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SWITCHING EXAMPLE—LECTURE 10

DATA: iin(t ) 2u(t ) 4u(t  1) 2u(t  2) , i L (0 ) 0 ,

Switch moves to A at t = 0; Switch moves back to B at t = 1;

Switch moves back to A at t = 2 and remains there.

Step 1: 0 ≤ t < 1 s

Equivalent circuit in s-domain:

 I in(s) 2

sand  I  L (s)

2

s

2

s 1

2

s 4

s(s 2) 2

s 2

s 2

  

  . Therefore

i L (t ) 2 1 e2t  u(t ), 0 t  1 .

Step 2: 1 ≤ t < 2 s or 0 ≤ t' < 1 where t' = t – 1.

Equivalent circuit in s-domain with respect to t':

One observes that the current source which accounts for the IC on the inductor at t  1  

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is part of the value of   I  L (s) in contrast to what Professor Ray did in class!!!!! He made the

same mistake he warned you about. Nevertheless the current moving up through the resistor 

(from bottom to top) is  I  L (s). Thus

 I  L (s) 0.5

0.5 2

s

i L (1 )

s i L (1 )

s 4  

Hence

i L (t ') i L (1 )e4t 'u(t ') 0 ≤ t' < 1 or equivalently, i L (t ) i L (1 )e4(t 1)

u(t  1) for 1 ≤ t < 1.

Step 3: 2 ≤ t < 3 s or 0 ≤ t' < 1 where t' = t – 2.

Equivalent circuit in s-domain with respect to t':

 I in(s) 4

s. By superposition,

 I  L (s)

2

s

2

s 1

4

s

1

12

s

i L (2 )

s

8

s(s 2)

i L (2 )

s 2

  4

s

4

s 2

  

 

i L (2 )

s 2

 

Therefore, i L (t ') 4 1 e2t ' u(t ') i L (2 )e2t '

u(t ') valid for 0 t ' or equivalently

i L (t ) 4 1 e2(t 2) u(t  2) i L (2 )e2(t 2)

u(t  2) valid for 2 t