Solution to ECE Test #3 Su05
1.
For each pole-zero plot, indicate which frequency response plot corresponds to it by
writing in its letter designation.
F
B
ω
6
[s]
4
4
2
2
σ
0
D
ω
6
[s]
2
σ
0
2
4
4
4
6
6
6
6
0
2
4
6
4
2
0
C
4
6
6
6
[s]
4
2
2
σ
0
[s]
σ
4
4
6
6
6
6
6
4
2
0
A
4
6
6
6
4
2
0
10
1.2
0.2
ω
-20
20
-20
ω
-20
−π
D
E
|H(jω)|
-20
ω
-20
ω
-20
Phase ofH(jω)
ω
-20
20
−π
−π
20
ω
20
ω
20
ω
1.2
ω
-20
Phase ofH(jω)
π
20
ω
F
Phase ofH(jω)
π
20
|H(jω)|
0.04
20
6
π
20
|H(jω)|
0.1
4
Phase ofH(jω)
π
−π
−π
ω
Phase ofH(jω)
20
2
C
|H(jω)|
π
-20
2
B
20
[s]
σ
|H(jω)|
Phase ofH(jω)
-20
ω
|H(jω)|
-20
6
0
4
4
4
2
3
0
2
2
2
4
2
0
0
6
2
2
2
A
ω
6
4
4
4
E
ω
6
2
σ
0
2
2
[s]
4
2
4
ω
6
π
20
ω
-20
−π
2.
Sketch a root locus for each of these pole-zero plots assuming that these are the
poles and zeros of the loop transfer function of a closed-loop system with an adjustable
gain.
A
B
ω
ω
8
8
[s]
6
[s]
6
4
4
2
2
σ
0
-2
-2
-4
-4
-6
-6
-8
σ
0
-8
-8
-6
-4
-2
0
2
4
6
8
-8
-6
-4
-2
C
0
2
4
6
8
D
ω
ω
8
8
[s]
6
[s]
6
4
4
2
2
σ
0
-2
-2
-4
-4
-6
-6
-8
σ
0
-8
-8
-6
-4
-2
0
2
4
6
8
-8
-6
-4
(continued on next page)
-2
0
2
4
6
8
For each system indicate whether it will definitely be unstable at some finite positive value
of K ( 0 < K < ∞ ) or definitely will be stable for all finite positive K or it is impossible to be
sure. Explain your answer.
A
Definitely unstable at some K, 0 < K < ∞
X
Definitely stable for all K, 0 < K < ∞
Impossible to be sure
Explanation Two branches of the root locus move into the right half-plane at a finite
positive value of K.
B
Definitely unstable at some K, 0 < K < ∞
Definitely stable for all K, 0 < K < ∞
X
Impossible to be sure
Explanation For any K > 0, the root locus is in the left half-plane and stays there because
the two branches that go to infinity go vertically up and down.
C
Definitely unstable at some K, 0 < K < ∞
Definitely stable for all K, 0 < K < ∞
X
Impossible to be sure
Explanation All root locus branches terminate on finite zeros, two of which are in the left
half-plane and one of which is at the origin. Therefore for any K less than infinity, the root
locus stays in the left half-plane.
D
Definitely unstable at some K, 0 < K < ∞
X
Definitely stable for all K, 0 < K < ∞
Impossible to be sure
Explanation One pole is in the RHP and the locus starts there so the system is unstable at
low positive values of K.
3.
Sketch a canonical realization of a system with the transfer function
H (s) = 3
H (s) =
s ( s − 2)
.
( s + 6) s2 + 4 s + 4
(
)
3s 2 − 6 s
s 3 + 10 s 2 + 28 s + 24
3
6
X(s)
1
s
1
s
1
s
10
28
24
Y(s)
4.
Sketch a cascade realization of a system with the transfer function
H (s) = 3
H (s) = 3
X(s)
s+5
.
s + 8s + 7
2
s+5
s+5
1
= 3×
×
s + 7 s +1
( s + 7 ) ( s + 1)
1
s
7
5
1
s
3
Y(s)
5.
A feedback system of the convential type has a forward-path transfer function
K
H1 ( s ) =
s+8
and a feedback-path transfer function
1
H2 ( s) =
.
s + 12
K can be any real number. (Not just positive numbers.) For what range of K’s is this
system stable?
K
K ( s + 12 )
K ( s + 12 )
s+8
H (s) =
= 2
= 2
K
1
s + 20 s + 96 + K s + 20 s + 96 + K
1+
s + 8 s + 12
The roots of the denominator are at
p1, 2 =
−20 ± 400 − 384 − 4 K
= −10 ± 4 − K
2
For any K greater than 4 the system is stable because the real part of the poles is fixed at
-10 while the imaginary part varies. For any K less than 4 the roots are both real. When
4 − K equals or exceeds 10, one root has a non-negative real part. This occurs when
K ≤ −96 . Therefore for any K ≤ −96 the system is unstable. So the range of K’s for
stability is K > −96 .