Digital Signal Processing Reference
In-Depth Information
The proof of the time-integration property is left as an exercise for the reader
(see Problem 6.7).
Example 6.13
Given the Laplace transform pair
cos(
ω
0
t
)
u
(
t
)
s
(
s
2
+
←→
with ROC: Re
{
s
}>
0
,
ω
0
)
derive the unilateral Laplace transform of sin(
ω
0
t
)
u
(
t
).
Solution
By applying the time-integration property to the aforementioned unilateral
Laplace transform pair yields
t
←→
1
s
s
cos(
ω
0
τ
)
u
(
τ
)d
τ
with ROC: Re
{
s
}>
0
,
ω
0
s
2
+
0
−
where the left-hand side of the transform pair is given by
t
t
0
t
=
sin(
ω
0
τ
)
ω
0
1
ω
0
cos(
ω
0
τ
)
u
(
τ
)d
τ
=
cos(
ω
0
τ
)d
τ
=
sin(
ω
0
t
)
.
0
−
0
Substituting the value of the integral in the transform pair, we obtain
sin(
ω
0
t
)
u
(
t
)
ω
0
(
s
2
+
←→
with ROC: Re
{
s
}>
0
,
ω
0
)
6.4.7 Time and s-plane convolution
If
x
1
(
t
) and
x
2
(
t
) are two arbitrary functions with the following Laplace trans-
form pairs:
x
1
(
t
)
←→
←→
X
1
(
s
)
with ROC:
R
1
and
x
2
(
t
)
X
2
(
s
)
with ROC:
R
2
,
then the convolution property states that
←→
X
1
(
s
)
X
2
(
s
)
containing at least ROC:
R
1
∩
R
2
;
time convolution
x
1
(
t
)
∗
x
2
(
t
)
(6.25)
←→
1
s-plane convolution
x
1
(
t
)
x
2
(
t
)
2
π
j
[
X
1
(
s
)
∗
X
2
(
s
)]
containing at least ROC:
R
1
∩
R
2
.
(6.26)
The convolution property is valid for both unilateral (for causal signals) and
bilateral (for non-causal signals), Laplace transforms. The overall ROC of the
convolved signals may be larger than the intersection of regions
R
1
and
R
2
because of possible cancellation of poles in the products.
Proof
Consider the Laplace transform of
x
1
(
t
)
∗
x
2
(
t
):
∞
∞
∞
−
st
d
t
−
st
d
t
.
L
x
1
(
t
)
∗
x
2
(
t
)
=
[
x
1
(
t
)
∗
x
2
(
t
)]e
=
x
1
(
τ
)
x
2
(
t
− τ
)d
τ
e
−∞
0
−
0
−
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