Solution: 7.9.5.5
R
+
C
G
c
G
p
Σ
-
Figure 1: Cascade Compensation with UnityFeedback
For the system of Figure 1 wehave
G
p
=
10
s(s +0:5)(s+10)
and G
c
(s)=
K
c
(s + z
1
)(s + z
2
)
(s +20)(s+ p)
:
Figure 2 shows the evaluation of G
c
G
p
(s)ats = ;4+j3incomponent
form. That is, eachofthe vectors in Figure 2 is the polar representation
of one of the factors in G
c
G
p
evaluated at s = ;4+j3. The vector V
1
, for
instance, is the polar representation of the factor s in the denominator of
G
c
G
p
, V
2
the polar representation of the factor s + p,andsoon. Notice
that the pole at s = ;0:5has been cancelled by the zero at s = ;0:5and
so those twoterms do not enter into the angle and magnitude calculations
to be made next. Thus,
G
c
G
p
j
s=;4+j3
=
K
c
jV
3
j
6
(jV
1
j
6
1
)(jV
2
j
6
2
)(jV
3
j
6
3
)(jV
4
j
6
4
)
=
K
c
jV
3
j
jV
1
jjV
2
jjV
3
jjV
4
j
6
( ;
1
;
2
;
3
;
4
)
The evaluation of G
c
G
p
at s = ;4+j3hasnowbeen broken down into
acomposite magnitude and a composite angle. For the root locus to pass
through this point, wemust have
;
1
;
2
;
3
;
4
= ;180
:
our tactic will be to iteratively choose p and then compute
=
1
+
2
+
3
4
; 180
=
h
180
; tan
;1
(3=4)
i
+
h
180
; tan
;1
[3=(4; p)]
i
+ tan
;1
(3=6)
+tan
;1
(3=16); 180
;;
1
θ
1
θ
2
α
-4 -z
Im(s)
Re(s)
3
V
5
=
s + z
V
2 =
s + p
V
4
=
s + 12 V
1 =
s
θ
3
V
3
=
s + 10
-10 -p
XXX
X
-20
Figure 2: Satisfaction of Angle Condition For Desired Closed Loop Poles
moving p closer and closer to the origin, evaluating K
v
and K
c
as wego.
At each iteration we will havetorecompute the location of the zero. The
MATLAB program shown belowdoesthis.
s=-4+j*3
p1 = 20
p2 = 10
pp = 0.1
p(1) = 0.1
dp = 0.001
i=1
while pp>0.001
p(i) = pp;;
alpha(i) = angle(s + p(i)) +angle(s) + angle(s + p1) + angle(s + p2) -pi;;
alphad(i) = alpha(i)*180/pi;;
z(i) = 3 + 2/tan(alpha(i));;
k(i) = (abs(s)*abs(s + p1)*abs(s + p2)*abs(s + p(i) ) )/(10 *abs(s + z(i) ));;
kv(i) = (10 * k(i) * z(i))/(p1*p2*p(i));;
pp = pp -dp;;
i=i+1;;
end
Figure 3 shows K
v
as a function of the pole being added near the origin.
The following MATLAB dialogue identies p, z,andK, that meet the spec-
ications
2
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
0
50
100
150
200
250
300
350
400
450
Figure 3: K
v
as a function of p
3
EDU>kv(93)
ans =
1.103282131050450e+02
EDU>k(93)
ans =
5.731420341078142e+02
EDU>z(93)
ans =
0.30799545394164
EDU>p(93)
ans =
0.00800000000000
EDU>
Then the compensator is
G
c
(s)=
573(s +0:308)(s+0:5)
(s +0:008)(s+20)
The following MATLAB dialogue generates the step response for the com-
pensated system, shown in Figure 4
EDU>g = zpk([-.308],[0 -0.008 -10 -20],573)
Zero/pole/gain:
573 (s+0.308)
-------------------------
s(s+0.008) (s+10) (s+20)
4
Time (sec.)
Amplitude
Step Response
0 1 2 3 4 5 6
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Figure 4: Step Response of Compensated System
EDU>h = 1
h=
1
EDU>tc = feedback(g,h)
Zero/pole/gain:
573 (s+0.308)
------------------------------------------
(s+22.11) (s+0.3469) (s^2 + 7.55s + 23.01)
EDU>step(tc,6)
EDU>print -deps lagld5step.eps
EDU>
The following dialogue generates the ramp response shown in Figure 5.
5
Time (sec.)
Amplitude
Linear Simulation Results
0 1 2 3 4 5 6
0
1
2
3
4
5
6
Figure 5: Ramp Response of Compensated System
EDU>step(tc,6)
EDU>print -deps lagld5step.eps
EDU>t = 0:0.1:6;;
EDU>u = t;;
EDU>lsim(tc,u,t)
EDU>print -deps lagld5ramp.eps
EDU>
6