Digital Signal Processing Reference
In-Depth Information
Taking the Laplace transform of Eq. (6.29) yields
(
s
+
4)
Y
zs
(
s
)
=
2
s
s
2
+
4
.
Using the partial fraction expansion, the above equation is expressed as follows:
2
s
(
s
+
4)(
s
2
+
4)
≡−
0
.
4
(
s
+
4)
+
0
.
4
s
+
0
.
4
(
s
2
+
4)
Y
zs
(
s
)
=
.
Taking the inverse Laplace transform, the zero-state response is given by
−
4
t
+
0
.
4 cos(2
t
)
+
0
.
2 sin(2
t
)]
u
(
t
)
,
y
(
t
)
=
[
−
0
.
4e
which is same as the result derived in Example 3.2.
We also know from Chapter 3 that the overall response
y
(
t
) is the sum of
the zero-input response
y
zi
(
t
) and the zero-state response
y
zs
(
t
). This is easily
verifiable for the above results.
Example 6.17
In Example 3.3, the following differential equation
d
2
w
d
t
2
+
7
d
w
d
t
+
12
w
(
t
)
=
12
x
(
t
)
(6.32)
was used to model the RLC series circuit shown in Fig. 3.1. Determine the
zero-input, zero-state, and overall
response
of the system produced by the input
x
(
t
)
=
2e
−
t
u
(
t
) given the initial conditions,
w
(0
−
−
)
=
5
V
and
w
(0
)
=
0.
Solution
Overall response
The Laplace transforms of the individual terms in Eq. (6.32)
are given by
2
s
+
1
,
−
t
u
(
t
)
=
X
(
s
)
=
L
x
(
t
)
=
L
2e
W
(
s
)
=
L
w
(
t
)
,
d
w
d
t
−
L
=
sW
(
s
)
−
w
(0
)
=
sW
(
s
)
−
5
,
and
d
2
w
d
t
2
=
s
2
W
(
s
)
−
sw
(0
−
−
)
=
s
2
W
(
s
)
−
5
s
.
L
)
−
w
(0
Taking the Laplace transform of both sides of Eq. (6.32) and substituting the
above values yields
24
s
+
1
[
s
2
W
(
s
)
−
5
s
]
+
7[
sW
(
s
)
−
5]
+
12
W
(
s
)
=
or
=
5
s
2
+
40
s
+
59
s
+
1
24
s
+
1
[
s
2
+
7
s
+
12]
W
(
s
)
=
5
s
+
35
+
,
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