|
13.7.2
frequency-shifting theorem
If
v
(
t
)
=
f
(
t
)
u
(
t
) has a Laplace transform
V
(
s
), then,
v
d
(
t
)
=
v t e
s t
o
( )
has a Laplace
transform
V s
V s s
d
o
( )
(
).
=
−
This theorem follows from the defining equation
for Laplace transforms.
V s
v t e dt
v t e e dt
v t e
d
d
st
s t
st
s s t
o
o
( )
( )
( )
( )
(
)
=
=
==
−
∞
−
∞
− −
∞
−
−
−
∫
∫
0
0
0
∫∫
=
−
dt V s s
o
(
)
13.7.3
time-differentiation theorem
If
v
(
t
)
=
f
(
t
)
u
(
t
) has a Laplace transform
V
(
s
), then,
v
d
(
t
)
=
dv t
dt
( )
has a Laplace transform
V s
sV s
v
d
( )
( )
( ).
=
−
−
0
Note that
dv t
dt
df t
dt
u t
( )
( )
( ).
≠
×
V s
dv t
dt
e dt
d v t e
dt
sv t e
dv t
dt
e
d
st
st
st
s
( )
( )
[ ( )
]
( )
( )
=
= −
+
−
∞
−
−
−
−
∫
0
tt
st
st
st
d
st
dv t
dt
e
d v t e
dt
sv t e
V s
dv t
dt
e
∴
=
+
∴
=
−
−
−
−
−
( )
[ ( )
]
( )
( )
( )
0
∞
∞
−
∞
−
∞
−
∫
∫
∫
=
+
=
+
−
−
−
dt s
v t e dt
d v t e
dt
dt
sV s
v t e
st
st
st
( )
[ ( )
]
( )
( )
0
0
0
∞
∞
The function
v t e
st
( )
-
will be a decaying function for any value of
s
in the ROC of
V
(
s
). Otherwise,
the Laplace transform will not converge for that value of
s
. Therefore, it will go to zero as
t
→
∞
.
∴
=
−
−
V s
sV s
v
d
( )
( )
( )
0
Now, by using mathematical induction, we may show that,
Laplace transform of
d v t
dt
s V s
sv
dv t
dt
2
2
2
0
0
( )
( )
( )
( )
(
)
=
−
−
−
−
Fig. 13.7-2
Output response and its
components in the circuit
in Example: 13.7-1
5
2
4
–5
10
(V)
v
S
(
t
)
v
(
t
)
(
s
)
Time
13.22
Analysis of Dynamic Circuits by Laplace Transforms
and that, in general, Laplace transform of
d v t
dt
s V s
s
v
s
dv t
dt
d
v t
dt
n
n
n
n
n
n
n
( )
( )
( )
( )
( )
(
)
=
−
−
−
−
−
−
−
−
−
−
1
2
0
1
0
11
0
(
)
−
13.7.4
time-Integration theorem
If
v
(
t
)
=
f
(
t
)
u
(
t
) has a Laplace transform
V
(
s
), then,
v
i
(
t
)
=
v t dt
t
( )
0
−
∫
has a Laplace transform
V s
V s
s
i
( )
( )
.
=
V s
v t dt e dt
d
v t dt e
dt
v t e
s
i
t
st
t
st
st
( )
( )
( )
( )
=
(
)
(
)
=
−
−
−
−
∫
∫
∫
−
∞
−
−
0
0
0
vv t dt e
V s
v t dt e dt
s
v t e
t
st
i
t
st
st
( )
( )
( )
( )
0
0
0
0
1
−
−
−
∫
∫
∫
(
)
∴
=
(
)
=
−
−
∞
−
−−
−
−
−
∞
−
∞
−
∞
∫
∫
∫
∫
+
(
)
=
+
dt
s
d
v t dt e
dt
dt
s
v t e dt
s
v t
t
st
st
1
1
1
0
0
0
( )
( )
( )
ddt e
t
st
0
0
−
−
∫
(
)
−
∞
The function
v
(
t
) is stated to possess a Laplace transform. This implies that there is an exponential
function
Me
a
t
with some positive value of
M
and some real value for
a
such that |
v
(
t
)| <
Me
a
t
.
Otherwise,
v
(
t
) would not have a Laplace transform. Therefore, the function
v t dt
t
( )
0
−
∫
will satisfy the
inequality |
v t dt
t
( )
0
−
∫
| < |
M/
a
(
e
a
t
– 1)| and therefore is bounded. Therefore, the Laplace transform of
v t dt
t
( )
0
−
∫
will exist. That is, it is possible to select a value for
s
such that the function
v t dt
t
( )
0
−
∫
×
e
-
st
is
a decaying function. For such an
s
,
i.e.,
for a value of
s
in the ROC of Laplace transform of
v t dt
t
( ) ,
0
−
∫
the value of
v t dt e
t
st
( )
0
−
∫
(
)
−
will go to zero as
t
→
∞
. And, the value of
v t dt e
t
st
( )
0
−
∫
(
)
−
at
t
=
0
-
is zero
in any case.
∴
=
=
−
∞
−
∫
V s
s
v t e dt
V s
s
i
st
( )
( )
( )
1
0
example: 13.7-2
Find the Laplace transform of
t
n
u
(
t
).
Solution
The function
tu
(
t
) is the integral of
u
(
t
). Therefore,
tu t
s
( )
/
⇔
1
2
. Now the function
t
2
u
(
t
) is
2 times the integral of
tu
(
t
). Therefore,
t u t
s
2
3
2
( )
/
⇔
. Proceeding similarly to the power
n
, we get,
t u t
n s
n
n
( )
!/
⇔
+
1
.
Some Theorems on Laplace Transforms
13.23
13.7.5
s
-domain-differentiation theorem
If
v
(
t
)
=
f
(
t
)
u
(
t
) has a Laplace transform
V
(
s
), then,
-
tv
(
t
) has a Laplace transform
dV s
ds
( )
.
We show this by determining the Laplace transform of
-
tv
(
t
) from the defining integral.
−
=
−
−
−
∞
−
∞
−
∫
∫
tv t e dt
v t te
dt
st
st
0
0
( )
( )[
]
We have repeatedly used the limit
lim
(
)
∆
∆
∆
a
a
a
a
a
a
→
+
−
=
0
e
e
te
t
t
t
many times before. We use this limit
again with
a
= -
s
within the integral.
∴
−
=
−
−
−
−
−
∞
−
∞
→
− +
−
∫
∫
tv t e dt
v t
e
e
s
d
st
s
s
s t
st
0
0
0
( )
( ) lim
(
)
∆
∆
∆
tt
Since the limiting operation is on
s
and integration is on
t
we may interchange the order of these
two operations.
∴
−
=
−
−
−
−
−
∞
−
→
∞
− +
−
∫
∫
tv t e dt
v t
e
e
s
d
st
s
s
s t
st
0
0 0
( )
lim
( )
(
)
∆
∆
∆
tt
s
v t e
e
dt
s
v t e
s
s
s t
st
s
=
−
=
→
− +
−
∞
→
−
−
∫
lim
( )
lim
( )
(
)
∆
∆
∆
∆
∆
0
0
0
1
1
((
)
( )
lim
(
)
( )
l
s
s t
st
s
dt
v t e
dt
s
V s
s
V s
+
∞
−
∞
→
−
−
∫
∫
−
(
)
=
+
−
(
)
=
∆
∆
∆
∆
0
0
0
1
iim
(
)
( )
( )
∆
∆
∆
s
V s
s
V s
s
dV s
ds
→
+
−
=
0
13.7.6
s
-domain-Integration theorem
If
v
(
t
)
=
f
(
t
)
u
(
t
) has a Laplace transform
V
(
s
) and
lim
( )
t
v t
t
→
0
is finite, then,
v t
t
( )
has a Laplace
transform
V s ds
s
( ) .
∞
∫
V s ds
v t e dt ds
v t
e ds
s
st
s
st
s
( )
( )
( )
∞
−
∞
∞
−
∞
∞
∫
∫
∫
∫
=
=
−
−
0
0
∫∫
∫
=
−
−
−∞
∞
−
dt
v t
e
e
t
dt
st
t
( )
0
If the integration
V s ds
s
( )
∞
∫
is carried out in the right-half of
s
-plane (ROC of right-sided functions
will have at least a part of right-half
s
-plane in it), then
e
t
−∞
in the last step in the equation above will
vanish. Then,
V s ds
v t
t
e
dt
v t
t
s
st
( )
( )
( )
∞
−
∞
∫
∫
=
=
−
0
Laplace Transform of
.
13.24
Analysis of Dynamic Circuits by Laplace Transforms
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