2260
F 09
HOMEWORK #6 prob 1 solution
E
X
:
a)
Find
L
e
−
5
t
(
t
−
2)
u
(
t
−
2)
{
}
.
b)
Find
v
(
t
) if
V
(
s
)
=
1
s
+
31
s
+
39
s
2
+
2
s
+
17
.
c)
Find
lim
t
→
0
+
v
(
t
)
if
V
(
s
)
=
23
s
2
−
4
29
s
3
+
5
s
2
+
37
.
d)
Plot the poles and zeros of
V
(
s
) in the
s
plane.
V
(
s
)
=
s
2
+
14
s
+
45
(
s
2
+
49)(
s
+
19)
S
OL
'
N
:
a)
We may transform the delayed signals first and, later, add the exponential.
We use the delay identity:
L
{
f
(
t
−
a
)
u
(
t
−
a
)}
=
e
−
as
L
{
f
(
t
)}
Here, this applies to the portion obtained by omitting the exponential:
L
e
−
5
t
(
t
−
2)
u
(
t
−
2)
{
}
=
e
−
2
s
L
tu
(
t
)
{
}
=
e
−
2
s
L
t
{ }
=
e
−
2
s
s
2
Now, we include the exponential and apply the corresponding identity:
L
{
e
−
at
f
(
t
)}
=
F
(
s
+
a
)
This means we replace
s
with
s
+ 5 in our previous result to obtain our
final answer:
L
e
−
5
t
(
t
−
2)
u
(
t
−
2)
{
}
=
e
−
2(
s
+
5)
(
s
+
5)
2
An alternative approach is to write the exponential as a delayed function
and use only the delay identity:
L
e
−
5
t
(
t
−
2)
u
(
t
−
2)
{
}
=
L
e
−
10
e
−
5(
t
−
2)
(
t
−
2)
u
(
t
−
2)
{
}
After taking the constant term,
e
−
10
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- Spring '12
- lim, constant term
-
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