Digital Signal Processing Reference
In-Depth Information
Solution
Taking the derivative of the CTFT pair for the cosine function yields
d
d
t
cos(
ω
0
t
)
CTFT
←−−→
(j
ω
)
π
[
δ
(
ω − ω
0
)
+ δ
(
ω + ω
0
)]
.
By rearranging terms, we obtain
CTFT
←−−→
−ω
0
sin(
ω
0
t
)
j
π
[
ω
0
δ
(
ω − ω
0
)
− ω
0
δ
(
ω + ω
0
)]
,
which can be expressed as follows:
←−−→
π
j
CTFT
ω
0
sin(
ω
0
t
)
[
ω
0
δ
(
ω − ω
0
)
− ω
0
δ
(
ω + ω
0
)]
,
obtained by using the multiplicative property of the impulse function,
x
(
t
)
δ
(
t
+
t
0
)
=
x
(
−
t
0
)
δ
(
t
+
t
0
). The CTFT of the periodic sine function is
therefore given by
←−−→
π
j
CTFT
sin(
ω
0
t
)
[
δ
(
ω − ω
0
)
− δ
(
ω + ω
0
)]
.
5.5.6 Time integration
The time-integration property expresses the CTFT of a time-integrated signal
∫
x
(
t
)d
t
in terms of the CTFT of the original signal
x
(
t
).
CTFT
←−−→
If x
(
t
)
X
(
ω
)
, then
t
X
(
ω
)
j
ω
CTFT
←−−→
x
(
τ
)d
τ
+ π
X
(0)
δ
(
ω
)
.
(5.53)
−∞
The proof of the time-integration property is left as an exercise for the reader
(see Problem 5.14).
Example 5.19
Given
δ
(
t
)
CTFT
←−−→
1, calculate the CTFT of the unit step function
u
(
t
) using
the time-integration property.
Solution
Integrating the CTFT pair for the unit impulse function yields
t
1
j
ω
CTFT
←−−→
δ
(
t
)d
t
+ πδ
(
ω
)
.
−∞
By noting that the left-hand side of the aforementioned CTFT pair represents
the unit step function, we obtain
1
j
ω
CTFT
←−−→
u
(
t
)
+ πδ
(
ω
)
.
The above CTFT pair can be verified from Table 5.2.
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