1
Solution 9.10.4.3
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)=
4
(s +4)
2
and wewant
T
c
(s)=
!
2
n
s
2
+2!
n
s + !
2
n
;;
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
:
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;;
whichcanberearranged as
C
d
=
D
1+G
;;
2
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
:
Since
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 +4)
2
4
=
(!
2
n
=4)(s+4)
2
s(s +2!
n
)
:
Thus, we see that the compensator cancels the twopoles of the plant and
replaces them with poles at s =0ands = ;2!
n
.
3
Wenow nd
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
:
If wechoose =0:8and!
n
=5rad/s, then
T
d
(s) =
s(s +2!
n
)
s
2
+2!
n
s + !
2
n
=
s(s +2(0:8)(5)
s
2
+2(0:8)(5)s+5
2
=
s[s +8]
s
2
+8s +5
2
=
s(s +8)
(s +4;j3)(s+4+j3)
G
c
(s) =
(5
2
=4)(s+4)
2
s[s +2(0:8)(5)]
=
(25=4)(s+4)
2
s(s +8)
:
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. Wecanimprove the situation
by increasing !
n
In fact, we can get anynoiseattentuation wewantat10
rad/s bymerely increasing !
n
.The price wewillpayisincreased 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, however, there is a
limitation on gain, then there will be a limitation on noise suppression. For
instance, suppose we let =0:8 and !
n
=1000rad/s. Then
T
d
(s) =
s(s +2!
n
)
s
2
+2!
n
s + !
2
n
=
s(s +2(0:8)(1000)
s
2
+2(0:8)(1000)s+1000
2
=
s(s +1600)
s
2
+1600ss+10
6
4
10
-1
10
0
10
1
10
2
10
3
-60
-50
-40
-30
-20
-10
0
10
20
Figure 2: Frequency Response of T
d
(j!)
.
5
10
-1
10
0
10
1
10
2
10
3
-60
-50
-40
-30
-20
-10
0
10
20
Figure 3: Frequency Response of T
d
(j!)
.
=
s(s +8)
(s +800;j600)(s+800+j600)
G
c
(s) =
(1000
2
=4)(s +4)
2
s[s +2(0:8)(1000)]
=
250;;000(s+4)
2
s(s +1600)
:
The Bode magnitude plot of T
d
is shown in Figure 3. Wehave plentyof
noise suppression around 10 rad/s, but weneedagain of 250,000 in our
compensator. If we use a more realistic value of !
n
=100rad/s, then wesee
from Figure 4 that wecanachieveabout 15 dB of noise suppression. This
is not great, but it is denitely an improvement. The gain requirementisa
more realistic 2500.
6
10
-1
10
0
10
1
10
2
10
3
-60
-50
-40
-30
-20
-10
0
10
20
Figure 4: Frequency Response of T
d
(j!)
.