Digital Signal Processing Reference
In-Depth Information
which reduces to
5
s
2
+
40
s
+
59
(
s
+
1) (
s
2
+
7
s
+
12)
=
5
s
2
+
40
s
+
59
(
s
+
1)(
s
+
3)(
s
+
4)
.
Taking the partial fraction expansion, we obtain
5
s
2
+
40
s
+
59
(
s
+
1)(
s
+
3)(
s
+
4)
W
(
s
)
=
k
1
(
s
+
1)
k
2
(
s
+
3)
k
3
(
s
+
4)
,
≡
+
+
where the partial fraction coefficients are given by
5
s
2
+
40
s
+
59
(
s
+
1)(
s
+
3)(
s
+
4)
=
5
−
40
+
59
(2)(3)
k
1
=
(
s
+
1)
=
4
,
s
=−
1
5
s
2
+
40
s
+
59
(
s
+
1)(
s
+
3)(
s
+
4)
=
45
−
120
+
59
(
−
2)(1)
k
2
=
(
s
+
3)
=
8
,
s
=−
3
and
5
s
2
+
40
s
+
59
(
s
+
1)(
s
+
3)(
s
+
4)
=
80
−
160
+
59
(
−
3)(
−
1)
k
3
=
(
s
+
4)
=−
7
.
s
=−
4
Substituting the values of the partial fraction coefficients
k
1
,
k
2
, and
k
3
,we
obtain
4
(
s
+
1)
8
(
s
+
3)
−
7
(
s
+
4)
.
Calculating the inverse Laplace transform of both sides, we obtain the output
signal as follows:
W
(
s
)
≡
+
−
t
−
3
t
−
4
t
]
u
(
t
)
.
w
(
t
)
≡
[4e
+
8e
−
7e
Zero-input response
To calculate the zero-input output, the input signal is
assumed to be zero. Equation (6.32) reduces to
d
2
w
zi
d
t
2
+
7
d
w
zi
d
t
+
12
w
zi
(
t
)
=
0
,
(6.33)
−
−
with initial conditions
w
(0
)
=
5 and
w
(0
)
=
0. Calculating the Laplace
transform of Eq. 6.33 yields
[
s
2
W
zi
(
s
)
−
5
s
]
+
7[
sW
zi
(
s
)
−
5]
+
12
W
zi
(
s
)
=
0
or
W
zi
(
s
)
=
5
s
+
35
s
2
+
7
s
+
12
.
Using the partial fraction expansion, the above equation is expressed as follows:
5
s
+
35
s
2
+
7
s
+
12
20
s
+
3
−
15
s
+
4
.
Taking the inverse Laplace transform, the zero-input response is given by
W
zi
(
s
)
=
≡
−
3
t
−
4
t
]
u
(
t
)
.
w
zi
(
t
)
≡
[20e
−
15e
Search WWH ::
Custom Search