ME2142/ME2142E Feedback Control Systems
Solutions to Tutorial Problems Set 2
5
th
Sept 2012
1.
G
1
G
2
G
3
G
4
G
5
G
6
+ +
+
+
+
+
-
-
R(s) C(s)
G
1
G
4
G
5
G
6
+ +
+
+
+
-
C(s) R(s)
2
2 3
1
G
G G +
G
1
G
4
G
6
+ +
+
+
+
-
C(s) R(s)
2
2 3
1
G
G G +
5 2 3
2
(1 ) G G G
G
+
G
1
+
-
C(s) R(s)
5 2 3
6
2
(1 ) G G G
G
G
+
+
2
2 3 2 4
1
G
G G G G +
1 2
2 3 2 4
6 2 5 2 3 1 2
2 3 2 4 2
1 ( )
( )
(1 )
1
1
GG
G G G G C s
R s
G G G G G GG
G G G G G
+
=
| || | + +
+
| |
+
\ . \ .
1 2
2 3 2 4 1 6 2 5 2 3
(1 ) ( (1 ))
GG
G G G G G G G G G G
=
+ + + +
2(a) (i) With D(s)=0 and N(s)=0,
) ( ) ( ) ( 1
) ( ) (
) (
) (
s H s G s G
s G s G
s R
s C
p c
p c
+
=
(ii) With R(s)=0 and N(s)=0, the block diagram can be re-drawn as
and the closed-loop transfer function is
) ( ) ( ) ( 1
) (
) (
) (
s H s G s G
s G
s D
s C
c p
p
+
=
(iii) With R(s)=0 and N(s)=0, the block diagram can be re-drawn as
and the closed-loop transfer function is
) ( ) ( ) ( 1
) ( ) ( ) (
)) ( )) ( )( ( 1
) ( ) ( ) (
) (
) (
s H s G s G
s G s G s H
s H s G s G
s G s G s H
s N
s C
c p
p c
c p
p c
+
=
[Note that for all three closed-loop transfer functions, the denominator remains
the same.]
2(b) Using the results obtained in 1(a) above,
(i)
K K s s
K K
s s
K K
s s
K
K
s H s G s G
s G s G
s R
s C
p
p
p
p
p c
p c
+ +
=
+
+
+
=
+
=
2
) 1 (
1
) 1 (
) ( ) ( ) ( 1
) ( ) (
) (
) (
t
t
t
For a unit step,
s
s R
1
) ( = ,
and so
s K K s s
K K
s C
p
p
1
) (
) (
2
+ +
=
t
and 1
1
) (
lim
2
0
=
+ +
=
s K K s s
K K
s c
p
p
s
ss
t
Since ) ( 1 ) ( t t r = , the steady-state error is then
0 1 1 1 ) ( ) ( = = = =
ss ss
c c r e
(ii)
K K s s
K
s s
K K
s s
K
s H s G s G
s G
s D
s C
p
p
p c
p
+ +
=
+
+
+
=
+
=
2
) 1 (
1
) 1 (
) ( ) ( ) ( 1
) (
) (
) (
t
t
t
For a unit step,
s
s D
1
) ( = ,
and so
s K K s s
K
s C
p
1
) (
) (
2
+ +
=
t
and
p p
s
ss
K s K K s s
K
s c
1 1
) (
lim
2
0
=
+ +
=
t
Since 0 ) ( = t r , the steady-state error is then
p
ss ss
K
c c r e
1
0 ) ( ) ( = = =
2(c) Because the system is linear, the principle of superposition applies. Thus,
) (
) (
) (
) (
) (
) (
) ( s D
s D
s C
s R
s R
s C
s C + =
Using the results obtained in (b) above,
p
ss
K
c
1
1+ =
2(d) The characteristic equation is obtained by equating the denominator of any of the
closed-loop transfer functions to zero. Thus
0 ) ( ) ( ) ( 1 = + s H s G s G
p c
Or 0
) 1 1 . 0 (
1
) 5 ( 1 =
+
+
s s
0 5 1 . 0
2
= + + s s
0 2 50 10
2 2 2
= + + + +
n n
s s s s e .e
giving 50
2
=
n
e and 10 2 =
n
.e
Therefore rad/s 2 5 =
n
e and 707 . 0
2
1
= = .
-5 -10
-5
5
-5 -10
-5
5
3. Let the output of the Controller be M(s).
v
v
K s
s
K
s
GH
G
s M
s
10 1 1 . 0
10
1 1 . 0
10
1
1 1 . 0
10
1 ) (
) (
+ +
=
+
+
+
=
+
=
u
and
p v
p
v
p
v
p
r
K s K s
K
K s s
K
K s s
K
GH
G
s
s
10 ) 10 1 ( 1 . 0
10
) 10 1 1 . 0 (
10
1
) 10 1 1 . 0 (
10
1 ) (
) (
2
+ + +
=
+ +
+
+ +
=
+
=
u
u
2 2
2
2
2 100 ) 100 10 (
100
n n
n
p v
p
s s K s K s
K
e .e
e
+ +
+ + +
=
with
p p n
K K 10 100 = = e and
p
v
p
v
K
K
K
K
2
10 1
20
100 10 +
=
+
= .
From the equations, it can be seen that increasing K
p
will increase
n
e and thus the
speed of response. Increasing K
v
will increase the damping ratio and thus the
damping in the system.
4 a). Pole-Zero locations:
(i) (ii)
Sys1 Sys2
-5 -10
-5
5
-0.5
-1.0
0.2
-0.2
Pole at -10
(iii)
The pole at s=-1 will dominate
the response. The response will
be close to that obtained for
Sys1.
(iv)
The response will
be dominated by
the pair of
complex
conjugate poles
Step Responses:
Octave Program:
sys1=zp([],-1,1);
sys2=zp([],-10,10);
sys3=zp([],[-1 -10],10);
step(sys1);
hold on;
step(sys2);
hold on;
step(sys3);
Sys3
Sys4
Octave Program: Sys4=tf(1,[0.1 1.11 1.12 0.111 0.01]);
Step(Sys4)
