EEE 480
HW # 6 SOLUTIONS
Problem 5.21
The rootlocus asymptotes for puregain compensation are at
¡
2
:
5.
Since the required closedloop pole
locations are
¡
1
§
j
, we need to introduce a lag compensator. Observe that asymptotes at
¡
1 are obtained for
poles at 0
;
¡
2. Hence, a possible controller would cancel the system pole at
¡
3 and put a pole at the origin.
That is,
C
(
s
) =
K
s
+ 3
s
To compute
K
, we can form the closedloop characteristic equation (simple enough), (
s
2
+ 2
s
+
K
)(
s
+ 3). One
of its roots is at
¡
3 due to the cancellation; the other two should be at
¡
1
§
j
and have magnitude
p
2, hence
K
= 2.
Problem 5.23
Here, the puregain asymptotes are at 0 and we need a lead compensator to achieve the desired speci¯cation.
At
¡
2
§
j
2, the system double integrator contributes 2
£
135
o
= 270
o
, so the compensator should contribute
90
o
phase lead.
Choosing the pole angle as 10
o
, the zero should contribute 100
o
. Computing the corresponding locations:
²
pole: tan 10
o
= 0
:
176
)
p
=
2
0
:
176
+ 2 = 13
:
3.
²
zero:
z
= 2
¡
(2)(0
:
176) = 1
:
65.
²
gain:
K
=
¯
¯
¯
s
2
(
s
+
p
)
s
+
z
¯
¯
¯
¯
¯
¯
s
=
¡
2+
j
2
=
(8)(
p
11
:
3
2
+2
2
)
p
0
:
35
2
+2
2
= 45
:
21.
The ¯nal compensator is
C
(
s
) = 45
:
21
s
+ 1
:
65
s
+ 13
:
3
Problem 5.25
Here, the puregain asymptotes are at 0.5 and we need a lead compensator to achieve the desired speci¯cation.
While the procedure used in Problem 5.23 is applicable, let us use a di®erent approach that can yield quick
results for simple problems.
Guided by the expected rootlocus of the compensated system, we select the desired closedloop characteristic
polynomial as
(
s
2
+ 2
s
+ 4)(
s
+ 10) =
s
3
+ 12
s
2
+ 24
s
+ 40
The ¯rst factor is from the speci¯ed dominant pair and the second factor is a reasonable choice for the third
closedloop pole.
Next, we write the closedloop characteristic polynomial in terms of the lead compensator
parameters
(
s
2
+
s
)(
s
+
p
) +
K
(
s
+
z
) =
s
3
+ (1 +
p
)
s
2
+ (
K
+
p
)
s
+
Kz
Equating the coe±cients of the two polynomials we ¯nd:
²
p
= 12
¡
1 = 11
²
K
= 24
¡
p
= 13
²
z
= 40
=K
= 3
:
08.
The ¯nal compensator is
C
(
s
) = 13
s
+ 3
:
08
s
+ 11
This approach is based on the socalled \poleplacement" design which computes the parameters of a general
(highorder) compensator to assign the closedloop poles to desired locations. The advantage of pole placement
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '10
 GU
 #, 36 K, compensators, ¡1, µlead

Click to edit the document details