file_down

S I X

Transient Response Stability

SOLUTIONS TO CASE STUDIES CHALLENGES

Antenna Control: Stability Design via Gain

From the antenna control challenge of Chapter 5,

( ) 3 2

76.39151.32 198 76.39

KT ss s s K

=+ + +

Make a Routh table:

s3 1 198

s2 151.32 76.39K

s1 29961.36 76.39

151.32K−

0

s0 76.39K 0

From the s1 row, 392.2.K < From the s0 row, 0 K< . Therefore, 0 392.2.K< <

UFSS Vehicle: Stability Design via Gain

( )( )( )( )

( )

( )

( ) ( )( )

( )

1

12

1

2 4 3 2

3 1 2

13 4 3 2

3 14 3 2

3 1

0.125 0.437 22 1.29 0.193

10.25 0.10925

3.483 3.465 0.60719

0.25 0.109253.483 3.465 0.60719

0.25 0.109251 3.483 3.465 0.25 2.42

sG

s s s sGG

G ssG

s s s sG K G

s KG

s s s sG s s K

T sG s s s s K

− + ⋅=

+ + +

=+ −

− −=

+ + += −

+=

+ + ++

= =+ + + + +( ) 188 0.10925s K+

6-2 Chapter 6: Transient Response Stability

s4 1 3.465 0.10925K1

s3 3.483 ( )10.25 2.4288K + 0

s2 ( )11 45.844

3.483

K− − 0.10925K1 0

s1 ( )( )1 1

1

4.2141 26.420.25

45.84K K

K+ −

− 0 0

s0 0.10925K1 0 0

1For stability :0 26.42K< <

ANSWERS TO REVIEW QUESTIONS

1. Natural response

2. It grows without bound

3. It would destroy itself or hit limit stops

4. Sinusoidal inputs of the same frequency as the natural response yield unbounded responses

even though the sinusoidal input is bounded.

5. Poles must be in the left-half-plane or on the jω axis.

6. The number of poles of the closed-loop transfer function that are in the left-half-plane, the

right-half-plane, and on the jω axis.

7. If there is an even polynomial of second order and the original polynomial is of fourth order,

the original polynomial can be easily factored.

8. Just the way the arithmetic works out

9. The presence of an even polynomial that is a factor of the original polynomial

10. For the ease of finding coefficients below that row

11. It would affect the number of sign changes

12. Seven

13. No; it could have quadrantal poles.

14. None; the even polynomial has 2 right-half-plane poles and two left-half-plane poles.

15. Yes

16. ( )Det 0s − =I A

Solutions to Problems 6-3

SOLUTIONS TO PROBLEMS

1.

s5 1 5 1

s4 3 4 3

s3 3.667 0 0

s2 4 3 0

s1 2.75− 0 0

s0 3 0 0

2 rhp; 3 lhp

2.

The Routh array for ( ) 5 3 26 5 8 20P s s s s s= + + + + is:

s5 1 6 8

s4 θ ε 5 20

s3 5ε

− 20ε

−

s2 5 20

s 10θ

1 20

The auxiliary polynomial for row 4 is ( ) 25 20Q s s= + , with ( ) 10Q s′ = , so there are two roots

half-plane. The balance, one root must be in the left half-plane.

on the jω -axis. The first column shows two sign changes so there are two roots on the right


3.

s5 1 4 3

s4 1− 4− 2−

s3 ε 1 0

s2 1 4εε− 2− 0

s1 22 1 41 4

ε εε

+ −−

0 0

s0 2− 0 0

3 rhp, 2 lhp

4.

s5 1 3 2

s4 1− 3− 2−

s3 2− 3− ROZ

s2 3− 4−

s1 1/ 3−

s0 4−

( ) ( ) ( )Even 4 :4 ;Rest 1 :1rhp;Total 5 :1rhp;4j jω ω

5.

s4 1 5 6

s3 4 8 0

s2 3 6 0

s1 6 0 0 ROZ

s0 6 0 0

( ) ( )Even 2 :2 ;Rest 2 :2 lhp;Total:2 ;2 lhpj jω ω


6.

s6 1 6− 1 6−

s5 1 0 1

s4 6− 0 6−

s3 24− 0 0 ROZ

s2 ε 6−

s1 144 /ε− 0

s0 6−

( ) ( )Even 4 :2 rhp;2lhp;Rest 2 :1rhp;1lhp;Total:3rhp;3lhp

7. Program: %–det ([si() si(); sj() sj()])/sj()

%Template for use in each cell.

syms e %Construct a symbolic object for

%epsilon.

%%%%%%%%%%%%%%%%%%%%%%%%%%$$$$$$$$$%%%%%%%%%%%

s5=[1 4 3 0 0] %Create s^5 row of Routh table.

%%%%%%%%%%%%%%%%%%%%%%%%%%%$$$$$$$$$%%%%%%%%%%

s4=[–1 –4 –2 0 0] %Create s^4 row of Routh table.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

if –det([s5(1) s5(2); s4(1) s4(2)])/s4(1)= =0

s3=[e...

–det([s5(1) s5(3); s4(1) s4(3)])/s4(1) 0 0];

%Create s^3 row of Routh table

%if 1st element is 0.

else

s3=[–det([s5(1) s5(2); s4(1) s4(2)])/s4(1). . .

–det([s5(1) s5(3); s4(1) s4(3)])/s4(1) 0 0];


%if 1st element is not zero.

end

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

if –det([s4(1) s4(2); s3(1) s3(2)])/s3(1)= =0

–det([s4(1) s4(3); s3(1) s3(3)])/s3(1) 0 0]; s2=[e...




else

s2=[–det([s4(1) s4(2); s3(1) s3(2)])/s3(1) . . .

–det([s4(1) s4(3); s3(1) s3(3)])/s3(1) 0 0];


%if 1st element is not zero.

end

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

if –det([s3(1) s3(2); s2(1) s2(2)])/s2(1)= =0

s1=[e . . .

–det([s3(1) s3(3); s2(1) s2(3)])/s2(1) 0 0];



else

s1=[–det([s3(1) s3(2); s2(1) s2(2)])/s2(1) . . .

–det([s3(1) s3(3); s2(1) s2(3)])/s2(1) 0 0];


%if 1st element is not zero

end

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

s0=[–det([s2(1) s2(2); s1(1) s1(2)])/s1(1) . . .

–det([s2(1) s2(3); s1(1) s1(3)])/s1(1) 0 0];

%Create s^0 row of Routh table.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

's3' %Display label.

s3=simplify(s3); %Simplify terms in s^3 row.

pretty(s3) %Pretty print s^3 row.









pretty(s0) %Pretty print s^0 row. Computer response: s5 =

1 4 3 0 0

s4 =

–1 –4 –2 0 0

ans =

s3

[e 1 0 0]


ans =

s2

[ –1 + 4 e ]

[– -------------- –2 0 0]

[ e ]

ans =

s1

[ 2 ]

[ 2 e + 1 - 4 e ]

[- -------------- 0 0 0]

[ -1 + 4 e ]

ans =

s0

[–2 0 0 0]

8.

( ) 4 3 2

24010 35 50 264

T ss s s s

=+ + + +

s4 1 35 264

s3 10 50 0

s2 30 264 0

s1 –38 0 0

s0 264 0 0

2 rhp, 2 lhp

9.

( ) 4 2

14 4 1

T ss s

=+ +

s4 4 4 1

s3 16 8 0 ROZ

s2 2 1 0

s1 4 0 0 ROZ

s0 1 0 0

( )Even 4 :4 jω


10.

The characteristic equation is:

( )( )( )

21 0

1 3s

Ks s s

++ =

− + or

( )( ) ( )1 3 2 0s s s K s− + + + = or

( )3 22 3 2 0s s K s K+ + − + =

The Routh array is:

s3 1 3K −

s2 2 2K

s 3−

1 2K

The first column will always have a sign change regardless of the value of K. There is no

value of K that will stabilize this system.

11.

( ) 8 7 6 5 4 3 2

845 12 25 45 50 82 60 84

T ss s s s s s s s

=+ + + + + + + +

S8 1 12 45 82 84

S7 1 5 10 12

s6 1 5 10 12

s5 3 10 10 ROZ

s4 5 20 36

s3 5− 29−

s2 1− 4

s1 49−

s0 4

( ) ( )Even 6 :2 rhp, 2 lhp, 2 ;Rest 2 :0 rhp, 2 lhp,0 ;Total:2 rhp, 4 lhp, 2j j jω ω ω


12.

( ) 4 3 2

12 5 2 1

T ss s s s

=+ + + +

s4 2 1 1

s3 5 2 0

s2 1 5

s1 23− 0

s0 5

Total: 2 lhp, 2 rhp

13.

( ) 7 6 5 4 3 2

82 2 4 8 4 8

T ss s s s s s s

=− − + + − − +

s7 1 1− 4 4−

s6 2− 2 8− 8

s5 12− 8 16− 0 ROZ

s4 0.6667 5.333− 8 0

s3 88− 128 0 0

s2 4.364− 8 0 0

s1 33.33− 0 0 0

s0 8 0 0 0

Even (6): 3 rhp, 3 lhp; Rest (1): 1 rhp; Total: 4 rhp, 3 lhp


14. Program: numg=8;

deng=[1 –2 –1 2 4 –8 –4 0];

'G(s)'

G=tf(numg,deng)

'T(s)'

T=feedback(G,1)

'Poles of T(s)'

pole(T)

Computer response: ans =

G(s)

Transfer function:

8

-----------------------------------------------

s^7 - 2 s^6 - s^5 + 2 s^4 + 4 s^3 - 8 s^2 - 4 s

ans =

T(s)

Transfer function:

8

---------------------------------------------------

s^7 - 2 s^6 - s^5 + 2 s^4 + 4 s^3 - 8 s^2 - 4 s + 8

ans =

Poles of T(s)

ans =

-1.0000 + 1.0000i

-1.0000 - 1.0000i

-1.0000

2.0000

1.0000 + 1.0000i

1.0000 - 1.0000i

1.0000


15. Thus, there are 4 rhp poles and 3 lhp poles.

16. Even (6): 1 rhp, 1 lhp, 4 jω ; Rest (1): 1 lhp; Total: 1 rhp, 2 lhp, 4 jω

( ) 5 4 3 2

64 2 2 6

T ss s s s s

=+ + − + −

s5 1 2 1

s4 4 –2 –6

s3 2.5 2.5 0

s2 –6 –6 0

s1 –12 0 0 ROZ

s0 –6 0 0

Even (2): 2 jω ; Rest (3): 1 rhp, 2lhp; Total: 2 lhp, 1 rhp, 2 jω

17.

( ) ( )4 3 2

507 1;3 10 30 169

G s H ss s s s s

= =+ + + +

. Therefore,

( ) 5 4 3 2

5071 3 10 30 169 507

G sT sG H s s s s s

= =+ + + + + +

s5 1 10 169

s4 3 30 507

s3 12 60 0 ROZ

s2 15 507 0

s1 345.6− 0 0

s0 507 0 0

Even (4): 2 rhp, 2 lhp, 0 jω ; Rest (1): 0 rhp, 1 lhp, 0 jω ; Total (5): 2 rhp, 3 lhp, 0 jω


18.

( ) ( )( ) ( )

2

2

11 3 2

K sT s

K s s K+

=+ + + +

. For a second-order system, if all coefficients are

positive, the roots will be in the lhp. Thus, 1K > − .

19.

( ) ( )( )3 2

64 3 6

K sT s

s s K s K+

=+ + + +

s3 1 3 K+

s2 4 6K

s1 132

K− 0

s0 6K 0

Stable for 0 6K< <

20.

The characteristic equation for all cases is ( )( )

1 0K s as s b

−+ =

− or ( )2 0s K b s Ka+ − − = . The Routh

array is

s2 1 Ka−

s K b−

1 Ka−

a) 0, 0 , 0 0a b K b K K< < ⇒ > > ⇒ >

b) 0, 0 , 0a b K b K K b< > ⇒ > > ⇒ >

c) 0, 0 , 0 0a b K b K b K> < ⇒ > < ⇒ < <

d) 0, 0 , 0 Nosolutiona b K b K> > ⇒ > < ⇒


21.

( ) ( )( )4 3 2

19 26 24

K sT s

s s s K s K+

=+ + + + +

s4 1 26 K

s3 9 0

s2 210 K− 9K 0

s1 2 105 5040

210K K

K− + +

− 0 0

s0 9K 0 0

Stable for 0 140.8K< <

22. Program: K=[0:0.2:200];

for i=1:length(K);

deng=poly ([0 –2 –3 –4]);

dent=deng+[0 0 0 K(i) K(i)];

R=roots (dent);

A=real (R);

B=max (A);

of B>0

R

K=K (i)

Break

end

end Computer response: R =

–8.0442

0.0000 + 4.2791i

0.0000 – 4.2791i

–0.9559

K =

140.8000

24 +K


23. Program: %-det([si() si();sj() sj()])/sj()

%Template for use in each cell.

syms K %Construct a symbolic object for

%gain, K.

s4=[1 26 K 0] %Create s^4 row of Routh table.

s3=[9 24+ K 0 0] %Create s^3 row of Routh table.

s2=[–det([s4(1) s4(2);s3(1) s3(2)])/s3(1). . .

–det([s4(1) s4(3);s3(1) s3(3)])/s3(1) 0 0];


s1=[–det([s3(1) s3(2);s2(1) s2(2)])/s2(1). . .


s0=[–det([s2(1) s2(2);s1(1) s1(2)])/s1(1). . .

–det([s2(1) s2(3);s2(1) s1(3)])/s1(1) 0 0];










pretty(s0) %Pretty print s^0 row. Computer response: s4 =

[ 1, 26, K, 0]

s3 =

[ 9, 24+K, 0, 0]

ans =

s2

[70/3 – 1/9 K K 0 0]

ans =

s1

[ 2 ]

[–105 K – 5040 + K ]

[------------------------ 0 0 0 ]

[ –210 + K ]

–det([s3(1) s3(3);s2(1) s2(3)])/s2(1) 0 0];


ans =

s0

[K 0 0 0]

Stable for 0 140.8K< <

24.

( ) ( )( )( ) ( )2

2 21 3 4

K s sT s

K s K+ −

=+ + −

. For positive coefficients in the denominator, 314

K− < < .

Hence marginal stability only for this range of K.

25.

( ) ( )5 4

12K s

T ss s Ks K

+=

+ + +. Always unstable since s3 and s2 terms are missing.

26.

( ) ( )( )( )( ) ( )3 2

2 4 57 1 2 3 40K s s s

T sKs K s Ks K

− + +=

+ + + + −

s3 K 2K

s1 254

7 1K KK

−+

0

s0 3 40K− 0

For stability, 1 354 40

K< <


27.

( ) ( )( ) ( )4 3 2

23 3 3 2 4

K sT s

s s s K s K+

=+ − + + + −

s4 1 3− 2 4K −

s3 3 3K + 0

s2 ( )123

K− + 2 4K − 0

s1 ( )3312

K KK

++

0 0

s0 2 4K − 0 0

Conditions state that 12, 2K K< − > , and 33K > − . These conditions cannot be met

simultaneously. System is not stable for any value of K.

28.

( ) ( )3 280 2001 15390KT s

s s s K=

+ + + +

s3 1 2001

s2 80 15390K +

s1 1 14469

80 8K− + 0

s0 15390K + 0

There will be a row of zeros at s1 row if 144690K = . The previous row, s2, yields the

auxiliary equation, ( )280 144690 15390 0s + + = . Thus, 44.73s j= ± . Hence, 144690K =

yields an oscillation of 44.73 rad/s.


29.

( ) ( ) ( ) ( )4 2

2

2 21 2 1 2 1Ks Ks Ks KT s

K s K s K− + +

=+ + − + +

Since all coefficients must be positive for stability in a second-order polynomial,

1 K− < < ∞ ; 1; 1 2K K−∞ < < − < < ∞ . Hence, 1 12

K− < < .

30.

( ) ( )( )( ) ( )4 3 2

2 711 31 8 21 12

s sT s

s s K s K s K+ +

=+ + + + + +

Making a Routh table,

s4 1 31K + 12K

s3 11 8 21K + 0

s2 3 320

11K + 12K 0

s1 224 1171 6720

3 320K K

K+ +

+0 0

s0 12K 0 0

31.

s2 row says 106.7 K− < . s1 row says 42.15K < − and 6.64 K− < . s0 row says 0 K< .

( ) ( )( )3 2

43 2 4

K sT s

s s K s K+

=+ + + +



s3 1 2 K+

s2 3 4K

s1 6 K− 0

s0 4K 0

a. For stability, 0 6K< < .

b. Oscillation for 6K = .

c. From previous row with 26,3 24 0K s= + = . Thus 8s j= ± , or 8ω = rad/s.

32.

a. ( ) ( )( )( )( )2

1 22 2 2

K s sG s

s s s− −

=+ + +

. Therefore, ( ) ( )( )( ) ( ) ( )3 2

2 14 6 3 2 2

s s KT s

s K s K s K− −

=+ + + − + +

.


s3 1 6 3K−

s2 4 K+ 4 2K+

s1 ( )23 8 204

K KK

− + −

+ 0

s0 4 2K+ 0

From s1 row: 1.57, 4.24K = − ; From s2 row: 4 K− < ; From s0 row: 2 K− < . Therefore,

2 1.57K− < < .

b. If 1.57K = , the previous row is 25.57 7.14s + . Thus, 1.13s j= ± .

c. From part b, 1.13rad/sω = .


33.

Applying the feedback formula on the inner loop and multiplying by K yields

( ) ( )2 5 7eKG s

s s s=

+ +

Thus,

( ) 3 25 7KT s

s s s K=

+ + +

Making a Routh table:

s3 1 7

s2 5 K

s1 35

5K− 0

s0 K 0

For oscillation, the s1 row must be a row of zeros. Thus, 35K = will make the system

oscillate. The previous row now becomes, 25 35s + . Thus, 2 7 0s + = , or 7s j= ± .

Hence, the frequency of oscillation is 7 rad/s .

34.

( ) ( ) ( )2

3 2

21 2 4 24Ks KsT s

s K s K s+

=+ − + − +

s3 1 2 4K −

s2 1K − 24

s1 22 6 20

1K K

K− −−

0

s0 24 0

For stability, 5K > ; Row of zeros if 5K = . Therefore, 24 24 0s + = . Hence, 6ω = for

oscillation.


35. Program:

K=[0:0.001:200];

for i=1:length(K);

deng=conv([1 –4 8],[1 3]);

numg=[0 K(i) 2*K(i) 0];

dent=numg+deng;

R=roots(dent);

A=real(R);

B=max(A);

if B<0

R

K=K(i)

break

end

end

Computer response:

R =

–4.0000

–0.0000 + 2.4495i

–0.0000 – 2.4495i

K =

5

a. From the computer response, (a) the range of K for stability is 0 5K< < .

b. The system oscillates at 5K = at a frequency of 2.4494 rad/s as seen from R, the poles

of the closed-loop system.


36. ( ) ( )( ) ( )4 3 2

23 3 3 2 4

K sT s

s s s K s K+

=+ − + + + −

s4 1 3− 2 4K −

s3 3 3K + 0

s2 12

3K +

− 2 4K − 0

s1 ( )3312

K KK

++

0 0

s0 2 4K − 0 0

For 33 :1K < − sign change; For 33 12 :1K− < < − sign change; For 12 0 :1K− < < sign

change; For 0 2 : 3K< < sign changes; For 2 : 2K > sign changes. Therefore, 2K >

yields two right-half-plane poles.

37.

( ) ( )4 3 27 15 13 4KT s

s s s s K=

+ + + + +

s4 1 15 4K +

s3 7 13 0

s2 927

4K + 0

s1 1000 49

92K−

0 0

s0 4K + 0 0

a. System is stable for 4 20.41K− < < .

b. Row of zeros when 20.41K = . Therefore, 292 24.417

s + . Thus, 1.3628s j= ± , or

1.3628rad/sω = .


38.

( ) ( )3 214 45 50KT s

s s s K=

+ + + +

s3 1 45

s2 14 50K +

s1 580

14K− 0

s0 50K + 0

a. System is stable for 50 580K− < < .

b. Row of zeros when 580K = . Therefore, 214 630s + . Thus, 45s j= ± , or

6.71rad/sω = .

39.

( ) 4 3 28 17 10KT s

s s s s K=

+ + + +

s4 1 17 K

s3 8 10 0

s2 126

8 K 0

s1 32 1063

K− + 0 0

s0 K 0 0

a. For stability 0 19.69K< < .


b. Row of zeros when 19.69K = . Therefore, 2126 19.698

s + . Thus, 1.25s j= ± , or

1.118rad/sω = .

c. Denominator of closed-loop transfer function is 4 3 28 17 10s s s s K+ + + + . Substituting

19.69K = and solving for the roots yield 1.118, 4.5s j= ± − , and 3.5− .

40.

( ) ( )( )

2

3 2

2 12 1K s s

T ss s K s K

+ +=

+ + + −

s3 1 1K +

s2 2 K−

s1 3 2

2K + 0

s0 K− 0

Stability if 2 03

K− < < .

41.

( ) ( )4 3 2

3 2

2 22

s K s KsT s

s s s K+ + +

=+ + +

s3 1 2

s2 1 K

s1 2 K− 0

s0 K 0

Row of zeros when 2K = . Therefore 2 2s + and 2s j= ± , or 1.414 rad/sω = . Thus

2K = will yield the even polynomial with 2 jω roots and no sign changes.


42.

1 K2 1

s3 K1 5 0

s2 1 2

1

5K KK− 1 0

s1 2

1 1 2

1 2

5 255

K K KK K

− +−

0 0

s0 1 0 0

For stability, 21 2 1 1 25; 25 5K K K K K> + < ; and 1 0K > . Thus 2

1 1 20 5 25K K K< < − , or

1 1 20 5 25K K K< < − .

43.

s4 1 1 1

s3 K1 K2 0

s2 1 2

1

K KK− 1 0

s1 2 2

1 1 2 2

2 1

K K K KK K− +

− 0 0

s0 1 0 0

For two 2 21 1 2 2poles, 0j K K K Kω − + = . However, there are no real roots. Therefore, there

is no relationship between K1 and K2 that will yield just two polesjω .


44.

s8 1 1.18 03E + 2.15 03E + 1.06 04E− + 415−

s7 103 4.04 03E + 8.96 03E− + 1.55 03E− + 0

s6 1140.7767 2236.99029 10584.951− 415− 0

s5 3838.02357 8004.2915− 1512.5299− 0 0

s4 4616.10784 10135.382− 415− 0 0

s3 422.685462 1167.4817− 0 0 0

s2 2614.57505 415− 0 0 0

s1 1100.3907− 0 0 0 0

s0 415− 0 0 0 0

a. From the first column, 1rhp,7 lhp,0 jω .

b. G(s) is not stable because of 1 rhp.

45.

( ) 2 31 2 6K s s s s= + + +

s3 1 1

s2 6 2

s 23

1 2

No RHP roots

( ) 2 32 2 2 6K s s s s= + + +

s3 1 2

s2 6 2

s 53

1 2

No RHP roots


( ) 2 33 4 4K s s s s= + + +

s3 1 1

s2 4 4

s 8

1 4

Auxiliary equation ( ) 24 4Q s s= + , no roots in RHP, but two roots in axisjω .

( ) 2 34 4 2 4K s s s s= + + +

s3 1 2

s2 4 4

s 1

1 4

No RHP roots

The interval polynomial has no roots in the RHP.

46.

The characteristic equation for this system is:

( )2 2 2

4 2 20 002 2 2

0

1 0or 0T T T

s KK Ks a sm m ms s a

ω ωωω

⎛ ⎞++ = + + + =⎜ ⎟

+ ⎝ ⎠

The Routh array is:

s4 1 20

T

Kam

ω⎛ ⎞

+⎜ ⎟⎝ ⎠

20

T

Kmω

s3 4θ 202

T

Kam

θ ω⎛ ⎞

+⎜ ⎟⎝ ⎠

s2 201/ 2

T

Kam

ω⎛ ⎞

+⎜ ⎟⎝ ⎠

20

T

Kmω

s

22200

220

4r r

r

K Kam m

Kam

ω ω

ω

⎛ ⎞− + +⎜ ⎟

⎝ ⎠⎛ ⎞

+⎜ ⎟⎝ ⎠

1 20

T

Kmω


The second row of zeros was substituted with the coefficients resulting from

differentiating the characteristic equation: ( )2

4 2 2 00a

T T

KKQ s s a sm m

ωω⎛ ⎞

= + + +⎜ ⎟⎝ ⎠

and

( ) 3 20' 4 2a

T

KQ s s am

ω⎛ ⎞

= + +⎜ ⎟⎝ ⎠

.

Since all the plant parameters are positive, there are two sign changes in the first column

of the Routh array. So there are two poles in the RHP, two must be in the LHP.

47.

The characteristic equation for the system is 21 0b

KI s

+ = or 2 0b

KsI

+ = . The system has two

complex conjugate poles at b

Ks jI

= ± . The arm will oscillate at a frequency rad/secb

KI

.

48.

( )( )( )

1 0 0 0 1 20 1 0 3 1 40 0 1 1 1 3

1.1427 2.5713 1.1245 2.5713 1.1245

ss

s s s i s i

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟− −⎜ ⎟ ⎜ ⎟− = ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠− = + − − − +

I A

I A

Hence, eigenvalues are 1.1427,2.5713 1.1245j− ± . Therefore, 2 rhp, 1 lhp, 0 jω .

49. Program: A=[0 1 0;0 1 –4;–1 1 3];

eig(A) Computer response: ans =

1.0000

1.5000 + 1.3229i

1.5000 – 1.3229I


50.

Writing the open-loop state and output equations we get,

1 2

2 2 3

3 1 2 3

2 3

33 4 5

x xx x xx x x x uy x x

== += − − − += +

&

&

&

Drawing the signal-flow diagram and including the unity feedback path yields,

3x x 12x

1 1

1

s 3

1

s

1

s 1

-5 1

-3

-1

r c = y

1

1

-4

Writing the closed-loop state and output equations from the signal-flow diagram,

( )

1 2

2 2 3

3 1 2 3

1 2 3 2 3

1 2 3

2 3

33 4 53 4 53 5 6

x xx x xx x x x r c

x x x r x xx x x r

y x x

== += − − − + −= − − − + − += − − − += +

&

&

&


In vector-matrix form,

[ ]

0 1 0 00 1 3 03 5 6 1

0 1 1

r

y

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= +⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥− − −⎣ ⎦ ⎣ ⎦

=

x X

x

&

Now, find the characteristic equation.

( )( )

3 2

0 0 0 1 0 1 00 0 0 1 3 0 1 30 0 3 5 6 3 5` 6

5 9 9

s ss s s

s s

s s s

−⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥⎢ ⎥ ⎢ ⎥− = − = − −⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥− − − +⎣ ⎦ ⎣ ⎦ ⎣ ⎦

= + + +

I A

Forming a Routh table to determine stability

s3 1 9

s2 5 9

s1 365

0

s0 9 0

Since there are no sign changes, the closed-loop system is stable.

51. Program: A=[0,1,0;0,1,3;–3,–4,–5];

B=[0;0;1];

C=[0,1,1];

D=0;

'G'

G=ss(A,B,C,D)

'T'

T=feedback(G,1)

'Eigenvalues of T'

ssdata(T);

eig(T)


Computer response:

ans =

G

a =

x1 x2 x3

x1 0 1 0

x2 0 1 3

x3 –3 –4 –5

b =

u1

x1 0

x2 0

x3 1

c =

x1 x2 x3

y1 0 1 1

d =

u1

y1 0

Continuous-time model.

ans =

T

a =

x1 x2 x3

x1 0 1 0

x2 0 1 3

x3 -3 -5 -6

b =

u1

x1 0

x2 0

x3 1


c =

x1 x2 x3

y1 0 1 1

d =

u1

y1 0

Continuous-time model.

ans =

Eigenvalues of T

ans =

–1.0000 + 1.4142i

–1.0000 – 1.4142i

–3.0000

52.

a. For ( )2

1 21, 1 0c

sn B sω

= = − = or 022 =+− cs ω . The Routh array is

s2 1− 2cω

s 2θ −

1 2cω

The auxiliary polynomial used in the second row is ( ) 2 2a cQ s s ω= − + , that row is

replaced with the coefficients of ( )' 2aQ s s= − .

The first column has one sign change, so there is one root I the RHP, one in the LHP.

b. For ( )4

2 42, 1 0c

sn B sω

= = + = or 4 4 0cs ω+ = .

The Routh array is


s4 1 0 4cω

s3 θ 4 0 0

s2 θ ε 4cω

s 44 cω

ε−

1 4cω

The second row was originally a row of zeros, the auxiliary equation used was

( ) 4 2a cQ s s ω= + , so its coefficients were substituted with the coefficients of

( ) 3' 4aQ s s= .

The first column in the array has two sign changes, so the polynomial has two roots in the

RHP and two must be in the LHP.

SOLUTIONS TO DESIGN PROBLEMS

53.

( ) ( )( )( ) ( ) ( )3 2

1 105.45 11.91 11 43.65 10

K s sT s

s K s K s K+ +

=+ + + + + +

s3 1 11.91 11K+

s2 5.45 K+ 43.65 10K+

s1 211 61.86 21.26

5.45K K

K+ +

+ 0

s0 43.65 10K+ 0

For stability, 0.36772 K− < < ∞ . Stable for all positive K.


54.

( ) ( )( )4 3 2

0.7 0.12.2 1.14 0.193 0.07 0.01

K sT s

s s s s K+

=+ + + + +

s4 1 1.14 0.07 0.01K +

s3 2.2 0.193 0

s2 1.0523 0.07 0.01K + 0

s1 0.17209 0.14635K− 0 0

s0 0.07 0.01K + 0 0

For stability, 0.1429 1.1759K− < <

55.

( ) ( ) ( )2

5 4 3 2

0.6 10 60.1130 3229 10 2348 60.1 58000 0.6

K K s K sT ss s s K s K s K

+ +=

+ + + + + + +

S s5 1 3229 60.1 58000K +

s4 130 10 23480K + 0.6K

s3 10 396290K− + 7812.4 7540000K + 0

s2 2100 2712488 8.3247 9

10 396290K K E

K− + +

− + 0.6K 0

s1 4 3 2

3 2

7813 3 5.1401 11 7.2469 15 3.3213 19 2.4874 221000 66753880 9.9168 11 3.299 15

E K E K E K E K EK K E K E

− + + +− + +

0 0

s0 0.6K 0 0


Note: s3 row was multiplied by 130

From s1 row after canceling common roots:

( )( )( )( )( )( )( )

7813000 39629 967.31586571671 2776.9294183336 29908.0706151651000 39629 2783.405672635 29908.285672635

K K K KK K K

− + + −− + −

0

3

2

1

From row : 0From row : 39629From row : 29908.29;39629 From row : 29908.29 ,or 29908.07;

s Ks Ks K Ks K K

><< <

< <

Therefore, for stability, 0 29908.07K< <

56.

s5 1 1311.2 ( )1000 100 1K +

s4 112.1 10130 60000K

s3 1220.8 99465 1000K + 0

s2 10038 9133.4K− 60000K 0

s1 ( )( )

( )0.010841 1.0192

994651.0991

K KK

+ −−

0 0

s0 60000K 0 0

2

1

0

From row : 1.099From row : 0.010841 1.0192; 1.0991From row : 0

s Ks K Ks K

<− < < ><

Therefore, 0 1.0192K< <


57.

Find the closed-loop transfer function.

( ) ( )( )( )

( ) ( )( ) ( ) ( )

6

6

3 2 6

63 1030 140 2.5

63 101 172.5 4625 10500 63 10

KG ss s s

G s KT sG s H s s s s K

×=

+ + +

×= =

+ + + + + ×

Make a Routh table.

s3 1 4625

s2 172.5 610500 63 10 K+ ×

s1 4564.13 365217.39K− 0

s0 610500 63 10 K+ × 0

The s1 line says 21.25 10K −< × for stability. The s0 line says 41.67 10K −> − × for

stability. Hence, 4 21.67 10 1.25 10K− −− × < < × for stability.

58.

Find the closed-loop transfer function.

( ) ( )( )( )( )

( ) ( )( ) ( )

( )( )( )3 2

7570 103 0.862.61 62.61

7570 103 0.81 7570 785766 3918.76 623768p p p

Kp s sG s

s s s

G s Kp s sT s

G s H s s K s K s K

+ +=

+ −

+ += =

+ + + − +

Make a Routh table:

s3 1 785766 3918.76pK −

s2 7570 623768Kp

s1 785766 4001.16pK − 0

s0 623768Kp 0


The s1 line says 35.09 10pK −> × for stability. The s0 line says 0pK > for stability.

Hence, 35.09 10pK −> × for stability.

59.

The characteristic equation is given by: 6 2 9 13

3 2 7 11

1 10 1.314 10 2.66 101 00.00163 5.272 10 3.538 10

s sKs s s

− − −

− −

× + × + ×+ =

+ + × + ×

Or

( ) ( ) ( )3 6 2 7 9 11 130.00163 1 10 5.272 10 1.314 10 3.538 10 2.66 10 0s K s K s K− − − − −+ + × + × + × + × + × =

The corresponding Routh array is:

s3 1 7 95.272 10 1.314 10 K− −× + ×

s2 60.00163 1 10 K−+ × 11 133.538 10 2.66 10 K− −× + ×

s ( )( )15

6

1.314 10 1371.6 457.80.00163 1 10

K KK

−

−

× + ++ ×

1 11 133.538 10 2.66 10 K− −× + ×

For stability row 2 requires 1630K > − and row 4 requires 133.008K > − . The dominant

requirement being the latter. It is clear also that when 133.008K > − , the first element on

row 3 is positive. So the overall requirement for stability is 133.008K > − .

60.

The characteristic equation of the system is given by:

2

11 01

C C

f

K Kms bs k K Ts

+ − =+ + +

or

( )( ) ( ) ( )2 21 1 0f f C CK ms bs k Ts K K Ts K ms bs k+ + + + + − + + = or

( ) ( )( ) ( ) ( )3 2 21 0f f C CK mTs bT m s kT b s k K K Ts K ms bs k+ + + + + + + − + + = or

( ) ( )3 2 0f f C f f C C f C CK mTs K bT m K m s K kT b K K T K b s k K K K k⎡ ⎤ ⎡ ⎤ ⎡ ⎤+ + − + + + − + + − =⎣ ⎦ ⎣ ⎦ ⎣ ⎦

Substituting numerical values the equation becomes:


6 3 2 65.28 10 0.03444 1.8 1584.78 11700 36000 31.5 10 0f f f fK s K s K s K− ⎡ ⎤⎡ ⎤ ⎡ ⎤× + − + − + − × =⎣ ⎦ ⎣ ⎦ ⎣ ⎦

The Routh array is given by

s3 65.28 10 fK−× 1584.78 11700fK −

s2 0.03444 1.8fK − 622500 31.5 10fK − ×

s

6 65.28 10 36000 31.5 10 1584.78 11700 0.03444 1.8

0.03444 1.8f f f

f

K K K

K

− ⎡ ⎤ ⎡ ⎤ ⎡ ⎤× − × − − −⎣ ⎦ ⎣ ⎦⎣ ⎦⎡ ⎤− −⎣ ⎦

1 636000 31.5 10fK − ×

To obtain positive quantities on the first column it is required: 6

6

6 6

5.28 10 0 00.03444 1.8 0 52.2636000 31.5 10 0 875

1584.78 11700 0.03444 1.8 5.28 10 36000 31.5 10 0

f f

f f

f f

f f f f

K KK K

K K

K K K K

−

−

× > ⇒ >− > ⇒ >

− × > ⇒ >

⎡ ⎤⎡ ⎤ ⎡ ⎤− − − × − × >⎣ ⎦ ⎣ ⎦ ⎣ ⎦

or 2 254.5798 3255.552 21060 0.19008 166.32f f f fK K K K− + > −

or 254.3897 3421.872 21060 0f fK K− + >

or 2 62.9139 387.206 0f fK K− + >

or

( )( )6.914 55.999 0f fK K− − >

So either 6.914fK < and 55.999 6.914f fK K< ⇒ <

or 6.914fK > and 55.999 6.914f fK K> ⇒ >

The most dominant requirement is given by the fourth row. We conclude requiring

875fK > .


61.

a. The Mesh equations obtained by defining clockwise mesh currents are given by

I1: 1 2 11 R I RI V

sC⎛ ⎞+ − =⎜ ⎟⎝ ⎠

I2: 1 2 312 0RI R I RI

sC⎛ ⎞− + + − =⎜ ⎟⎝ ⎠

I3: 2 312 0RI R I

sC⎛ ⎞− + + =⎜ ⎟⎝ ⎠

Solving for I3,

1

21

3 2 23

1

12 0

0 0

1 1 1 20 2 3

12

10 2

R R VsC

R RsC

RR VI

RR R R R RsC sC sC sC

R R RsC

R RsC

+ −

− +

−−

= =⎛ ⎞⎛ ⎞+ − + + − −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

− + −

− +

3

12 3 2 2

31 1 22 3

R VV RIRR R R

sC sC sC

−= =

⎛ ⎞⎛ ⎞+ + − −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

22

1

11 1 21 2 3

VV

sRC sRC sRC

−=⎛ ⎞⎛ ⎞+ + − −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

b. The gain of the inverting amplifier is given by: 1 2

2 1

V RV R

= − or 2 1

1 2

1V RV R K

= − = − .

Equating to the transfer function obtained in Part a


21 1

1 1 21 2 3K

sRC sRC sRC

−− =

⎛ ⎞⎛ ⎞+ + − −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

. Equivalently

21 01 1 21 2 3

K

sRC sRC sRC

− =⎛ ⎞⎛ ⎞+ + − −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

c. The characteristic equation can be written as: 21 1 21 2 3 0K

sRC sRC sRC⎛ ⎞⎛ ⎞+ + − − − =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

or

3 3 3 2 2 2

1 5 6 1 0Ks R C s R C sRC

+ + + − = or

( ) 3 3 3 2 2 21 6 5 1 0K s R C s R C sRC− + + + =

The Routh array is given by

s3 ( ) 3 31 K R C− 5RC

s2 6R2C2 1

s ( )298K RC−

1 1

So for oscillation it is required to have 29K = . The resulting auxiliary equation is

( ) 2 2 26 1Q s R C s= + . Solving the latter for the jω -axis poles we obtain 16

s jRC

= ± .

The oscillation frequency is 12 6

fRCπ

= .


62.

( ) ( )

11.7 6.8 61.6 7.73.5 24 66.9 8.40 1 01 0 10

24 66.9 8.4 3.5 66.9 8.4det 11.7 1 0 6.8 0 0

0 10 1 10

3.5 24 8.4 3.5 24 66.961.6 0 1 0 7.7 0 1

1 0 1 0 10

s K Ks K K

ss

s

s K K K Ks s s s

s s

s K s KK K s

s

+ − − −⎡ ⎤⎢ ⎥+ −⎢ ⎥− =⎢ ⎥−⎢ ⎥−⎣ ⎦

+ − −− = + − +

−

+ − +− − + −

− −

I A

I A

( ) ( )

( )

( )4 3 2

0 1 0 011.7 24 66.9 8.4

10 0 1 10

0 0 0 16.8 3.5 66.9 8.4

10 1 0 10

1 0 0 0 0 161.6 3.5 66.9 8.4

0 1 1 0

1 0 0 17.7 3.5 24 66.9

0 10 1 10 1 0

35.7 304.6 59.2 840.41

s ss s K K

s s

s sK K

s s

K K Ks s

s sK s K

s s K s

⎧ − ⎫= + + − −⎨ ⎬−⎩ ⎭

⎧ − ⎫+ − −⎨ ⎬−⎩ ⎭

⎧ − − ⎫− − −⎨ ⎬− −⎩ ⎭

⎧ − − ⎫+ − + +⎨ ⎬− −⎩ ⎭= + + + + 2713.3 1032.57 0Ks K K+ − =

The Routh array is:

s4 1 304.6 59.2K+ 713.3K

s3 35.7 21032.57K−

s2 35.66 304.6K + 228.92 713.3K K+

s 228936.58 230524.08

35.66 304.6K K

K++

1 228.92 713.3K K+

Row 3 is positive if 8.54K > −

Rows 4 and 5 are positive if 0K >

So the system is closed loop stable if 0K > .

840.4 1 K


63.

Sensor

+

-Input

transducer

Desired force

Input voltage

Controller Actuator Pantograph dynamics

Spring

Fup

Yh-Ycat Spring

displacement

Fout1

100K 1

1000

0.7883( s + 53.85)

( s2 + 15.47 s + 9283 )( s2 + 8.119 s+ 376 .3)82300

1

100

+

-

Desired force

Controller Actuator Pantograph dynamics

Spring

Fup

Yh-Ycat Spring

displacement

Fout1

1000

0.7883( s + 53.85)

(s2 + 15.47s + 9283 )(s 2 + 8.119 s + 376 .3)82300

K

100

( ) ( ) ( )( )

( )( )( )

( ) ( ) ( ) ( )

( ) ( )( )( )

2 2

2 2

0.7883 53.8515.47 9283 8.119 376.3

/100 * 1/1000 * *82.3 30.6488 53.85

8.119 376.3 15.47 9283

h cat

up

e

e

Y s Y s sG s

F s s s s s

G s K G s eK s

G ss s s s

− += =

+ + + +

=

+=

+ + + +

( ) ( )

( )4 3 2

0.6488 53.85

23.589 9784.90093 0.6488 81190.038

K sT s

s s s K

+=

+ + + +

s + (34.94 K + 0.3493192910)


s4 1 9785 ( )0.3493 7 34.94e K+ +

s3 23.59 ( )0.6488 81190K + 0 +

s2 ( )0.0275 6343K− + ( )0.3493 7 34.94e K+ 0 230654K <

s1 20.0178 1058.7 432.59 6.0275 6343

K K eK

− + +− +

0 128966 188444K− < <

s0 ( )0.3493 7 34.94e K+ 0 99971 K− <

The last column evaluates the range of K for stability for each row. Therefore

99971 188444K− < < .

64.

The characteristic equation is given by

3 2

520 10.38441 02.6817 0.11 0.0126

sKs s s

− −+ =

+ + +

or

( )3 22.6817 0.11 0.0126 520 10.3844 0s s s K s+ + + − + =

or

( ) ( )3 22.6817 0.11 520 0.0126 10.3844 0s s K s K+ + − + − =

The Routh array is:

s3 1 0.11 520K−

s2 2.6817 0.0126 10.3844K−

s 0.2824 1384.1

2.6817K−

1 0.0126 10.3844K−


Thus for stability

40.2824 1384.1 0or 2.04 102.6817

K K −−> < ×

and 30.0126 10.3844 0or 1.21 10K K −− > < ×

The intersection of both requirements gives 42.04 10K −< × .

file_down

Documents

Transcript of file_down