Digital Signal Processing Reference
In-Depth Information
Solution
Substituting
L
=
1
/
12 H,
R
=
7
/
12
, and
C
=
1 F into Eq. (3.6) and multi-
plying both sides of the equation by 12 yields
d
2
w
d
t
2
+
7
d
w
d
t
+
12
w
(
t
)
=
12
x
(
t
)
,
(3.18)
−
−
with initial conditions,
w
(0
)
=
5 and
w
(0
)
=
0, and the input signal is given
−
t
u
(
t
).
by
x
(
t
)
=
2e
(i) Zero-input response of the system
Based on Eq. (3.18), the characteristic
equation of the LTIC system is given by
s
2
+
7
s
+
12
=
0
,
which has roots at
s
=−
4
, −
3. The zero-input response is therefore given by
−
4
t
−
3
t
)
u
(
t
)
,
w
zi
(
t
)
=
(
A
e
+
B
e
where
A
and
B
are constants. To calculate the value of the constants, we sub-
stitute the initial conditions
w
(0
−
−
)
=
0 in the above equation.
The resulting simultaneous equations are as follows:
)
=
5 and
w
(0
A
+
B
=
5
,
4
A
+
3
B
=
0
,
which have the solution
A
=−
15 and
B
=
20. The zero-input response is
therefore given by
−
3
t
−
4
t
)
u
(
t
)
.
w
zi
(
t
)
=
(20e
−
15e
(ii) Zero-state response of the system
To calculate the zero-state response
of the system, the initial conditions are assumed to be zero, i.e. the capaci-
tor is assumed to be uncharged. Hence, the zero-state response
w
zs
(
t
) can be
calculated by solving the following differential equation:
d
2
w
d
t
2
+
7
d
w
d
t
+
12
w
(
t
)
=
12
x
(
t
)
,
(3.19)
−
=
−
=
=
with initial conditions,
w
(0
)
0 and
w
(0
)
0, and input
x
(
t
)
2 exp(
−
t
)
u
(
t
).
The homogeneous solution of Eq. (3.18) has the same form as the zero-input
response and is given by
w
(h)
−
4
t
−
3
t
,
zs
(
t
)
=
C
1
e
+
C
2
e
where
C
1
and
C
2
are constants. The particular solution for input
x
(
t
)
=
2e
−
t
u
(
t
)
is of the form
w
(p)
−
t
u
(
t
). Substituting the particular solution into
zs
(
t
)
=
K
e
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