8.Consider the case where
( )
cos4 ,
0
u t
t t
=
≥
and
( )
0
0
x
=
while
( )
0
0
x
=
dotnosp
.
Obtain the solution
for
( )
x t
utilizing the phasor method.
Repeat the problem but this time utilize the Laplace
transform method.
Is the answer the same?
Should it be?
From Problem 1,
( )
( )
(
)
0
0
2
2
2
2
5
2
5
U s
s x x
X s
s s
s s
+
+
=
+
+
+
+
+
dotnosp
.
As the initial conditions are both zero
( )
( )
2
2
5
U s
X s
s s
=
+
+
Phasor
Method
The phasor method, as shown in Lecture 35, gives the steady state portion of the
particular solution when the input is sinusoidal.
For this problem, as the initial conditions are
zero, the homogeneous solution is zero for all points in time.
Thus the phasor solution gives the
steady state portion of the total solution.
The abrupt start of the input causes a transient solution
to occur which is not captured by this method.
First find
(
)
(
)
(
)
2
1
2
5
X S
G S
U S S S
=
=
+
+
Recall, for a sinusoidal input, the phasor method gives the general solution,
( )
(
)
(
)
0
cos
p
x t u G j
t
ω
ω
β
φ
=
+
+
, where
(
)
(
)
Im
tan
Re
G j
G j
ω
φ
ω
=
, for the case where the input is
(
)
0
cos
u
t
ω
β
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 Spring '08
 Perreira
 Algebra, Steady State, Laplace, Partial fractions in complex analysis, Transient response, Laplace Transform Method The Laplace Transform Table

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