1
Solution 9.10.4.2
Σ Σc
(s)G G
p
(s)
R(s)
C(s)
+
_
+
+
D(s)
Figure 1: UnityFeedbackwith Disturbance at the Output
For the system of Figure 1,
G
p
(s)=
2
s(s +2)
and wewant
T
c
(s)=
!
2
n
s
2
+2!
n
s + !
2
n
;;
With =0:8 and !
n
=10.
Then
T
c
(s)=
100
(s +8;j6)(s+8+j6)
TondG
c
wesetD =0andnotethat
T
c
=
G
c
G
p
1+G
c
G
p
:
Solving rst for G
c
G
p
,wehave
G
c
G
p
=
T
c
1;T
c
:
Then
G
c
=
1
G
p
T
c
1;T
c
:
In the presentcase
T
c
1;T
c
=
100
(s +8;j6)(s+8+j6)
1;
100
(s +8;j6)(s+8+j6)
2
=
100
s
2
+16s +100;100
=
100
s(s +16)
:
Then
G
c
(s) =
100
s(s +16)
s(s +2)
2
=
50(s+2)
(s +16)
:
Thus, G
c
simply cancels the pole at s = ;2 and replaces it byanother
pole at s = ;16.
The closed loop transfer function between D and C with R =0,canbe
found as follows. With R =0,
C
d
= D;C
d
G;;
whichcan be rearranged as
C
d
=
D
1+G
;;
or
T
d
=
C
d
D
=
1
1+G
=
1
1+G
c
G
p
=
1
1+
T
c
1;T
c
= 1;T
c
:
Wethus havetwoexpressions for T
d
:
T
d
=
1
1+G
c
G
p
=1;T
c
:
Wenow compute T
d
bybothmethods.
T
d
(s) =
1
1+G
c
G
p
3
10
-2
10
-1
10
0
10
1
10
2
10
3
10
4
-100
-80
-60
-40
-20
0
20
40
Figure 2: Frequency Response of T
d
(j!)
.
=
1
1+
50(s+ 2)(2)
s(s +2)(s+16)
=
s(s +16)
s
2
+16s +100
= 1;T
c
= 1;
100
(s +8;j6)(s+8+j6)
=
s
2
+16s+100;100
(s +8;j6)(s+8+j6)
=
s(s +16)
(s +8;j6)(s+8+j6)
b.
The frequency response of T
d
(j!)isshown in Figure 2 The disturbance
rejection at 10 r./s. is ver poor, about 0 db. To see howtoimproveitwe
4
generalize the analysis somewhat. Wehavechosen
T
c
(s)=
!
2
n
s
2
+2!
n
s + !
2
n
:
Then
T
c
1;T
c
=
!
2
n
s
2
+2!
n
s + !
2
n
1;
!
2
n
s
2
+2!
n
s + !
2
n
=
!
2
n
s
2
+2!
n
s
:
Then
G
c
(s) =
T
c
1;T
c
1
G
p
=
!
2
n
s
2
+2!
n
s
s(s +2)
2
=
(!
2
n
=2)(s+2)
s +2!
n
T
d
(s) = 1;T
c
(s)
= 1;
!
2
n
s
2
+2!
n
s + !
2
n
=
s(s +2!
n
)
s
2
+2!
n
s + !
2
n
The situation should nowbeclear. Wecanget anynoise attentuation we
wantat10rad/sbymerelyincreasing !
n
.Theprice we will payisincreased
gain in our compensator, since the gain of the compensator is !
n
=2. As
long as gain is not a problem, noise suppression is not a problem. If, how-
ever, there is a limitation on gain, then there will be a limitation on noise
suppression.