04/23/08
ECE 483 Homework #7 Solution
Problem 82
Solution:
First, we obtain
G
(
z
) as
G
(
z
) =
z
−
1
z
Z
b
1
s
2
(
s
+ 1)
B
=
(
T
−
1 +
e
−
T
)
z
+ 1
−
e
−
T
−
Te
−
T
(
z
−
1)(
z
−
e
−
T
)
=
0
.
3679
z
+ 0
.
2642
(
z
−
1)(
z
−
0
.
3679)
.
Transforming it to the
w
plane, we get
G
(
w
) =
G
(
z
)
v
v
z
=
2+
w
2

w
=
−
0
.
1037
w
2
−
1
.
0568
w
+ 2
.
5284
2
.
7358
w
2
+ 2
.
5284
w
,
whose frequency response is plotted in Fig. 1. Alternatively, the data in Table P82 can be
used to plot approximately the frequency response.
40
30
20
10
0
10
20
30
40
50
60
Magnitude (dB)
10
2
10
1
10
0
10
1
10
2
10
3
135
180
225
270
Phase (deg)
Bode Diagram
Frequency
(rad/sec)
Figure 1: Frequency response of the uncompensated system in Problem 82.
(a) With
D
(
z
) = 1, from the frequency response of the uncompensated system in Table
P82, we ±nd that the 0 dB crossover frequency of

G
(
jω
w
)

occurs at approximately
ω
w
= 0
.
8, and at this frequency,
∠
G
(
jω
w
) =
−
149
◦
(or 211
◦
in Fig. 1). Thus, the
phase margin is approximately
φ
m
=
−
149
◦
−
(
−
180
◦
)
≃
31
◦
.
(b) If
D
(
z
) =
K
, then as we adjust the gain
K >
0, the phase plot of the compensated
frequency response
D
(
jω
w
)
G
(
jω
w
) will remain the same, while the magnitude plot
1
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View Full Documentwill shift by 20 log
K
dB. To have a 45
◦
phase margin, we need to place the new 0
dB crossover frequency at
ω
w
= 0
.
56 rad/s where
∠
D
(
jω
w
)
G
(
jω
w
) =
∠
G
(
jω
w
) =
180
◦
+ 45
◦
= 225
◦
, i.e.
20 log

D
(
jω
w
)
G
(
jω
w
)

ω
w
=0
.
56
= 20 log
K
+ 20 log

G
(
j
0
.
56)

= 0
.
Since 20 log

G
(
j
0
.
56)

= 3
.
98 dB, we must have 20 log
K
=
−
3
.
98 dB, i.e.,
K
= 0
.
8195.
0
5
10
15
20
25
0
0.2
0.4
0.6
0.8
1
1.2
1.4
n (samples)
Amplitude
Step Response
Figure 2: Step response of part (b) of Problem 82 of Problem 82.
(c) See Fig. 2 for the unit response of part (b). The rise time is approximately 1.5 seconds,
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 Spring '08
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