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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 ï70 ï60 ï50 ï40 ï30 ï20 ï10 Magnitude (dB) 10 ï2 10 ï1 10 0 10 1 10 2 ï180 ï135 ï90 ï45 0 45 Phase (deg) Bode Diagram Frequency (rad/sec)

Solution Assignment #8 - Engineering and …skoskie/ECE382/ECE382_s12/ECE382_s12_hw8soln...Solution Assignment #8 True or False: (1) True (2) ... Wcp = 2.5586 (Cross over ... Solution_homework8.docx

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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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Mag

nitu

de (d

B)

10 2 10 1 100 101 102180

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eg)

Bode Diagram

Frequency (rad/sec)

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)

10 1 100 101 1020

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eg)

Bode Diagram

Frequency (rad/sec)

80

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0

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

Mag

nitu

de (d

B)

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

Figure a)

Figure b)

Figure c)

CP9.4

CP9.10

Figure CP9.10

E10.3

Figure E10.3

E10.14

Figure E10.14

E10.20

Figure E10.20