Page 119 - Schaum's Outline of Theory and Problems of Electric Circuits

WAVEFORMS AND SIGNALS
[CHAP. 6
108
EXAMPLE 6.12 Compute the average power dissipated from 0 to T in a resistor connected to a voltage vðtÞ.
Replace vðtÞ by a constant voltage V dc . Find V dc such that the average power during the period remains the same.
2
p ¼ vi ¼ v =R
ð 2
1 T 2 1 2 V dc
P avg ¼ v ðtÞ dt ¼ V eff ¼ or V dc ¼ V eff
RT 0 R R

EXAMPLE 6.13 The current iðtÞ shown in Fig. 6-6 passes through a 1-mF capacitor. Find (a) v ac the voltage
across the capacitor at t ¼ 5k ms (k ¼ 0; 1; 2; 3; ...Þ and (b) the value of a constant current source I dc which can
produce the same voltage across the above capacitor at t ¼ 5k ms when applied at t > 0. Compare I dc with hiðtÞi, the
average of iðtÞ in Fig. 6-6, for a period of 5 ms after t > 0.

Fig. 6-6
(a)At t ¼ 5ms
ð 3 " ð 3 ð 3 #
1 5 10 6 3 3 10 5 10
v ac ¼ iðtÞ dt ¼ 10 ð10 Þ 4 dt 2 dt ¼ 12 4 ¼ 8V
C 0 0 3 10 3
This is the net charging effect of iðtÞ during each 5-ms interval. Every 5 ms the above amount is added to the
capacitor voltage. Therefore, at t ¼ 5k ms, v ¼ 8k (V).
(b) With a constant current I dc , the capacitor voltage v dc at t ¼ 5k ms is
ð 5k 10 3
1 6 3 3
v dc ¼ I dc dt ¼ 10 ðI dc Þð5k 10 Þ¼ 10 ð5kÞðI dc Þ ðVÞ
C 0
Since v dc ¼ v ac at 5k ms, we obtain
3
3
10 ð5kÞðI dc Þ¼ 8k or I dc ¼ 8k=ð5k 10 Þ¼ 1:6 10 3 A ¼ 1:6mA
Note that I dc ¼hiðtÞi of Fig. 6-6 for any period of 5 ms at t > 0.

6.7 NONPERIODIC FUNCTIONS
A nonperiodic function cannot be specified for all times by simply knowing a finite segment.
Examples of nonperiodic functions are

0 for t < 0
(a) v 1 ðtÞ¼ (21)
1 for t > 0
8
< 0 for t < 0
(b) v 2 ðtÞ¼ 1=T for 0 < t < T (22)
:
0 for t > T

0 for t < 0
(c) v 3 ðtÞ¼ t= (23)
e for t > 0