PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

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.