Digital Signal Processing Reference
In-Depth Information
Calculating the CTFT of both sides of Eq. (5.66) and applying the time-
differentiation property, we obtain
n
m
a
k
(j
ω
)
k
Y
(
ω
)
=
b
k
(j
ω
)
k
X
(
ω
)
k
=
0
k
=
0
or
n
b
k
(j
ω
)
k
H
(
ω
)
=
Y
(
ω
)
X
(
ω
)
k
=
0
=
.
(5.67)
m
a
k
(j
ω
)
k
k
=
0
Given one representation for an LTIC system, it is straightforward to derive
the remaining two representations based on the CTFT and its properties. We
illustrate the procedure through the following examples.
Example 5.27
Consider an LTIC system whose input-output relationship is modeled by the
following third-order differential equation:
d
3
y
d
t
3
+
6
d
2
y
d
t
2
+
11
d
y
d
t
+
6
y
(
t
)
=
2
d
x
d
t
+
3
x
(
t
)
.
(5.68)
Calculate the transfer function
H
(
ω
) and the impulse response
h
(
t
) for the LTIC
system.
Solution
Using the time-differentiation property for the CTFT, we know that
d
n
x
d
t
n
CTFT
←−−→
(j
ω
)
n
X
(
ω
)
.
Taking
the
CTFT
of
both
sides
of
Eq.
(5.47)
and
applying
the
time-
differentiation property yields
(j
ω
)
3
Y
(
ω
)
+
6( j
ω
)
2
Y
(
ω
)
+
11( j
ω
)
Y
(
ω
)
+
6
Y
(
ω
)
=
2( j
ω
)
X
(
ω
)
+
3
X
(
ω
)
.
Making
Y
(
ω
) common on the left-hand side of the above expression, we obtain
[( j
ω
)
3
+
6( j
ω
)
2
+
11( j
ω
)
+
6]
Y
(
ω
)
=
[2( j
ω
)
+
3]
X
(
ω
)
.
Based on Eq. (5.46), the transfer function is therefore given by
H
(
ω
)
=
Y
(
ω
)
X
(
ω
)
2( j
ω
)
+
3
(j
ω
)
3
+
6( j
ω
)
2
+
11( j
ω
)
+
6
.
=
(5.69)
The impulse response
h
(
t
) is obtained by taking the inverse CTFT of Eq. (5.69).
Factorizing the denominator, Eq. (5.69) is expressed as
2( j
ω
)
+
3
(1
+
j
ω
)(2
+
j
ω
)(3
+
j
ω
)
,
H
(
ω
)
=
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