Homework 2 Solution
Solutions by Jiachen Xue and Ramamurthy
Mani
Problem 5 (a)
H
(
S
) can be expressed as
H
(
s
)
=
K
Q
m
j
=1
(
s

z
j
)
Q
n
i
=1
(
s

p
i
)
=
n
X
i
=1
K
i
s

p
i
(1)
where
z
j
is the
j
th
zero and
p
i
is the
i
th
pole of the transfer function.
K
i
is the coefficient. Take
the inverse Laplace transformation of
H
(
s
) to get the impulse response.
h
(
t
)
=
L

1
[
H
(
s
)]
=
L

1
"
n
X
i
=1
K
i
s

p
i
#
=
n
X
i
=1
K
i
e
p
i
t
(2)
If
p
i
<
0, that is to say there is a pole existing on the right half plane, then
lim
t
→∞

K
i
e
p
i
t
 → ∞
.
(3)
Thus
Z
∞
∞

h
(
t
)

dt
→ ∞
.
(4)
which makes the system unstable.
(b)
Assume
H
(
s
) has zeros
z
1
, z
2
on the right plan:
z
1
=
σ
j
+
jω
j
z
2
=
σ
j

jω
j
in which
σ
j
>
0 and
ω
j
6
= 0.
H
(
s
) can be further expended as
H
(
s
)
=
H
min
(
s
) (
s

σ
j

jω
j
) (
s

σ
j
+
jω
j
)
=
H
min
(
s
)
£
(
s

σ
2
j
) +
ω
2
j
/
(5)
where
H
min
(
s
) has all its zeros on the left plane. Thus,
H
(
s
) can be written as
H
(
s
)
=
£
H
min
(
s
)
(
(
s
+
σ
2
j
) +
ω
2
j
)/
(
s

σ
2
j
) +
ω
2
j
(
s
+
σ
2
j
) +
ω
2
j
=
H
1
(
s
)
A
(
s
)
(6)
in which
H
1
(
s
) =
H
min
(
s
)
(
(
s
+
σ
2
j
) +
ω
2
j
)
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 Fall '07
 Chakrabarti
 Hmin

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