Digital Signal Processing Reference
In-Depth Information
Solution
Using the above Laplace transform pair and
s
-shifting property, the Laplace
transforms of exp( j
ω
0
t
)
u
(
t
) and exp(
−
j
ω
0
t
)
u
(
t
) are given by
1
(
s
−
j
ω
0
)
←→
e
j
ω
0
t
u
(
t
)
with ROC: Re
s
>
0
and
1
(
s
+
j
ω
0
)
←→
−
j
ω
0
t
u
(
t
)
e
with ROC: Re
s
>
0
.
(i) To calculate the Laplace transform of
x
1
(
t
)
=
cos(
ω
0
t
)
u
(
t
), we add the
above transform pairs to obtain
1
(
s
−
j
ω
0
)
1
(
s
+
j
ω
0
)
←→
e
j
ω
0
t
u
(
t
)
+
e
−
j
ω
0
t
u
(
t
)
+
with ROC: Re
s
>
0
,
which reduces to
s
s
2
+ ω
0
←→
cos(
ω
0
t
)
u
(
t
)
with ROC: Re
s
>
0
.
(ii) To evaluate the Laplace transform of
x
2
(
t
)
=
sin(
ω
0
t
)
u
(
t
), we take the
difference of the above transform pairs to obtain
1
(
s
−
j
ω
0
)
1
(
s
+
j
ω
0
)
←→
e
j
ω
0
t
u
(
t
)
−
e
−
j
ω
0
t
u
(
t
)
−
with ROC: Re
s
>
0
,
which simplifies to
ω
0
s
2
+ ω
0
←→
sin(
ω
0
t
)
u
(
t
)
with ROC: Re
s
>
0
.
6.4.5 Time differentiation
←→
If
x
(
t
)
X
(
s
) with ROC:
R
, then the Laplace transform of
d
x
d
t
←→
−
sX
(
s
)
−
x
(0
)
with ROC:
R
.
(6.21)
−
Note that if the function
x
(
t
) is causal,
x
(0
)
=
0.
Proof
By Eq. (6.9), the Laplace transform of the derivative d
x
/
d
t
is given by
∞
d
x
d
t
d
x
d
t
−
st
d
t
.
L
=
e
0
−
Applying integration by parts on the right-hand side of the equation yields
∞
∞
d
x
d
t
−
st
−
st
d
t
.
L
=
x
(
t
)e
−
(
−
s
)
x
(
t
)e
A
0
−
0
−
X
(
s
)
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