ECE 483 Midterm 2 Solution
Problem 1 (25 points)
In the above system,
T
= 0
.
5 s, and
D
(
z
) = 1
−
z
−
1
,
G
(
z
) =
Z
bracketleftbigg
1
−
e
−
sT
s
G
p
(
s
)
bracketrightbigg
=
0
.
125
(
z
−
1)(
z
−
0
.
5)
.
(a)
(5 pts)
Is the closedloop system stable?
Solution:
The characteristic equation is
1 +
D
(
z
)
G
(
z
) = 1 + (1
−
z
−
1
)
0
.
125
(
z
−
1)(
z
−
0
.
5)
= 1 +
0
.
125
z
(
z
−
0
.
5)
= 0
,
which has two solutions 0
.
25
±
j
0
.
25. These are the closedloop poles and are both inside the
unit circle:

0
.
25
±
j
0
.
25

= 0
.
3536
<
1. Therefore, the system is stable.
(b)
(10 pts)
What are the steadystate errors of the system for tracking the unit step and the
unit ramp input, respectively?
Solution:
Since the system is stable, from the openloop transfer function
D
(
z
)
G
(
z
) =
0
.
125
z
(
z
−
0
.
5)
,
we conclude that the system is of type 0, with
K
dc
= lim
z
→
1
D
(
z
)
G
(
z
) = 0
.
25
.
Thus, the steady state tracking error for unit step input is
e
ss
=
1
1 +
K
dc
= 0
.
8
,
and the steady state tracking error for unit ramp input is infinity.
(c)
(10 pts)
What are the approximate settling time (2% criterion) and maximum overshoot of
the system’s step response?
Solution:
Pick any of the two closedloop pole, say,
p
= 0
.
25 +
j
0
.
25. We have

p

= 0
.
3536
and
negationslash
p
= 0
.
7854 rad.
Thus, the corresponding pole on the
s
plane is
q
=
ln

p

T
+
j
negationslash
p
T
=
−
2
.
0794 +
j
1
.
5708.
This pole
q
has the damping ratio
ζ
= 2
.
0794
/
√
2
.
0794
2
+ 1
.
5708
2
=
0
.
7979. Therefore, the approximate settling time and maximum overshot are
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 Spring '08
 evens
 0 k, KDC, 1.56%, 0.7854 rad, 1.17 Pole

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