202 lec 10
TRANSCRIPT
8/7/2019 202 Lec 10
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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