360 Practical MATLAB® Applications for Engineers
60 eut 2 di t 6
dt
t()() it()
Step b
60
1
26
s
sI s I s
() () (Taking the LT)
observe that the initial current iL(0) = 0 , then
60
1
26
s
Is s
()[ ]
Step c
Is
ss ss
()
()( )()( )
60
12 6
60
21 3
Is
ss
A
s
B
s
()
()( )
(
30
13 1 3
by partial fractions expansion))
A
s s
30
3
30
2
15
(^1)
A
s s
30
1
30
2
15
3
then
Is
ss
()
15
1
15
3
taking the ILT
Step d
it e ut e ut
() 15 tt() 15 3 ()
V(t) = 60 e− t
L = 2 H
R = 6 Ω R = 6 Ω
t = 0
- −
Switch moves upwards at t = 0 i(t)
FIGURE 4.18
Network of R.4.112.