ECE 421, Spring 2006
Solutions to HW Assignment #5
Problem #B523
From the given
G
(
s
)
,
the closedloop transfer function and closedloop characteristic equation are
T
CL
(
s
)=
C
(
s
)
R
(
s
)
=
G
(
s
)
1+
G
(
s
)
=
K
s
(
s
+1)(
s
+2)+
K
=
K
s
3
+3
s
2
+2
s
+
K
(1)
∆
(
s
s
3
s
2
s
+
K
(2)
and the Routh array is
s
3
:
1
2
s
2
:
3
K
s
1
:
6
−
K
3
s
0
:
K
For closedloop stability, there must be no sign changes in the
f
rst column of the array. Checking the
s
0
and
s
1
rows (the only ones involving
K
)
produces the following limits on
K
for stability.
s
0
:
K>
0
,s
1
:6
−
0
⇒
K<
6
(3)
0
<K<
6
(4)
Problem #B524
G
(
s
)
,
the closedloop transfer function and closedloop characteristic equation are
T
(
s
C
(
s
)
R
(
s
)
=
G
(
s
)
G
(
s
)
=
10
s
(
s
−
1) (2
s
+3)+10
=
10
2
s
3
+
s
2
−
3
s
+10
(5)
∆
(
s
)=2
s
3
+
s
2
−
3
s
(6)
It is obvious by inspection of
∆
(
s
)
that the closedloop system is unstable since not all the coe
ﬃ
cients
have the same sign. Forming the Routh array to check this conclusion gives the following result.
s
3
:
2
−
3
s
2
:
1
10
s
1
:
−
23
s
0
:
10
Since there are 2 sign changes in the
f
rst column of the array, there are 2 closedloop poles in the righthalf
plane. Therefore, the closedloop system is unstable.
Problem #B525
From the given closedloop characteristic
∆
(
s
s
4
s
3
+(4+
K
)
s
2
+9
s
+25=0
(7)
the Routh array is
1
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4
:
1
(4 +
K
)
25
s
3
:
2
9
s
2
:
2
K
−
1
2
25
s
1
:
18
K
−
109
2
K
−
1
s
0
:
25
For closedloop stability, there must be no sign changes in the
f
rst column, so the
s
2
and
s
1
rows must
be checked.
s
2
:2
K
−
1
>
0
⇒
K>
0
.
5
(8)
s
1
:1
8
K
−
109
>
0
⇒
109
/
18 = 6
.
056
(9)
Since the constraint from the
s
1
row is more restrictive than the costraint from the
s
2
row, the limit on
K
for closedloop stability is
6
.
056
.
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 Spring '08
 Cook,G
 closedloop characteristic equation

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