Solutions to Homework 5
ECE 460: Automatic Control
Due: Monday, February 22, at 2:10pm
Note 1.
Your solutions should be neat and legible; always show your work.
Problem 1.
(30 points) Given the system shown in Figure 1, find the following.
(a)
The closedloop transfer function
(b)
The system type
(c)
The steadystate error for an input of 5
u
(
t
).
(d)
The steadystate error for an input of 5
tu
(
t
).
(e)
The closedloop poles.
(f)
Discuss the validity of your answers to
part (c)
and
(d)
.
Figure 1
Solution:
(a)
For the inner loop:
G
1
(
s
) =
1
s
2
(
s
+1)
1 +
1
s
3
(
s
+1)
=
s
s
4
+
s
3
+ 1
The equivalent openloop transfer function
G
e
(
s
) is then given by
G
e
(
s
) =
1
s
2
(
s
+ 3)
G
1
(
s
) =
1
s
(
s
5
+ 4
s
4
+ 3
s
3
+
s
+ 3)
The closedloop transfer function is given by
T
(
s
) =
G
e
(
s
)
1 +
G
e
(
s
)
=
1
s
6
+ 4
s
5
+ 3
s
4
+
s
2
+ 3
s
+ 1
(b)
From inspection of
G
e
(
s
), we can say that
K
p
= 1;
K
v
= constant and
K
a
= 0. Thus
the system is Type 1.
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
(c)
Since the system is type 1, for
r
(
t
) = 5
u
(
t
),
e
ss
=
e
(1) = 0.
(d)
From
G
e
(
s
),
K
v
= lim
s
→
0
sG
e
(
s
) = 1
/
3. Therefore,
e
ss
= 5
/K
v
= 15.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Winter '14

Click to edit the document details