Solution Assignment #8et.engr.iupui.edu/~skoskie/ECE382/ECE382_s12/ECE382_s12_hw8soln_rough.pdf ·...

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Transcript of Solution Assignment #8et.engr.iupui.edu/~skoskie/ECE382/ECE382_s12/ECE382_s12_hw8soln_rough.pdf ·...

Solution Assignment #8 True or False: (1) True (2) True (3) True (4) True (5) False 6.) b %Skill check 6 p=3.25*[1/6 1]; q=[1/24 11/24 1 0]; sys=tf(p,q); [mag,phase,w]=bode(sys); [Gm,Pm,Wcg,Wcp]=margin(mag,phase,w) Script Run Gm = 5.2850e+03 Pm = 54.8907 (Phase margin) Wcg = 262.3778 Wcp = 2.5586 (Cross over frequency) 7.) a p=[1 0.2]; q=[1/8 11/8 19/4 5]; sys=tf(p,q); bode(sys) The gain margin is ∞ so the system is stable

Figure Skill check 7

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8.) d

! ! =9  !

(!" + 1)( !" ! + 3!" + 9) =9  ![ 9− 4!! − !(12! − !!)]

9− 4!! − !(12! − !!)

At limiting K, imaginary of L(s)=0 => ω=0 or ! = 2 3 rad/s

!"#$ ! ! = −1 =>9  ! 9− 4!! − ! 12! − !!

9− 4!! − ! 12! − !!!!! !

= −1     =>              − 351 !1521

= −1 => ! = 4.33 9.) a %Skill check 9 p=4.3*9*[0.2 1]; q=[1 4 12 9]; sys=tf(p,q); [mag,phase,w]=bode(sys); [Gm,Pm,Wcg,Wcp]=margin(mag,phase,w) Script run Gm = 5.1077 Pm = 28.1321 Wcg = 7.1834 Wcp = 3.7548 10.) d %Skill check 10 p=[1 1]; q=[4 1 0 0]; sys=tf(p,q); [mag,phase,w]=bode(sys); [Gm,Pm,Wcg,Wcp]=margin(mag,phase,w) Script run Gm = 1.3693e-05 Pm = -35.7368 Wcg = 0.0037 Wcp = 0.6537

The phase margin is negative so the system is unstable 11.) b !!" = 180+ !"#!!

!4 − 180 => ! = 3.3564  !"#/!

20 log ! + 20log  ( !" + 4 )+20log( !! )=0 (eq11) using ! = 3.3564  !"#/!  in (eq11) we get K=2.15 12.) a Replace !!!.!!  !"   !!.!!!!

!.!!!! and repeat the procedure of question #11

14.) c p=[-0.3 1]; q=[3/50 1/2 1 0]; sys=tf(p,q); bode(sys) grid on

15.) a %Skill check 15 p=[1 4]; q=[1 6 5 0]; sys=tf(p,q); [mag,phase,w]=bode(sys); [Gm,Pm,Wcg,Wcp]=margin(mag,phase,w)

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40System: sysFrequency (rad/sec): 1.37Magnitude (dB): 3.05

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Gm = 7.2196e+04 Pm = 58.1073 Wcg = 268.7613 Wcp = 0.6678 Word Match (in or, top to bottom): f, e, k, b, j, a, i, d, h, c, g E9.16

The phase approximation is

E9.24 Using the Nyquist criterion, we have P=1 and N=0 which implies Z=N+P=1. Hence the system has root in the right half-plane. E9.25 p=[11.7]; q=[1/200 3/20 1 0]; sys=tf(p,q); bode(sys) grid on p=[11.7]; q=[1/200 3/20 1 0]; sys=tf(p,q); bode(sys) grid on Using the bode plot of the loop transfer function

CP9.2

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Figure b)

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CP9.10

Figure CP9.10

E10.3

Figure E10.3

E10.14

Figure E10.14

E10.20

Figure E10.20