Page 118 - Circuit Analysis II with MATLAB Applications

Chapter 3 Elementary Signals

vt = e – 2t u t – e – 2t u t – 2 + 10tu t – 2 – 30u t – 2 – 10tu t – 3 + 30u t – 3

0
0
0
0
0
0
– 10tu t – 0 3 + 50u t – 3 + 10tu t – 5 – 50u t – 5 + 10tu t – 5

0
0
0
0
70u t – – 0 5 – 10tu t – 7 + 70u t – 7

0
0
e – 2t u t + e – = – 2t + 10t – 30 u t – 2 + – 20t + 80 u t – 3 + 20t – 120 u t – 5

0
0
0
0
+ – 10t + 70 u t – 0 7

b.
dv – 2t – 2t – 2t – 2t
------ = – 2e u t + e G t + 2e + 10 u t – 2 + e – + 10t – 30 G t – 2

dt 0 0
20u t – – 0 3 + – 20t + 80 G t – 3 + 20u t – 5 + 20t – 120 G t – 5 (1)

0
10u t – – 0 7 + – 10t + 70 G t – 7

Referring to the given waveform we observe that discontinuities occur only at t = 2 , t = , 3
and t = 5 . Therefore, G t = 0 and G t – 7 = 0 . Also, by the sampling property of the delta

function
– 2t – 2t
–

e – + 10t – 30 G t – 2 = e – + 10t – 30 G t – 2 | 10G t – 2
t = 2

– 20t + 80 G t – 3 = – 20t + 80 G t – 3 = 20G t – 3

t = 3

20t – 120 G t – 5 = 20t – 120 G t – 5 = – 20G t – 5

t = 5
and with these simplifications (1) above reduces to
dv dt = – 2e – 2t u t + 2e – 2t u t – 2 + 10u t – 2 – 10G t – 2

e

0
0
0
20u t – – 0 3 + 20G t – 3 + 20u t – 5 – 20G t – 5 – 10u t – 7

0
0
+
–
= – 2e – 2t > u t – u t – 2 @ 10G t – 2 + 10 u t – 2 – u t – 3 @ 20G t – 3

>

0
0
0
0
– 10 u t – > 0 3 – u t – 5 @ 20G t – 5 + 10 u t – 5 – u t – 7 @
–

>

0
0
0
The waveform for dv dte is shown below.
dv dt Vs
e
e
–

20 20G t 3
10
1 2 3 4 5 6 ts
– 10 7
10G t – 2
–
– 20
–
–
– 2e – 2t 20G t 5
3-22 Circuit Analysis II with MATLAB Applications
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