0
which is also perfect.We also have
ζ =c
2√
20KI= 1
which gives
KI =c2
80Since ζ = 1 we can write
τ =2(20)
c=
40c
So, if c = 10,
KI =10080
= 1.25
τ =4010
= 4
If c = 0.2,
KI =0.0480
= 0.0005
τ =400.2
= 200
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10.22 (a) The command transfer function is
Ω(s)Ωr(s)
=KP s + KI
4s2 + (4 + KP )s + KI(1)
For both cases (1) and (2), the time constant is
τ =8
4 + KP= 0.2
Therefore, KP = 36 for both cases. For case (1),
ζ =40
2√
4KI= 0.707
Thus, KI = 200. For case (2), the same procedure gives KI = 100.(b) Because ζ > 1 is required, there will be two real roots, and the dominant root must
be s = −1/τ = −5. For this to be the dominant root, the second root must be to the leftof s = −5. Choosing an arbitrary separation factor of 10, we place the secondary root ats = −50. The characteristic polynomial must therefore be
(s + 5)(s + 50) = s2 + 55s + 250 = 0
Multiply this equation by 4 before comparing it with the denominator of equation (1),so that the highest coefficients will be the same in each polynomial. Thus the requiredcharacteristic polynomial must be
4s2 + 220s + 1000
Comparing its coefficients with those of the transfer function denominator gives the equa-tions for the required gain values.
4 + KP = 220
KI = 1000
Thus, KP = 216.
(continued on the next page)
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Problem 10.22 continued:
For a unit-step command input, a MATLAB file to compute the unit-step response forthe first case, where KP = 216 and KI = 1000, is
KP = 216; KI = 1000;sys = tf(4*[KP,KI],[4,4+KP,KI]);
Right-click on the resulting plot, select “Characteristics”, then “Peak Response” to deter-mine the maximum percent overshoot, and the rise time. The following table summarizesthe important response characteristics. The overshoot decreases as ζ increases, as expected.But the 10-90% rise time is the smallest for ζ = 1.74. A common misconception is thatresponse is sluggish for ζ > 1, but this is clearly not the case here. The gains KP = 216,KI = 1000 give a good response with a very small rise time and a small overshoot. However,high gain values are required, and this might make the physical system expensive.
The fast response with ζ = 1.74 is due to the numerator dynamics, whose effect isincreased by the high gain values used in the third case. This effect can be seen by comparingthe actual overshoot and rise time with those predicted by the second-order model withoutnumerator dynamics. These values are given in the following table.
Without Numerator DynamicsOvershoot Rise Time Overshoot Rise Time
Case KP KI ζ % 10 − 90% % 10 − 90%1 9 50 0.707 16 0.14 4 0.362 9 25 1 8 0.18 0 0.663 54 250 1.74 5 0.03 0 0.44
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10.23 With the disturbance torque absent, the motor torque expression can be found asfollows. From the block diagram,
T (s) =(
KPKI
s
)E(s) =
(KPs + KI
s
)[Ωr(s) − Ω(s)]
Using the transfer function Ω(s)/Ωr(s), we have
T (s) =(
KP s + KI
s
)Ωr(s)
(1 − KP s + KI
4s2 + (4 + KP )s + KI
)
or
T (s) = 4KP s2 + (KP + KI)s + KI
4s2 + (4 + KP )s + KIΩr(s) (1)
This can be used to obtain the unit-step responses for the various gain values computedabove.
For a unit-step command input, a MATLAB file to compute the actuator response forthe first case, where KP = 36 and KI = 200, is
KP = 36; KI = 200;sys = tf(4*[KP,KP+KI,KI],[4,4+KP,KI]);
Right-click on the resulting plot, select “Characteristics”, then “Peak Response” to deter-mine the peak value. For Case 1 it is 36. Repeating this for the other cases, we obtain peakvalues of 36 for case 2 and 216 for case 3.
Note that the maximum torque occurs at t = 0 for all three cases, that the maximumtorque equals KP , and thus is much greater for the case 3, which has the largest gain values.So we see that the fast rise time of case 3 is obtained at the expense of having to provide amotor with a larger torque capability.
(continued on the next page)
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Problem 10.23 continued:
In the absence of a disturbance, the maximum torque occurs at t = 0 for both theproportional control system and the PI control system, if the command is a step function.The maximum torque is KP M , where M is the step magnitude. This is because the stepinput is a sudden command, and the proportional term responds instantaneously to try toreduce the error, whereas the integral term takes time to build up. Thus the maximumerror, and maximum torque, occur at t = 0. At t = 0, the speed is zero and the torque is
T (0) = KP e(0) = KP [ωr(0)− ω(0)] = KP ωr(0) = KP M
This reasoning assumes that the maximum error occurs at t = 0. This is not true for somesystems, such as the system shown in Figure 10.6.7. For such systems, there is no simpleformula to compute the maximum torque.
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10.24 The error equation is
E(s) =4s2 + 4s
4s2(4 + Kp)s + KIΩr(s) +
s
4s2(4 + Kp)s + KITd(s)
Apply the final value theorem using Ωr(s) = Td(s) = 1/s2. For the command error,
ess =4
KI
• For ζ = 0.707, KI = 200, and ess = 4/200
• For ζ = 1, KI = 100, and ess = 4/100
• For ζ = 1.74, KI = 1000, and ess = 4/1000
For the disturbance error,
ess =1
KI
• For ζ = 0.707, KI = 200, and ess = 1/200
• For ζ = 1, KI = 100, and ess = 1/100
• For ζ = 1.74, KI = 1000, and ess = 1/1000
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10.25 From the figure given in the problem statement ,
Ω(s)Ωr(s)
=KP s + KI
2s2 + (2 + KP )s + KI
So the characteristic polynomial is
2s2 + (2 + KP )s + KI (1)
a) 1. The characteristic polynomial must be factored as
2(s + 10)(s + 8) = 2s2 + 36s + 160
Comparing coefficients with the characteristic polynomial (1), we obtain
KP = 34 KI = 160
2. The characteristic polynomial must be factored as
2(s + 10)(s + 20) = 2s2 + 60s + 400
Comparing coefficients with the characteristic polynomial (1), we obtain
KP = 58 KI = 400
3. The characteristic polynomial must be factored as
2(s + 10)(s + 50) = 2s2 + 120s + 1000
Comparing coefficients with the characteristic polynomial (1), we obtain
KP = 118 KI = 1000
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10.26 From the figure given in the problem statement ,
Ω(s)Ωr(s)
=KI
4s2 + (4 + K2)s + KI
So the characteristic polynomial is
4s2 + (4 + K2)s + KI
and the damping ratio is
ζ =4 + K2
2√
4KI=
4 + K2
4√
KI
a) 1. Since ζ ≤ 1, we can write an expression for the time constant:
τ =8
4 + K2= 0.2
which gives K2 = 36. Thus
ζ =4 + K2
4√
KI=
10√KI
= 0.707
which gives KI = 200.2. Since ζ ≤ 1, we can write an expression for the time constant:
τ =8
4 + K2= 0.2
which gives K2 = 36. Thus
ζ =4 + K2
4√
KI=
10√KI
= 1
which gives KI = 100.3. The characteristic polynomial must be factored as
4(s + 5)(s + 50) = 4s2 + 220s + 1000
Comparing coefficients with the characteristic polynomial, we obtain
K2 = 216 KI = 1000
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10.27 The actuator equation is
T (s) =(4s + 4)KI
4s2 + (4 + K2)s + KIΩr(s) +
K2s + KI
4s2 + (4 + K2)s + KITd(s)
For a unit-step command input, a MATLAB file to compute the actuator response for thefirst case, where K2 = 36 and KI = 200, is
K2 = 36; KI = 200;sys = tf(KI*[4,4],[4,4+K2,KI]);
Right-click on the resulting plot, select “Characteristics”, then “Peak Response” to deter-mine the peak value. For Case 1 it is 14.5. Repeating this for the other cases, we obtainpeak values of 8.58 for case 2 and 16.1 for case 3.
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10.28 The error equation is
E(s) =4s2 + (4 + K2)s
4s2(4 + K2)s + KIΩr(s) +
s
4s2(4 + K2)s + KITd(s)
Apply the final value theorem using Ωr(s) = Td(s) = 1/s2. For the command error,
ess =4 + K2
KI
• For ζ = 0.707, K2 = 36, KI = 200, and ess = 40/200
• For ζ = 1, K2 = 36, KI = 100, and ess = 40/100
• For ζ = 1.74, K2 = 216, KI = 1000, and ess = 220/1000
For the disturbance error,
ess =1
KI
• For ζ = 0.707, KI = 200, and ess = 1/200
• For ζ = 1, KI = 100, and ess = 1/100
• For ζ = 1.74, KI = 1000, and ess = 1/1000
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10.29 From the figure given in the problem statement ,
Ω(s)Ωr(s)
=KI
2s2 + (2 + K2)s + KI
So the characteristic polynomial is
2s2 + (2 + K2)s + KI (1)
a) 1. The characteristic polynomial must be factored as
2(s + 10)(s + 8) = 2s2 + 36s + 160
Comparing coefficients with the characteristic polynomial (1), we obtain
K2 = 34 KI = 160
2. The characteristic polynomial must be factored as
2(s + 10)(s + 20) = 2s2 + 60s + 400
Comparing coefficients with the characteristic polynomial (1), we obtain
K2 = 58 KI = 400
3. The characteristic polynomial must be factored as
2(s + 10)(s + 50) = 2s2 + 120s + 1000
Comparing coefficients with the characteristic polynomial (1), we obtain
K2 = 118 KI = 1000
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10.30 The compensated system diagram is shown in the following figure.
Figure : for Problem 10.30
Command compensation cannot affect the characteristic roots or the disturbance re-sponse, so we set D(s) = 0 here. The error equation is
E(s) =s(Is + c− Kf)
Is2 + (c + KP )s + KI)R(s)
For a unit-step input, the final value theorem gives ess = 0 as long as the system is stable.Thus ess = 0 even if there is no command compensation (Kf = 0). For a unit-ramp input,the final value theorem gives
ess =c − Kf
KI= 0
if Kf = c. Thus the command compensation improves the ramp response without affectingthe step response.
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10.31 The requirements are (for unit step inputs)
1. the steady-state command error must be zero.
2. the magnitude of the steady-state disturbance error must be ≤ 0.1
3. ζ = 0.707
Part (a). We can do P control and PD control at the same time by finding the errorequation for PD control, and then setting KD = 0 to obtain the results for P control. FromFigure 10.7.4 with I = 20 and c = 10, the error equation can be written as follows (afterdoing some algebra):
(1) E(s) = Θr(s)−Θ(s) =20s2 + 10s
20s2 + (10 + KD)s + KPΘr(s)+
120s2 + (10 + KD)s + KP
Td(s)
The characteristic equation is obtained from the denominator:
(2) 20s2 + (10 + KD)s + KP = 0
For a unit step command, the steady state error is found by setting Θr(s) = 1/s andTd(s) = 0 in equation (1), and using the final value theorem.
ess = lims→0
sE(s) = lims→0
s
[20s2 + 10s
20s2 + (10 + KD)s + KP
]1s
= 0
if the system is stable (that is, if 10 + KD > 0 and KP > 0).For a unit step disturbance, the steady state error is found by setting Td(s) = 1/s and
Θr(s) = 0 in equation (1), and using the final value theorem.
ess = lims→0
sE(s) = lims→0
s
[1
20s2 + (10 + KD)s + KP
]1s
=1
KP
if the system is stable.Thus the steady state errors are the same for P and PD control. So they both satisfy
specification #1, and specification #2 if KP ≥ 10. The difference between the two lies intheir transient performance.
(continued on the next page)
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Problem 10.31 continued:
P control: For P control, the characteristic equation is (set KD = 0 in equation (2)):
20s2 + 10s + KP = 0
Thusζ =
102√
20KP
To satisfy specification #3 (ζ = 0.707), KP must be 2.5, which is smaller than the valuerequired to satisfy specification #2. Thus P control cannot satisfy simultaneously bothspecification #2 and specification #3.
PD control. For PD control, since ζ < 1 we can write the following formula for the timeconstant [the real part of the roots of equation (1) is −(10 + KD)/40]:
(3) τ =40
10 + KD
The formula for ζ is
(4) ζ =10 + KD
2√
20KP= 0.707
A solution that exactly satisfies specification #3 is KP = 10. From equation (4) we findthat KD = 10. From equation (3) we see that τ = 40/20 = 2. Other solutions are possiblefor values of KP greater than 10.
Part (b). In summary
1. With P control, the requirements on damping ratio and disturbance error cannot bemet simultaneously.
2. With PD control, one solution is KP = 10, KD = 10
Thus P control does not work. PD control works but has numerator dynamics that increasesthe overshoot.
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10.32 Because ζ = cos β, the value ζ = 0.707 corresponds to β = 45, which is a 45 line onthe complex plane. Thus ζ = 0.707 corresponds to a pair of roots whose real and imaginaryparts have the same magnitude; that is, s = −a ± ja. The time constant of these roots isτ = 1/a. Thus the roots must be s = −1±j if τ = 1 and ζ = 0.707. These roots correspondto the polynomial equation
(s + 1− j)(s + 1 + j) = (s + 1)2 + 1 = s2 + 2s + 2 = 0
or10s2 + 20s + 20 = 0
Compare this with the system’s characteristic equation obtained from the denominator ofthe transfer function:
10s2 + (3 + KD)s + KP = 0
Thus KP = 20 and 3 + KD = 20, or KD = 17.
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10.33 From the figure given in the problem statement we obtain
Θ(s) =KP
20s2 + (10 + K2)s + KPΘr(s)−
120s2 + (10 + K2)s + KP
Td(s)
The steady-state unit-step response for the command input is
θss =KP
KP= 1
which is perfect. The steady-state deviation caused by a unit-step disturbance is
∆θss = − 1KP
Thus KP ≥ 10 is required to meet the specification.If ζ ≤ 1,
τ =40
10 + K2= 0.1
which gives K2 = 390.Thus
ζ =10 + K2
2√
20KP=
200√20KP
≤ 1
if KP ≥ 2000.So an acceptable design is KP = 2000, K2 = 390, which gives ζ = 1, τ = 0.1, and
∆θss = − 12000
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10.34 (a) With the specific parameter and gain values used in Problem 10.33, the resultingcommand transfer function is
Θ(s)Θr(s)
=20 + 17s
10s2 + 20s + 20(1)
A MATLAB file to compute the unit-ramp response is the following.
systheta = tf([17,20],[10,20,20]);t = [0:0.001:4];input = t;lsim(systheta,input,t)
Even though the system is underdamped, we do not see oscillations in the response to a unitramp command because the oscillation period is 2π = 6.28, which is greater than 4τ = 1.Thus the oscillations die out before one period has occurred. The steady-state error canbe found using the final value theorem. From the block diagram in Figure 10.7.4, with thespecific parameter and gain values used here, we obtain
E(s) = Θr(s) − Θ(s) = Θr(s)(1 − 20 + 17s
10s2 + 20s + 20
)= Θr(s)
(10s2 + 3s
10s2 + 20s + 20
)
With Θr(s) = 1/s2, we have
ess = lims→0
sE(s) = lims→0
s1s2
(10s2 + 3s
10s2 + 20s + 20
)=
320
From Figure 10.7.4 with the specific parameter and gain values used here and Θr(s) =1/s2, we obtain the following actuator equation:
T (s) =170s3 + 251s2 + 60s
10s2 + 20s + 20Θr(s)
Note that the MATLAB tf function cannot be used to compute this transfer functionbecause the order of the numerator is greater than that of the denominator. We can,however, work around this difficulty as follows.
(continued on the next page)
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Problem 10.34 continued:
With Θr(s) = 1/s2, and canceling an s term from the numerator and denominator, weobtain
T (s) =170s2 + 251s + 6010s2 + 20s + 20
1s
This is equivalent to the unit-step response of the following transfer function
T (s)Ωr(s)
=170s2 + 251s + 6010s2 + 20s + 20
A MATLAB file to compute the unit-ramp response is the following.
systorque = tf([170,251,60],[10,20,20]);step(systorque)
The torque approaches the constant value of 3 needed to counteract the damping torque3θ = 3(1) = 3.
(continued on the next page)
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Problem 10.34 continued:
(b) Actual disturbance inputs are not always “clean” functions like steps, ramps, andsine waves. We often do not know the exact functional form of the disturbance. It isoften a random function, such as the disturbance torque due to wind gusts of a rotatingradar antenna. The analysis of random inputs is beyond the scope of this text. However,although frequency response plots strictly speaking describe only the steady-state responsefor periodic inputs, the plots are used to obtain a rough idea of the system’s transientresponse to fluctuating inputs that are not periodic. The disturbance transfer function forthe system in question is
Θ(s)Td(s)
=−1
10s2 + 20s + 20(1)
The frequency response plot can be obtained with the following MATLAB file.
sys=tf(1,[10, 20,20]);bodemag(sys)
Right-click on the plot, select “Characteristics”, then “Peak Response”, to see the peakvalue and the corresponding frequency. They are m = −26 dB at ω = 0 rad/s. The plotshows that the system responds more to slowly varying disturbances, and tends to rejectdisturbances whose frequencies lie above the frequency ω = 1.4 radians per unit time. Fordisturbance frequencies within the bandwidth of 0 ≤ ω ≤ 1.4, the system attenuates theinput by approximately m = −26 db, or by a multiplicative factor of M = 10−26/20 = 0.05.
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10.35 a) Because this system is third order, we do not have formulas to use for the dampingratio and the time constant. In addition, we must interpret the specifications to apply tothe dominant root or root pair, and must choose the secondary root somewhat arbitrarily.The values ζ = 0.707 and τ = 1 correspond to the dominant root pair s = −1 ± j. Thethird root must be less than −1 so that it will not be dominant. We arbitrarily select thethird root to be s = −2. This choice can be investigated later if necessary. These threeroots correspond to the polynomial equation
(s + 2)(s + 1 − j)(s + 1 + j) = (s + 2)[(s + 1)2 + 1
]= s3 + 4s2 + 6s + 4 = 0
To compare this with the system’s characteristic equation, we multiply it by 10.
10s3 + 40s2 + 60s + 40 = 0
Compare this with the system’s characteristic equation:
10s3 + (3 + KD)s2 + KP s + KI
Thus KP = 60, KI = 40, and 3 + KD = 40, or KD = 37.b) The resulting disturbance transfer function for this system is
Θ(s)Td(s)
=−s
10s3 + 40s2 + 60s + 40
The frequency response plot can be obtained with the following MATLAB file.
sys=tf([1,0],[10, 40,60,40]);bodemag(sys)
Right-click on the plot, select “Characteristics”, then “Peak Response”, to see the peakvalue and the corresponding frequency. They are m = −33.7 dB at ω = 1.14 rad/s. Theplot shows that the system attenuates disturbance inputs by a factor of m = −33.7 dB ormore. This corresponds to an amplitude reduction of M = 10−33.7/20 = 0.0207. The systemrejects by an even greater amount any disturbances whose frequencies lie below or abovethe frequency ω = 1.14. Compare this performance with the PD control system of Problem10.35, which does not have as great an attenuation and responds more to low frequencydisturbances than high frequency ones.
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10.36 The requirements are (for unit step inputs)
1. the steady-state command error must be zero.
2. the magnitude of the steady-state disturbance error must be ≤ 0.1
3. the time constant must be 0.1
We can do P control and PD control at the same time by finding the error equationfor PD control, and then setting KD = 0 to obtain the results for P control. From Figure10.7.4 with I = 20 and c = 10, the error equation can be written as follows (after doingsome algebra):
(1) E(s) = Θr(s)−Θ(s) =20s2 + 10s
20s2 + (10 + KD)s + KPΘr(s)+
120s2 + (10 + KD)s + KP
Td(s)
The characteristic equation is obtained from the denominator:
(2) 20s2 + (10 + KD)s + KP = 0
For a unit step command, the steady state error is found by setting Θr(s) = 1/s andTd(s) = 0 in equation (1), and using the final value theorem.
ess = lims→0
sE(s) = lims→0
s
[20s2 + 10s
20s2 + (10 + KD)s + KP
]1s
= 0
if the system is stable (that is, if 10 + KD > 0 and KP > 0).For a unit step disturbance, the steady state error is found by setting Td(s) = 1/s and
Θr(s) = 0 in equation (1), and using the final value theorem.
ess = lims→0
sE(s) = lims→0
s
[1
20s2 + (10 + KD)s + KP
]1s
=1
KP
if the system is stable.Thus the steady state errors are the same for P and PD control. So they both satisfy
specification #1, and specification #2 if KP ≥ 10. The difference between the two lies intheir transient performance.
(continued on the next page)
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Problem 10.36 continued:
P control. For P control, we have
ζ =10
2√
20KP
Thus, if KP ≥ 10, ζ ≤ 0.35. Because ζ < 1, the time constant gives
τ =4010
= 4
Thus specification #3 is not satisfied.
PD control. For PD control, specifications #1 and #2 are met if KP ≥ 10. There are twoways to meet specification #3: i) set ζ ≤ 1 or ii) set ζ > 1.
Case (i). For case (i), (ζ ≤ 1), the time constant is given by
τ =40
10 + KD= 0.1
if KD = 390. For this value, the damping ratio is found from equation (4) to be
ζ =10 + 3902√
20KP=
400√20KP
Thus ζ ≤ 1 only if KP ≥ 8000. Thus one solution is KP = 8000 and KD = 390. This givesa damping ratio ζ = 1.Case (ii). For case (ii), (ζ > 1), the dominant root must have a real part equal to −10 forthe time constant to be 0.1. The other root must be placed to the left of s = −10. Thusthere are an infinite number of solutions, depending on where the second root is placed.For example, if the second root is placed at s = −b, the characteristic equation will factoras 20(s + 10)(s + b) = 20s2 + (200 + 20b)s + 200b. Comparing this to the characteristicequation shows that KP = 200b and KD = 190 + 20b. Thus the larger b is, the greaterthe gains KP and KD. For example, using an arbitrary root separation factor of 10, onesolution is b = 100, KP = 20, 000 and KD = 2190. Generally, placing the second root far tothe left requires higher gains, which is not desirable. In this case, additional specificationsare needed to arrive at a unique solution.
(continued on the next page)
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Problem 10.36 continued:
Part (b). In summary,
1. Using P control, specifications #2 and #3 cannot be met simultaneously.
2. Using PD control, a solution with ζ ≤ 1 is: KP = 8000, KD = 390. A solution withζ > 1 is: KP = 20, 000, KD = 2190. Other solutions are possible.
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10.37 From the following diagram with Td(s) = 0 we obtain
Θ(s) =1
s(Is + c)KfΘr + (KP + KDs) [Θr(s) − Θ(s)]
ThusΘ(s)Θr(s)
=Kf + KP + KDs
Is2 + (c + KD)s + KP
The steady-state unit-step response for the command input is
θss =Kf + KP
KP= 1
only if Kf = 0. The error equation is
E(s) = Θr(s) − Θ(s) =Is2 + cs − Kf
Is2 + (c + KD)s + KPΘr(s)
For a unit-ramp command,ess = lim
s→0sE(s) = ∞
regardless of the value of Kf . So Kf does not improve the steady-state response for a stepor a ramp input.
Figure : for Problem 10.37
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10.38 a) Try I-action: G(s) = KI/s. Thus
C(s)R(s)
=KI
20s2 + 0.2s + KI
and the error equation with a unit-ramp input is
E(s) = R(s)− C(s) =1s2
(20s2 + 0.2s
20s2 + 0.2s + KI
)
Thus ess = 0.2/KI = 0.01 if KI = 20. This gives ζ = 0.2/(2√
20KI) = 0.005 and τ =2(20)/0.2 = 200.
b) Try PI-action: G(s) = KP + KI/s. Thus
C(s)R(s)
=KP s + KI
20s2 + (0.2 + KP )s + KI
and the error equation with a unit-ramp input is
E(s) = R(s) − C(s) =1s2
(20s2 + 0.2s
20s2 + (0.2 + KP )s + KI
)
Thus ess = 0.2/KI = 0.01 if KI = 20. This gives
ζ =0.2 + KP
2√
20KI=
0.2 + KP
40= 1
if KP = 38.8. This gives τ = 2(20)/(0.2+ KP ) = 1.
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Problem 10.38 continued:
c) Try PID-action: G(s) = KP + KI/s + KDs. Thus
C(s)R(s)
=KDs2 + KP s + KI
(20 + KD)s2 + (0.2 + KP )s + KI
and the error equation with a unit-ramp input is
E(s) = R(s)− C(s) =1s2
(20s2 + 0.2s
(20 + KD)s2 + (0.2 + KP )s + KI
)
Thus ess = 0.2/KI = 0.01 if KI = 20. The damping ratio is
ζ =0.2 + KP
2√
(20 + KD)KI=
0.2 + KP
2√
20(20 + KD)= 1
Because ζ = 1, the expression for the time constant is
τ =2(20 + KD)0.2 + KP
= 0.1
The solution is KP = 3.8 and KD = −19.8.
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10.39 The error equation is:
(1) E(s) =s2 + s
s2 + (1 + KD)s + KPΘr(s) +
1s2 + (1 + KD)s + KP
Td(s)
For a unit ramp command, the steady state error is found by setting Θr(s) = 1/s2 andTd(s) = 0 in equation (1), and using the final value theorem.
ess = lims→0
sE(s) = lims→0
s
(s2 + s
s2 + (1 + KD)s + KP
)1s2
=1
KP
if the system is stable (that is, if 1 + KD > 0 and KP > 0).For a unit ramp disturbance, the steady state error is found by setting Td(s) = 1/s2 and
Θr(s) = 0 in equation (1), and using the final value theorem.
ess = lims→0
sE(s) = lims→0
s
(1
s2 + (1 + KD)s + KP
)1s2
= ∞
if the system is stable.
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10.40 The system’s closed-loop transfer function can be found from the block diagram. Itis
C(s)R(s)
=KP + KDs
2s2 + (2 + KD)s + KP
Thus the characteristic equation is 2s2 + (2 + KD)s + KP = 0. Because ζ is specified to beless than 1, we can write the following formula for the time constant.
τ =2(2)
2 + KD= 1
This gives KD = 2. The damping ratio is given by
ζ =2 + KD
2√
2KP=
2 + 22√
2KP=
2√2KP
= 0.9
This gives KP = 2.469.If the command is a step input with a magnitude m, then R(s) = m/s, and the Final
Value Theorem gives
css = lims→0
sC(s) = lims→0
s
(KP + KDs
2s2 + (2 + KD)s + KP
)m
s= m
Thus the steady state error is zero, so all three specifications are satisfied.
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10.41 a) The output equation is
Θ(s) =KDs2 + KP s + KI
10s3 + (2 + KD)s2 + KP s + KIΘr(s)−
s
10s3 + (2 + KD)s2 + KP s + KITd(s)
The characteristic equation
10s3 + (2 + KD)s2 + KP s + KI = 0 (1)
For case 1, the desired characteristic equation can be expressed as
10(s + 0.5)[(s + 5)2 + 25] = 10s3 + 105s2 + 550s + 250 = 0 (2)
Comparing coefficients in equations (1) and (2), we obtain
KP = 550 KI = 250 KD = 103
For case 2, the desired characteristic equation can be expressed as
10(s + 0.5)(s + 1)(s + 2) = 10s3 + 35s2 + 35s + 10 = 0 (3)
Comparing coefficients in equations (1) and (3), we obtain
KP = 35 KI = 10 KD = 33
b) The MATLAB file is
KP1 = 550; KD1 = 103; KI1 = 250;systheta1 = tf([KD1,KP1,KI1],[10,2+KD1,KP1,KI1]);KP2 = 35; KD2 = 33; KI2 = 10;systheta2 = tf([KD2,KP2,KI2],[10,2+KD2,KP2,KI2]);step(systheta1,systheta2)
Right-click on the plot, select “Characteristics”, then “Peak Response” to determine themaximum percent overshoot and the peak time. For Case 1, the maximum percent overshootis 22%; for Case 2, it is15%. For the gains given in Example 10.7.7, it is 10%. The separationfactors are 10, 2, and 10, respectively. On the basis of this example we would conclude thata large separation factor is better, provide that the dominant root is complex.
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Problem 10.41 continued:
c) The disturbance transfer function is
Θ(s)Td(s)
= − s
10s3 + (2 + KD)s2 + KP s + KI
The MATLAB file, which is a continuation of the previous file, is
sysdist1 = tf([-1,0],[10,2+KD1,KP1,KI1]);sysdist2 = tf([-1,0],[10,2+KD2,KP2,KI2]);bodemag(sysdist1,sysdist2)
Right-click on the plot, select “Characteristics”, then “Peak Response” to determine thepeak response and the corresponding frequency. For Case 1, the maximum response ism = −54.2 dB at ω = 2.44; for Case 2, it is −30 dB at ω = 0.585. For the gains given inExample 10.7.7, it is m = −34.1 dB at ω = 0.7. The separation factors are 10, 2, and 10,respectively. On the basis of this example we would conclude that a large separation factorgives more attenuation, provided that the dominant root is real.
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10.42 Combining equations (1) and (2) in Example 10.7.5, with Td(s) = 0, we obtain
T (s) =KIs(10s + 2)
10s3 + (2 + K2)s2 + K1s + KIΘr(s)
Using the values KI = 25, K1 = 55, and K2 = 58 from the example, we can use theMATLAB code shown below to obtain the plot of T (t).
KI=25;K1=55;K2=58;sys=tf([10*KI,2*KI,0],[10,2+K2,K1,KI])
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10.43 With I action only, the transfer function is
C(s)R(s)
=KIK
τs2 + s + KIK
andζ =
12√
τKIK
With DI action,C(s)R(s)
=K(KDs2 + KI)
(τ + KKD)s2 + s + KIK
andζ =
12√
(τ + KKD)KIK
If KD > 0 the derivative action makes ζ smaller, and thus the response is more oscillatory.So D action does not improve the response.
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10.44 The compensated system diagram is shown in the following figure.
Figure : for Problem 10.44
Command compensation cannot affect the characteristic roots or the disturbance re-sponse, so we set D(s) = 0 here. The error equation is
E(s) =Is2 + cs − Kf
Is2 + cs + KPR(s)
For a unit-step input, the final value theorem gives ess = −Kf/KP as long as the system isstable. Thus ess = 0 only if there is no command compensation (Kf = 0)! For a unit-rampinput, the final value theorem gives ess = ∞. Thus the command compensation does notimprove the ramp response. Therefore, command compensation for this system degradesthe step response, and does not improve the ramp response.
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10.45 a) The closed-loop transfer function is
C(s)R(s)
=3KP + 3KDs
s2 + 3KDs + 3KP − 4
The damping ratio is
ζ =3KD
2√
3KP − 4= 0.707
The time constant isτ =
23KD
= 0.1
These two conditions give KP = 68 and KD = 20/3.b) The closed-loop transfer function with the negative rate feedback gain K1 is
C(s)R(s)
=3KP
s2 + 3K1s + 3KP − 4
This denominator has the same form as the transfer function in part (a) with KD replacedby K1. Thus KP = 68 and K1 = 20/3.
c) For part (a),C(s)R(s)
=20s + 204
s2 + 20s + 200
This has numerator dynamics and will overshoot more than the design in part (b), for which
C(s)R(s)
=204
s2 + 20s + 200
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10.46 Since the desired value of θ is 0, the error signal is e = θr − θ = 0 − θ = −θ. TryingPD control we use f = KP e + KDe = −KP θ −KDθ. Substituting this into the equation ofmotion we obtain
MLθ − (M + m)gθ = f = −KP θ − KDθ
The characteristic equation is
MLs2 + KDs + KP − (M + m)g = 0
Using M = 40, m = 8, L = 20, and g = 32.2 this becomes
800s2 + KDs + KP − 1545.6 = 0
A 2% settling time of 10 seconds corresponds to 4τ = 10 or τ = 2.5 seconds. Because ζ < 1,we can use the following formula for the time constant.
τ =2(800)KD
= 2.5
This gives KD = 640 lb/rad/sec.The formula for the damping ratio is
ζ =KD
2√
800(KP − 1545.6)=
6402√
800(KP − 1545.6)= 0.707
Solve this for KP to obtain KP = 1801.6 lb/rad.
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10.47 The closed-loop transfer function is
Θ(s)Θr(s)
=2412 + 1200s
1500τs3 + 1500s2 + (1200− 1932τ)s + 480
The characteristic equation is
1500τs3 + 1500s2 + (1200− 1932τ)s + 480 = 0
From the Routh-Hurwitz criterion, we can tell that the system is stable if 0 ≤ τ ≤ 0.498,so the case where τ = 1 is unstable.
If τ = 0.1, the roots ares = −0.3418± 0.4761j
The time constant is 1/0.3418 = 2.926, and the damping ratio is 0.583. The specificationscall for a dominant time constant of 2.5 and a damping ratio of 0.707.
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10.48 The closed-loop transfer function is
Θ(s)Θr(s)
=(2412 + 1200s)(τs + 1)
1500τs3 + 1500s2 + (1200− 1932τ)s + 480
The characteristic equation is
1500τs3 + 1500s2 + (1200− 1932τ)s + 480 = 0
From the Routh-Hurwitz criterion, we can tell that the system is stable if 0 ≤ τ ≤ 0.498,so the case where τ = 1 is unstable.
If τ = 0.1, the roots ares = −0.3418± 0.4761j
The time constant is 1/0.3418 = 2.926, and the damping ratio is 0.583. The specificationscall for a dominant time constant of 2.5 and a damping ratio of 0.707.
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10.49 a) From Figure P10.49,
(ms2+cs+k)X(s) = Fd(s)+(ces+ke)X(s)+me
s2Xr(s) +
KDs2 + KP s + KI
s[Xr(s) − X(s)]
If me = m, ce = c, and ke = k, this equation becomes
(ms3 + mKDs2 + mKP s + mKI)X(s) = sFd(s) + (ms3 + mKDs2 + mKP s + mKI)Xr(s)
ThusX(s)Xr(s)
= 1
which is perfect, and
X(s)Fd(s)
=s
ms3 + mKDs2 + mKPs + mKI
For a step disturbance, Fd(s) = 1/s, Xss = 0. For a unit-ramp disturbance, Fd(s) = 1/s2,Xss = 1/mKI .
Since the characteristic equation is third order, there is no expression for the dampingratio or time constant. We can, however, obtain a dominant root pair that meets therequirements that ζ = 1 and τ = τd. This means that the dominant root pair must be twoidentical roots at s = −1/τd. Thus the third root will be real. Denote it by s = −b. Thenthe characteristic equation can be factored as
ms3+mKDs2+mKP s+mKI = m
(2 +
1τd
)2
(s+b) = m
[s3 +
(b +
2τd
)s2 +
(1τ2d
+2b
τd
)s +
b
τ2d
]
Comparing coefficients, we see that the gains must be chosen so that
KP = b +2τd
KP =1τ2d
+2b
τd
KI =b
τ2d
The value of b must be selected such that the root at s = −b is not the dominant root. Thisrequires that b > 1/τd.
(continued on the next page)
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Problem 10.49 continued:
b) In addition to requiring accurate estimates of m, c, and k, this scheme requires thatthe second derivative, xr, of the command input must be computed in real time. This canbe difficult to do for noisy inputs or inputs having discontinuities in xr or xr, such as stepsand ramps.
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10.50 a) From the diagram,
ΩL(s) =1
N(Ies + ce)KT
Las + Ra[Vm(s) − KbΩm(s)]
and Ωm(s) = NΩL(s). Thus
ΩL(s)Vm(s)
=KT
NIeLas2 + N(ceLa + RaIe)s + ceRaN + NKTKb
With the given values,
ΩL(s)Vm(s)
=1.4 × 105
4.368s2 + 894.4s + 2.8015× 104=
4.9971.5592× 10−4s2 + 0.0319s + 1
Since the two roots are real, this can be expressed as
ΩL(s)Vm(s)
=K
(τ1s + 1)(τ2s + 1)
where K = 4.997, τ1 = 2.591× 10−2 and τ2 = 6.02× 10−3.
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Problem 10.50 continued:
b) Using PI control with the above transfer function results in a third-order system,for which there are no convenient design formulas. Since τ2 τ1, we will neglect τ2 andexpress the transfer function as
ΩL(s)Vm(s)
=K
τ1s + 1
With PI control, the closed-loop transfer function is
ΩL(s)Ωr(s)
=K(KPs + KI)
τ1s2 + (1 + KKP )s + KKI
Since we will set ζ < 1, we can express the closed-loop time constant as
τ =2τ1
1 + KKP= 0.05
ThusKP =
2τ1 − 0.050.05K
= 7.284× 10−3
The damping ratio is
ζ =1 + KKP
2√
τ1KKI=
20τ1√τ1KKI
Thus
KI =400τ2
1
ζ2τ1K
Choosing ζ = 0.707, we obtainKI = 4.1481
The resulting third-order transfer function is
ΩL(s)Ωr(s)
=0.0364s + 20.73
0.0001559s3 + 0.03193s2 + 1.036s + 20.73
The roots of the third-order system are s = −170.39 and s = −17.21 ± 22j. Thus thedominant time constant is 1/17.21 = 0.058, which is close to the desired value. The dampingratio of the dominant root pair is ζ = cos[tan−1(22/17.21)] = 0.616, which is within therequired range.
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10.51 The value from Problem 10.19 is KP = 3.6772. From Example 10.8.6,
Tm(s)Ωr(s)
=NKP KT (Ies + ce)
D(s)
and thus, since ia = TM/KT ,
Ia(s)Ωr(s)
=NKP (Ies + ce)
D(s)
whereD(s) = N
[LaIes
2 + (RaIe + ceLa)s + Race + KT Kb + KP KT
]
Note that the term N in D(s) cancels N in the numerator of Ia(s)/Ωr(s).The MATLAB program is the following.
KT = 0.2; Kb = 0.2;Ie = 1.775e-3;ce = 1e-3;Ra = 0.8;La = 4e-3;KP = 3.672;num = KP*[Ie, ce];den = [La*Ie, Ra*Ie+ce*La, Ra*ce+KT*Kb+KP*KT];sys = tf(num, den);step(209.4*sys)
Note that we can use the step function by multiplying the transfer function by magnitudeof the step input. The plot is shown on the next page. The maximum current is 389 A!This unattainable value can be reduced by using a modified step input that increases slowly(see Problem 10.53 for a discussion of this type of input).
(continued on the next page)
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Problem 10.51 continued:
0 0.01 0.02 0.03 0.04 0.05 0.06−200
−100
0
100
200
300
400
Step Response
Time (sec)
Cur
rent
i a (A
)
Figure : for Problem 10.51.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
10.52 The new transfer functions are
Ω(s)Ωr(s)
=KP s + KI
0.5s3 + 5.2s2 + (2 + KP )s + KI
T (s)Ωr(s)
=KP s + KI
0.1s + 10.5s2 + 5.2s + 2
0.5s3 + 5.2s2 + (2 + KP )s + KI
Ω(s)Td(s)
=−s(0.1s + 1)
0.5s3 + 5.2s2 + (2 + KP )s + KI
The three cases are:
1. KP = 18, KI = 40;
2. KP = 18, KI = 20;
3. KP = 108, KI = 200;
a) The MATLAB program is
KP = [18, 18, 108];KI = [40, 20, 200];% Command transfer functionk = 1; sysa1 = tf([KP(k), KI(k)],[0.5, 5.2, 2 + KP(k), KI(k)]);k = 2; sysa2 = tf([KP(k), KI(k)],[0.5, 5.2, 2 + KP(k), KI(k)]);k = 3; sysa3 = tf([KP(k), KI(k)],[0.5, 5.2, 2 + KP(k), KI(k)]);% Actuator transfer functionk = 1; numb1 = conv([KP(k), KI(k)],[0.5, 5.2, 2]);denb1 = conv([0.1, 1],[0.5, 5.2, 2 + KP(k), KI(k)]);sysb1 = tf(numb1, denb1)k = 2; numb2 = conv([KP(k), KI(k)],[0.5, 5.2, 2]);denb2 = conv([0.1, 1],[0.5, 5.2, 2 + KP(k), KI(k)]);sysb2 = tf(numb2, denb2)k = 3; numb3 = conv([KP(k), KI(k)],[0.5, 5.2, 2]);denb3 = conv([0.1, 1],[0.5, 5.2, 2 + KP(k), KI(k)]);sysb3 = tf(numb3, denb3)
(continued on the next page)
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Problem 10.52 continued:
subplot(2,1,1), step(sysa1, sysa2, sysa3)subplot(2,1,2), step(sysb1, sysb2, sysb3)
The characteristic roots are as follows:KP KI Intended ζ Intended Roots Actual Roots18 40 0.707 −2 ± 2j (dominant) −2.2353± 2.9147j (ζ = 0.6086), −5.929418 20 1 −2, −2 (dominant) −1.5016, −4.4492± 2.6159j108 200 1.74 −2, −20 (dominant) −1.9664, −4.2168± 13.6248j
The plots are shown in the following figure. The peak actuator values are 16.2, 14.2,and 52.
0 0.5 1 1.50
0.5
1
1.5
0 0.5 1 1.5−20
0
20
40
60
Step Response
Time (sec)
Spe
ed (r
ad/s
ec)
Step Response
Time (sec)
Act
uato
t Res
pons
e
1 2
3
1 2
3
Figure : for Problem 10.52a.
(continued on the next page)
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
Problem 10.52 continued:
b) The MATLAB program is
KP = [18, 18, 108];KI = [40, 20, 200];% Disturbance transfer functionk = 1; sysc1 = tf([-0.1, -1, 0],[0.5, 5.2, 2 + KP(k), KI(k)]);k = 2; sysc2 = tf([-0.1, -1, 0],[0.5, 5.2, 2 + KP(k), KI(k)]);k = 3; sysc3 = tf([-0.1, -1, 0],[0.5, 5.2, 2 + KP(k), KI(k)]);bodemag(sysc1, sysc2, sysc3)
The plots are shown in the figure on the following page. The peak values are −23.3,−24.7, and −30.8 dB.
c) The peak actuator values are less than in Example 10.6.4 because the actuator re-sponds more slowly in this problem (it has a time constant of 0.1, whereas in the examplethe actuator responded instantaneously). The peak disturbance response values are slightlyhigher than in the example.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
10−1
100
101
102
−70
−65
−60
−55
−50
−45
−40
−35
−30
−25
−20
Mag
nitu
de (d
B)
Bode Diagram
Frequency (rad/sec)
1
2
3
Figure : for Problem 10.52b.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
10.53 a) The transfer functions are
Ω(s)Ωr(s)
=KP s + KI
5s2 + (2 + KP )s + KI
T (s)Ωr(s)
=(KP s + KI)(5s + 2)
5s3 + (2 + KP )s + KI
The three cases are:
1. KP = 18, KI = 40;
2. KP = 18, KI = 20;
3. KP = 108, KI = 200;
a) The MATLAB program is
KP = [18, 18, 108];KI = [40, 20, 200];% Command transfer functionk = 1; sysa1 = tf([KP(k), KI(k)],[5, 2 + KP(k), KI(k)]);k = 2; sysa2 = tf([KP(k), KI(k)],[5, 2 + KP(k), KI(k)]);k = 3; sysa3 = tf([KP(k), KI(k)],[5, 2 + KP(k), KI(k)]);% Actuator transfer functionk = 1; sysb1 = tf(conv([KP(k), KI(k)],[5, 2]),[5, 2 + KP(k), KI(k)])k = 2; sysb2 = tf(conv([KP(k), KI(k)],[5, 2]),[5, 2 + KP(k), KI(k)])k = 3; sysb3 = tf(conv([KP(k), KI(k)],[5, 2]),[5, 2 + KP(k), KI(k)])tc = [0:.001:2];rc = 1 - exp(-20*tc);subplot(2,1,1),lsim(sysa1,sysa2,sysa3,rc,tc)ta = [0:.001:0.8];ra = 1 - exp(-20*ta);subplot(2,1,2),lsim(sysb1,sysb2,sysb3,ra,ta)
(continued on the next page)
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
Problem 10.53 continued:
The plots are shown in the following figure. The peak actuator values are 14.5, 13.4,and 40.1. The peak actuator values are less than in Example 10.6.4 because the commandinput does not act as fast as the step input used in the example. This gives the systemmore time to respond, and thus makes less demand on the actuator.
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20
0.2
0.4
0.6
0.8
1
1.2
1.4
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.80
10
20
30
40
Linear Simulation Results
Time (sec)
Am
plitu
de
Linear Simulation Results
Time (sec)
Am
plitu
de
Command
1
2
3
1
2
3
Figure : for Problem 10.53.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
10.54 The new transfer functions are
Ω(s)Ωr(s)
=KI
0.5s3 + 5.2s2 + (2 + K2)s + KI
T (s)Ωr(s)
=KI(5s + 2)
0.5s3 + 5.2s2 + (2 + K2)s + KI
Ω(s)Td(s)
=−s(0.1s + 1)
0.5s3 + 5.2s2 + (2 + K2)s + KI
The three cases are:
1. K2 = 18, KI = 40, ζ = 0.707;
2. K2 = 18, KI = 20, ζ = 1;
3. K2 = 108, KI = 200, ζ = 1.74;
a) The MATLAB program is
K2 = [18, 18, 108];KI = [40, 20, 200];% Command transfer functionk = 1; sysa1 = tf(KI(k),[0.5, 5.2, 2 + K2(k), KI(k)]);k = 2; sysa2 = tf(KI(k),[0.5, 5.2, 2 + K2(k), KI(k)]);k = 3; sysa3 = tf(KI(k),[0.5, 5.2, 2 + K2(k), KI(k)]);% Actuator transfer functionk = 1; sysb1 = tf(KI(k)*[5, 2],[0.5, 5.2, 2 + K2(k), KI(k)]);k = 2; sysb2 = tf(KI(k)*[5, 2],[0.5, 5.2, 2 + K2(k), KI(k)]);k = 2; sysb3 = tf(KI(k)*[5, 2],[0.5, 5.2, 2 + K2(k), KI(k)]);subplot(2,1,1), step(sysa1, sysa2, sysa3)subplot(2,1,2), step(sysb1, sysb2, sysb3)
(continued on the next page)
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Problem 10.54 continued:
The characteristic roots are as follows:K2 KI Intended ζ Intended Roots Actual Roots18 40 0.707 −2 ± 2j (dominant) −2.2353± 2.9147j (ζ = 0.6086), −5.929418 20 1 −2, −2 (dominant) −1.5016, −4.4492± 2.6159j108 200 1.74 −2, −20 (dominant) −1.9664, −4.2168± 13.6248j
The plots are shown in the following figure. The peak actuator values are 8.64, 4.87,and 11.3.
0 0.5 1 1.5 2 2.5 3 3.50
0.2
0.4
0.6
0.8
1
1.2
1.4
0 0.5 1 1.5 2 2.5 3 3.50
2
4
6
8
10
12
Step Response
Time (sec)
ω(t)
Step Response
Time (sec)
T(t)
ζ = 0.707
ζ = 1.74
ζ = 1
ζ = 0.707
ζ = 1.74
ζ = 1
Figure : for Problem 10.54a.
(continued on the next page)
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
Problem 10.54 continued:
b) The MATLAB program is
K2 = [18, 18, 108];KI = [40, 20, 200];% Disturbance transfer functionk = 1; sysc1 = tf([-0.1, -1, 0],[0.5, 5.2, 2 + K2(k), KI(k)]);k = 2; sysc2 = tf([-0.1, -1, 0],[0.5, 5.2, 2 + K2(k), KI(k)]);k = 3; sysc3 = tf([-0.1, -1, 0],[0.5, 5.2, 2 + K2(k), KI(k)]);bodemag(sysc1, sysc2, sysc3)
The plots are shown in the following figure. The peak values are −23.3, −24.7, and−30.8 dB.
c) The peak actuator values are greater than in Example 10.6.5. The peak disturbanceresponse values are slightly higher than in the example.
10−2
10−1
100
101
102
−110
−100
−90
−80
−70
−60
−50
−40
−30
−20
Mag
nitu
de (d
B)
Bode Diagram
Frequency (rad/sec)
ζ = 1.74
ζ = 0.707
ζ = 1
Figure : for Problem 10.54b.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
10.55 LetK =
KpotKaKT
6Ie
whereIe = I1 +
14
[I2 +
19I3
]= 0.0157
and K = 12.7547.The transfer function is
Θ(s)Θr(s)
=K(KDs + KP )
Ls3 + Rs2 + KKDs + KKP
a) For L = 0,Θ(s)Θr(s)
=K(KDs + KP )
Rs2 + KKDs + KKP
To obtain τ = 0.5 and ζ = 1,
τ = 0.5 =2R
KKD
andζ = 1 =
KKD
2√
RKKP
These give
KD = KP =4R
K= 0.0941
(continued on the next page)
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Problem 10.55 continued:
Parts (b), (c), and (d) are done with the following MATLAB program.
I1 = 0.01; I2 = 5e-4; I3 = 0.2;Ka = 1; KT = 0.6; Kpot = 2;R = 0.3;Ie = I1+(1/4)*(I2+(1/9)*I3)K = Kpot*Ka*KT/(6*Ie)KD = 4*R/KKP = 4*R/Ksys1 = tf([K*KD,K*KP],[R,K*KD,K*KP])L = 0.015;sys2 = tf([K*KD,K*KP],[L,R,K*KD,K*KP])step(sys1,sys2)
The resulting plot is shown below. With L = 0, the overshoot is 13.5% and the settlingtime is 2.7 s. With L = 0.015, the overshoot is 16.2% and the settling time is 2.59 s. Sothe response predicted by the second order model, which makes the calculation of the gainsKP and KD much easier, is close to that of the third order system. This will not be true ifL is much larger than 0.015.
0 0.5 1 1.5 2 2.5 3 3.50
0.2
0.4
0.6
0.8
1
1.2
1.4
Step Response
Time (sec)
Ampl
itude
L = 0.015
L = 0
Figure : for Problem 10.55.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
10.56 Since the acceleration time is 4 s, we set τ = 1 so that the controllers can follow theramp input. The time constant expressions for the three controllers are:
1. τ = I/(c + KP ) = 10/(3 + KP ) for P control.
2. τ = I/(c + KP ) = 10/(3 + KP ) for PI control, if ζ ≤ 1.
3. τ = I/(c + K2) = 10/(3 + K2) for Modified I control, if ζ ≤ 1.
Requiring, for example, a maximum steady state error of 10% of the slew speed meansthat ess ≤ 0.1. The slope of the input is 1/4, and the steady state errors for a rampcommand of slope 1/4 is
1. ess = 0.25/(c + KP ) = 0.25/(3 + KP ) for P control.
2. ess = 0.25c/KI = (0.25)3/KI for PI control.
3. ess = 0.25(c + K2)/KI = 0.25(3 + K2)/KI for Modified I control.
Combining these requirements and satisfying the requirement on τ exactly, we obtainthe following gains:
1. KP = 7 for P control.
2. KP = 7 , KI = 7.5 for PI control.
3. K2 = 8 , KI = 110 for Modified I control.
(continued on the next page)
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Problem 10.56 continued:
The following program is a modification of the programs command_tf.m, actuator_tf.m,and trap1.m given in Section 10.9. The program calls on the files named plot_commandand plot_actuator listed in Table 10.9.3.
KPa = 120;KPb = 7; KIb = 7.5;K2 = 8;KIc = 110;I = 10;c = 3;% Create the command transfer functions.sysa = tf(KPa,[I,c+KPa]);sysb = tf([KPb,KIb],[I,c+KPb,KIb]);sysc = tf(KIc,[I,c+K2,KIc]);% Create the actuator transfer functionssysaACT = tf(KPa*[I,c],[I,c+KPa]);sysbACT = tf(conv([KPb,KIb],[I,c]),[I,c+KPb,KIb]);syscACT = tf(KIc*[I,c],[I,c+K2,KIc]);% A specific trapezoidal profilet = [0:0.01:19];for k = 1:length(t)if t(k) <= 4r(k) = (1/4)*t(k);elseif t(k) <= 10r(k) = 1;elseif t(k) <= 14r(k) = 14/4-(1/4)*t(k);elser(k) = 0;endendsubplot(2,1,1),plot_commandsubplot(2,1,2),plot_actuator
The value KP = 7 for P control was found to be too small, so KP was increased to 120for P control only. This improved response is shown in the figure on the following page.
(continued on the next page)
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Problem 10.56 continued:
0 2 4 6 8 10 12 14 16 18 20−0.2
0
0.2
0.4
0.6
0.8
1
1.2
Time
Com
man
d R
espo
nse
0 2 4 6 8 10 12 14 16 18 20−4
−2
0
2
4
6
Time
Act
uato
r O
utpu
t
P
PI I Modified
I Modified
P
PI
Figure : for Problem 10.56.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
10.57 a) First we reflect the load and tachometer inertias back to the motor shaft to obtainthe equivalent inertia Ie:
Ie = Im +1
N2(IL + It)
= 2 × 10−5 +1
1.52(4× 10−3 + 10−5) = 1.802× 10−3 kg · m2
Next do the same for the load damping coefficient:
ce = cm +cL
N2= 0 +
10−3
1.52= 4.444× 10−4 N · m · s/rad
The transfer functions derived in Example 10.3-5 and are repeated here.
ΩL(s)Ωr(s)
=KPKT
D(s)(1)
ΩL(s)TL(s)
= −(Las + Ra)/ND(s)
(2)
where the denominator is
D(s) = NLaIes2 + N(RaIe + ceLa)s + NRace + NKTKb + KP KT (3)
Thus the characteristic equation is
D(s) = 5.407× 10−6s2 + 1.623× 10−3s + 2.8× 10−3 + 0.04KP = 0
The Routh-Hurwitz criterion shows that the system is stable if and only if KP > 0.The desired speed ωr is a constant and so can be modeled as a step function of magnitude
1000(2π)/60 = 104.7 rad/s. Applying the Final Value Theorem to equation (1) we have
ωLss =KP KT
NRace + NKTKb + KP KTωr =
0.04KP
2.8× 10−3 + 0.04KPωr (4)
Solve for KP as follows.
KP =0.07ωLss/ωr
1 − ωLss/ωr(5)
For the steady-state speed ωLss to be within 10% of ωr, ωLss/ωr = 0.9 or 1.1. These valuesgive KP = 0.63 and KP = −0.77 respectively.
(continued on the next page)
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Problem 10.57 continued:
Choosing the positive solution for stability, we obtain the roots s = −282 and −18.4.The system is stable and the result obtained from the Final Value Theorem is thus valid.Thus the solution is KP = 0.63 for which ωLss = 900 rpm. The largest time constant of theclosed-loop system is 1/18.4 = 0.054 s. The speed will reach its steady-state value of 900rpm after approximately 4(0.054) = 0.217 s. It will not oscillate because the roots are real.With KP = 0.63, we can then determine the value of the gain K1, which is given by
K1 =KP
KtachKa=
0.6310(5)
= 0.0126
as discussed in Example 10.3-5.b) Applying the Final Value Theorem to equation (2), with KP = 0.63 and TL = 1 N-m,
we obtain the steady-state deviation caused by the load torque.
ωLss = − Ra/N
NRace + NKTKb + KP KTTL = −14.3 rad/s (6)
or −136 rpm. This is not the actual load speed, but only the deviation in the load speedcaused by the load torque. Thus the load torque causes the steady-state speed to decreaseby 136 rpm.
The actual load speed is the sum of the speeds obtained from equations (1) and (2), forthe given inputs. Here, at steady-state, the actual load speed is ωLss = 900 − 136 = 764rpm.
You should note that because the model is linear, we can deduce that for KP = 0.63,the steady-state speed will be 90% of the desired speed, regardless of the magnitude of ωr,as long as it is a step function and if the load torque is zero. Similarly, linearity enables usto see from equation (6) that the steady-state deviation caused by a constant load torqueis proportional to TL; that is, ωLss = −14.3TL rad/s.
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10.58 The maximum speed error occurs at t = 0 and is 1000 rpm, or 104.7 rad/s.From (6.5.7), the energy consumption from t = 0 to t = T is
E =∫ T
0Rai
2a(t) dt +
∫ T
0ceω
2L(t) dt (1)
The rms current is
irms =
√1T
∫ T
0i2a(t) dt (2)
The rms speed error is
ωrms =
√1T
∫ T
0[104.7− ωL(t)]2 (t) dt (3)
We can perform these computations either 1) by obtaining the step response numericallyand then performing the integrations numerically, or 2) by first obtaining the closed-formexpressions for the current and speed, and then evaluating the integrals. These integralscan be evaluated in closed form.
Both methods require the transfer functions. From the example, we have the expressionfor the speed:
ΩL(s)Ωr(s)
=KP KT
D(s)
whereD(s) = NLaIes
2 + N(RaIe + ceLa)s + NRace + NKTKb + KP KT
The expression for the current is
Ia(s)Ωr(s)
=KP (Ies + ce)
D(s)
(continued on the next page)
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Problem 10.58 continued:
The following MATLAB program does the computations.
N = 1.5;La = 2e-3;Ra = 0.6;Ie = 1.802e-3;ce = 4.444e-4;KP = 0.63;KT = 0.04;Kb = 0.04;D = [N*La*Ie,N*(Ra*Ie+ce*La),N*Ra*ce+N*KT*Kb+KP*KT];sysia = tf(KP*[Ie,ce],D);syswL = tf(KP*KT,D);[ia,t1] = step(104.7*sysia);[wL,t2] = step(104.7*syswL);integrand1 = Ra*ia.^2;integrand2 = ce*wL.^2;E = trapz(t1,integrand1) + trapz(t2,integrand2)max_current = max(ia)rms_current = sqrt((1/max(t1))*trapz(t1,ia.^2))rms_speed_error = sqrt((1/max(t2))*trapz(t2,(104.7-wL).^2))
The results are:Energy consumed = E = 97.0283 J
The maximum current is 64.2584 A, and the rms current is 23.1163 A. The rms speed erroris 34.9 rad/s, or 333.271 rpm.
For those not wishing to use MATLAB, we give the following expressions for the speedand current.
ωL(t) = 94.041 + 6.5643e−282t − 100.6053e−18.4t
ia(t) = 1.0448− 83.3224e−282t + 82.2776e−18.4t
The integrals in equations (1), (2), and (3) can be evaluated in closed form or numeri-cally.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
10.59 From Figure 10.6.4, we see that the only case where the saturation block with anupper limit of 20 will have any effect is the case for which ζ = 1.74. This corresponds toKP = 108 and KI = 200.
The Simulink model is shown in the following figure. In the To Workspace block,specify the Save format as Array. Type KP = 108; KI = 200; in MATLAB and then runthe model. Then in MATLAB type
subplot(2,1,1),plot(tout,simout(:,2)),subplot(2,1,2),plot(tout,simout(:,1))
The resulting plot, which has been edited, is shown on the next page. As compared to thecase with no limit on the actuator, the overshoot in ω(t) is larger (18% vs approximately5%).
Figure : for Problem 10.59.
(continued on the next page)
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
Problem 10.59 continued:
0 0.5 1 1.50
0.2
0.4
0.6
0.8
1
1.2
t
ω(t)
0 0.5 1 1.50
5
10
15
20
25
t
T(t)
Figure : for Problem 10.59.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
10.60 The Simulink model is shown in the following figure. We must use the PID Controllerblock instead of a Transfer Function block, because the latter does not allow the denominatorto have an order greater than the numerator. In the Block Parameters window of the PIDController block, enter 55 for Proportional, 25 for Integral, and 58 for Derivative.
We must use the Transfer Function (with initial outputs) block because the initial po-sition is not zero. In the To Workspace block, specify the Save format as Array.
Figure : for Problem 10.60.
Run the model and then in MATLAB type
plot(tout,simout)
The resulting plot, which has been edited, is shown on the following page.
(continued on the next page)
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
Problem 10.60 continued:
0 1 2 3 4 5 6 7 8 9 100.5
1
1.5
2
2.5
3
t
θ(t)
Figure : for Problem 10.60.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
10.61 a) The Simulink model and the responses are shown in the following figures.
Figure : for Problem 10.61.
(continued on the next page)
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
Problem 10.61 continued:
0 5 10 15 20 25 300
2
4
6
t
θ(t)
0 5 10 15 20 25 30
0
10
20
t
T(t
)
0 5 10 15 20 25 30−5000
0
5000
10000
t
T(t
)
Figure : for Problem 10.61.
(continued on the next page)
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
Problem 10.61 continued:
b) Comparison with Figure 10.7.4 shows that the response with a limited actuator ismuch slower. However, the response shown in Figure 10.7.7 is misleading because it iscaused by an actuator response that contains an impulse, something that is not physicallypossible. To see this, derive the expression for the actuator output, as follows. From thediagram in Figure 10.7.3, we can derive the following expression.
T (s) =IKDs4 + (IKP + cKD)s3 + (IKI + cKP )s2 + cKIs
Is3 + (c + KD)s2 + KP s + KIΘr(s)
If the command is a unit step, then Θr(s) = 1/s, and, after canceling the s terms in thenumerator and denominator, we obtain
T (s) =IKDs3 + (IKP + cKD)s2 + (IKI + cKP )s + cKI
Is3 + (c + KD)s2 + KP s + KI
Using synthetic division, this can be expressed as
T (s) = KD +N(s)
Is3 + (c + KD)s2 + KP s + KI
where N(s) is a second-order polynomial. Thus the constant term KD will produce animpulse of strength KD in T (t).
The top graph shows the command response, which is much slower than when theactuator output is unlimited. The second of the three graphs shows the limited actuatoroutput. The third of the three graphs shows the actuator output if unlimited. It has avery large positive value initially and then goes negative. So the limiter causes the veryslow command response. Note that Simulink cannot deal with an impulse in the actuatorresponse, but rather computes the response numerically. This effect causes the simulated,unlimited actuator response to have some approximation error.
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10.62 a) The Simulink model and the responses are shown in the following figures. Wemust use the PID Controller block instead of a Transfer Function block, because the latterdoes not allow the denominator to have an order greater than the numerator. In the BlockParameters window of the PID Controller block, enter 55 for Proportional, 25 for Integral,and 58 for Derivative.
Figure : for Problem 10.62.
(continued on the next page)
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
Problem 10.62 continued:
0 0.5 1 1.5 2 2.5 30
0.5
1
1.5
t
θ(t)
0 0.5 1 1.5 2 2.5 3−200
−100
0
100
200
300
t
T(t
)
Figure : for Problem 10.62.
b) The maximum overshoot in the command response is 42%, as compared to 10% inthe example. The 2% settling time is 2.5 as compared with 2.89 in the example.
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10.63 a) The Simulink model is shown in the following figure. Note that since N = 1, wedo not need a gain block for the gain 1/N .
Figure : for Problem 10.63a.
(continued on the next page)
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Problem 10.63 continued:
In MATLAB, type
Kb = 0.199; KT = 0.14;Ie = 2.08e-3; ce = 3.6e-4;Ra = 0.43; La=2.1e-3;KP = 7.284e-3; KI = 4.1481;
Then run the model and type the following in MATLAB.
subplot(2,1,1),plot(tout,simout(:,1)),subplot(2,1,2),plot(tout,simout(:,2))
This produces the following plots, which have been edited.
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5−10
−8
−6
−4
−2
0
2
t (sec)
ω L(t) (r
ad/s
ec)
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5−2
0
2
4
6
8
10
t (sec)
i a(t) (A
)
Figure : for Problem 10.63a.
(continued on the next page)
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
Problem 10.63 continued:
b) The Simulink model is shown in the following figure. Note that since N = 1, we donot need a gain block for the gain 1/N .
Figure : for Problem 10.63b.
In MATLAB, type
Kb = 0.199; KT = 0.14;Ie = 2.08e-3; ce = 3.6e-4;Ra = 0.43; La=2.1e-3;KP = 7.284e-3; KI = 4.1481;
Then run the model and type the following in MATLAB.
subplot(2,1,1),plot(tout,simout(:,1)),subplot(2,1,2),plot(tout,simout(:,2))
This produces the plots shown on the following page, which have been edited.
(continued on the next page)
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
Problem 10.63 continued:
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20
0.01
0.02
0.03
0.04
0.05
0.06
t (sec)
Spe
ed E
rror
, ωL(t
) −
ω(t
) (
rad/
sec)
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20
0.005
0.01
0.015
0.02
t (sec)
i a(t)
(A)
Figure : for Problem 10.63b.
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10.64 Modify the Simulink model from Problem 10.63a as shown below.
Figure : for Problem 10.64.
(continued on the next page)
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
Problem 10.64 continued:
In MATLAB, type
Kb = 0.199; KT = 0.14;Ie = 2.08e-3; ce = 3.6e-4;Ra = 0.43; La=2.1e-3;KP = 7.284e-3; KI = 4.1481;
Then run the model and type the following in MATLAB.
subplot(2,1,1),plot(tout,simout(:,1)),subplot(2,1,2),plot(tout,simout(:,2))
This produces the following plots, which have been edited.
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50
50
100
150
t (s)
ωL(t
) (r
ad/s
ec)
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
−10
−5
0
5
10
t (sec)
i a(t)
(A)
Figure : for Problem 10.64.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
Solutions Manual c©
to accompany
System Dynamics, Second Edition
by
William J. Palm III
University of Rhode Island
Solutions to Problems in Chapter Eleven
c©Solutions Manual Copyright 2010 The McGraw-Hill Companies. All rightsreserved. No part of this manual may be displayed, reproduced, or distributedin any form or by any means without the written permission of the publisheror used beyond the limited distribution to teachers or educators permitted byMcGraw-Hill for their individual course preparation. Any other reproductionor translation of this work is unlawful.
11.1 From the quadratic formula,
s =−12 ±
√144− 12k
6= −2 ±
√4− k/3
For k ≤ 12 the roots are real, the dominant root lies between 0 and −2, and the dominanttime constant lies between 1/2 and ∞. For k > 12 the roots are complex, and the real partis −2. Thus the dominant time constant is 1/2. The root locus plot is identical to Figure11.1.2. The smallest possible dominant time constant is 1/2 and is obtained for any k ≥ 12.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
11.2 From the quadratic formula,
s =−c ±
√c2 − 1446
For c ≥ 12 the roots are real, the dominant root lies between 0 and −2, and the dominanttime constant lies between 1/2 and ∞. For c < 12 the roots are complex, and the undampednatural frequency is ωn =
√12/3 = 2. This means that the complex roots lie on a semicircle
of radius 2 centered at the origin. The root locus plot is identical to Figure 11.1.3. Thesmallest possible dominant time constant is 1/2 and is obtained for c = 12.
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11.3 Dividing by the highest coefficient and factoring out c gives the standard form (11.1.4):
s2 + 666.7s + 2.78× 104 + 8333c(s + 666.7) = 0
where α = β = 666.7, γ = 2.78 × 104, and µ = 8333c. From the results of (11.1.4) and(11.1.5) we see that the root locus is a circle centered at s = −666.7 with a radius of 166.7,and is similar to Figure 11.1.4(a) with the ×’s at s = −44.7 and s = −622, and the © ats = −667. The starting points with c = 0 are at the ×’s. As c → ∞ one root approachess = −667 and the other root approaches s = −∞.
The plot illustrates the sometimes counterintuitive behavior of dynamic systems. Onewould think that increasing the damping c would increase the time constant and thus slowthe response. However, as c is increased up to the value where the two roots meet ats = −833, the dominant time constant decreases, and we see that the smallest possibledominant time constant is τ = 1/833 s.
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11.4 Let µ = (m− 2)/2. Thus m = 2µ + 2 and the characteristic equation becomes
(2µ + 2)s2 + 12s + 10 = 0
Divide by 2:(µ + 1)s2 + 6s + 5 = 0
From the quadratic formula,
s =−3 ±
√4− 5µ
µ + 1
For 0 ≤ µ ≤ 4/5, the roots are real. For µ = 0, the roots are s = −1 and s = −5. Forµ = 4/5, the roots are equal at s = −1.67. For µ > 4/5 the roots are complex and approachs = 0 as µ → ∞. The plot is similar to that in Figure 11.1.5 with the ×’s located at s = −1and s = −5, and with the circle intersecting the real axis at s = −1.67.
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11.5 a) The equation can be expressed as
1 +3p
61
s(s + 4/3)= 0
so K = 3p/6 = p/2.b) The equation can be expressed as
1 +p
3s + 2
s2 + 2s + 5/3= 0
so K = p/3.c) The equation can be expressed as
1 +4p
4s2 + 1/4
s(s2 + 0.5)= 0
so K = 4p/4 = p.
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11.6 a) Write Gc(s)Gp(s) as
Gc(s)Gp(s) = KP TD
(s2 + 1
TDs + 1
TDTI
)(s + 10)
s(s + 1)(s + 2)= 0.5KP
(s2 + 2s + 10)(s + 10)s(s + 1)(s + 2)
The poles are s = 0, −1, and −2. The zeros are s = −10 and −1 ± 3j.b) K = TDKP = 0.5KP
c) The closed-loop transfer function is
C(s)R(s)
=Gc(s)Gp(s)
1 + Gc(s)Gp(s)
When KP = 10, this becomes
C(s)R(s)
=5s3 + 60s2 + 150s + 5006s3 + 63s2 + 152s + 500
The closed-loop poles are the roots of the denominator, and are s = −8.688 and −0.9059±2.962j. The closed-loop zeros are the roots of the numerator, and are s = −10 and −1±3j.
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11.7Part (a)
• Poles: s = 0 and s = −5
• Zeros: None
• Two paths must leave the plot, so there will be two asymptotes.
• The locus exists on the real axis in the following interval: −5 < Re(s) < 0.
• To find the breakaway and breakin points:
K = −(s2 + 5s)
dK
ds= −(2s + 5) = 0
The solution is s = −2.5, which must be a breakaway point. The value of K at thebreakaway point is
K = − (s2 + 5s)∣∣∣s=−2.5
= 6.25
• Angles of the asymptotes:
θ =(2n + 1)180
2 − 0= 90, 270
• Intersection point of the asymptotes:
σ =∑
sp −∑
sz
2 − 0=
0 − 5 − 02
= −2.5
• Crossover point: None for K ≥ 0
• The root locus plot is shown in the figure on the following page.
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−5 −4.5 −4 −3.5 −3 −2.5 −2 −1.5 −1 −0.5 0−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
Root Locus
Real Axis
Imag
inar
y A
xis
Figure : for Problem 11.7a
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
11.7 Part (b)
• Poles: s = 0, s = −7, and s = −9. Zeros: None
• Three paths must leave the plot, so there will be three asymptotes.
• The locus exists on the real axis in the following intervals:
Re(s) < −9 − 7 < Re(s) < 0
• To find the breakaway and breakin points:K = −(s3 + 16s2 + 63s) and
dK
ds= −(3s2 + 32s + 63) = 0
The solutions are s = −8.0618 and s = −2.6049. The point s = −8.0618 correspondsto a breakaway or breakin point for K < 0. The value of K at the breakaway points = −2.6049 is
K = − (s3 + 16s2 + 63s)∣∣∣s=−2.6049
= 73.22
• Angles of the asymptotes:
θ =(2n + 1)180
3 − 0= 60, 180, 300
• Intersection point of the asymptotes:
σ =∑
sp −∑
sz
3 − 0=
0 − 7− 9 − 03
= −163
= −5.333
• The crossover point is found as follows. Substitute s = jω into the characteristicequation s3 + 16s2 + 63s + K = 0.
−jω3 − 16ω2 + 63ωj + K = 0
which gives(63− ω2)ω = 0 and K − 16ω2 = 0
The first equation gives ω = 0, which is not of interest, and ω = ±√
63. From thesecond equation, K = 16(63) = 1008 at the crossover points. The root locus plot isshown in the figure on the following page.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
−35 −30 −25 −20 −15 −10 −5 0 5 10 15−20
−15
−10
−5
0
5
10
15
20
Root Locus
Real Axis
Imag
inar
y A
xis
Figure : for Problem 11.7b
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
11.7 Part (c)
• Poles: s = −1.5± 1.6583j
• Zero: s = −3
• One path must leave the plot, so there will be one asymptote, which is 180.
• The locus exists on the real axis in the following interval: Re(s) < −3
• To find the breakaway and breakin points:
K = −s2 + 3s + 5s + 3
dK
ds= −(s + 3)(2s + 3) − (s2 + 3s + 5)
(s + 3)2=
s2 + 6s + 4(s + 3)2
= 0
The solutions are s = −5.2361 and s = −0.7639. The point s = −0.7639 correspondsto a breakaway or breakin point for K < 0. The value of K at the breakaway points = −5.2361 is
K = − s2 + 3s + 5s + 3
∣∣∣∣∣s=−5.2361
= 7.47
• There is no crossover point for K ≥ 0. This can be proved as follows. Substitutes = jω into the characteristic equation s2 + 3s + 5 + K(s + 3) = 0.
−ω2 + 3jω + 5 + K(jω + 3) = 0
which gives(K + 3)ω = 0 and 5 + 3K − ω2 = 0
The first equation gives two possibilities: 1) ω = 0, which is not of interest becauseit does not lie on the root locus for K > 0, and 2) K = −3. With K = −3, thesecond equation gives ω = ±
√5− 9 = ±4j, which means that s = −4 and is thus not
a solution of interest.
• The root locus plot is shown in the figure on the following page.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
−9 −8 −7 −6 −5 −4 −3 −2 −1 0
−3
−2
−1
0
1
2
3
Root Locus
Real Axis
Imag
inar
y A
xis
Figure : for Problem 11.7c
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
11.7 Part (d)
• Poles: s = 0 and s = −4. Zero: s = −5
• One path must leave the plot, so there will be one asymptote, which is 180.
• The locus exists on the real axis in the following intervals
Re(s) < −5 and − 4 < Re(s) < 0
• To find the breakaway and breakin points:
K = −s2 + 4s
s + 5
dK
ds= −(s + 5)(2s + 4)− (s2 + 4s)
(s + 5)2=
s2 + 10s + 20(s + 5)2
= 0
The solutions are s = −7.2361 and s = −2.7639. The point s = −2.7639 correspondsto the breakaway point, and s = −7.2361 corresponds to the breakin point. Becauseeach point is symmetrically placed a distance 2.2361 from the zero, this indicates thatthe locus is a circle of radius 2.2361 off the real axis.
• The value of K at the breakaway point is
K = − s2 + 4s
s + 5
∣∣∣∣∣s=−2.7639
= 1.5279
The value of K at the breakin point is
K = − s2 + 4s
s + 5
∣∣∣∣∣s=−7.2361
= 10.4721
(continued on the next page)
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
Problem 11.7d continued:
• There is no crossover point for K ≥ 0. This can be proved as follows. Substitutes = jω into the characteristic equation s2 + 4s + K(s + 5) = 0.
−ω2 + 4jω + K(jω + 5) = 0
which gives(K + 4)ω = 0 and 5K − ω2 = 0
The first equation gives two possibilities: 1) ω = 0, which is not of interest because itdoes not lie on the root locus for K > 0, and 2) K = −4. With K = −4, the secondequation gives ω = ±j
√20, which means that s = −
√20 and is thus not a solution of
interest. The root locus plot is shown in the following figure.
−10 −9 −8 −7 −6 −5 −4 −3 −2 −1 0
−3
−2
−1
0
1
2
3
Root Locus
Real Axis
Imag
inar
y A
xis
Figure : for Problem 11.7d
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
11.7 Part (e)
• Poles: s = 0 and s = −1.5± 1.6583j
• Zeros: None
• Three paths must leave the plot, so there will be three asymptotes.
• The locus exists on the real axis in the following interval: Re(s) < 0
• To find the breakaway and breakin points:
K = −(s3 + 3s2 + 5s)
dK
ds= −(3s2 + 6s + 5) = 0
The solutions s = −1 ± 0.8165j are complex, so there are no breakaway or breakinpoints.
• Angles of the asymptotes:
θ =(2n + 1)180
3 − 0= 60, 180, 300
• Intersection point of the asymptotes:
σ =∑
sp −∑
sz
3 − 0=
0− 1.5− 1.6583j − 1.5 + 1.6583j − 03
= −33
= −1
• The crossover point is found as follows. Substitute s = jω into the characteristicequation s3 + 3s2 + 5s + K = 0.
−jω3 − 3ω2 + 5ωj + K = 0
which gives(5 − ω2)ω = 0 and K − 3ω2 = 0
The first equation gives ω = 0, which is not of interest, and ω = ±√
5. From thesecond equation, K = 3(5) = 15 at the crossover points.
• The root locus plot is shown in the figure on the following page.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
−8 −6 −4 −2 0 2 4−5
−4
−3
−2
−1
0
1
2
3
4
5
Root Locus
Real Axis
Imag
inar
y A
xis
Figure : for Problem 11.7e
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
11.7 Part (f)
• Poles: s = 0, s = −3 and s = −7. Zero: s = −4
• Two paths must leave the plot, so there will be two asymptotes. Angles of the asymp-totes:
θ =(2n + 1)180
3 − 1= 90, 270
• Intersection point of the asymptotes:
σ =∑
sp −∑
sz
3 − 1=
0 − 3 − 7 + 42
= −3
• The locus exists on the real axis in the following intervals:
−7 < Re(s) < −4 and − 3 < Re(s) < 0
• To find the breakaway and breakin points:
K = −s3 + 10s2 + 21s
s + 4
dK
ds= −(s + 4)(3s2 + 20s + 21)− (s3 + 10s2 + 21s)
(s + 4)2= −2s3 + 22s2 + 80s + 84
(s + 4)2= 0
The only real solution is s = −1.7812. This point corresponds to a breakaway point.The value of K at the breakaway point is
K = − s3 + 10s2 + 21s
s + 4
∣∣∣∣∣s=−1.7812
= 5.1062
(continued on the next page)
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
Problem 11.7f continued:
• There is no crossover point for K ≥ 0. This can be proved as follows. Substitutes = jω into the characteristic equation s3 + 10s2 + 21s + K(s + 4) = 0.
−ω3 − 10ω2 + 21jω + K(jω + 4) = 0
which gives(K + 21− ω2)ω = 0 and 4K − 10ω2 = 0
The first equation gives two possibilities: 1) ω = 0, which is not of interest because itdoes not lie on the root locus for K > 0, and 2) K = ω2 − 21. Substituting this intothe second equation gives ω = ±j
√7/3, which means that s = −
√7/3 and is thus
not a solution of interest. The root locus plot is shown in the following figure.
−15 −10 −5 0 5−10
−8
−6
−4
−2
0
2
4
6
8
10
Root Locus
Real Axis
Imag
inar
y A
xis
Figure : for Problem 11.7f
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
11.8 The root locus equation is
1 + Gc(s)Gp(s) = 1 + KP
(1 +
1TIs
+ TDs
)s + 10
(s + 2)(s + 5)= 0
or
1 + KP TD
s2 + 1TD
s + 1TITD
s
s + 10(s + 2)(s + 5)
= 0
Substitute the given values to obtain
1 + 0.5KP(s2 + 2s + 10)(s + 10)
s(s + 2)(s + 5)= 0
The root locus parameter is K = 0.5KP .
• Poles: s = 0, s = −2 and s = −5
• Zeros: s = −10 and s = −1 ± 3j
• No paths must leave the plot, so there will be no asymptotes.
• The locus exists on the real axis in the following intervals:
−10 < Re(s) < −5 and − 2 < Re(s) < 0
• To find the breakaway and breakin points:
K = − s3 + 7s2 + 10s
s3 + 12s2 + 30s + 100
dK
ds= −5s4 + 60s3 + 390s2 + 1400s + 1000
(s3 + 12s2 + 30s + 100)2= 0
The only relevant solution is s = −0.9187, which is a breakin point for K > 0. Thevalue of K at this breakin point is
K = − s3 + 7s2 + 10s
s3 + 12s2 + 30s + 100
∣∣∣∣∣s=−0.9187
= 0.0496
(continued on the next page)
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
Problem 11.8 continued:
• There is no crossover point for K ≥ 0. This can be proved as follows. Substitutes = jω into the characteristic equation s3 + 10s2 + 21s + K(s + 4) = 0.
−ω3 − 10ω2 + 21jω + K(jω + 4) = 0
which gives(K + 21− ω2)ω = 0 and 4K − 10ω2 = 0
The first equation gives two possibilities: 1) ω = 0, which is not of interest because itdoes not lie on the root locus for K > 0, and 2) K = ω2 − 21. Substituting this intothe second equation gives ω = ±j
√7/3, which means that s = −sqrt7/3 and is thus
not a solution of interest. The root locus plot is shown in the following figure.
−15 −10 −5 0 5−10
−8
−6
−4
−2
0
2
4
6
8
10
Root Locus
Real Axis
Imag
inar
y A
xis
Figure : for Problem 11.8
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
11.9 a) The equation can be expressed as
1 +5p
4s2 + 6s + 8
s[s2 + (25/4)s + 4]= 0
so K = 5p/4.b) The poles are the roots of s[s2 + (25/4)s + 4] = 0, which are s = 0, −5.5262, and
−0.7238. The zeros are the roots of s2 + 6s + 8 = 0, which are s = −2 and −4.One path must leave the plot, so there will be one asymptote, which is 180.The locus exists on the real axis in the following intervals:
Re(s) < −5.5262 − 4 < Re(s) < −2 − 0.7238 < Re(s) < 0
To find the breakaway and breakin points:
K = −s[s2 + (25/4)s + 4]s2 + 6s + 8
= −s3 + (25/4)s2 + 4s
s2 + 6s + 8
dK
ds= −(s2 + 6s + 8)(3s2 + (25/2)s + 4)− (s3 + (25/4)s2 + 4s)(2s + 6)
(s2 + 6s + 8)2
=s4 + 12s3 + 57.5s2 + 100s + 32
(s2 + 6s + 8)2= 0
The solutions are s = −4.4115 ± 2.9797j, s = −2.7693, and s = −0.4077. The breakawaypoint is at s = −0.4077 and the breakin point is at s = −2.7693. The value of K at thebreakaway point is
K = − s3 + (25/4)s2 + 4s
s2 + 6s + 8
∣∣∣∣∣s=−0.4077
= 0.1153
The value of K at the breakin point is
K = − s3 + (25/4)s2 + 4s
s2 + 6s + 8
∣∣∣∣∣s=−2.7693
= 16.49
There is no crossover point for K ≥ 0.The root locus plot is shown in the figure on the following page.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
−10 −9 −8 −7 −6 −5 −4 −3 −2 −1 0
−3
−2
−1
0
1
2
3
Root Locus
Real Axis
Imag
inar
y A
xis
Figure : for Problem 11.9
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
11.10 a) The poles are s = 0, 0, and −9. The zero is s = −1. The plot is shown in thefollowing figure.
−9 −8 −7 −6 −5 −4 −3 −2 −1 0 1
−3
−2
−1
0
1
2
3
4
Root Locus
Real Axis
Imag
inar
y A
xis
Figure : for Problem 11.10a
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
11.10 b)
K = −s2(s + 9)s + 1
dK
ds= −(s + 1)(3s2 + 18s) − (s3 + 9s2)
(s + 1)2= 0
which gives s = 0 and s = −3 for the breakaway points. For the breakaway point of interest,s = −3,
K = − s2(s + 9)s + 1
∣∣∣∣∣s=−3
= 27
From Guide 10, ∑roots = −3 − 3 + r3 = −9
Thus the third root is r3 = −3. So when K = 27, all three roots are at s = −3.c) The smallest possible dominant time constant occurs at the breakaway point s = −3
and is τ = 1/3.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
11.11 a) The root locus equation is
1 + Ks(s + 0.5)(s + 1.5)
(s + 2.5)(s2 + 7s + 24.5)
The poles are s = −2.5, and −3.5± 3.5j. The zeros are s = 0, −0.5, and −1.5. The plot isshown in the figure on the following page.
b) The two points on the root locus that have a time constant of τ = 0.5 can be read offthe plot. They are s = −2 and s = −2 ± 3.35j. You can use the graphical method to findthe value of K at each point, and then use Guide 10 to determine the locations of the otherroots. For the first point, s = −2, K = 4.92, and the other roots are s = −0.64 ± 2.19j,which are the dominant roots. Thus the dominant time constant is not 0.5. For the secondpoint, s = −2 ± 3.35j, K = 0.73, and the third root is s = −2.32, which is not dominant.Thus the solution is K = 0.73.
You can solve this problem very easily by using the rlocus and rlocfind functions ofMATLAB to find the values of K and the root locations.
−6 −5 −4 −3 −2 −1 0 1 2 3−4
−3
−2
−1
0
1
2
3
4
Root Locus
Real Axis
Imag
inar
y A
xis
Figure : for Problem 11.11
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
11.12 a) Separate the equation as follows:
9s3 + 6s2 + 2 − 5ps = 0
The root locus form of the equation is
1 +(−5p
9
)s
s3 + (2/3)s2 + 2/9= 0
The root locus parameter is K = −5p/9 ≤ 0.b) Separate the equation as follows:
4s3 + 2s + 7 − ps2 = 0
The root locus form of the equation is
1 +(−p
4
)s2
s3 + (1/2)s2 + 7/4= 0
The root locus parameter is K = −p/4 ≤ 0.c) Separate the equation as follows:
s2 + 3s + 4 − p(s − 4) = 0
The root locus form of the equation is
1 + (−p)s − 4
s2 + 3s + 4= 0
The root locus parameter is K = −p ≤ 0.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
11.13Part (a)
• Poles: s = 0 and s = −5
• Zeros: None
• Two paths must leave the plot, so there will be two asymptotes.
• The locus exists on the real axis in the following intervals:
Re(s) < −5 and Re(s) > 0
• Angles of the asymptotes:
θ =m360
0 − 2= ±180
• The two previous results lead us to conclude that there are no breakaway or breakinpoints.
• Crossover point: None for K ≤ 0
• The root locus plot is shown in the figure on the following page.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
−12 −10 −8 −6 −4 −2 0 2 4 6 8
−6
−4
−2
0
2
4
6
Root Locus
Real Axis
Imag
inar
y A
xis
Figure : for Problem 11.13a
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
11.13 (b)
• Poles: s = −1.5± 0.866j
• Zero: s = −3
• One path must leave the plot, so there will be one asymptote.
• The locus exists on the real axis in the following interval: Re(s) > −3
• Angles of the asymptote:
θ =m360
1 − 2= 0
• The breakaway and breakin points are found as follows.
K = −s2 + 3s + 3s + 3
dK
ds= −(s + 3)(2s + 3)− (s2 + 3s + 3
(s + 3)2=
s2 + 6s + 6(s + 3)2
= 0
The solutions are s = −4.7321 and s = −1.2679. The point is at s = −4.7321 is abreakaway or breakin point for K > 0. The point s = −1.2679 is a breakin point forK < 0. The value of K at the breakaway point is
K = − s2 + 3s + 3s + 3
∣∣∣∣∣s=−1.2679
= −0.4641
• The crossover point obviously occurs at s = 0, at which K = −1.
• The root locus plot is shown in the figure on the following page.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
−3 −2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5
−1.5
−1
−0.5
0
0.5
1
1.5
Root Locus
Real Axis
Imag
inar
y A
xis
Figure : for Problem 11.13b
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
11.13 (c)
• Poles: s = 0, s = −1.5± 0.866j
• Zeros: None
• Three paths must leave the plot, so there will be three asymptotes.
• The locus exists on the real axis in the following interval: Re(s) > 0
• Angles of the asymptotes:
θ =m360
0 − 3= 0, −120, −240
which are equivalent to θ = 0, −120, 60.
σ =∑
sp −∑
sz
2 − 0=
0− 1.5− 0.866j − 1.5 + 0.866j − 03
= −1
• The breakaway and breakin points are found as follows.
K = −(s3 + 3s2 + 3s)
dK
ds= −(3s2 + 6s + 3) = 0
The solutions are s = −1 and s = −1, which do not lie on the locus for K < 0. Thusthere are no breakaway or breakin points for K < 0.
• There is no crossover point for K ≤ 0. This can be proved as follows. Substitutes = jω into the characteristic equation s3 + 3s2 + 3s + K = 0.
−jω3 − 3ω2 + 3jω + K = 0
which gives(3 − ω2)ω = 0 and K − 3ω2 = 0
The first equation gives two possibilities: 1) ω = 0, which is not of interest because itdoes not lie on the root locus for K < 0, and 2) ω = ±
√3. Substituting this into the
second equation gives K = 9, which corresponds to the locus for K > 0.
• The root locus plot is shown in the figure on the following page.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
−3 −2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
Root Locus
Real Axis
Imag
inar
y A
xis
Figure : for Problem 11.13c
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
11.13 (d)
• Poles: s = 0, s = −5, s = −7. Zeros: None
• Three paths must leave the plot, so there will be three asymptotes.
• The locus exists on the real axis in the following intervals:
−7 < Re(s) < −5 and Re(s) > 0
• Angles of the asymptotes:
θ =m360
0 − 3= 0, −120, −240
which are equivalent to θ = 0, −120, 60.
• Intersection point of the asymptotes:
σ =∑
sp −∑
sz
2 − 0=
0 − 5 − 7 − 03
= −4
• The breakaway and breakin points are found as follows: K = −(s3 + 12s2 + 35s) and
dK
ds= −(3s2 + 24s + 35) = 0
The solutions are s = −6.0817 and s = −1.9183, which does not lie on the locus forK < 0. The value of K at the breakaway point is
K = − (s3 + 12s2 + 35s)∣∣∣s=−6.0817
= −6.0411
(continued on the next page)
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
Problem 11.13d continued:
• There is no crossover point for K ≥ 0. This can be proved as follows. Substitutes = jω into the characteristic equation s3 + 12s2 + 35s + K = 0.
−jω3 − 12ω2 + 35jω + K = 0
which gives(35− ω2)ω = 0 and K − 12ω2 = 0
The first equation gives two possibilities: 1) ω = 0, which is not of interest because itdoes not lie on the root locus for K < 0, and 2) ω = ±
√35. Substituting this into the
second equation gives K = 12(35), which corresponds to the locus for K > 0. Theroot locus plot is shown in the following figure.
−25 −20 −15 −10 −5 0 5 10 15 20 25−20
−15
−10
−5
0
5
10
15
20
Root Locus
Real Axis
Imag
inar
y A
xis
Figure : for Problem 11.13d
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
11.13 (e)
• Poles: s = 0, s = −3
• Zero: s = −4
• One path must leave the plot, so there will be one asymptote.
• The locus exists on the real axis in the following intervals:
−4 < Re(s) < −3 and Re(s) > 0
• Angle of the asymptote:
θ =m360
1 − 2= 0
• From the two previous items, it is obvious that there are no breakaway or breakinpoints for K < 0.
• There is no crossover point for K ≤ 0. This can be proved as follows. Substitutes = jω into the characteristic equation s2 + 3s + K = 0.
−ω2 + 3jω + K = 0
which givesω = 0 and K − ω2 = 0
The first equation gives ω = 0, which is not of interest because it does not lie on theroot locus for K < 0. The second equation gives K = 0.
• The root locus plot is shown in the figure on the following page.
c©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
−4 −3 −2 −1 0 1 2 3 4 5 6
−3
−2
−1
0
1
2
3
Root Locus
Real Axis
Imag
inar
y A
xis
Figure : for Problem 11.13e
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
11.13 (f)
• Poles: s = and s = −6. Zero: s = 4. One path must leave the plot, so therewill be one asymptote. The locus exists on the real axis in the following intervals:−6 < Re(s) < 0 and Re(s) > 4. The angle of the asymptote is θ = m360/(1−2) = 0
• The breakaway and breakin points are found as follows.
K = −s2 + 6s
s − 4
dK
ds= −(s − 4)(2s + 6)− (s2 + 6s
(s − 4)2=
s2 − 8s − 24(s − 4)2
= 0
The solutions are s = 10.325 and s = −2.325. The point is at s = 10.325 is a breakinpoint. The point s = −2.325 is a breakaway point. The value of K at the breakinpoint is
K = − s2 + 6s
s − 4
∣∣∣∣∣s=10.325
= −26.65
The value of K at the breakaway point is
K = − s2 + 6s
s − 4
∣∣∣∣∣s=−2.325
= −1.35
The breakin and breakaway points are symmetrically located each a distance 6.325from the zero at s = 4. The locus off the real axis is a circle of radius 6.325 centeredat the zero.
• The crossover points can be found as follows. Substitute s = jω into the characteristicequation s2 + 6s + K(s − 4) = 0.
−ω2 + 6jω + K(jω − 4) = 0
which gives(6 + K)ω = 0 and 4K + ω2 = 0
The first equation gives 1) ω = 0, which is not of interest because it does not lie onthe root locus for K < 0, and 2) K = −6. Substituting K = −6 into the secondequation gives ω = ±
√24 = ±2
√6. The root locus plot is shown in the figure on the
following page.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
−6 −4 −2 0 2 4 6 8 10 12−8
−6
−4
−2
0
2
4
6
8
Root Locus
Real Axis
Imag
inar
y A
xis
Figure : for Problem 11.13f
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
11.14 a) The closed-loop transfer function is
C(s)R(s)
=KP (8− s)
s2 + 2s + 3 + KP (8− s)=
KP (8− s)s2 + (2 − KP )s + 3 + 8KP
The root locus equation is
1 + KP8 − s
s2 + 2s + 3= 1 + K
s − 8s2 + 2s + 3
= 0
where K = −KP ≤ 0. The poles are s = −1± 1.414j, and the zero is s = 8. The root locusplot is shown in the figure on the following page.
The breakaway and breakin points are found as follows.
K = −s2 + 2s + 3s − 8
dK
ds= −(s − 8)(2s + 2)− (s2 + 2s + 3)
(s − 8)2= −s2 − 16s− 19
(s − 8)2= 0
The solutions are s = 17.11, which is the breakin point, and s = −1.11, which is a breakawayor breakin point for K > 0. From the location of the breakin point we can tell that the rootlocus off the real axis is a circle of radius 17.11− 8 = 9.11 centered on the zero at s = 8.
From the Routh-Hurwitz criterion applied to the characteristic equation
s2 + (2− KP )s + 3 + 8KP = 0
the system is stable if 2 − KP > 0 and 3 + 8KP > 0; that is, if −3/8 < KP < 2. Thereforethe crossover points correspond to KP = 2 and can be located by solving the characteristicequation for this value of KP . The result is
s2 + 19 = 0
or s = ±√
19.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
−5 0 5 10 15 20−10
−8
−6
−4
−2
0
2
4
6
8
10
Root Locus
Real Axis
Imag
inar
y A
xis
Figure : for Problem 11.14a
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
11.14 b) To achieve τ = 2/3, either 1) the two roots must be real with the dominant rootat s = −3/2, or 2) the roots must be complex with the real part equal to −3/2. Since thelocus does not exist on the real axis for Re(s) < 0, it is impossible for case (1) to occur.Case (2) is also impossible because the circular locus lies entirely to the right of the verticalline passing through s = −3/2. Thus we see that the smallest time constant possible forthis system is 1/1 = 1, which corresponds to KP = 0.
c) With KP = 1 the closed-loop transfer function becomes
C(s)R(s)
=4 − s
s2 + s + 3 + 11
The characteristic roots are s = −0.5± 3.2787j. The unit step response can be found withthe Laplace transform.
C(s) =4 − s
(ss2 + s + 3 + 11)=
C1
s+ C2
3.2787s2 + s + 3 + 11
+ C3s + 0.5
s2 + s + 3 + 11
whereC1 + C3 = 0 C1 + 3.2787C2 + 0.5C3 = −1 11C1 = 8
The solution is C1 = 8/11 = 0.7273, C2 = −0.4159, and C3 = −0.7273. Thus
c(t) = 0.7273− 0.4149e−0.5t sin 3.2787t− 0.7273e−0.5t cos 3.2787t
The response is plotted in the figure on the following page. The negative sign in thenumerator causes the negative initial slope in the response.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
0 2 4 6 8 10 12−0.1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
Step Response
Time (sec)
Am
plitu
de
Figure : For Problem 11.14c
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
11.15 a) The root locus equation is
1 + KP−2s2 + s + 26s(s + 2)(s + 3)
= 1 + Ks2 − 0.5s − 13s(s + 2)(s + 3)
= 0
where K = −2KP ≤ 0. The poles are s = 0, −2, −3, and the zeros are s = 3.86 and −3.36.The root locus plot is shown in the figure on the following page.
The characteristic equation is
s(s + 2)(s + 3) + K(s2 − 0.5s − 13) = s3 + (5 + K)s2 + (6− 0.5K)s− 13K = 0
From the Routh-Hurwitz criterion, the system is stable if
5 + K > 0 6 − 0.5K > 0 − 13K > 0
and(5 + K)(6− 0.5K) > −13K
These reduce to −1.7277 < K < 0, or 0 < KP < 0.8639.At a breakaway or breakin point, ζ = 1. These points are found as follows.
K = − s3 + 5s2 + 6s
s2 − 0.5s− 13
dK
ds= −(s2 − 0.5s − 13)(3s2 + 10s + 6)− (s3 + 5s2 + 6s)(2s − 0.5)
(s2 − 0.5s − 13)2= 0
This is true if s4 − s3 − 4.75s2 − 130s − 78 = 0, which gives s = 8.5, −3.96, −2.68, and−0.8628. The first solution is an unstable point, the second and third solutions correspondto KP < 0, which is unstable, and the fourth is the desired solution. At this point,
K|s=−0.8628 =
(− s3 + 5s2 + 6s
s2 − 0.5s− 13
)∣∣∣∣∣s=−0.8628
= −0.177
which gives KP = 0.177/2 = 0.088.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
−4 −2 0 2 4 6 8 10
−5
−4
−3
−2
−1
0
1
2
3
4
5
Root Locus
Real Axis
Imag
inar
y A
xis
Figure : for Problem 11.15a
c©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
11.15 b) The closed-loop transfer function is
C(s)R(s)
=KP (−2s2 + 1s + 26)
s3 + (5 − 2KP )s2 + (KP + 6)s + 26KP
The step response with KP = 0.088 is shown in the figure on the following page. Thenegative sign in the numerator causes the negative initial slope in the response.
0 1 2 3 4 5 6 7 8−0.2
0
0.2
0.4
0.6
0.8
1
1.2
Step Response
Time (sec)
Am
plitu
de
Figure : for Problem 11.15b
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
11.16 For ∆ = 0, the closed-loop transfer function is
C(s)R(s)
=KP + KDs
s2 + KDs + KP − 5
Assuming the system is stable, the steady-state response for a unit-step input is
css = lims→0
KP + KDs
s2 + KDs + KP − 51s
=KP
KP − 56= 1
Thus we cannot achieve the specification that css = 1.For ζ < 1, the time constant expression is
τ =2
KD= 0.1
Thus KD = 20.For the damping ratio,
ζ =KD
2√
KP − 5= 0.707
which gives KP = 205. Thus the steady-state response will be css = 205/200 = 1.025,which is close to the desired value.
With KP = 205 and KD = 20, for ∆ 6= 0, the closed-loop transfer function is
C(s)R(s)
=205 + 20s
s2 + 20s + 200− ∆
The root locus equation is
1 + K1
s2 + 20s + 200= 0
where K = −∆ and −1 ≤ K ≤ 0. The root locus for K ≤ 0 is shown in the figure onthe following page. The breakin point is at s = −0.1 when K = −100. It shows that thetime constant remains fixed at τ = 0.1 for −100 ≤ K ≤ 0, that is for 0 ≤ ∆ ≤ 100. Thedamping ratio variation as K varies from 0 to −1 is too small to be seen on the plot, butwe can calculate the sensitivity of ζ to changes in K as follows.
ζ =20
2√
200 + K
From this we see that ζ varies from ζ = 0.707 to ζ = 0.709 as K varies from 0 to −1. Thusthe system is insensitive to K and therefore to ∆.
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−20 −15 −10 −5 0−10
−8
−6
−4
−2
0
2
4
6
8
10
Root Locus
Real Axis
Imag
inar
y A
xis
Figure : for Problem 11.16
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
11.17 a) With proportional control, the root locus equation is
1 + KP6
s(2s + 2)(3s + 24)= 0
or1 + K
1s(s + 1)(s + 8)
= 0
where K = KP . The root locus is similar in shape to that shown in Figure 11.1.20.The breakaway point is found as follows.
K = −s(s + 1)(s + 8) = −(s3 + 9s2 + 8s
)
dK
ds= −(3s2 + 18s + 8) = 0
which gives s = −5.517 and s = −0.4833. The first solution corresponds to a breakin pointfor K < 0. The second solution is the breakaway point for K > 0. For this plot, the farthestto the left the dominant root can lie is at the breakaway point. Thus the smallest possibledominant time constant is 1/0.4833 = 2.07.
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11.17 b) PD action will add a zero that will pull the locus to the left, stabilize the system,and allow a smaller time constant to be obtained, if the zero is placed properly. With PDaction, the open-loop transfer function becomes
KP (1 + TDs)s(s + 1)(s + 8)
=KP TD
(s + 1
TD
)
s(s + 1)(s + 8)
The poles are still at s = 0, s = −1, and s = −8, and the zero is at s = −1/TD. The rootlocus parameter is K = KP TD.
We want τ = 0.5. Thus the zero at s = −1/TD must be placed far enough to the left topull the locus to the left of s = −2. Thus we choose TD = 0.5 to place the zero at s = −2.The resulting root locus is shown in the figure on the following page.
Next we choose ζ = 0.707, which along with τ = 0.5, specifies the point s = −2 ± 2j.From the plot we can see that the line corresponding to ζ = 0.707 passes through thelocus, so we know that a solution exists. At s = −2 ± 2j, we find that K = 19.7. ThusKP = 19.7/TD = 2(19.7) = 39.4. The third root lies at s = −5 and thus is not dominant.Therefore, one solution is TD = 0.5, KP = 39.4. Other solutions are possible, depending onthe choices made for TD and ζ.
−5 −4 −3 −2 −1 0 1 2 3 4
−2
−1
0
1
2
3
4
0.707
0.707
Root Locus
Real Axis
Imag
inar
y A
xis
s = −2 + 2j
Figure : for Problem 11.17
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11.18 The root locus equation is
1 + KPs + 10
(s + 2)(s + 3)= 0
The root locus plot is shown in the following figure.The line tangent to the circle gives the smallest ζ. This line has an angle of 49. Thus
the smallest ζ is ζ = cos 49 = 0.66.The 45 line corresponds to ζ = 0.707. This line intersects the circle at two points. The
left-most point gives the smallest time constant. This point, which can be found graphicallyor analytically, is s = −6.73 + 6.73j, which corresponds to
KP = − (s + 2)(s + 3)s + 10
∣∣∣∣s=−6.73+6.73j
= 8.46
The time constant is τ = 1/6.73 = 0.149.
−18 −16 −14 −12 −10 −8 −6 −4 −2 0−8
−6
−4
−2
0
2
4
6
8
Root Locus
Real Axis
Imag
inar
y A
xis
ζ = 0.66 ζ = 0.707
Figure : for Problem 11.18
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
11.19 The characteristic equation is found from
1 + KP Gp(s) = 0
which becomes1 + 2
5(s + 4)(s + 3)(s + p)
= 0
ors2 + 13s + 40 + p(s + 3) = 0
Let K = p − 7. Then the characteristic equation becomes
s2 + 20s + 61 + K(s + 3) = 0
where K ≥ 0. The poles are s = −3.76 and s = −16.2. The zero is s = −3. The root locusplot is sketched in the figure on the following page. As p increases above 7, the dominantroot moves from −3.76 to −3, and the other root moves from −16.2 to −∞. The roots arealways real, so there will be no oscillations in the free response.
The dominant time constant varies from τ = 1/3.76 = 0.266 to τ = 0.333 as p increasesabove p = 7.
−25 −20 −15 −10 −5 0
−8
−6
−4
−2
0
2
4
6
8
Root Locus
Real Axis
Imag
inar
y A
xis
Figure : for Problem 11.19
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11.20 Express the controller transfer function as
Gc(s) =KP (s + KI
KP)
s=
KP (s + b)s
where b = KI/KP . The root locus equation is
1 + KPs + b
s(s + 1)(s + 2)= 0
The poles are s = 0, s = −1, and s = −2. The zero is at s = −b. The asymptotic anglesare θ = ±90. The asymptotes intersect at
σ =b − 3
2
Sketches of the root locus plots for three cases are shown in the figure on the followingpages. These three cases correspond to
a) 0 < b < 1, using b = 0.5b) 1 < b < 2, using b = 1.5, andc) b > 2, using b = 3.Case (a) does not satisfy the specification that the dominant root have a damping ratio
of ζ = 0.707. Case (b) allows a smaller dominant time constant than Case (c).
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−6 −5 −4 −3 −2 −1 0 1 2 3 4−4
−3
−2
−1
0
1
2
3
4
Root Locus
Real Axis
Imag
inar
y A
xis
Figure : for Problem 11.20, case (a)
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
−6 −4 −2 0 2 4−5
−4
−3
−2
−1
0
1
2
3
4
5
Root Locus
Real Axis
Imag
inar
y A
xis
Figure : for Problem 11.20, case (b)
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
−8 −6 −4 −2 0 2 4−5
−4
−3
−2
−1
0
1
2
3
4
5
Root Locus
Real Axis
Imag
inar
y A
xis
Figure : for Problem 11.20, case (c)
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
Problem 11.20 continued:
The characteristic equation is
s3 + 3s2 + (KP + 2)s + bKP = 0
Because we require that ζ = 0.707, the dominant root pair has the form s = −c ± cj. Thecharacteristic equation can be factored as
[(s + c)2 + c2](s− s3) = s3 + (2c− s3)s2 + (2c2 − 2cs3)s − 2c2s3 = 0
where the third root is s3 and must be real. Comparing coefficients, we see that
2c− s3 = 3 2c2 − 2cs3 = KP + 2 − 2c2s3 = bKP
Thus the third root is s3 = 2c− 3. In order for s = −c ± cj to be dominant, s3 < −c, andthus we must choose c < 1. To minimize the dominant time constant τ = 1/c, we shouldchoose c as large as possible, subject to the restriction that c < 1. However, the closer c
is to 1, the closer the secondary root s3 is to the dominant root. If s3 is too close to thedominant root, the response expected for ζ = 0.707 might not be obtained. Thus we shouldtry a value for c and check the effects of the secondary root by analysis or simulation.
For example, one solution is c = 0.9. This gives s3 = −1.2, b = 1.09, and KP = 1.78.
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11.21 The characteristic equation is
1 + Gc(s)Gp(s) = 0
which becomes1 +
(KP +
KI
s+ KDs
)4
3s2 + 3= 0
or3s3 + 4KDs2 + (3 + 4KP )s + 4KI = 0 (1)
To achieve τ = 1 and ζ = 0.5, the desired dominant roots are s = −1 ± j√
3. The thirdroot is s = −b, where b has some arbitrary value such that b > 1 (so that s = −b will notbe the dominant root). The polynomial corresponding to these three roots is
[(s + 1)2 + 3](s + b) = 0
ors3 + (b + 2)s2 + (4 + 2b)s + 4b = 0
To compare this with equation (1), we must multiply by 3:
3s3 + 3(b + 2)s2 + 3(4 + 2b)s + 12b = 0 (2)
Comparing the coefficients of (1) and (2), we obtain
KP =9 + 6b
4
KI = 3b
KD =3(b + 2)
4The gains can be computed once a value for b > 1 has been selected.
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11.22 a) Assuming that sin φ ≈ φ, cosφ ≈ 1, and sinφ φ2 ≈ 0, we obtain from (2.4.8) and(2.4.11),
(IG + mL2)φ− mLx = T + mgLφ (1)
(m + M)x− mLφ = −f (2)
Solve (2) for x and substitute into (1) to obtain[(m + M)IG + mML2
]φ − (m + M)mgLφ = (m + M)T − mLf (3)
Note that (3) implies that the arm dynamics (φ) are independent of the base dynamics (x),unless either T or f are dependent on x through feedback control.
b) Substituting the given values into (3) gives
500φ− 29 430φ = 60T − 50f (4)
c) For the personal transporter discussed in Chapter 2, T = 0 since the device is con-trolled entirely with the base force f . So we will imitate this approach. Noting that (4)represents an unstable system because the φ term is missing and because the coefficient ofφ is negative, we try a feedback control law of the form
f = KP φ + KDφ (5)
This is a PD control algorithm.Setting T = 0 in (4), substituting (5) into (4), and cleaning up, we obtain
φ − 58.86φ = −0.1KPφ − 0.1KDφ
orφ + 0.1KDφ + (0.1KP − 58.86)φ = 0
A settling time of 4 s implies a time constant of 1 s. If ζ ≤ 1, the time constant is givenby
τ =2
0.1KD= 1
which gives KD = 20.Choosing ζ = 1 gives
ζ =0.1KD
2√
0.1KP − 58.86=
1√0.1KP − 58.86
= 1
which gives KP = 598.6.
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11.23 From Example 11.3.1,
LIs3 + I(R + K3)s2 + +KT K2s + KTK1 = 0
Here,IL = 1.2 × 10−7 IR = 3.6× 10−6
Thus
1.2× 10−7s3 + (3.6× 10−6 + 6 × 10−5K3)s2 + 0.04K2s + 0.04K1 = 0 (1)
We wants = − 1
0.5= −2 and s =
10.05
(−1 ± j) = −20 ± 20j
This requires a characteristic equation of the form
(s + 2)[(s + 20)2 + 202
]= 0
ors3 + 42s2 + 880s + 1600 = 0
Multiply by the leading coefficient of Eq. (1) to obtain
1.2×10−7(s3+42s2+880s+1600) = 1.2×10−7s3+50.4×10−7s2+1056×10−7s+1920×10−7 = 0 (2)
Comparing the coefficients of Eqs. (1) and (2) gives
3.6× 10−6 + 6× 10−5K3 = 50.4× 10−7
0.04K2 = 1056× 10−7
0.04K1 = 1920× 10−7
These giveK1 = 0.0048 V/rad
K2 = 0.00264 V · rad
K3 = 0.024 V/rad
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11.24 Let x1 = θ and x2 = θ. Thenx1 = x2
x2 =1
mL(mgθ − mu) =
g
Lx1 −
1L
u
But u = K1x1 + K2x2, sox1 = x2
x2 =g
Lx1 −
1L
(K1x1 + K2x2)
Taking the Laplace transform of each equation with zero initial conditions and collectingterms, we obtain
sX1(s)− X2(s) = 0
(K1 − g)X1(s) + (Ls + K2)X2(s) = 0
The determinant of these equations must be zero. This gives∣∣∣∣∣
s −1(K1 − g) Ls + K2
∣∣∣∣∣ = 0
This givess(Ls + K2) + (K1 − g) = 0
orLs2 + K2s + K1 − g = 0
The control system will be stable if and only if K2 > 0 and K1 > g.
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11.25 a) Let the state variables be
z1 = x1 z2 = ˙ x1 z3 = x2 z4 = ˙ x2
The input is f . The state equations are
z1 = z2 z2 =1
m1[k1y − (k1 + k2)z1 − cz2 + k2z3 + cz4 − f ]
z3 = z4 z4 =1
m2[k2z1 + cz2 − k2z3 − cz4 + f ]
where the disturbance input is the road displacement y. The matrices are
A =
0 1 0 0−k1+k2
m1− c
m1
k2m1
cm1
0 0 0 1k2m2
cm2
− k2m2
− cm2
=
0 1 0 0−3240 −2 240 2
0 0 0 148 0.2 −48 −0.2
B =
0− 1
m1
01
m2
=
0−0.02
00.004
b) The MATLAB script file is
A = [0,1,0,0;-3240,-2,240,2;0,0,0,1;48,0.2,-48,-.2];B = [0;-.02;0;0.004];p = [-1.397+69.94j,-1.397-69.94j,-0.168+7.779j,-0.168-7.779j];K = acker(A,B,p)
The result is K = [-80 794, -22, 12 688, 124, 124]. For this application, r(t) =0, and the control algorithm is f(t) = K1(0 − x1) − K2x2 − K3x3 − K4x4 = 80 794x1 +22x2 − 12 688x3 − 124x4.
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11.26 There are two circuits and one inertia in this system, and we must write an equationfor each. For the inertia I , Newton’s law gives:
Idω
dt= T − cω − TL
where T = KT ia. For the generator circuit, Kirchhoff’s voltage law gives
vf = Rf if + Lfdifdt
For the motor circuit, Kirchhoff’s voltage law gives
va = Raia + Ladiadt
+ Kbω
where va = Kf if . Substitute for va and T , and rearrange to obtain:
Idω
dt+ cω = KT ia − TL
Lfdifdt
+ Rf if = vf
Ladiadt
+ Raia = Kf if − Kbω
Let the state variable vector be x = [ω, if , ia]. Let the input vector be u = [vf , TL]. Thestate equations are
x = Ax + Bu
where
A =
−c/I 0 KT/I
0 −Rf/Lf 0−Kb/La Kf/La −Ra/La
=
−2 0 0.050 −10 0
−2.5 250 −5
B =
0 −1/I−1/Lf 0
0 0
=
0 −0.1−5 00 0
(continued on the next page)
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
Problem 11.26 continued:
State variable feedback with proportional control has the form
vf = KP (ωr − ω) − K2if − K3ia
To determine the gain vector K = [KP , K2, K3] we use the acker function with the inputmatrix B1 consisting of the first column of B: B1 = [0, 1/Lf , 0]T .
First we must specify the three desired roots. Since the speed time constant is I/c = 0.5,the field time constant is Lf/Rf = 0.1, and the armature time constant is La/Ra = 0.2, wemay reasonably choose a dominant closed-loop time constant of 1 s. Choosing ζ = 0.707,gives the two dominant roots to be s = −1± j. Choosing the third root to be s = −2 givesthe root vector p = [-1 + j, -1-j, -2]. In MATLAB type
A=[-2, 0, 0.05; 0, -10, 0; -2.5, 250, -5];B1 = [0;5;0];p = [-1 + j,-1-j,-2];K = acker(A,B1,p)K =
0.0100 -2.6000 0.0135
Thus the control algorithm is
vf = 0.01(ωr − ω) + 2.6if − 0.0135ia
The closed-loop dynamics are described by
x = (A− B1K)x +
0 −1/IKP 00 0
[
ωr
TL
]
Let
B2 =
0 −1/IKP 00 0
(continued on the next page)
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
Problem 11.26 continued:
To find the transfer functions, first create the state space model Mss as follows:
KP = K(1);B2 = [0,-1/10; KP,0;0,0]C = [1, 0, 0]; D = [0];Mss = ss(A -B1*K,B2,C,D)
Then find the two transfer functions as follows:
Mtf = tf(Mss)
The first transfer function displayed on the screen is Ω(s)/Ωr(s), which is
Ω(s)Ωr(s)
=0.125
s3 + 4s2 + 6s + 4
The second transfer function displayed on the screen is Ω(s)/TL(s), which is
Ω(s)TL(s)
= −0.1s2 + 0.2s + 0.1875s3 + 4s2 + 6s + 4
For a unit-step command ωr, the steady-state response is
ωss = 0.0313
which is a 97% error. For a unit-step disturbance TL, the steady-state response is
ωss = −0.0469
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
11.27 With b = 10, τ = 1, and u = −K1δ − K2ω − K3φ, the equations become
δ = u = −K1δ − K2ω − K3φ
ω = −ω + 10δ
φ = ω
Taking the Laplace transform of each equation with zero initial conditions and collectingterms, we obtain
(s + K1)∆(s) + K2Ω(s) + K3Φ(s) = 0
−10∆(s) + (S + 1)Ω(s) + k3Φ(s) = 0
0∆(s)− Ω(s) + sΦ(s) = 0
The determinant of these equations must be zero. This gives∣∣∣∣∣∣∣
(s + K1) K2 K3
−10 (s + 1) 00 −1 s
∣∣∣∣∣∣∣= 0
This gives
(s + K1)
∣∣∣∣∣(s + 1) 0−1 s
∣∣∣∣∣+ 10
∣∣∣∣∣K2 K3
−1 s
∣∣∣∣∣ = 0
This reduces tos3 + (K1 + 1)s2 + (K1 + 10K2)s + 10K3 = 0 (1)
The desired roots are s = −20 and s = −10 ± 10j. These correspond to the factoredcharacteristic equation:
(s + 20)[(s + 10)2 + 102
]= s3 + 40s2 + 600s + 4000 = 0 (2)
Comparing the coefficents of Eqs. (1) and (2), we obtain
K1 + 1 = 40
K1 + 10K2 = 600
10K3 = 4000
The solution isK1 = 39 K2 = 56.1 K3 = 400
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
11.28 The basic equations for the motor circuit is
e = Raia + Ladiadt
+ Keωm (1)
Tm = KT ia (2)
The dynamics of the windup roll give
d(Iωr)dt
= nTm − cωr − TR (3)
The rate of change of paper tension is given as
dT
dt= k(vr − vp) (4)
The linear and rotational roll velocities are related as
Rvr = ωr (5)
and the effect of the gear pair isωm = nωr (6)
From calculus,d(Iωr)
dt= Iωr + Iωr (7)
andI =
12ρπW (4RR3) = 2ρπWR3
(12dW
)= ρπdW 2R3 (8)
a) If R ≈ 0, then I ≈ 0 andd(Iωr)
dt= Iωr
The three state variables are ia, ωm, and T . Thus the state equations are
diadt
=1La
(e − Raia − Keωm) (9)
dωm
dt=
1I
(n2KT ia − cωm − nRT
)(10)
dT
dt=
k
nRωm − kvp (11)
(continued on the next page)
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
Problem 11.28 continued:
b) In this case, I 6= 0 and
R =d
2Wt + R(0)
From (3), (7), and (8),
ρπdW 2R3
nωm +
12
ρπWR4
nωm = nKT ia −
c
nωm − RT
or
ωm =2n
ρπWR4
[−ρπdW 2R3
nωm + nKT ia −
c
nωm − RT
](12)
The state model consists of (9), (11), and (12).
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
11.29 Change a) For the given values,
Y (s)Ic(s)
=KaK1K2
s(τs + 1)(s2 + 2ζωns + ω2n)
=3× 106
s4 + 260s3 + 26000s2 + 106s
which can be expressed as
Y (s)Ic(s)
=3 × 106/s4
1 + 260/s + 26000/s2 + 106/s3
Let x1 = y and u = ic. Then the state model is
x =
−260 1 0 0−2600 0 1 0−106 0 0 1
0 0 0 0
x +
000
3 × 106
u
(continued on the next page)
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Problem 11.29 continued:
b) Note that if Gc(s) = KP , then
ic = KPe = KP (K3yr − K3y) = 20KPyr − 20KPy
Substitute this into the matrix equation in part (a) to obtain
x = Ax + Bu ==
−260 1 0 0−2600 0 1 0−106 0 0 1
−60 × 106KP 0 0 0
x +
000
60× 106KP
u
The charactersitic polynomial of A is found from the detrminant |sI− A| = 0, and is
s4 + 260s3 + 2600s2 + 106s + 6 × 107KP = 0
The root locus equation is
1 + K1
s4 + 260s3 + 2600s2 + 106s= 0
where K = 6 × 107KP . The root locus plot is shown in the figure on the following page.Since two root always lie in the right-half plane, it is impossible for the proportional controlsystem to be stable.(continued on the next page)
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−1000 −800 −600 −400 −200 0 200 400 600 800−800
−600
−400
−200
0
200
400
600
800
Root Locus
Real Axis
Imag
inar
y A
xis
Figure : For Problem 11.29
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
11.30 a) Assuming that sin φ ≈ φ, cosφ ≈ 1, and sinφ φ2 ≈ 0, we obtain from (2.4.8) and(2.4.11),
(IG + mL2)φ− mLx = T + mgLφ (1)
(m + M)x− mLφ = −f (2)
Solve (2) for x and substitute into (1) to obtain[(m + M)IG + mML2
]φ − (m + M)mgLφ = (m + M)T − mLf (3)
Note that (3) implies that the arm dynamics (φ) are independent of the base dynamics (x),unless either T or f are dependent on x through feedback control.
b) Substituting the given values into (3) gives
500φ− 29 430φ = 60T − 50f (4)
c) For the personal transporter discussed in Chapter 2, T = 0 since the device is con-trolled entirely with the base force f . So we will imitate this approach. Set T = 0 anddivide (4) by 500 to obtain
φ − 58.86φ = −0.1f (5)
Let x1 = φ and x2 = φ to put the model into the following state variable form:
x1 = x2
x2 = 58.86x1 − 0.1f
For linear state variable feedback, f = K1x1 + K2x2, and
x1 = x2
x2 = (58.86− 0.1K1)x1 − 0.1K2x2
Transforming both equations and rearranging gives
sX1(s) − X2(s) = 0 (6)
(0.1K1 − 58.86)X1(s) + (s + 0.1K2)X2(s) = 0 (7)
(continued on the next page)
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Problem 11.30 continued:
Solve (6) for X2(s) and substitute into (7) to obtain
(0.1K1 − 58.86)X1(s) + (s + 0.1K2)sX1(s) = 0 (7)
or (s2 + 0.1K2s + 0.1K1 − 58.86
)X1(s) = 0
Thus the characteristic equation is
s2 + 0.1K2s + 0.1K1 − 58.86 = 0 (8)
A settling time of 4 s implies a time constant of 1 s. If ζ ≤ 1, the time constant is givenby
τ =2
0.1K2= 1
which gives K2 = 20.Choosing ζ = 1 gives
ζ =0.1K2
2√
0.1K1 − 58.86=
1√0.1K1 − 58.86
= 1
which gives K1 = 598.6.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
11.31 a) Using P action, the closed-loop transfer function is
T (s) = T (s) =0.5KP
s3 + 1.5s2 + 1.5s + 0.5 + 0.5KP
Substitute s = jω into the characteristic equation:
j(−ω3 + 1.5ω) + (−1.5ω2 + 0.5 + 0.5KP ) = 0
Thus ωu = 1.225, KPu = 3.5, Pu = 2π/ωu = 5.129.From Table 11.4-1,
For P action : KP = 0.5KPu = 1.75
For PI action : KP = 0.45KPu = 1.575 TI = 0.83Pu = 4.2572
KI =KP
TI= 0.37
For PID action : KP = 0.6KPu = 2.1 TI = 0.5Pu = 2.5646
KI =KP
TI= 0.8189
TD = 0.125Pu = 0.6411 KD = KP TD = 1.3464
b) The responses are shown in the figure on the following page.
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0 5 10 15 20 25 300
0.5
1
t
θProportional
0 5 10 15 20 25 300
0.5
1
1.5
t
θ
PI control
0 5 10 15 20 25 300
0.5
1
1.5
t
θ
PID control
Figure : For Problem 11.31. Responses of the P, PI, and PID controllers.
(continued on the next page)
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Problem 11.31 continued:
The overshoot in the PID response can be reduced by using KP = 1, and reducingthe other gains by a factor of KP = 1.75, that KP = 1, KI = 0.8189/1.75, and KD =1.3464/1.75. The value of KP = 1 was found by simulation. The response of the improvedPID is shown in the following figure.
0 5 10 15 20 25 300
0.2
0.4
0.6
0.8
1
1.2
1.4
t
θ
PID
Improved PID
Figure : For Problem 11.31. Response of the improved PID controller.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
11.32 a) Using P action, the closed-loop transfer function is
T (s) =5KP
2s3 + 10s2 + 2s + 4 + 5KP
Substituting s = jω into the characteristic equation gives ωu = 1, KPu = 1.2, Pu = 2π/ωu =2π.
From Table 11.4-1,For P action : KP = 0.5KPu = 0.6
For PI action : KP = 0.45KPu = 0.54 TI = 0.83Pu = 5.215
KI =KP
TI= 0.1035
For PID action : KP = 0.6KPu = 0.72 TI = 0.5Pu = 3.1416
KI =KP
TI= 0.2292
TD = 0.125Pu = 0.7854 KD = KP TD = 0.5655
b) The responses are shown in the figure on the following page.
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0 10 20 30 40 50 60 700
0.5
1Proportional
t
y
0 10 20 30 40 50 60 700
0.5
1PI control
y
t
0 10 20 30 40 50 60 700
1
2
t
y
PID control
Figure : For Problem 11.32. Responses of the P, PI, and PID controllers.
(continued on the next page)
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Problem 11.32 continued:
The oscillations in the PID response can be reduced by increasing the derivative gain.The value of KD = 2 was found by simulation. The response of the improved PID is shownin the following figure.
0 5 10 15 20 25 30 35 40 45 500
0.2
0.4
0.6
0.8
1
1.2
1.4
t
y
PIDImproved PID
Figure : For Problem 11.32. Response of the improved PID controller.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
11.33 a) Set p = 1 and ζ = 0.5. Using P action, the closed-loop transfer function is
C(s)R(s)
=4KP
s3 + 3s2 + 6s + 4 + 4KP
Substituting s = jω into the characteristic equation gives ωu =√
6 = 2.45, KPu = 3.5,Pu = 2π/ωu = 2.57.
From Table 11.4-1,
For PID action : KP = 0.6KPu = 2.1 TI = 0.5Pu = 1.2825
KI =KP
TI= 1.6374
TD = 0.125Pu = 0.3206 KD = KP TD = 0.6733
The PID control law is
G(s) = 2.1(1 +
11.285s
+ 0.3206s
)=
0.6733s2 + 2.1s + 1.637s
and the resulting closed-loop transfer function is
C(s)R(s)
=2.693s2 + 8.4s + 6.55
s4 + 3s3 + 8.693s2 + 12.4s + 6.55
The roots are s = −0.4848± 2.3038j and s = −1.0152± 0.3886j.In terms of p and ζ, the closed-loop transfer function is
C(s)R(s)
=4p(0.6733s2 + 2.1s + 1.637)
s4 + (4ζ + p)s3 + (4 + pζ + 2.6932p)s2 + 12.4ps + 6.548p
b) With p = 1 the root locus equation in terms of ζ is
1 + 4ζs3 + s2
s4 + s3 + 6.932s2 + 12.4s + 6.548= 0
So the root locus parameter is K = 4ζ. The plot is shown in the figure on the followingpage. For 0.4 ≤ ζ ≤ 0.6, the dominant roots vary from s = −0.313 ± 2.42j, which have adamping ratio of 0.129 and a time constant of 3.19, to s = −0.641 ± 2.16j, which have adamping ratio of 0.284 and a time constant of 1.56.
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−3.5 −3 −2.5 −2 −1.5 −1 −0.5 0 0.5−3
−2
−1
0
1
2
3
Root Locus
Real Axis
Imag
inar
y A
xis
Figure : For Problem 11.33b.
(continued on the next page)
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Problem 11.33 continued:
c) With ζ = 0.5 the root locus equation in terms of p is
1 + ps3 + 4.6932s2 + 12.4s + 6.548
s4 + 2s3 + 4s2= 0
So the root locus parameter is K = p. The plot is shown in the following figure. For0.5 ≤ p ≤ 1.5, the dominant roots vary from s = −0.589±1.9j, which have a damping ratioof 0.296 and a time constant of 1.69, to s = −0.542± 2.6j, which have a damping ratio of0.204 and a time constant of 1.845.
Root Locus
Real Axis
Imag
inar
y A
xis
−4 −3.5 −3 −2.5 −2 −1.5 −1 −0.5 0 0.5−4
−3
−2
−1
0
1
2
3
4
Figure : For Problem 11.33c.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
11.34 a) Using P action, the closed-loop transfer function is
C(s)R(s)
=KP
s3 + 6s2 + 11s + 6 + KP
Substituting s = jω into the characteristic equation gives ωu =√
11 = 3.317, KPu = 60,Pu = 2π/ωu = 1.8945. From Table 11.4-1,
For PID action : KP = 0.6KPu = 36 TI = 0.5Pu = 0.9472
KI =KP
TI= 38.006
TD = 0.125Pu = 0.2368 KD = KP TD = 8.525
b) The response is shown in the figure on the following page, and is labeled ”OriginalPID”. The overshoot can be lowered by decreasing KP , for example, from 36 to 5, anddecreasing KI and KD by the same ratio. The response labeled ”Improved PID” wasobtained with the gain values: KP = 5, KI = (5/36)38.0065.2786, and KD = (5/36)8.525 =1.184.
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0 2 4 6 8 100
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
t
Res
pons
e
Original PID
Improved PID
Figure : For Problem 11.34.
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11.35 In the linear region,C(s)R(s)
=KP s + KI
5s2 + KP s + KI
To avoid saturation,
KP ≤ mmax
rmax=
202
= 10
If we set ζ ≤ 1,
τ =10KP
= 0.1 which gives KP = 100 > 10
If we set KP = 10, the characteristic equation becomes 5s2 + 10s + KI = 0. If ζ ≤ 1, thenτ = 10/10 = 1 > 0.1. If ζ > 1 and τ = 0.1, the two roots will be real and the equation canbe factored as 5(s + 10)(s + a) = 5s2 + (50 + 4a)s + 50a = 5s2 + 10s + KI . Thus a = −8and KI = 50a = −40, which gives an unstable system.
Thus all the possibilities have been exhausted, and we cannot achieve τ = 0.1 becauseof the limit on KP .
So one possible solution is KP = 10 with ζ = 1. This gives
ζ =KP
2√
5KI= 1 which gives KI =
K2P
20= 5
but then τ = 10/5 = 2 > 0.1, so all the specifications have not been achieved.Using the Saturation block with Simulink, simulation with a step input of magnitude
2 shows that the actuator is not saturated except at t = 0+, and that the steady stateresponse is css = 2 but it takes about 8 seconds to reach steady state.
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11.36 a) The closed-loop transfer function is
C(s)R(s)
=KP s + KI
10s2 + (20 + KP )s + KI
The damping ratio is
ζ =20 + KP
2√
10KI= 1
The time constant isτ =
2020 + KP
= 0.1
These two conditions give KP = 180 and KI = 1000.b) Because ζ ≤ 1,
τ =20
20 + KP= 0.1
Thus, to minimize τ , we maximize KP . The largest we can make KP without causingsaturation is
KP =mmax
rmax=
11
= 1
Thus the damping ratio is
ζ =21
2√
10KI
To obtain ζ = 1 requires that KI = 11.025. The solution is KP = 1, KI = 11.025.c) The easiest way to compare these two designs is to use the Saturation block with
Simulink. The unit-step responses show that saturation occurs for both designs and thatthe steady state response is css = 0.05. Examining the plant’s equation of motion, whichis 10c + 20c = m(t), we see that if the maximum value of m is 1, then the steady stateresponse will be css = 1/20 = 0.05. So with the given value of mmax = 1, it is impossibleto achieve a steady state response of css = 1.
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11.37 In the linear region,
C(s)R(s)
=KP s + KI
7s2 + (5 + KP )s + KI
To avoid saturation,
KP ≤ mmax
rmax=
205
= 4
If ζ > 1 and τ = 0.2, the two roots will be real and the equation can be factored as7(s + 5)(s + a) = 7s2 + (35 + 5a)s + 35a. Thus 5 + KP = 35 + 5a and KI = 35a. ThusKI = 5KP − 150 and the system is stable KI > 0, which is true only if KP > 30.
So the only choice is ζ ≤ 1, which gives
τ =14
5 + KP≥ 14
9> 0.2 because KP ≤ 4
With KP = 4, τ = 14/9, and KI can be selected to achieve any value of ζ ≤ 1. However,the time constant specification cannot be satisfied because of the limit on KP .
For example, KI = 2.8929 gives ζ = 1. Using the Saturation block with Simulink,simulation with a step input of magnitude 5 shows that the actuator is always saturated,and that the response reaches c = 4.5 at t = 4 and takes more than 10 seconds to reachsteady state at css = 5. Examining the plant’s equation of motion, which is 7c+4c = m(t),we see that since mmax is 20, then the steady state response will be css = 20/4 = 5. So withthe given value of mmax = 20, it is possible to achieve a steady state response of css = 5,but with a time constant much larger than the desired value of 0.2.
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11.38 a) For PI control with ζ = 1
KP =mmax
rmax=
63
= 2
KI =1
4Iζ2
(mmax
rmax
)2
=1
4ζ2= 0.25
For modified I controlK2 = 5.44
mmax
rmax= 10.88
KI =29.64I
(mmax
rmax
)2
= 7.4
b) For a unit-step command, the PI controller has an overshoot of about 15% andthe modified I controller has no overshoot. Neither controller saturates the actuator. Ifthe magnitude of the step command is rmax = 3, the PI controller has an overshoot ofabout 13% and the modified I controller has no overshoot. Neither controller saturates theactuator.
c) For a unit-ramp command, neither controller saturates the actuator. The responseof the modified I controller lags behind that of the PI controller by about 1.5 seconds atsteady-state.
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11.39 The closed-loop transfer functions for the PI controller are
C(s)R(s)
=KP s + KI
Is2 + KP s + KI
C(s)D(s)
=−s
Is2 + KP s + KI
Assuming the controller is critically damped (ζ = 1), the roots are repeated at s = −KP /2I .The closed-loop transfer functions for the modified I controller are
C(s)R(s)
=KI
Is2 + K2s + KI
C(s)D(s)
=−s
Is2 + K2s + KI
Assuming the controller is critically damped (ζ = 1), the roots are repeated at s = −K2/2I .a) For a unit ramp disturbance, the response of the PI controller is
c(t) =4I
K2P
[1 −
(1 +
KP
2It
)e−KP t/2I
]
and the response of the modified I controller is
c(t) =4I
K22
[1 −
(1 +
K2
2It
)e−K2t/2I
]
For both controllers, css = 1/KI , but note that the value of KI for the modified I controlleris 29.6 times lager than for PI. Thus the steady-state response of the modified I controlleris 29.6 times smaller than that of the PI controller.
b) For a sinusoidal disturbance of unit amplitude and frequency ω, D(s) = ω/(s2 +ω2),and the magnitude ratio of the PI controller is
M(ω) =∣∣∣∣C(iω)D(iω)
∣∣∣∣ =ω√
(KI − Iω2)2 + (KPω)2
Its maximum value occurs at ω =√
KI/I and is 1/KP .
(continued on the next page)
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Problem 11.39 continued:
The magnitude ratio of the modified I controller is
M(ω) =∣∣∣∣C(iω)D(iω)
∣∣∣∣ =ω√
(KI − Iω2)2 + (K2ω)2
Its maximum value occurs at ω =√
KI/I and is 1/K2.Because K2 = 5.44KP (see (11.5.3)), the maximum magnitude ratio of the PI controller
is 5.44 times larger than that of the modified I controller. This means that the deviationcaused by the disturbance is 5.44 times greater for the PI controller.
c) For a sinusoidal command of unit amplitude and frequency ω, R(s) = ω/(s2 + ω2),and the magnitude ratio of the PI controller is
M(ω) =∣∣∣∣C(iω)R(iω)
∣∣∣∣ =KI√
(KI − Iω2)2 + (KP ω)2
The magnitude ratio of the modified I controller is
M(ω) =∣∣∣∣C(iω)R(iω)
∣∣∣∣ =
√K2
I + (K2ω)2√
(KI − Iω2)2 + (K2ω)2
The ratio for PI is greater than 1 at low frequencies. The ratio for modified I is alwaysless than 1. The PI controller has a high- frequency slope of -20 db/decade, whereas themodified I controller has a slope of -40 db/decade.
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11.40 A plot of the response data yields an eyeball estimate of R = 58 and L = 1. If othersalso estimate this line by eye, some variations can be expected in the values of R and L. Abetter way is the use a least squares fit. A Matlab file to do this is shown below.
% Times at which measurements were made.tm=[0:0.5:6,7];% Measured Response.resp=[0,4,20,32,56,84,116,140,160,172,184,190,194,196];% Fit a 5th order polynomial to the data.coef=polyfit(tm,resp,5);% Evaluate the polynomial at a large number of points.dt = .01;t=[0:dt:7];resppoly=polyval(coef,t)’;% Find the coefficients of the polynomial’s derivative.coef2=[5*coef(1),4*coef(2),3*coef(3),2*coef(4),coef(5)];% der is the derivative polynomial.der=polyval(coef2,t)’;% Find the maximum slope R, and the time tdmax at which it occurs.[R,i]=max(der);tdmax = i*dt;% Compute the response where the slope is the maximum.slopemax=resppoly(i);% Compute the intercept L on the time axis.L=(R*tdmax-slopemax)/R;% Compute the points on the straight line having the slope R% and intercept L.stline=R*(t-L);plot(t,resppoly,t,stline,tm,resp,’+’), ...axis([0 7 0 200]),ylabel(’Response’),xlabel(’Time (min)’)RL% Gains For P Control:KP=1/(R*L)
(continued on the next page)
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Problem 11.40 continued:
% Gains For PI Control:KP=0.9/(R*L)KI=KP/(3.3*L)% Gains For PID Control:KP=1.2/(R*L)KI=KP/(2*L)KD=.5*L*KP
The resulting plot is shown in the figure on the following page. The tangent line and thedata are shown along with the polynomial for visual checking. The results are R = 54.5612and L = 0.9422. The resulting values of the gains are
For P action : KP =1
RL= 0.0195
For PI action : KP =0.9RL
= 0.0175 KI =KP
TI= 0.0056
For PID action : KP =1.2RL
= 0.0233 KI =KP
TI= 0.0124 KD = KP TD = 0.0110
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0 1 2 3 4 5 6 70
20
40
60
80
100
120
140
160
180
200
Re
spo
nse
Time (min)
Figure : For Problem 11.40
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11.41 A MATLAB file to fit a 5th degree polynomial to the data is given below. Thepolynomial is then used to compute R, L, and the PID gains.
% Raw Data.% Times at which measurements were made.tm=[0:1:12];% Measured Temperatures (deg C).tempm=[156,157,159,162,167,172,175,179,181,182,183,184,184];% Change in steam pressure.deltap = 3;% Shift the data to indicate relative temperature change.tstart = tempm(1);deltatempm = tempm-tstart;% Scale the data for a unit step input.scaledtempm=deltatempm/deltap;% Fit a 5th order polynomial to the data.coef=polyfit(tm,scaledtempm,5);% Evaluate the polynomial at a large number of points.dt = .01;t=[0:dt:12];scaledtemp=polyval(coef,t)’;% Find the coefficients of the polynomial’s derivative.coef2=[5*coef(1),4*coef(2),3*coef(3),2*coef(4),coef(5)];% der is the derivative polynomial.der=polyval(coef2,t)’;% Find the maximum slope R, and the time tdmax at which it occurs.[R,i]=max(der);tdmax = i*dt;% Compute the scaled temperature where the slope is the maximum.scaledtdmax=scaledtemp(i);% Compute the intercept L on the time axis.L=(R*tdmax-scaledtdmax)/R;% Compute the points on the straight line having the slope R% and intercept L.
(continued on the next page)
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Problem 11.41 continued:
stline=R*(t-L);plot(t,scaledtemp,t,stline,tm,scaledtempm,’+’), ...axis([0 12 0 15]), title(’Unit Step Response’),...ylabel(’Change in Temperature, deg C’),xlabel(’Time (min)’)RL% Gains For PID Control:KP=1.2/(R*L)KI=KP/(2*L)KD=0.5*L*KP
The resulting plot is shown in Figure 11.41. The tangent line and the data are shown alongwith the polynomial for visual checking. The computed values are R = 1.4842, L = 1.5396,KP = 0.5251, KI = 0.1705, and KD = 0.4043.
0 2 4 6 8 10 120
1
2
3
4
5
6
7
8
9
10Unit Step Response
Cha
nge
in T
empe
ratu
re, d
eg C
Time (min)
Figure : For Problem 11.41
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
11.42 The standard form is1 +
c
5s
s2 + 9= 0
where the locus parameter is K = c/5. Type rlocus([1,0],[1,0,9]) to plot the rootlocus.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
11.43 The equation of motion is
4x + 8x + kx = f(t)
and the characteristic equation is
4s2 + 8s + k = 0
The standard form is1 +
k
41
s2 + 2s= 0
where the locus parameter is K = k/4. Type rlocus(1,[1,2,0]) to plot the root locus.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
11.44 The equation of motion is
4x + cx + 64x = f(t)
and the characteristic equation is
4s2 + cs + 64 = 0
The standard form is1 +
c
4s
s2 + 16= 0
where the locus parameter is K = c/4. Type rlocus([1,0],[1,0,16]) to plot the rootlocus.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
11.45 The equation of motion is
2x + 8x + (26 + k2)x = k2y(t)
and the characteristic equation is
2s2 + 8s + 26 + k2 = 0
The standard form is1 +
k2
21
s2 + 4s + 13= 0
where the locus parameter is K = k2/2. Type rlocus(1,[1,4,13]) to plot the root locus.The plot shows that the largest attainable value of ζ for k2 > 0 is ζ = cos[tan−1(3/2)] = 0.55.
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11.46 The equation of motion is
2x + (8 + c2)x + 26x = c2vi(t)
and the characteristic equation is
2s2 + (8 + c2)s + 26 = 0
The standard form is1 +
c2
2s
s2 + 4s + 13= 0
where the locus parameter is K = c2/2. Type rlocus([1,0],[1,4,13]) to plot the rootlocus. The plot shows that the smallest attainable value of τ for c2 > 0 occurs whenK = 3.21 at s = −3.61. This gives τ = 1/3.61 and c2 = 2K = 6.42.
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11.47 The characteristic equation in standard form is
1 + K1
s3 + 13s2 + 52s + 60= 0
Type
rlocus(1,[1,13,52,60]),sgrid([0.5,0.707],[3,5])
to plot the root locus with a (ζ, ωn) grid. The plot shows that the specifications are achiev-able with K in the range 38.6 ≤ K ≤ 83.8.
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11.48 a) The characteristic equation in standard form is
1 +K
21
s3 + 6s2 + 8s= 0
where the locus parameter is K/2. Type rlocus(1,[1,6,8,0]),sgrid(0.707,[]) to plotthe root locus with a (ζ, ωn) grid.
b) The plot shows that the specifications are achievable with K/2 ≈ 5.18; that is, withK = 10.36.
c) The step response with K = 10.36 is obtained by typing
sys = tf(10.36,[2,12,16,10.36]);step(sys)
Right click on the plot and select “Characteristics”, then “Peak response”. The maximumovershoot is 4.03% at t = 4.4.
With K = 10.36, the roots are s = −4.4694 and s = −0.7653± 0.7572j. The dominanttime constant is 1/0.7653 = 1.31. The secondary time constant is 1/4.4694 = 0.2237. It ismuch smaller than the dominant time constant and thus has little effect on the response.This is illustrated by the fact that the overshoot predicted from the dominant roots withζ = 0.707 is 4.33%, which is almost identical to that of the third-order system.
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11.49 Isolate the parameter c as follows.
LaIs2 + RaIs + KbKT + c(Las + +Ra) = 0
Put this into the form of (11.6.1):
1 +c
I
s + Ra/La
s2 + (Ra/La)s + KbKT/LaI= 0
where K = c/I . The zero is s = −Ra/La = −666.67. The poles are the roots ofs2 + (Ra/La)s + KbKT/LaI = s2 + 666.67s + 8.3333 × 104 = 0 and are s = −500 ands = −166.7. Type rlocus([1, 666.67], [1, 666.67, 8.3333e+4]), axis equal toplot the root locus. The locus is a circle centered at s = −666.67 with a radius of 288.5.The smallest possible time constant occurs when both roots are repeated at s = −955. Thisoccurs when K = 1240, or when c = 1240I = 0.0496.
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11.50 a) The charactersitic polynomial is
D(s) = s4 + (16 + c1)s3 + (9 + 8c1)s2 + (8 + 4c1)s + 4 = 0
Isolate the parameter c1 as follows.
s4 + 16s3 + 9s2 + 8s + 4 + c1(s3 + 8s2 + 8s) = 0
Put this into the standard root locus form:
1 + c1s3 + 8s2 + 8s
s4 + 16s3 + 9s2 + 8s + 4= 0
where K = c1. The zeros are the roots of s3 +8s2 +8s = 0 and are s = 0, s = −6.8284, ands = −1.1716. The poles are the roots of s4 + 16s3 + 9s2 + 8s + 4 = 0 and are s = −15.4499,s = −0.0107 ± 0.6997j, and s = −0.5287. Type rlocus([1, 8,8,0], [1,16,9,8,4]),sgrid(0.707,[]) to plot the root locus with a gridline corresponding to ζ = 0.707.
b) By moving the cursor along the plot we can determine that a root exists at s =−0.973 + 0.966j, which gives ζ = 0.71, which is close enough to 0.707. The value of c1
required to give this root is determined as we move the cursor, and is the “gain”, which is3.69. However, this root is not the dominant root. This can be seen by typing K = 3.69;roots([1,(16+K),(9+8*K),(8+8*K), 4]). The roots are s = −17.6245, s = −0.9725 ±0.9673j, and s = −0.1206. The latter is the dominant root.
So it is not possible to achieve a dominant root with a damping ratio of 0.707.
(continued on the next page)
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Problem 11.50 continued:
c) Type rlocus([1,8,8,0], [1,16,9,8,4]),sgrid([],0.25) to plot the root locuswith a gridline corresponding to ωn = 0.25. This will give a circular gridline that in-tersects the real axis at the point corresponding to τ = 4. By moving the cursor alongthe plot we can determine that a root exists at s = −0.252, which gives τ close enoughto 4. The value of c1 required to give this root is determined as we move the cur-sor, and is the “gain”, which is 1.51. To see if this is the dominant root, type K =1.51; roots([1,(16+K),(9+8*K),(8+8*K), 4]). The roots are s = −16.2908, s =−0.4835 ± 0.8601j, and s = −0.2522. The latter is the dominant root. So it is possi-ble to achieve a dominant real root with a time constant of 4.
d) With c1 = 1.51 the two transfer functions are:
X1(s)D(s)
=8s + 4
s4 + (16 + c1)s3 + (9 + 8c1)s2 + (8 + 4c1)s + 4=
8s + 4s4 + 17.51s3 + 21.08s2 + 14.04s + 4
X2(s)D(s)
=s2 + 9.51s + 5
s4 + 17.51s3 + 21.08s2 + 14.04s + 4
The unit-step response plot for x1 is found by typing
sys1 = tf([8,4],[1,17.51,21.08,14.04,4]);step(sys1)
The unit-step response plot for x2 is found by typing
sys2 = tf([1,9.51,5],[1,17.51,21.08,14.04,4]);step(sys2)
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11.51 The root locus plots are shown in the solutions to Problem 11.7.
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11.52 The root locus plots are shown in the solutions to Problem 11.13.
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11.53 a) K = 27.7. The three roots are s = −1.37± 1.37j, and s = −7.2.b) K = 16.9. The three roots are s = −1.5, s = −1.65, and s = −6.87.
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11.54 a) K = 1.27. The three roots are s = −0.569, s = −0.569, and s = −7.86.b) When K = 1.14, the dominant root is s = −0.402. When K = 1.4, the dominant
roots are s = −0.576±).156j.
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11.55 It is possible. With K = 3.96, the dominant roots are s = −0.724± 0.724j.
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11.56 The two candidate solutions are on the boundaries of the ζ-ωn lines. One candidateis ζ = 0.5, which gives ωn = 2.39 and ζωn = 1.7. The other candidate is ωn = 2, which givesζ = 0.677 and ζωn = 1.35. So the solution is ζ = 0.5 and ωn = 2.39, for which K = 43.6.
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Problem 11.57 The characteristic equation is 10s3 + (2 + KD)s2 + KP s + KI = 0. WithKP = 55 and KD = 58, we have 10s3 + 60s2 + 55s + KI = 0, or
1 +KI
101
s3 + 6s2 + 5.5s= 0
The root locus gain is K = KI/10. In MATLAB type
sys = tf(1, [1, 6, 5.5, 0]);rlocus(sys), axis equal
The plot follows. In Example 10.7.4, KI = 25. This gives ζ = 0.707 and τ = 2. The plotshows that to reduce the error by increasing KI , the dominant time constant will becomelarger and the damping ratio will decrease, making the system slower and more oscillatory.Moving the cursor along the plot shows that if we decrease the error by half by doublingKI to 50, the new damping ratio will be approximately 0.44 and the new time constant willbe approximately 1/0.43 = 2.3.
−20 −15 −10 −5 0 5 10−15
−10
−5
0
5
10
15
Root Locus
Real Axis
Ima
gin
ary
Axi
s
Figure : for Problem 11.57.
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11.58 The characteristic equation is s3 +KDs2 +(KP − 4)s+KI = 0. With KP = 604 andKD = 40, we have s3 + 40s2 + 600s + KI = 0 or
1 + KI1
s3 + 40s2 + 600s= 0
The root locus gain is K = KI . In MATLAB type
sys = tf(1, [1, 40, 600, 0]);rlocus(sys)
The plot follows. In Example 10.8.3, KI = 4000. This gives ζ = 0.707 and τ = 0.1. The plotshows that to reduce the error by increasing KI , the dominant time constant will becomelarger and the damping ratio will decrease, making the system slower and more oscillatory.If we decrease the error by half by doubling KI to 8000, the plot shows that the newdamping ratio will be approximately 0.3 and the new time constant will be approximately1/5.68 = 0.18.
−60 −50 −40 −30 −20 −10 0 10−40
−30
−20
−10
0
10
20
30
40
Root Locus
Real Axis
Imag
inar
y A
xis
Figure : for Problem 11.58.
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11.59 a) The model is shown in the figure. The outputs of the proportional and integralterms are stored in the variables simout(:,1) and simout(:,2). To plot them, typeplot(tout,simout(:,1),tout,simout(:,2)
Figure : for Problem 11.59a
(continued on the next page)
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Problem 11.59 continued:
b) The model is shown in the figure. With KA = 0, the overshoot id about 18% andthe actuator remains saturated until abot t = 0.6. There are a range of values for KA thatwill reduce the overshoot. For example, with KA = 50, the overshoot is about 3% and theactuator remains saturated until about t = 0.3.
Figure : for Problem 11.59b
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11.60 The model is shown in the figure. Without the rate limiter, the overshoot is about9%, but with the rate limiter, the overshoot is about 50%.
Figure : for Problem 11.60
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11.61 a) By reducing the loops in the block diagram we obtain the following transferfunction:
Θ(s)Θr(s)
=KT KI
ILs4 + (IR + IK3)s3 + K2KTs2 + K1KTs + KIKT
To place the four roots at s = −2, −20, and −20 ± 20j, requires the following polynomial:
s4 + 62s3 + 1720s2 + 19 200s + 32 000
This polynomial was obtained using the MATLAB conv function as follows. Note that theroot pair −20 ± 20j can be written as (s + 20)2 + 400 = s2 + 40s + 400.
step1 = conv([1,2],[1,40,800];)step2 = conv(step1,[1,20])step2 =
1 62 1720 19200 32000
Comparing the denominator of the transfer function with this polynomial, and account-ing for the leading coefficient IL, we obtain
KI =32 000IL
KT= 0.0960
K1 =19 200IL
KT= 0.0576
K2 =1720IL
KT= 0.0052
K3 = 62L− R = −0.4760
(continued on the next page)
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Problem 11.61 continued:
b) The Simulink model is shown on the next page, using a unit-step command startingat t = 0 and a step disturbance of magnitude 0.1 starting at t = 1. Before running thismodel, in the Command window enter the values of the parameters as follows.
L=0.002;R=0.6;KT=0.04;I=6e-5;KI=32000*I*L/KT;K1=19200*I*L/KT;K2=1720*I*L/KT;K3=62*L-R;
The second figure shows the response.
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Figure : for Problem 11.61b. The Simulink model.
0 0.5 1 1.5 2 2.5 3 3.5 4−5
−4
−3
−2
−1
0
1
Time (s)
Posit
ion (r
ad)
Figure : for Problem 11.61b. The response.
c) The model is shown on the following page, using a unit-step command starting att = 0 and a step disturbance of magnitude 0.1 starting at t = 1. Be sure to first enter theparameters as discussed in part (b). The second figure shows the response. Obviously thesaturation limit prevents the system from counteracting the disturbance.
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Figure : for Problem 11.61c. The Simulink model.
0 0.5 1 1.5 2 2.5 3 3.5 4−1800
−1600
−1400
−1200
−1000
−800
−600
−400
−200
0
200
Time (s)
Posit
ion (r
ad)
Figure : for Problem 11.61c. The response.
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11.62 The model without rate limiting is shown in the figure on the following page. Thecorresponding response is shown in the following plot. The disturbance acts at t = 20.Saturation blocks 1 and 2 were added to prevent the liquid heights from becoming negative,which of course is physically impossible.
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Figure : for Problem 11.62. Model without rate limiting.
0 10 20 30 40 50 60 700
5
10
15
20
25
30
35
Time
Flow
Rat
e
With no limit on rate
0 10 20 30 40 50 60 700
1
2
3
4
5
6
7
Time
Heigh
t h2
Figure : for Problem 11.62. Response with no rate limiting.
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Problem 11.62 continued:
The model with rate limiting is shown in the figure on the following page. The cor-responding response is shown in the following plot. As we would expect, the flow rateresponds slower with rate limiting, and the liquid height also responds slower. The steady-state response is the same.
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Figure : for Problem 11.62. Model with rate limiting.
0 10 20 30 40 50 60 700
10
20
30
40
50
60
Time
Flow
Rat
e
With rate limited
0 10 20 30 40 50 60 700
2
4
6
8
10
12
14
16
Time
Heigh
t h2
Figure : for Problem 11.62. Response with rate limiting.
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Solutions Manual c©
to accompany
System Dynamics, Second Edition
by
William J. Palm III
University of Rhode Island
Solutions to Problems in Chapter Twelve
c©Solutions Manual Copyright 2010 The McGraw-Hill Companies. All rightsreserved. No part of this manual may be displayed, reproduced, or distributedin any form or by any means without the written permission of the publisheror used beyond the limited distribution to teachers or educators permitted byMcGraw-Hill for their individual course preparation. Any other reproductionor translation of this work is unlawful.
12.1 The closed-loop transfer function is
C(s)R(s)
=KP + KDs
s2 + KDs + KP − 5
For ωn = 0.5,
ωn =
√KP − 6
5= 0.5
Thus KP = 6.25.For the damping ratio,
ζ =KD
2√
5(KP − 6=
KD
2√
5(0.25= 0.707
which gives KD = 2.5.
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12.2 a) For K = 4 the roots are s = −1 ±√
3j. For ωn = 4 and ζ = 0.5, the roots mustbe s = −2 ± 2
√3j. A simple gain adjustment will not give the desired roots. The angle
deficiency is6
4s(s + 2)
∣∣∣∣s=−2+2
√3j
= 6 4 − 6 s − 6 (s + 2) = −210
Thus the compensator must add (210 − 180) = 30 to the system. This implies a leadcompensator. Place the compensator’s pole and zero to the left of s = −2. Using the sameprocedure as in Example 12.1.2, with µ = 10, we obtain T = 0.0138 and T = 0.293. ChooseT = 0.0138 because the second value would place the zero to the right of s = −2, and thusviolate the assumed geometry. The compensator’s pole is s = −1/T = −72.6, and its zerois s = −1/aT = −7.26. The compensated open-loop transfer function is
Gc(s)G(s) =s + 7.26s + 72.6
K
s(s + 2)
The value of K required to place the root at s = −2± 2√
3j is K = 155.6. Other solutionsare possible, depending on the choice for µ.
For C = 1µF ,
T = 0.0138 =R1R2
R1 + R210−6
(continued on the next page)
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Problem 12.2 continued:
b) A ramp error of 0.05 implies that Cv = 20. With K = 4, s = −1 ± j√
3 and Cv = 2.To obtain Cv = 20 we need to increase the gain by a factor of 10, but without changing theroot locations appreciably. Therefore we use a lag compensator. Because the PM and GMare not specified, we use a root locus design method.
Step 1: With no compensation, K = 4 will place the roots at the desired locations.Thus KP1 = 4.
Step 2: To obtain Cv = 20, K must be increased from 4 to 40. Thus KP2 = 40 andµ = KP1KP2 = 0.1.
Step 3: Choose T large. Note that the plant has a pole at s = −2. Select the compen-sator’s pole and zero to be well to the right of s = −2. So try s = −0.02 and s = −0.2.This gives T = 100.
Step 4: The open-loop compensated transfer function is
Gc(s)G(s) =110
2 + 0.2s + 0.02
K
s(s + 2)
The precise value of the desired roots s = −1±j√
3 probably will not lie exactly on the newroot locus, but if T is chosen large enough, they should be close. The two specifications areζ = 0.5 and ωn = 2, so we can look for a root that satisfies one specification exactly, andhope that the other is close to being satisfied.
For T = 100, the root locus shows that s = −0.8989+1.56j gives ζ = 0.5, but ωn = 1.8,when K = 36. If ωn is close enough to 2, we can stop. Otherwise a larger value of T canbe chosen. Note that the roots s = −0.8989± 1.56j are “dominant” only if we consider thepole at s = −0.2 to be canceled by the zero at s = −0.2.
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12.3 Try a PID compensator. The transfer function is
C(s)R(s)
=5(KDs2 + KP s + KI)
s3 + 5KDs2 + 5KP s + 5KI
The steady-state error for a step command input will be zero, as required. The characteristicequation is
s3 + 5KDs2 + 5KPs + 5KI = 0 (1)
To achieve τ = 1 and ζ = 0.45, the desired dominant roots are s = −1 ± j1.985. The thirdroot is s = −b, where b has some arbitrary value such that b > 1 (so that s = −b will notbe the dominant root). The polynomial corresponding to these three roots is
[(s + 1)2 + (1.985)2](s + b) = 0
ors3 + (b + 2)s2 + (4.94 + 2b)s + 4.94b = 0 (2)
Comparing the coefficients of (1) and (2), we obtain
KP =4.94 + 2b
5
KI =4.94b
5
KD =b + 2
5The gains can be computed once a value for b > 1 has been selected.
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12.4 Because the desired roots are obtainable with a simple gain selection, we need a lagcompensator to achieve the desired Cv value. The lag compensator has the form
Gc(s) = Ks + z
s + p
ThusCv =
Kz
2p
To achieve Cv = 5 with K = 1, we need z/p = 10. Choosing z = 0.05 and p = 0.005satisfies this requirement, and places the compensator’s pole and zero near the origin, faraway from the desired root locations of s = −0.338 ± 0.562j. Using the root locus plot,we can determine the value of K required to place the closed-loop roots near the desiredlocation. This value is K = 1.024. Thus one solution is the compensator transfer function
Gc(s) = 1.024s + 0.05s + 0.005
The resulting closed-loop roots are s = −2.326, −0.055, and −0.312 ± 0.55j, which areclose to the desired locations. The closed-loop zero at s = 0.05 approximately cancels theclosed-loop pole at s = −0.055. The resulting Cv value is Cv = 5.12, which is close to thedesired value.
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12.5 The values ζ = 0.5 and ωn = 2 correspond to a dominant root location of s =−0.5 ± j
√3. These roots can be obtained by setting K = 1. So the gain KP required to
achieve the desired transient performance has already been established as KP1 = K = 1.A lag compensator is indicated, because the steady-state error is too large. The second
step is to determine the value of the parameter µ. For this system, the coefficient Cv is
Cv = lims→0
sKP
s(s + 1)= KP
and KP2 is the value of KP that gives Cv = 10. Thus, KP2 = 10, and the parameterµ = KP /KP2 = 1/10. The compensator’s pole and zero must be placed close to theimaginary axis, with the ratio of their distances being 1/10. Noting that the plant has apole at s = −1, we select locations well to the right of this pole, say, at s = −0.01 ands = −0.1 for the pole and zero, respectively. This gives T = 100.
The open-loop transfer function of the compensated system is thus
Gc(s)G(s)H(s) =0.1Kc(s + 0.1)
s(s + 1)(s + 0.01)
The root locus is shown in the figure on the following page.(continued on the next page)
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Problem 12.5 continued:
−1 −0.9 −0.8 −0.7 −0.6 −0.5 −0.4 −0.3 −0.2 −0.1 0−0.4
−0.3
−0.2
−0.1
0
0.1
0.2
0.3
0.4
Root Locus
Real Axis
Imag
inar
y A
xis
Figure : for Problem 12.5
For the desired damping ratio of 0.5, the locus essentially lies on the asymptote thatpasses through
σ =∑
sp −∑
sz
P − Z=
0 − 1 − 0.01 + 0.13 − 1
= −0.455
So the real part of the root is approximately −0.455, and the imaginary part will be0.455 tan 60 = 0.788 to obtain ζ = 0.5. Thus the root is approximately s = −0.455 +0.788j. The value of Kc required to obtain this root is found from
0.1Kc = −∣∣∣∣s(s + 1)(s + 0.01)
s + 0.1
∣∣∣∣s=−0.455+0.788j
= 0.9128
or Kc = 9.128. This is only a tentative estimate because the root does not lie exactly on thelocus. Using this value of Kc in the characteristic equation, we find that the actual rootsare s = −0.4496 ± 0.788j and s = −0.1109. So our estimate turns out to be very close tothe true value.(continued on the next page)
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Problem 12.5 continued:
The error coefficient is thus Cv = 9.128 and less than the desired value of 10. Also,the dominant roots differ somewhat from the desired locations at s = −0.5 + j0.866. Ifthese differences are too large, the compensator’s pole and zero can be placed closer to theimaginary axis, say, at s = −0.01 and s = −0.001, respectively, with T = 1000. This willdecrease the compensator’s influence on the locus near the desired root locations.
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12.6Step 1: Adjust KP to meet Cv = 20. Cv = K/2 = 20. Thus KP = 40. Use KP = 40 in
the following steps.Step 2: Check phase margin and gain margin with KP = 40, for
G(s) =20
s(0.5s + 1)
The phase margin is 17 and the gain margin is infinite. Thus we must add 40 − 17 = 23
to obtain a phase margin of 40. This requires a lead compensator.Following the method of Example 12.2.3 we add a 5 safety factor and choose φm to be
23 + 5 = 28. Then
µ =1 + sin 28
1 − sin 28= 2.7698
Now find the frequency at which the uncompensated gain equals −20 log√
µ = −4.42 db.This occurs at about ω = 8.2 rad/sec. Thus choose ωm = 8.2 and ωm = 1/T
õ = 8.2,
which gives T = 0.073. Thus the compensator’s parameters are µ = 2.7698 and T = 0.073.The pole and zero are s = −1/T = −13.699 and s = −4.946. The open-loop transferfunction of the compensated system is
Gc(s)G(s) =s + 4.946s + 13.699
40s(s + 2)
As s → 0, the compensator’s gain is seen to introduce a gain factor of 4.946/13/699 =0.361 = 1/µ. Thus, in order to preserve Cv = 20, we must increase KP by a factor of1/0.361 = 2.7698. Therefore we set KP = 40(2.7698) = 110.79. The compensated open-loop transfer function is
Gc(s)G(s) = 110.79s + 4.946s + 13.699
1s(s + 2)
=20(0.2022s + 1)
s(0.073s + 1)(0.5s + 1)
The frequency response plots show that the phase margin is 42, the gain crossoverfrequency is ωg = 8, and the gain margin is infinite. Thus the specifications are satisfied.The closed-loop transfer function is
Gc(s)G(s)1 + Gc(s)G(s)
=20(0.2022s + 1)
0.0365s3 + 0.573s2 + 5.044s + 20
The roots are s = −7.104 and s = −4.297±7.659j, with a dominant time constant of 0.233,a dominant damping ratio of 0.489, and natural frequency of 8.78.
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12.7 With no compensation, Cv = 0.625, which is much less than the desired value ofCv = 80. Thus a lag compensator is needed to increase the gain by 80/0.625 = 128. Leadcompensation is needed because it is not possible to place the roots at the desired locationwith a simple gain adjustment. Thus a simple lead or a simple lag compensator will notwork.
We therefore will design a lag-lead compensator. The angle deficiency is found from
6 G(s)|s=−2+2√
3j = 610
s(s + 2)(s + 8)
∣∣∣∣s=−2+2
√3j
= 120− 90− 30 = −240
Thus, the angle deficiency is 180− 240 = 60. For the lead compensator G1(s),
6 G1(s)|s=−2+2√
3j = 61 + µT1s
1 + T1s
∣∣∣∣s=−2+2
√3j
= 60
This is satisfied if the lead compensator has a pole at s = −53 and a zero at s = −3.65.Thus 1/µT1 = 3.65 and 1/T1 = 53. These give T1 = 0.0189 and µ = 14.5.
For the lag compensator, the choice of T2 = 100 gives∣∣∣∣∣s + µ
T2
s + 1T2
∣∣∣∣∣ =∣∣∣∣s + 0.145s + 0.01
∣∣∣∣s=−2+2
√3j
= 0.967
which we take to be close enough to 1 to indicate adequate pole-zero cancelation.The lag-lead compensator is thus given by
Gc(s) = 128s + 3.65s + 53
s + 0.145s + 0.01
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12.8 We try a compensator of the form
Gc(s) = Ks + a
s + b
The characteristic equation is found from
1 + Gc(s)Gp(s) = 0
which becomes1 + K
s + a
s + b
10s2(0.1s + 1)
= 0
or0.1s4 + (1 + 0.1b)s3 + bs2 + 10Ks + 10Ka = 0 (1)
For τ = 1 and ζ = 0.5, the required dominant root pair is s = −1 ± 1.732j. So thecharacteristic equation must be factored as
0.1[(s + 1)2 + (1.732)2](s2 + cs + d) = 0
or
0.1(s2+2s+4)(s2+cs+d) = 0.1s4+0.1(2+c)s3+0.1(4+2c+d)s2+0.1(4c+2d)s+0.4d = 0 (2)
Comparing the coefficients of equations (1) and (2), we obtain
1 + 0.1b = 0.1(2 + c) (3)
b = 0.1(4 + 2c + d) (4)
10K = 0.1(4c + 2d) (5)
10Ka = 0.4d (6)
We can choose c and solve (3) for b and (4) for d. The choice of c = 14 gives b = 6 andd = 28. So the secondary roots from s2 + cs + d = 0 are s = −2.417 and s = −11.58.
Then solve (5) for K to obtain K = 1.12. Finally, solve (6) for a to obtain a = 1. Theresulting compensator is
Gc(s) = 1.12s + 1s + 6
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12.9 The root locus equation is
1 +K
5s + a
s2(s + b)= 0
The poles are s = 0, s = 0, and s = −b. The zero is s = −a. The asymptotic angles areθ = ±90. The asymptotes intersect at
σ =a − b
2
For stability, the intersection point should be negative; thus, we should select b > a.For a 5% overshoot, the damping ratio of the dominant root must be ζ = 0.69, which
corresponds to a complex root. For a time constant of τ = 0.5, the dominant roots mustbe s = −2 ± 2.096j to achieve ζ = 0.69. The corresponding factor is (s + 2)2 + (2.096)2 =s2 + 4s + 8.393. Thus the characteristic equation can be factored as
(s2 + 4s + 8.393)(s− s3) = s3 + (4− s3)s2 + (8.393− 4s3)s − 8.393s3 = 0
The characteristic equation is
s3 + bs2 +K
5s + a
K
5= 0
Comparing coefficients, we see that
s3 = 4 − bK
5= 8.393− 4s3 a = −8.393s3
K/5
The third root must lie to the left of the dominant root, whose real part is −2. Therefore,b must be chosen so that b > 6. The other restriction is that b > a.
One solution is b = 10, which gives s3 = −6, K = 161.965, and a = 1.55.If we try to cancel one of the poles at s = 0 by letting a → 0, the root locus equation
becomes1 +
K
51
s(s + b)= 0
and the characteristic equation is
s2 + bs +K
5= 0
The specifications that ζ = 0.69 and τ = 0.5 require that s = −2± 2.096j. This is achievedwith b = 4 and K = 41.95.
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12.10 The open-loop poles are s = 0 and −2 ± 3j.a) The root locus equation for the lead compensator is
1 +s + 1
µT
s + 1T
Kc
s(s2 + 4s + 13)= 0
where µ > 1 and the root locus parameter is K = Kc ≥ 0.b) The root locus equation for the lag compensator is
1 +s + 1
µT
s + 1T
µKc
s(s2 + 4s + 13)= 0
where µ < 1 and the root locus parameter is K = µKc ≥ 0.c) The root locus equation for the reverse-reaction compensator is
1 +s − 1
T1
s + 1T2
−T1T2
Kc
s(s2 + 4s + 13)= 0
where the root locus parameter is K = −T1Kc/T2 ≤ 0.The root-locus plots are shown in the following figures. All three compensators can
produce an unstable system if the gain Kc is too large. However, only the reverse-reactioncompensator can pull the complex-root paths starting at s = −2±3j to the left. This enablesa dominant root to be obtained that is not close to the imaginary axis. The optimum rootlocations are shown on the plot on the next page.(continued on the next page)
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Problem 12.10 continued:
Figure : For Problem 12.10
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12.11 a) Assuming that sin φ ≈ φ, cosφ ≈ 1, and sinφ φ2 ≈ 0, we obtain from (2.4.8) and(2.4.11),
(IG + mL2)φ− mLx = T + mgLφ (1)
(m + M)x− mLφ = −f (2)
Solve (2) for x and substitute into (1) to obtain[(m + M)IG + mML2
]φ − (m + M)mgLφ = (m + M)T − mLf (3)
Note that (3) implies that the arm dynamics (φ) are independent of the base dynamics (x),unless either T or f are dependent on x through feedback control.
b) Substituting the given values into (3) gives
500φ− 29 430φ = 60T − 50f (4)
c) For the personal transporter discussed in Chapter 2, T = 0 since the device is con-trolled entirely with the base force f . So we will imitate this approach. Noting that (4)represents an unstable system because the φ term is missing and because the coefficient ofφ is negative, we try a feedback control law of the form
f = KP φ + KDφ (5)
This is equivalent to a series compensator of the PD type.Setting T = 0 in (4), substituting (5) into (4), and cleaning up, we obtain
φ − 58.86φ = −0.1KPφ − 0.1KDφ
orφ + 0.1KDφ + (0.1KP − 58.86)φ = 0
A settling time of 4 s implies a time constant of 1 s. If ζ ≤ 1, the time constant is givenby
τ =2
0.1KD= 1
which gives KD = 20.Choosing ζ = 1 gives
ζ =0.1KD
2√
0.1KP − 58.86=
1√0.1KP − 58.86
= 1
which gives KP = 598.6.
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12.12 a) The plot for K = 2 is shown in figure (a). It can be obtained with the followingMATLAB file.
K = 2;sys = tf(5*K,[1,6,5,0])margin(sys)
The results are that the gain margin is 9.54 dB; the phase crossover frequency is 2.24rad/sec; the phase margin is 25.4; and the gain crossover frequency is 1.23 rad/sec.
b) The plot for K = 20 is shown in figure (b). It can be obtained in a manner similar tothat used for part (a). The results are that the gain margin is −10.5 dB (which means thatthe system is unstable); the phase crossover frequency is 2.24 rad/sec; the phase margin is−23.7 (another indication of instability); and the gain crossover frequency is 3.91 rad/sec.
c) For K = 6 both the gain and phase margins are 0. Thus they both are a limitingfactor for stability.
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12.13 The open-loop transfer function is
G(s) = Gc(s)Gp(s) =2 + 19s
100s2 + s
The Bode plots are shown in following figure. They can be obtained with the MATLABmargin function. The gain margin is infinite because the phase curve is always above the−180 line. The phase margin is 66. The system is stable.
−50
0
50
100
150
Mag
nitu
de (d
B)
10−4
10−3
10−2
10−1
100
101
−150
−120
−90
Pha
se (d
eg)
Bode DiagramGm = Inf , Pm = 66.3 deg (at 0.212 rad/sec)
Frequency (rad/sec)
Figure : For Problem 11.13
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12.14 The open-loop transfer function is
G(s) = Gc(s)Gp(s) =25(7s + 64)
5s3 + 6s2 + 5s=
175s + 16005s3 + 6s2 + 5s
The Bode plots are shown in following figure. They can be obtained with the MATLABmargin function. Both the gain and phase margins are negative, so the system is unstable.
−100
−50
0
50
100
150
Mag
nitu
de (d
B)
10−2
10−1
100
101
102
103
−270
−225
−180
−135
−90
Pha
se (d
eg)
Bode DiagramGm = −47.3 dB (at 1.07 rad/sec) , Pm = −41.5 deg (at 7.46 rad/sec)
Frequency (rad/sec)
Figure : For Problem 12.14
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12.15 The open-loop transfer function is
G(s) =0.1KP (10)
(0.2s + 1)(s2 + s + 10)=
KP
0.2s3 + 1.2s2 + 3s + 10
The Bode plots, phase margins, and gain margins can be obtained with the MATLABmargin function. The results are as follows:
a) KP = 1: Gain margin = 20.9 dB, Phase margin = 88.3
b) KP = 10: Gain margin = 0.916 dB, Phase margin = 70.4
c) KP = 100: Gain margin = −19.1 dB, Phase margin = −115
Cases (a) and (b) are stable, but Case (c) is unstable. This means that any initial rollangle will continue to increase.
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12.16 A solution is K = 0.376.
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12.17 The answers are given in the following table.
Case Type No. Cp Cv Ca Step Error Ramp Error(a) 1 ∞ 20 0 0 1/20(b) 0 20 0 0 1/21 ∞(c) 2 ∞ ∞ 7 0 0
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12.18 The open-loop transfer function is
G(s) = KP e−(D1+D2)s 1100s + 1
= KP e−100s 1100s + 1
The delay of 100 s lowers the phase curve by 100ω rad, or 100ω(180/π) degrees. A plotof m and φ for KP = 1 shows that instability is caused by the negative gain margin.The phase crossover frequency is approximately 0.02 rad/sec, and m at this frequency isapproximately −7 dB. To achieve a positive gain margin, we must therefore increase KP toat least 107/20 = 2.239. For this value of KP both the phase margin and the gain marginare zero.
The calculations for KP = 1 can be done in MATLAB as follows.
sys = tf(1,[100,1]);w = [0.001:0.001:0.02]’;[mag, phase] = bode(sys,w);m = 20*log10(mag(:));phasetotal = phase(:)-100*w*(180/pi);semilogx(w,m,w,phasetotal),grid
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12.19 With D = 0, the open-loop frequency response plot of 10/(0.1s + 1) has a gaincrossover at ω ≈ 100, and a phase margin of 95. For PM = 40, the dead time D canreduce the phase curve at ω = 100 by no more than 95− 40 = 55. Because
6 P (iω)e−iωD = 6 P (iω) − ωD
we have100D ≤ 55
π
180rad
or D ≤ 0.0096.
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12.20 One solution is the compensator transfer function
Gc(s) =s + 0.1s + 0.01
This gives a gain margin of 14.3 dB and a phase margin of 41.6.
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12.21 One solution is the compensator transfer function
Gc(s) = 42s + 4.4s + 18
This gives an infinite gain margin and a phase margin of approximately 50.
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12.22 One solution is the compensator transfer function
Gc(s) = 206.66s + 166.6s + 1
This gives an infinite gain margin and a phase margin of approximately 57.
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12.23 One solution is the compensator transfer function
Gc(s) = 200.34s + 10.07s + 1
This gives an infinite gain margin and a phase margin of approximately 50.
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12.24 One solution is the compensator transfer function
Gc(s) = 50(s + 1)(s + 0.16)(s + 0.01)(s + 16)
The closed-loop transfer function is
C(s)R(s)
=50s2 + 58s + 8
0.2s5 + 4.402s4 + 20.24s3 + 66.2s2 + 58.16s + 8
There is pole-zero cancelation of the factor s + 1, so the transfer function becomes
C(s)R(s)
= 250s + 0.16
(s + 17.185)(s + 0.168)(s2 + 3.656s + 13.846)
The roots are s = −17.1837, s = −0.1681, and s = −1.8291 ± 3.2401j. If we take theroots s = −0.1681 to be canceled approximately by the compensator zero at s = −0.16,the root pair s = −1.8291 ± 3.2401j is the dominant root. It has a damping ratio ofζ = cos(tan−1(3.2401/1.8291)) = 0.4916, which is approximately the desired value of 0.5.
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12.25 One solution is the compensator transfer function
Gc(s) = 20(s + 0.7)(s + 0.15)(s + 7)(s + 0.015)
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12.26 The magnitude ratio is
M =ω2
n
ω√
4ζ2ω2n + ω2
A gain margin of 0 implies that M = 1. So, setting M = 1 in the above equation results inthe following polynomial:
ω4 + 4ζ2ω2nω2 − ω4
n = 0
The positive solution is
ω = ωn
√−2ζ2 +
√4ζ4 + 1 (1)
This is the gain crossover frequency.The phase angle is
φ = − 6 jω − 6 (2ζωn + jω) = −90 − tan−1 ω
2ζωn
The phase margin is
PM = φ(ω) − (−180) = φ(ω) + 180 (2)
where ω is given by (1). With this substitution, we obtain
PM = 90 − tan−1 ω
2ζωn= 90 − tan−1
√−2ζ2 +
√4ζ4 + 1
2ζ(3)
Using the identitiescot−1 y
x= tan−1 x
y
tan−1 y
x+ cot−1 y
x= 90
we can show that90 − tan−1 y
x= tan−1 x
y(4)
Letting y =√−2ζ2 +
√4ζ4 + 1 and x = 2ζ in (3), identity (4) gives
PM = tan−1 2ζ√−2ζ2 +
√4ζ4 + 1
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12.27 Proportional control is sufficient here, with KP = 912.01. The resulting maximumpercent overshoot is 1.03% and a 2% settling time of 0.866 s.
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12.28 A closed-loop damping ratio of ζ = 0.4559 corresponds to an overshoot of 20%,and a phase margin of 48.1. Try PD control action with Gc(s) = KP (TDs + 1), with aguessed value of TD = 0.2. Using the margin function with different values of KP givesGc(s) = KP (TDs + 1) = 605(0.2s + 1), which gives an infinite gain margin and a phasemargin of 48.
So the open-loop transfer function is
2.42s + 12.10.02s3 + 0.3s2 + s
The closed-loop transfer function is
2.42s + 12.10.02s3 + 0.3s2 + 3.42s + 12.1
The roots are s = −5 and −5.0000 ± 9.7980j. So the time constant is 0.2, which gives asettling time of 0.8, less than the required value of 1 s.
The error transfer function is
0.02s3 + 0.3s2 + s
0.02s3 + 0.3s2 + 3.42s + 12.1
Using the final value theorem with a unit-ramp input, we find that the steady-state erroris 1/12.1 = 0.0826, which is less than 0.1 as required.
So one solution is Gc(s) = KP (TDs + 1) = 605(0.2s + 1). Other solutions are possible.
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12.29 a) Since the open-loop transfer function is
KP s + KI
s(s + 4)
the margins are calculated in MATLAB as follows:
KP = 6;KI = 50;num = [KP,KI];den = [1,4,0];sys = tf(num,den);[GM, PM, Wg, Wp] = margin(sys)
The results are GM = ∞ and PM = 70.26.b) In MATLAB, type
sysd = tf(num,den,’iodelay’,0.1);[GM, PM, Wg, Wp] = margin(sysd)
The results are GM = 1.9068 and PM = 25.5. So the dead time has reduced the stabilityof the system quite a bit.
To use the feedback function, first convert the continuous-time function sysd to thediscrete-time function discrete using the c2d function with a sampling time equal to 1/100of the time constant. Then use the feedback and step functions.
discrete = c2d(sysd, 0.002);sysCL = feedback(discrete,1);step(sysCL)
The step response is very oscillatory.c) Iterate using the following commands, increasing the dead time until the margin
function indicates an unstable system.
sysd = tf(num,den,’iodelay’,0.157);[GM, PM, Wg, Wp]=margin(sysd)
To three decimal places, the upper limit on the dead time is 0.157. Any value greater thanthat results in a negative phase margin.
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12.30 The new open-loop transfer function with KP = 101 is
101(s + 3)(s + 0.0578)(s + 3.2)(s + 0.01)(s + 11.9)(s(s + 5)
=101s2 + 308.8s + 17.51
s5 + 20.11s4 + 113.8s3 + 191.5s2 + 1.904s
The new closed-loop transfer function is
101s2 + 308.8s + 17.51s5 + 20.11s4 + 113.8s3 + 292.5s2 + 310.7s + 17.51
which has the roots −13.2, −3.66± 3.25j, −0.056, and 0.206± 0.973j, which indicate thatthe system is unstable.
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12.31 The Simulink diagram is shown below. The PID controller block has the gains:KP = KI = 16, KD = 6. Set the command step to start at t = 0 and the disturbance stepto start at t = 5. You will see an overshoot of about 27% for the command response, andan undershoot of 5% for the disturbance response. In both cases the steady-state error iszero.
Figure : For Problem 12.31
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Solutions Manual c©
to accompany
System Dynamics, Second Edition
by
William J. Palm III
University of Rhode Island
Solutions to Problems in Chapter Thirteen
c©Solutions Manual Copyright 2010 The McGraw-Hill Companies. All rightsreserved. No part of this manual may be displayed, reproduced, or distributedin any form or by any means without the written permission of the publisheror used beyond the limited distribution to teachers or educators permitted byMcGraw-Hill for their individual course preparation. Any other reproductionor translation of this work is unlawful.
13.1 The equation of motion ismx + kx = ky
where y(t) = Y sin 6πt = 4× 10−3 sin 6πt. Also,
ω2n =
k
m=
5000.5
= 1000 rad/s
From (13.1.9), with ζ = 0 and r2 = (6π)2/1000,
X = Y
∣∣∣∣1
1 − r2
∣∣∣∣ = 6.2× 10−3 m or 6.2 mm
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13.2 For a period of 20 feet and a vehicle speed of v (mph), the frequency ω is
ω =(
528020
)(1
3600
)(2π) = 0.4608v rad/sec
Thus ω = 0.4608(20) = 9.216 rad/sec for v = 20 mph, and ω = 0.4608(50) = 23.04 rad/secfor v = 50 mph.
For a car weighing 2000 lb, the quarter-car mass is m = 500/32.2 slugs. Its naturalfrequency is ωn =
√k/m =
√2000/(500/32.2) = 11.35 rad/sec. Its frequency ratio at 20
mph is r = ω/ωn = 9.216/11.35 = 0.812, and at 50 mph it is r = 23.04/11.35 = 2.03. Itsdamping ratio is
ζ =360
2√
2000(500/32.2)= 1.02
Now substitute these values of r and ζ and Y = 0.03 ft into the following expressions,obtained from (13.1.9) and (13.1.12).
X = Y
√1 + 4ζ2r2
(1 − r2)2 + 4ζ2r2
Ft = r2kX
This gives the following table.
2000 lb Car(ζ = 1.02)
v (mph) r X (ft) Ft (lb)20 0.812 0.034 45.1250 2.03 0.025 203.4
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13.3 We are given m = 1500 kg, k = 20, 000 N/m, ζ = 0.04, and Y = 0.01 m. At resonance,
r =ωr
ωn=
ωn
√1 − 2ζ2
ωn= 0.998
and
Ft = r2kY
√4ζ2r2 + 1
(1 − r2)2 + 4ζ2r2= 2500 N
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13.4 The static deflection is δ = mg/k. Thus k = mg/δ = 200/0.003 = 66 667 N/m. Also,
ω = 40 Hz = 215 rad/s ωn =
√k
m=
√66 667
(200/9.81)= 57.2 rad/s
r2 =(
ω
ωn
)2
=(
25157.2
)2
= 19.26
From (13.1.9) with c = 0,X
Y=
1|1 − r2|
= 0.055
Thus 5.5% of the airframe motion is transmitted to the module.
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13.5 a) Neglect damping in the isolator, and determine its required stiffness k. From(13.1.9),
X
Y=
1|1− r2| = 0.1
which gives r2 = 11. Thus
ω2n =
ω2
r2=
[3000(2π)/60]2
11=
(314)2
11
and
k = mω2n =
232.2
(314)2
11= 556.7lb/ft
b) r = ω/ωn and ωn = 314/√
11. Thus
r1 =2500(2π)/60
314/√
11= 2.77
r2 =3500(2π)/60
314/√
11= 3.87
From (13.1.9)X
Y=
1|1 − r2|
Thus the highest percentage of motion will be transmitted at the lowest r value, whichcorresponds to 2500 rpm. For 2500 rpm,
X
Y=
1|1− r2
1|= 0.15
For 3500 rpm,X
Y=
1|1− r2
2|= 0.07
Thus at most, 15% of the crane motion will be transmitted to the module.
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13.6 We are given m = 5/32.2 slug and
r2 =(
ω
ωn
)2
=(30(2π)/60)2
32.2k/5
From (13.1.9) with ζ = 0,X
Y=
1|1− r2|
= 0.1
which gives r2 = 11. Thus(30(2π)/60)2
32.2k/5= 11
Solve for k to obtain k = 0.139 lb/ft. The transmitted force is
Ft =(r2 X
Y
)kY = 11(0.1)0.139(0.003) = 4.59× 10−4 lb
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13.7 a) From Newton’s law,mx = c(y − x) − kx
ormx + cx + kx = cy
T (s) =X(s)Y (s)
=cs
ms2 + cs + k
ThusT (jω) =
cωj
−mω2 + cωj + k=
cωj/k
1 − r2 + cωj/k
This gives
T (jω) =cωmmk j
1 − r2 + cωmj/mk=
2ζωnωj/ω2n
1 − r2 + 2ζωnωj/ω2n
=2ζrj
1 − r2 + 2ζrj
where we have used the fact that ωn =√
k/m, r = ω/ωn, and c/m = 2ζωn. The magnitudeis
X =2ζr√
(1− r2)2 + (2ζr)2Y
b) Thus
Ft = kX =2ζrk√
(1− r2)2 + (2ζr)2Y
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13.8 The displacement transmissibility with r = 2 is
X
Y=
√4ζ2r2 + 1
(1 − r2)2 + 4ζ2r2=
√1 + 16ζ2
9 + 16ζ2
This has a minimum of 1/3 when ζ = 0. This is the best choice for ζ if the velocity increases(r > 2). However, if the velocity decreases (so that r → 1, and the system approachesresonance), an non-zero value of ζ would be a better choice to limit the resonant response.
With ζ = 0 and a 20% increase in r (r = 2.4), we have X/Y = 0.21. With ζ = 0 and a20% decrease in r (r = 1.6), we have X/Y = 0.64.
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13.9 The relations areFt = r2kX
X = Y
√1 + 4ζ2r2
(1 − r2)2 + 4ζ2r2
whereζ =
c
2√
mkr =
ω
ωn
At 40 mph, the forcing frequency is
ω =(
528020
)(1
3600
)2π(4) = 18.432 rad/sec
and r = 18.432/ωn, where
ωn =
√k
m
To minimize Ft, choose either
1. r small (with ζ near 1 and ωn large; this means k large and c large), or
2. r > 2 (this means ωn small).
For example, with m = 800/32.2, choosing ζ = 1 and r = 0.5, implies that
ωn = 2(18.432) = 36.864 k = m(36.864)2 = 33 763 lb/ft
c = 2√
mkζ = 1832 lb sec/ft
This gives
X = Y
√4ζ2r2 + 1
(1 − r2)2 + 4ζ2r2= 0.0589 ft
andFt = r2kX = 497 lb
(continued on the next page)
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Problem 13.9 continued:
On the other hand, choosing ζ = 1 and r = 3 for example, we obtain
k = mω2n =
80032.2
(18.432
3
)2
= 937.9 lb/ft
c = 2√
80032.2
937.9 = 305 lb sec/ft
This gives
X = Y
√4ζ2r2 + 1
(1 − r2)2 + 4ζ2r2= 0.0317 ft
andFt = r2kX = 268 lb
Note that the plot in Figure 13.1.3 is a plot of Ft/kY . Thus the curves in the plot mustbe interpreted with the value of k kept in mind. Even though the curve for r = 0.5 is lowerthat the curve for r = 3, for ζ = 1, the corresponding values of k are different, and thus thecase r = 3 gives a lower value of Ft.
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13.10 For the lumped-mass equivalent system, we have the following spring constant andequivalent mass (after converting inches to feet):
k =Ewh3
4L3=
(4.32× 109)(0.333)(0.03125)3
4(0.5)3
= 8.78× 104 lb/ft
We compute the equivalent mass of the beam as follows. Using 15.2 slug/ft3 for the densityof steel, and including 23% of the beam’s mass, we obtain
me =20
32.17+ 0.23(15.2)(0.333)(0.5)(0.03125) = 0.640 slug
The unbalanced mass is m = 1/32.17 = 0.0311 slug.The model for the system is
mex + cx + kx = f(t) = muεω2 sin ωt
This gives the transfer function
T (s) =X(s)F (s)
=1
mes2 + cs + k
=1/k
mek s2 + c
ks + 1
Thus,
T (jω) =1/k
1 −(
ωωn
)2+ 2ζω
ωnj
where 1/k = 1.139×10−5 and ωn =√
k/me = 370 rad/sec = 3537 rpm. The log magnituderatio is.
m(ω) = 20 log(1/k)− 10 log
(
1 − ω2
ω2n
)2
+(
2ζω
ωn
)2
The motor speed of 1750 rpm gives a forcing frequency of ω = 183 rad/sec. Usingζ = 0.1 and ω/ωn = 183/370, we see that m = 2.36 − 98.9 = −96.5 db, which correspondsto a magnitude ratio of 1.5 × 10−5. The amplitude of the forcing function is muεω2 =0.0311(0.01)(183)2 = 10.4 lb. Thus, the steady-state amplitude is 10.4(1.5 × 10−5) =15.6×10−5 ft. On the downward oscillation, the total amplitude as measured from horizontalis the preceding value plus the static deflection, or 15.6× 10−5 +20/87800 = 3.84× 10−4 ft.(continued on the next page)
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Problem 13.10 continued:
Because ω is not close to ωn in this problem, the preceding results are not very sensitiveto the assumed value of ζ = 0.1. For example, the calculated amplitudes of vibration forζ = 0.05 and 0.2 are close to the amplitude for ζ = 0.1 (the amplitudes are 15.7 × 10−5
ft and 15.2 × 10−5 ft, respectively). However, if we had used a motor with a speed of3500 rpm = 366 rad/sec, this choice would put the forcing frequency very close to thenatural frequency. In this region, the assumed value of ζ would be critical in the amplitudecalculation. In practice, such a design would be avoided. That is, in vibration analysis,the most important quantity to know is the natural frequency ωn. If the damping is slight,the resonant frequency is near ωn. If ωn is designed so that it is not close to the forcingfrequency, it is often not necessary to know the precise amount of damping.
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13.11 We are given that the natural frequency is ωn1 = 900 rpm. Thus, at 1750 rpm,
r1 =1750900
= 1.94
If we decrease the stiffness k by 1/2, then the new natural frequency will be
ωn2 =
√k/2m
=1√2ωn1 =
900√2
rpm
Therfore,
r2 =1750
900/√
2= 2.75
From the rotating unbalance equation (13.2.6) with ζ = 0,
X =muε
m
r2
|1− r2| (1)
We are given that at r1, X1 = 8 mm. We need to compute X2. From (1),
X2
X1=(
r2
r1
)2 1 − r21
1 − r22
= 2−2.7636−6.5625
= 0.842
ThusX2 = 0.842X1 = 6.73 mm
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13.12 We are given that
k =500
0.25/12= 24, 000 lb/ft
Thus
ωn1 =
√k
m=
√24, 000
500/32.2= 39.31 rad/sec
From the rotating unbalance equation (13.2.6) with ζ = 0,
X =muε
m
r2
|1 − r2|
and thusX2
X1=(
r2
r1
)2 1 − r21
1 − r22
wherer1 =
1750(2π)/6039.31
= 4.66
After the block is added, the new mass is m2 = 4m1 and the new natural frequency is
ωn2 =
√k
4m1=
12ωn1
Therefore,r2 =
ω
ωn2= 2
ω
ωn1= 2r1
Thus,X2
X1= (2)2
1 − r21
1 − r22
= 4−20.7156−85.8624
= 0.9651
and given that X1 = 0.1 in, we have
X2 = 0.9651(0.1) = 0.0965 in
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13.13 The transmitted force is given by
Ft = muεω2
√1 + 4ζ2r2
(1− r2)2 + 4ζ2r2
We are given that mu = 0.05/32.2 = 0.00155 slug, m = 50/32.2 = 1.55 slug. R = 0.1/12 =0.0083 ft, and ω = 1000(2π)/60 = 104.7 rad/sec. Thus
r =ω
ωn=
104.7√500/1.55
= 5.83
Thus
Ft = (0.00155)(0.0083)(104.7)2√
1 + 136ζ2
1088 + 136ζ2= 0.141
√1 + 136ζ2
1088 + 136ζ2
a) For ζ = 0.05, Ft = 0.0049 lb.b) For ζ = 0.7, Ft = 0.034 lb.
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13.14 We haveFt = muεω2Tr
where Ft = 15, ω = 200(2π)/60 = 20.9 rad/s, and because c = 0,
Tr =1
r2 − 1
We haver =
ω
ωn=
20.9√k/M
=20.9√
2500/75= 3.62
Thus Tr = 0.0826 andFt = muε(20.9)2(0.0826) = 15
Solve for muε to obtain muε = 0.416 kg m.
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13.15 We want Tr = 0.1. Neglecting damping, we have from (13.2.12)
r2 =1 + Tr
Tr= 11
Butr =
3000(2π)/60ωn
=314ωn
=√
11
Thus ωn = 314/√
11 = 94.7. But
ωn =
√k
m=
√k
3
from which we obtain k = 3ω2n = 3(94.7)2 = 26 904 N/m.
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13.16 Damping is assumed to be negligible. Thus for the vertical motion,
Ft
F= Tr =
1r2 − 1
wherer =
ω
ωn=
ω√4k/m
=1750(2π)/60√
8000/25= 10.24
Thus Tr = 0.0096.The equation of motion for rotation is
Iθ = −(4kRθ)D/2
Thus the natural frequency for rotation is ωn =√
4kD2/4I =√
80/0.2 = 20. For therotational motion,
Tt
T= Tr =
1r2 − 1
wherer =
ω
ωn=
1750(2π)/6020
= 9.16
Thus Tr = 0.012.
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13.17 We are given m = 20/g slug, mu = 1/g slug, ε = 0.01 ft, and ω = 3500(2π)/60 =366.5 rad/sec. Thus the unbalance force amplitude is
muεω2 =(
132.2
)(0.01)(366.5)2 = 41.719 lb
The beam stiffness is
k =Ewh3
4L3=
4.32× 109(1/3)(3/96)3
4(1/2)3= 8.79× 104 lb/ft
The beam mass is m = ρV = 15.2(1/3)(3/96)(1/2) = 0.079 slug.The first design equation for the absorber is r2 = 1, or
r2 =ω
ωn2=
366.5√k2/m2
= 1
This implies that k2/m2 = (366.5)2. The second design equation is X2 = 0.25/12 = 1/48ft, where
X2 =1k2
F =41.719
k2
Thus k2 = 41.719(48) = 2002.5 lb/ft. Substitute this value into the first design equation toobtain
m2 =k2
(366.5)22002.5
(366.5)2= 0.0149 slug
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13.18 We are given that X2 ≤ 0.08/12 ft, ω = 6000(2π)/60 = 628.3 rad/sec, and that theunbalance force amplitude is muεω2 = 60 lb. Thus
muε =60ω2
=60
628.3= 0.0955
The first design equation for the absorber is
k2 =F
X2=
600.08/12
= 9000 lb/ft
The second design equation is √k2
m2= ω = 628.3
Thusm2 =
k2
(628.3)2= 0.023 slug
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13.19 We are given that ω = 200(2π)/60 = 20.94 rad/sec, the amplitude of the unbalanceforce is muεω2 = 4 lb, and that X2 ≤ 1/12 ft. Assume that the table legs are rigid.
The first design equation for the absorber is√
k2
m2= ω = 20.94
The second design equation is
k2 =F
X2=
41/12
= 48 lb/ft
Thusm2 =
48(20.94)2
= 0.109 slug
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13.20 The machine mass is m1 = 8 kg. a) We are given that ωn = 2π(6) = 12π rad/sec,muεω2 = 50 N, ω = 4(2π) = 8π rad/sec, and X2 ≤ 0.1 m. The design equation for theabsorber is √
k2
m2= ω = 8π
Thus k2 = 50/0.1 = 500 N/m, and
m2 =k2
(8π)2=
500(8π)2
= 0.792 kg
b) We have that k1 = m1ω2n = (12π)2m1 = 1421m1 = 11368 N/m. From (13.3.7),
T1(jω) =X1(jω)F (jω)
=1
11368
∣∣∣∣∣1 − r4
2b2r4
2 − [1 + (1 + µ)b2] r22 + 1
∣∣∣∣∣
whereb =
ωn2
ωn1=
8π
12π=
23
µ =m2
m1=
0.7928
= 6.336
1 + (1 + µ)b2 =139
+3.168
8= 1.84
The amplitude of F (jω) is muεω2, where r2 = ωωn2 = ω/8π. Thus
X1
muε=
111368
∣∣∣∣∣ω2(1 − r2
2
)
49r4
2 − 1.84r22 + 1
∣∣∣∣∣
orX1
muε=
111368
∣∣∣∣∣∣
ω2(1 − ω2
64π2
)
[49
ω4
4096π4 − 1.84 ω2
64π2 + 1]
∣∣∣∣∣∣
The plot is shown in the figure on the following page.(continued on the next page)
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Problem 13.20 continued:
0 10 20 30 40 50 60 70 80 90 1000
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Forcing frequency ω (rad/s)
X1/m
uε (s
lug−1
)
Figure : for Problem 13.20.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
13.21 An absorber tuned to 2100 rpm requires that√
k2
m2=
√k1
m1=
2100(2π)60
= 219.9 rad/sec
Because we are given that m2 = 5/32.2 = 0.155, we have that k2 = (219.9)2m2 = 9747.8.Also,
k1 = (219.9)2m1 (1)
The characteristic equation of the combined system is given by the denominator of (13.3.1),and is
m1m2s4 + (m2k1 + m2k2 + m1k2)s2 + k1k2 = 0 (1)
One of the observed resonances of the combined system is 2850 rpm, or 2850(2π)/60 =298.45 rad/sec. Thus substituting s = 298.45j and relation (1) into (2), along with thevalues m2 = 0.155 and k2 = 9747, we can determine the value of m1, which is m1 = 0.4slug. Thus gives k1 = 19, 344 lb/ft.
With m1 and k1 determined, we now calculate the required values of m2 and k2. Supposewe choose to put the resonances just outside the operating range, say at 1400 and 3100 rpm(146.6 and 324.6 rad/sec). Let λ = s2, and let λ1 and λ2 denote the desired values of s2.We can factor the characteristic equation (2) as follows:
m1m2(λ− λ1)(λ− λ2) = m1m2λ2 − m1m2(λ1 + λ2)λ + m1m2λ1λ2 = 0 (3)
Comparing the coefficients of (3) with (2), we see that
(m2k1 + m2k2 + m1k2 = −m1m2(λ1 + λ2) (4)
andm1m2λ1λ2 = k1k2 (5)
We can solve (5) for k2 as follows.
k2 =m1m2λ1λ2
k1= Dm2 (6)
whereD =
m1λ1λ2
k1
(continued on the next page)
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Problem 13.21 continued:
Substitute (6) into (4) and solve for m2.
m2 = −m1(λ1 + λ2) + k1 + m1D
D
The desired values are λ1 = −(146.6)2 and λ2 = −(324.6)2. These give the absorber valuesm2 = 0.2706 slug and k2 = 12, 672 lb/ft.
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13.22 a) From Newton’s law,
m1x1 = f − kx1 + c(x2 − x1)
m2x2 = −c(x2 − x1)
Transform these equations with zero initial conditions to obtain
(m1s2 + cs + k)X1(s) − csX2(s) = F (s)
−csX1(s) + (m2s2 + cs)X2(s) = 0
The solutions obtained with Cramer’s rule are
X1(s) =m2s
2 + cs
D(s)F (s)
X2(s) =cs
D(s)F (s)
where Cramer’s determinant is
D(s) = s(m1m2s3 + c(m1 + m2)s2 + km2s + ck)
Define the following parameters:
µ =m2
m1ω2
1 =k
m1
ζ =c
2√
m1kr =
ω
ω1
Then D(jω) can be written as
D(jω) =∣∣∣ωj[−m1m2ω
3j − c(m1 + m2)ω2 + km2ωj + kc]∣∣∣
orD(jω) = m2
1ω41r√
(2ζr)2[1 − (1 + µ)r2]2 + µ2r2(1 − r2)2
(continued on the next page)
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Problem 13.22 continued:
The numerator of kX1(jω)/F (jω) is | − m2ω2 + cωj| = m1rω
21
√4ζ2 + µ2r2. The nu-
merator of X2(jω)/F (jω) is |cωj|. Thus,
kX(jω)F (jω)
=√
4ζ2 + µ2r2
√(2ζ)2[1− (1 + µ)r2]2 + µ2r2(1− r2)2
andX2(jω)F (jω)
=1ω2
1
2ζ√(2ζ)2[1− (1 + µ)r2]2 + µ2r2(1− r2)2
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13.23 a) From Newton’s law,
m1x1 = f − kx1 + k2(x2 − x1) + c(x2 − x1)
m2x2 = −k2(x2 − x1) − c(x2 − x1)
Transform these equations with zero initial conditions to obtain
(m1s2 + cs + k1 + k2)X1(s)− (cs + k2)X2(s) = F (s)
−(cs + k2)X1(s) + (m2s2 + cs + k2)X2(s) = 0
The solutions obtained with Cramer’s rule are
X1(s) =m2s
2 + cs + k2
D(s)F (s)
X2(s) =cs + k2
D(s)F (s)
where Cramer’s determinant is
D(s) = m1m2s4 + c(m1 + m2)s3 + (m2k1 + m2k2 + m1k2)s2 + ck1s + k1k2
Define the following parameters:
µ =m2
m1ω2
1 =k1
m1ω2
2 =k2
m2
α =ω1
ω2r =
ω
ω1
ζ =c
2√
m1k1λ =
k2
k1
Then D(jω) can be written as
D(jω) =∣∣∣m1m2ω
4 − (m2k1 + m2k2 + m1k2)ω2 + k1k2 + [ck1ω − (m1 + m2)cω3]j∣∣∣
orD(jω) =
∣∣∣m21ω
4[µr4 − (1 + λ + µλ)r2 + µα2] + [2ζr − 2(1 + µ)r3]j∣∣∣
(continued on the next page)
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Problem 13.23 continued:
The numerator of k1X1(jω)/F (jω) is |k2 − m2ω2 + cωj| = k1
∣∣[λ − µr2 + 2ζri]∣∣. The
numerator of k1X2(jω)/F (jω) is |cωj + k2| = k1|(λ + 2ζrj)|. Thus,
k1X(jω)F (jω)
=√
(λ − µr2)2 + (2ζr)2√[µr4 − (1 + λ + µλ)r2 + µα2]2 + (2ζr)2 [1− (1 + µ)r2]2
andk1X2(jω)
F (jω)=
√λ2 + (2ζr)2√
[µr4 − (1 + λ + µλ)r2 + µα2]2 + (2ζr)2 [1− (1 + µ)r2]2
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13.24 The equations of motion are
m1x1 = −k1x1 − k2(x1 − x2)
m2x2 = k2(x1 − x2)
From these equations we can write the modal amplitude equations by substituting x1 =A1e
st and x2 = A2est, and using the given parameter values. The result is
(10s2 + 30, 000)A1 − 20, 000A2 = 0
−20, 000A1 + (30, 000s2 + 20, 000)A2 = 0
These give the solution
A2 =s2 + 3000
2000A1 (1)
The roots are found from Cramer’s determinant of the modal equations, which is
3s4 + 11, 000s2 + 2 × 106 = 0
The roots are s2 = −3475 and s2 = −192. Substitute s2 = −3475 into (1) to obtainA2 = −0.2375A1. Substitute s2 = −192 into (1) to obtain A2 = 1.404A1.
In the first mode, the masses oscillate in opposite directions with a radian frequencyof
√3475. The displacement amplitude of mass 2 is 0.2375 times that of mass 1. In the
second mode, the masses oscillate in the same direction with a radian frequency of√
192.The displacement amplitude of mass 2 is 1.404 times that of mass 1.
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13.25 The equations of motion are
m1L2θ1 = −mg1Lθ1 − kd(dθ1 − dθ2)
m2L2θ2 = −m2gLθ2 + kd(dθ1 − dθ2)
From these equations we can write the modal amplitude equations by substituting θ1 =A1e
st and θ2 = A2est. The result is
(m1L2s2 + m1gL + kd2)A1 − kd2A2 = 0
−kd2A1 + (m2L2s2 + m2gL + kd2)A2 = 0
Using the given parameter values and g = 9.81 m/s2, these equations give the solution
A2 = (3.13s2 + 7.13)A1 (1)
The roots are found from Cramer’s determinant of the modal equations, which is
2500s4 + 10810s2 + 11 586 = 0
The roots are s2 = −1.96 and s2 = −2.36. Substitute s2 = −1.96 into (1) to obtainA2 = 0.995A1. Substitute s2 = −2.36 into (1) to obtain A2 = −0.257A1.
In the first mode, the masses oscillate in the same direction with a radian frequency of√1.96. The displacement amplitude of mass 2 is 0.995 times that of mass 1. In the second
mode, the masses oscillate in the opposite direction with a radian frequency of√
2.36. Thedisplacement amplitude of mass 2 is 0.257 times that of mass 1.
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13.26 The equations of motion are
I1θ1 = k2(θ2 − θ1) − k1θ1
I2θ2 = −k2(θ2 − θ1)
From these equations we can write the modal amplitude equations by substituting θ1 =A1e
st and θ2 = A2est. The result is
(I1s2 + k1 + k2)A1 − k2A2 = 0
−k2A1 + (I2s2 + k2)A2 = 0
Using the given parameter values, these equations give
A2 =I1s
2 + k1 + k2
k2θ1 =
s2 + 43
A1 (1)
The roots are found from Cramer’s determinant of the modal equations, which is
5s4 + 23s2 + 3 = 0
The roots are s2 = −0.134 and s2 = −4.47. Substitute s2 = −0.134 into (1) to obtainA2 = 1.29A1. Substitute s2 = −4.47 into (1) to obtain A2 = −0.157A1.
In the first mode, the masses oscillate in the same direction with a radian frequency of√0.134. The displacement amplitude of mass 2 is 1.29 times that of mass 1. In the second
mode, the masses oscillate in the opposite direction with a radian frequency of√
4.47. Thedisplacement amplitude of mass 2 is 0.157 times that of mass 1.
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13.27 The equations of motion are
mx = −2k sin 45 x = −1.41kx
my = −ky − 2k sin 45 x = −2.41kx
In the first mode, the mass oscillates in the x direction with a radian frequency of√
1.41k/m.In the second mode, the mass oscillates in the y direction with a radian frequency of√
2.41k/m.
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13.28 From (13.4.8):
(750)(1350)s4 +730[(1.95× 104)(1.5)2 + 2.3× 104(1.1)2] + 1350(4.25× 104)
s2
+ 1.95(2.3)× 108(2.6) = 0
ors4 + 111.329s2 + 1.166× 104 = 0
This givess2 = −99.43 and s2 = −11.9
ors = ±9.971j and s = ±3.45j
These correspond to frequencies of 1.587 Hz and 0.549 Hz.
A1
A2=
x
θ=
k1L1 − k2L2
ms2 + k1 + k2=
3.95× 103
730s2 + 4.25× 104=
5.411s2 + 58.2192
For mode 1 (s2 = −99.43),
x
θ= −0.131 m ahead of the mass center
For mode 2 (s2 = −11.9),
x
θ= 0.1168 m behind the mass center
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13.29 Referring to Figure 13.4.4, we are given that m1g = 1000 lb, c1 = 0, and k2 = 1300lb/in. The ride rate should be
ke =m1g
∆=
10009.8
= 102 lb/in
Thus the suspension stiffness should be
k1 =kek2
k2 − ke=
102(1300)1300− 102
= 111 lb/in
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13.30 Define the following (refer to Figure 13.4.2):k1 = rear quarter-car suspension stiffness.k2 = front quarter-car suspension stiffness.kr = total rear suspension stiffness = 2k1.kf = total front suspension stiffness = 2k2.ker = total rear ride rate (including suspension and tire stiffness).kef = total front ride rate (including suspension and tire stiffness).ke1 =quarter-car rear ride rate.ke2 =quarter-car front ride rate.kt = individual tire stiffness.Referring to the guidelines on page 851, and using equations (4) and (5) of Example
13.4.3 as approximations for the bounce and pitch dynamics, we have
kef = 0.7ker (1)
ωbounce
2π=
12π
√kf + kr
m≤ 1.3 Hz (2)
ωpitch
2π=
12π
√kfL2
1 + krL22
IG≤ 1.3 Hz (3)
ke1 =(m/4)g
∆1(4a)
ke2 =(m/4)g
∆2(4b)
From the above definitions,
kef = 2ke2 ker = 2ke1 (5)
Because the tire stiffness is in parallel with the suspension stiffness,
ke1 =k1kt
k1 + ktke2 =
k2kt
k2 + kt(6)
(continued on the next page)
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Problem 13.30 continued:
These relations must be satisfied by k1, k2, and kt. Using the given values m =4800/32.2, IG = 1800, L1 = 3.5, and L2 = 2.5, these equations can be rearranged asfollows:
0.02019√
k1 + k2 ≤ 1.3 (7)
0.0375√
24.5k1 + 12.5k2 ≤ 1.3 (8)
k1 =1200∆1
kt
kt − 1200∆1
(9)
k2 =1200∆2
kt
kt − 1200∆2
(10)
∆1 = 0.7∆2 (11)
The procedure is to select suitable values for kt and ∆2, solve (9) and (10) for k1 and k2,and see if (7) and (8) are satisfied. Trying ∆2 = 9.8/12 ft, as suggested on page 851, andkt = 1200(12) = 14400 lb/ft, we obtain from (9) and (10) k1 = 2457 lb/ft and k2 = 1636lb/ft. With these values, the left-hand sides of (7) and (8) are 1.29 and 1.07. Thus therequirements have been met.
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Problem 13.31 The equation of motion with the controller is
125x + (5000 + KD)x + (7 × 106 + KP )x = 100 sin ωt
where ω = 2500(2π)/60 = 261.8. It is desired that
ωn = 100 =
√7 × 106 + KP
125
which gives KP = (1.25)2 × 108 − 7 × 106 = 1.4925× 108 N/m.Also
ζ =5000 + KD
2√
125[(125)2 × 108]= 0.5
This gives KD = 6.25√
5 × 104 − 5000 = 1.3475× 105 N·s/m.Substituting these values into the equation of motion and simplifying, we obtain
x + 500√
5x + 1.25× 106x = 0.8 sin ωt
The steady state amplitude is
X(ω) =0.8∣∣∣−ω2 + 1.25× 106 + 500
√5ωi
∣∣∣
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Problem 13.32 From the given information, the stiffness is
k = mω2n = 20(13.2)2 = 3485
and the damping isc = 2(0.28)
√20(3485) = 147.8
It is desired that ωn = 141. Thus
ωn = 141 =
√k + KP
20
andKP = (141)220− k = 3.9414× 105 N/m
The damping ratio is
ζ =c + KD
2√
20(k + KP )= 0.707
which givesKD = 2(0.707)
√20(k + KP ) = 3987 N · s/m
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Problem 13.33 Assuming that xs, xw, and xr are measured from the equilibrium positions,the equation of motion for the sprung mass is
msxs = c(xw − xs) + ks(xw − xs) + f
where f is the actuator force. Since we are neglecting the tire mass, a force balance at thatpoint gives
mwxw = 0 = kt(xr − xw)− ks(xw − xs)− c(xw − xs) − f
The actuator force isf = −Kp(xs − xw) − KD(xs − xw)
Taking the Laplace transform of these equations, using zero initial conditions, eliminat-ing the variables f and xw , and reverting back to the time domain, we obtain
ms(c + KD)d3xs
dt3+ ms(ks + kt + KP )
d2xs
dt2+ kt(c + KD)
dxs
dt+ kt(ks + KP )xs
= kt(c + KD)dxr
dt+ kt(ks + KP )xr
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Problem 13.34 Set the derivatives to zero to find the equilibrium solution ye.
1700y3 + 900y − 5g = 0
which has the roots y = 0.0542 and s = −0.027 ± 0.729i. Discard the complex valuedsolution. Near y = 0.0542,
y3 ≈ (0.0542)3 + 3y2e (y − ye) = (0.0542)2 + 0.00881x
where x = y − 0.0542. The linearized model is 5x = −900x − 1200(0.000881)x, or 5x +914.977x = 0. The roots are s = ±13.5i, and the natural frequency is ωn = 13.5 rad/sec.
b) Because the roots are s = ±13.5i, the solution of 5x + 914.977x = 0 has the formx(t) = A sin 13.5t + B cos 13.5t. Note that x(0) = B = 0.002 and x(0) = 13.5A = 0.005.Thus A = 3.704× 10−4, and
x(t) = 3.704× 10−4 sin 13.5t + 0.002 cos 13.5t
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Problem 13.35dv
dt= − c
mv3
∫ v(t)
v(0)
dv
v3= − c
m
∫ t
0dt = − c
mt
− v−2
2
∣∣∣∣∣
v(t)
v(0)
= − c
mt
v−2(t) − v−2(0) = 2c
mt
Thus
v(t) = v(0)
√1
1 + 2cv2(0)t/m
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Problem 13.36 The equilibria are the roots of
16ye − 4y3e = 0
and are ye = 0 and ye = ±2.The linearized model is
x + 12x + 16x − 12(ye)2x = 0
For ye = 0,x + 12x + 16x = 0
which is locally stable.For ye = ±2,
x + 12x− 32x = 0
which is locally unstable.The state variable form of the model is
y1 = y2
y2 = −16y1 + 4y31 − 12y2
where y1 = y and y2 = ˙ y.Create the following function file.
function ydot = prob13p36(t,y)ydot = [y(2); -16*y(1)+4*y(1)^3-12*y(2)];
For y(0) = 1, type the following in MATLAB.
y0 = 1;[t, y] = ode45(′prob13p36′, [0, 1], [y0, 0]);plot(t, y(:,1))
Do the same for the other initial conditions. The simulations show that for y(0) = ±2.1,the model is unstable (|y(t)| → ∞). For y(0) = ±1 and y(0) = ±1.9, y(t) → 0. Thus theequilibria at ye = ±2 are globally unstable, and the equilibrium at ye = ±0 is locally stable.
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Problem 13.37 The state variable form of the model is
y1 = y2
y2 = −2y1 − 0.1y31
where y1 = y and y2 = ˙ y.Create the following function file.
function ydot = prob13p37(t,y)ydot = [y(2); -2*y(1)-0.1*y(1)^3];
For y(0) = 10, type the following in MATLAB.
y0 = 10;[t, y] = ode45(′prob13p37′, [0, 10], [y0, 0]);plot(t, y(:,1))
Do the same for the other initial condition. The simulations show that for y(0) = 10, theresponse oscillates with a frequency of about 3 rad/s. For y(0) = 40, the response oscillateswith a frequency of about 11 rad/s. So the oscillation frequency depends greatly on theinitial condition.
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Problem 13.38 The three cases are of the form
x + cx + 2x = 0
The state variable form isx1 = x2
x2 = −2x1 − cx2
In MATLAB create the following file.
function xdot = prob13p38(t,y)global cxdot = [x(2); -2*x(1)-c*x(2)];
For part (a), where c = 0.1, type the following in MATLAB.
global cc = 0.1;[t, x] = ode45(′prob13p38′, [0, 50], [1, 0]);plot(x(:,1), x(:,2))
The phase plane plot is a tight spiral ending at (0, 0).For parts (b) and (c) change the second line to c = 2 and c = 4, respectively. The
phase plane plots head almost directly toward the point (0, 0) without much of a spiral.
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Problem 13.39 The state variable form of the model is
y1 = y2
y2 = −2y1 − 3y31 − 2y2
where y1 = y and y2 = ˙ y.Create the following function file.
function ydot = prob13p39(t,y)ydot = [y(2); -2*y(1)-3*y(1)^3-2*y(2)];
Then type the following in MATLAB.
[t, y] = ode45(′prob13p39′, [0, 10], [1, 0]);plot(y(:,1), y(:,2))
The phase plane plot heads almost directly toward the point (0, 0) without much of aspiral.
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Problem 13.40 The state variable form of the model is
y1 = y2
y2 = −y1 + 5(1− y21)y2
where y1 = y and y2 = ˙ y.Create the following function file.
function ydot = prob13p40(t,y)ydot = [y(2); -y(1)+5*(1-y(1)^2)*y(2)];
Then type the following in MATLAB.
[t, y] = ode45(′prob13p40′, [0, 50], [1, 0]);plot(y(:,1), y(:,2))
The phase plane plot is a limit cycle.
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Solutions Manual c©
to accompany
System Dynamics, Second Edition
by
William J. Palm III
University of Rhode Island
Solutions to Problems in Appendix C
c©Solutions Manual Copyright 2010 The McGraw-Hill Companies. All rightsreserved. No part of this manual may be displayed, reproduced, or distributedin any form or by any means without the written permission of the publisheror used beyond the limited distribution to teachers or educators permitted byMcGraw-Hill for their individual course preparation. Any other reproductionor translation of this work is unlawful.
C.1 The session is:
a = 1.12; b = 2.34; c = 0.72;d = 0.81;f = 19.83;x = 1 + a/b + c/f^2x =
1.4805s = (b-a)/(d-c)s =
13.5556r = 1/(1/a + 1/b + 1/c + 1/d)r =
0.2536y = a*b/c*f^2/2y =
715.6766
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C.2 The session is
x = -7-5i;y = 4+3i;x+yans =-3.0000 - 2.0000ix*yans =-13.0000 -41.0000ix/yans =-1.7200 + 0.0400i(3/2)*jans =0 + 1.5000i3/(2j)ans =0 - 1.5000i
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C.3 The session is:
x=[1:0.2:5];y = 7*sin(4*x);length(y)ans =
21y(3)ans =
-4.4189
There are 21 elements. The third element is −4.4189.
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C.4 The session is
wage = [5,5.5,6.5,6,6.25]; hours = [40,43,37,50,45];output = [1000,1100,1000,1200,1100];earnings = wage.*hoursearnings =
200.0000 236.5000 240.5000 300.0000 281.2500total_salary = sum(earnings)total_salary =
1.2582e+003total_widgets = sum(output)total_widgets =
5400average_cost = total_salary/total_widgetsaverage_cost =
0.2330average_hours = sum(hours)/total_widgetsaverage_hours =
0.0398[maximum,most_efficient] = max(output./earnings)maximum =
5most_efficient =
1[minimum,least_efficient] = min(output./earnings)minimum =
3.9111least_efficient =
5
The workers earned $200, $236.50, $240.50, $300, and $281.25 respectively. The total salarypaid out was $1258.20, and 5400 widgets were made. The average cost to produce one widgetwas 23.3 cents, and it took an average of 0.0398 hr to produce one widget. The first worker,who produced 5 widgets per dollar of earnings, was the most efficient. The fifth worker,who produced 3.911 widgets per dollar of earnings, was the least efficient.
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C.5 The MATLAB expressions are:
f = 1./sqrt(2*pi*c./x)E = (x + w./(y + z))./(x + w./(y - z))A = exp(-c./(2*x))./(log(y).*sqrt(d*z))S = x.*(2.15 + 0.35*y).^1.8./(z.*(1-x).^y)
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C.6 For part (a) note that the first quarter material cost is computed by
7(16) + 3(12) + 9(8) + 2(14) + 5(13) = 326
and the second quarter material cost is computed by
7(14) + 3(15) + 9(9) + 2(13) + 6(16) = 346
and so on. Thus the quarterly costs can be computed by multiplying the transpose of thematrix of unit costs by the matrix of quarterly production volumes. The resulting 3 × 4matrix is quarterly_costs. Its first row contains the material costs, its second row containsthe labor costs, and the third row contains the transportation costs. The four columnsof quarterly_costs correspond to the four quarters. For part (b) the yearly costs formaterials, labor, and transportation are found by summing the rows of quarterly_costs,or equivalently, by summing the columns of the transpose of quarterly_costs. For part(c) the total quarterly costs are found by summing the columns of quarterly_costs. Thesession is
unit_cost = [7,3,2;3,1,3;9,4,5;2,5,4;6,2,1];quarterly_volume = [16,14,10,12;12,15,11,13;8,9,7,11;14,13,15,17;13,16,12,18];quarterly_costs = unit_cost′*quarterly_volumequarterly_costs =
326 346 268 364188 190 168 214177 186 160 204
yearly_costs = sum(quarterly_costs′)yearly_costs =
1304 760 727total_quarter_cost = sum(quarterly_costs)total_quarter_cost =
691 722 596 782
(continued on the next page)
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Problem C.6 continued:
The table C.6 was created from the matrix quarterly_costs. All costs are in thousandsof dollars.
Table C.6Quarterly Costs
Category Quarter 1 Quarter 2 Quarter 3 Quarter 4Materials 326 346 268 364
Labor 188 190 168 214Transportation 177 186 160 204
From the vector yearly_costs we obtain the following information: yearly materialscost = $1,304,000, yearly labor cost = $760,000, and yearly transportation cost = $727,000.From the vector total_quarter_cost we find that the total costs in each quarter are$691,000, $722,000, $596,000, and $782,000 respectively.
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C.7 The session is
roots([13,182,-184,2503])ans =
-15.68500.8425 + 3.4008i0.8425 - 3.4008i
poly(ans)ans =
1.0000 14.0000 -14.1538 192.538513*ans
ans =1.0e+003 *0.0130 0.1820 -0.1840 2.5030
This gives the original coefficients.
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C.8 The session is
roots([36,12,-5,10])ans =
-0.86510.2659 + 0.5004i0.2659 - 0.5004i
polyval([36,12,-5,10],ans)ans =
1.0e-013 *0.01780.1243 + 0.0178i0.1243 - 0.0178i
Since the last result is essentially zero, the polynomial evaluates to zero, showing that theroots are correct.
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C.9 The session is
poly([3+6j,3-6j,8,8,20])ans =
1 -42 645 -5204 24960 -57600roots(ans)ans =
20.00003.0000 + 6.0000i3.0000 - 6.0000i8.00008.0000
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
C.10 The session is
conv([10,-9,-6,12],[5,-4,-12,8])ans =
50 -85 -114 272 -48 -192 96
The answer is 50s6 − 85s5 − 114s4 + 272s3 − 48s2 − 192s + 96.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
C.11 The session is
[q,r] = deconv([14,-6,3,9],[5,7,-4])q =
2.8000 -5.1200r =
0.0000 0.0000 50.0400 -11.4800
The answer is 2.8s− 5.12 with a remainder of 50.04s− 11.48.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
C.12 The session is:
x = [0:0.01:2];u = 2*log10(60*x+1);v = 3*cos(6*x);plot(x,u,x,v,′--′),ylabel(′Speed (mi/hr)′),...xlabel(′Distance x (mi)′)
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
C.13 The session is
p1 = [3,-6,8,4,90];p2 = [3, 5, -8, 70];x=[-3:0.01:3];y = polyval(p1,x);z = polyval(p2,x);plot(x,y,x,z,′--′),xlabel(′x′),ylabel(′y and z′),gtext(′y′),gtext(′z′)
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
C.14 The script file is:
x = -5;if x < -1
y = exp(x + 1)elseif x < 5
y = 2 + cos(pi*x)else
y = 10*(x - 5) + 1end
The answer for x = −5 is y = 0.0183. Change the first line to x = 3 to obtain y = 1. Thenchange the first line to x = 15 to obtain y = 101.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
C.15 a) z = 0 1 0 0 1b) z = 1 0 1 0 0c) z = 1 1 1 0 1d) z = 0 0 0 1 0e) z = 0 1 0 1 1
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
C.16 (a) The session is
price_A = [19,18,22,21,25,19,17,21,27,29];price_B = [22,17,20,19,24,18,16,25,28,27];price_C = [17,13,22,23,19,17,20,21,24,28];length(find(price_A>price_B&price_A>price_C))ans =
4
Thus the price of stock A was above both B and C on four days.(b) Replace the fourth line in the above session with
length(find(price_A>price_B|price_A>price_C))ans =
9
The answer is nine days.(c) Replace the fourth line in the session in part (a) with
length(find(xor(price_A>price_B,price_A>price_C)))ans =
5
The answer is five days.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
C.17 The script file is
price = [19,18,22,21,25,19,17,21,27,29];cost_new_shares = 100*sum(price.*(price<20))income_selling = 100*sum(price.*(price>25))shares_change = 100*(sum(price<20)-sum(price>25));total_shares = 1000 + shares_changenet_increase = price(10)*total_shares - price(1)*1000
The results are: cost_new_shares = 7300, income_selling = 5600, total_shares =1200, and net_increase = 15800. Thus you spent $7300 in buying shares. You received$5600 from the sale of shares. After the tenth day you own 1200 shares. The net increasein the worth of your portfolio is $15,800.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
C.18 The script file is:
sum = 0;for k = 1:10
sum = sum + 5*k^3;endsum
The answer is sum = 15125.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
C.19 The script file is:
sum = 0;k = 0;while sum <= 2000
k = k + 1;sum = sum + 2^k;
endksum
The answers are k = 10 and sum = 2046.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
C.20 The script file is:
amt1 = 1000;amt2 = 1000;k1 = 0;k2 = 0;while amt1 < 50000
k1 = k1 + 1;amt1 = amt1*1.055 + 1000;
endwhile amt2 < 50000
k2 = k2 + 1;amt2 = amt2*1.045 + 1000;enddiff = k2 - k1
The answer is diff = 2. Thus it takes 2 more years in the second bank.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
C.21 The function is
function theta = asine(x,q)if abs(x) <= 1
if q==1theta=asin(x);
elseif q==2theta = asin(x)+pi;
elseif q==3theta = asin(x)+pi;
elsetheta = asin(x);
endtheta = theta*(180/pi)
elsedisp(’|x| >1’)
end
The function does not protect against incorrect values of the quadrant number q, but theproblem statement did not require this.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
C.22 The function file is
function y = f6(x)y = 1 + exp(-0.2*x).*sin(x+2);
You can plot the function to obtain solution estimates to use with fminbnd, or you cansimply try values of x between 0 and 10. The session is
fminbnd(′f6′,0)ans =2.5150f6(ans)ans =
0.4070fminbnd(′f6′,10)ans =
8.7982f6(ans)ans =
0.8312
So the solutions are (x, y) = (2.5150, 0.4070) and (x, y) = (8.7982, 0.8312).
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
C.23 The function file is
function t = time(h,v0,g)% Computes time t to reach a specified height h, with initial speed v0.roots([0.5*g,-v0,h])
A test session follows.
time(100,50,9.81)ans =
7.46122.7324
The smaller value is the time to reach the height while ascending; the larger value is thetime to reach the height while descending.
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
Solutions Manual c©
to accompany
System Dynamics, Second Edition
by
William J. Palm III
University of Rhode Island
Solutions to Problems in Appendix D
c©Solutions Manual Copyright 2010 The McGraw-Hill Companies. All rightsreserved. No part of this manual may be displayed, reproduced, or distributedin any form or by any means without the written permission of the publisheror used beyond the limited distribution to teachers or educators permitted byMcGraw-Hill for their individual course preparation. Any other reproductionor translation of this work is unlawful.
D.1 The closed-form solution isy(t) = 6e−5t
After five steps, t = 0.1 and the exact solution is
y(0.1) = 3.6392
c©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
D.2 The closed-form solution isy(t) = 6 + sin t
After five steps, t = 1.5 and the exact solution is
y(1.5) = 6.9975
c©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
D.3 The closed-form solution is
y(t) = 12 − 2 cos 3t
After five steps, t = 0.5 and the exact solution is
y(0.5) = 11.8585
c©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
D.4 The closed-form solution is
y(t) = 3.25− 12.5e−4t
After five steps, t = 0.125 and the exact solution is
y(0.125) = −4.3316
c©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
D.5 The closed-form solution is
y(t) =657
e−3t +57e4t
After five steps, t = 0.05 and the exact solution is
y(0.05) = 8.8647
c©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
D.6 The closed-form solution isy(t) = 6e−5t
After five steps, t = 0.1 and the exact solution is
y(0.1) = 3.6392
c©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
D.7 The closed-form solution isy(t) = 6 + sin t
After five steps, t = 1.5 and the exact solution is
y(1.5) = 6.9975
c©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
D.8 The closed-form solution is
y(t) = 12 − 2 cos 3t
After five steps, t = 0.5 and the exact solution is
y(0.5) = 11.8585
c©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
D.9 The closed-form solution is
y(t) = 3.25− 12.5e−4t
After five steps, t = 0.125 and the exact solution is
y(0.125) = −4.3316
c©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
D.10 The closed-form solution is
y(t) =657
e−3t +57e4t
After five steps, t = 0.05 and the exact solution is
y(0.05) = 8.8647
c©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
D.11 The equations of motion are given by equations (3) and (4) in Example 2.2.4. Theseequations are programmed in the following m file.
function ydot = mast(t,y)Q=sqrt(2020+1650*cos(1.33+y(1)));ydot(1) = y(2);ydot(2) = (1/25400)*(-17500*cos(y(1))+(626000./Q).*sin(1.33+y(1)));ydot = [ydot(1);ydot(2)];
This file is called as follows. Since we do not know how long it will take for the mast to reach90, we must experiment with the stop time. Since the geometry on which the equationsare based breaks down if θ ≥ 90, we start with a low value of the stop time, say 2 seconds,and gradually increase the stop time uintil the plot shows that 90 has been reached. Totwo decimal places, the answer is 8.62 seconds.
[t,y] = ode45(@mast,[0, 8.62],[pi/6,0]);plot(t,(180/pi)*y(:,1))
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
D.12 The closed-form solution isy(t) = 6e−5t
After five steps, t = 0.1 and the exact solution is
y(0.1) = 3.6392
c©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
D.13 The closed-form solution is
y(t) = 6 + sin t
After five steps, t = 1.5 and the exact solution is
y(1.5) = 6.9975
c©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
D.14 The closed-form solution is
y(t) = 3.25− 12.5e−4t
After five steps, t = 0.125 and the exact solution is
y(0.125) = −4.3316
c©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
D.15 The closed-form solution isy(t) = 6e−5t
After five steps, t = 0.1 and the exact solution is
y(0.1) = 3.6392
c©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
D.16 The closed-form solution is
y(t) = 6 + sin t
After five steps, t = 1.5 and the exact solution is
y(1.5) = 6.9975
c©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
D.17 The closed-form solution is
y(t) = 12 − 2 cos 3t
After five steps, t = 0.5 and the exact solution is
y(0.5) = 11.8585
c©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
D.18 The closed-form solution is
y(t) = 6 + sin t
After five steps, t = 1.5 and the exact solution is
y(1.5) = 6.9975
c©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
D.19 The closed-form solution is
y(t) = 12 − 2 cos 3t
After five steps, t = 0.5 and the exact solution is
y(0.5) = 11.8585
c©2010 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.