360
Practical MATLAB
® Applications for Engineers
60
2
6
e u t
di t
dt
i t
t
Ϫ ( )
( )
( )
ϭ
ϩ
Step b
60
1
2
6
s
sI s
I s
ϩ
ϭ
ϩ
( )
( ) (
)
Taking the LT
observe that the initial current i L (0) = 0, then
60
1
2
6
s
I s s
ϩ
ϭ
ϩ
( )[
]
Step c
I s
s
s
s
s
( )
(
)(
)
(
)(
)
ϭ ϩ
ϩ
ϭ
ϩ
ϩ
60
1 2
6
60
2
1
3
I s
s
s
A
s
B
s
( )
(
)(
)
(
ϭ ϩ
ϩ
ϭ ϩ
ϩ ϩ
30
1
3
1
3
by partial fractions expansion) )
A
s
s
ϭ ϩ
ϭ
ϭ
ϭ
30
3
30
2
15
1
Ϫ
A
s
s
ϭ ϩ
ϭ
ϭ
ϭ
30
1
30
2
15
3
Ϫ
Ϫ
Ϫ
then
I s
s
s
( ) ϭ ϩ
ϩ ϩ
15
1
15
3
Ϫ
taking the ILT
Step d
i t
e u t
e u t
t
t
( )
( )
( )
ϭ
Ϫ
Ϫ
Ϫ
15
15
3
V(t ) = 60 e − t
L = 2 H
R = 6 Ω
R = 6 Ω
t =
0
+
−
i (t )
Switch moves
upwards at t = 0
FIGURE 4.18
Network of R.4.112.
CRC_47760_CH004.indd 360
CRC_47760_CH004.indd 360
7/28/2008 12:26:02 PM
7/28/2008 12:26:02 PM