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
Bode Diagram
ï10
Magnitude (dB)
ï20
ï30
ï40
ï50
ï60
ï70
45
Phase (deg)
0
ï45
ï90
ï135
ï180
ï2
10
ï1
10
0
10
Frequency (rad/sec)
Figure Skill check 7
1
10
2
10
8.) d
9 !
9 ![ 9 − 4!! − !(12! − !! )]
! ! =
=
(!" + 1)( !" ! + 3!" + 9)
9 − 4! ! − !(12! − ! ! )
At limiting K, imaginary of L(s)=0 => ω=0 or ! = 2 3 rad/s
9 ! 9 − 4!! − ! 12! − !!
− 351 !
!"#$ ! ! = −1 =>
= −1 => !
!
9 − 4! − ! 12! − !
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 => ! = 3.3564 !"#/!
4
20 log ! + 20log ( !" + 4 )+20log( !! )=0 (eq11)
using ! = 3.3564 !"#/! in (eq11) we get K=2.15
!!" = 180 + !"#!!
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
Bode Diagram
40
System: sys
Frequency (rad/sec): 1.37
Magnitude (dB): ï3.05
Magnitude (dB)
20
0
ï20
ï40
ï60
ï80
270
Phase (deg)
225
180
135
90
45
0
ï1
10
0
10
Frequency (rad/sec)
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)
1
10
2
10
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