Page 127 - Schaum's Outline of Theory and Problems of Electric Circuits

WAVEFORMS AND SIGNALS
116
Table 6-2 [CHAP. 6
n 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
xðnÞ 2 4 11 5 7 6 9 10 3 6 8 4 1 3 5 12

2
The time averages of xðtÞ and x ðtÞ may be approximated from xðnÞ.
X avg ¼ð2 þ 4 þ 11 þ 5 þ 7 þ 6 þ 9 þ 10 þ 3 þ 6 þ 8 þ 4 þ 1 þ 3 þ 5 þ 12Þ=16 ¼ 6
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
3
2
X eff ¼ð2 þ 4 þ 11 þ 5 þ 7 þ 6 þ 9 þ 10 þ 3 þ 6 þ 8 þ 4 þ 1 þ 3 þ 5 þ 12 Þ=16 ¼ 46
X eff ¼ 6:78
EXAMPLE 6.26 A binary signal vðtÞ is either at 0.5 or 0:5 V. It can change its sign at 1-ms intervals. The sign
change is not known a priori, but it has an equal chance for positive or negative values. Therefore, if measured for a
long time, it spends an equal amount of time at the 0.5-V and 0:5-V levels. Determine its average and effective
values over a period of 10 s.
During the 10-s period, there are 10,000 intervals, each of 1-ms duration, which on average are equally divided
between the 0.5-V and 0:5-V levels. Therefore, the average of vðtÞ can be approximated as
v avg ¼ð0:5 5000 0:5 5000Þ=10,000 ¼ 0
The effective value of vðtÞ is
2
2
2
V eff ¼½ð0:5Þ 5000 þð 0:5Þ 5000Š=10,000 ¼ð0:5Þ 2 or V eff ¼ 0:5V
The value of V eff is exact and independent of the number of intervals.
Solved Problems
6.1 Find the maximum and minimum values of v ¼ 1 þ 2 sinð!t þ Þ, given ! ¼ 1000 rad/s and ¼ 3
rad. Determine if the function v is periodic, and find its frequency f and period T. Specify the
phase angle in degrees.

V max ¼ 1 þ 2 ¼ 3 V min ¼ 1 2 ¼ 1
The function v is periodic. To find the frequency and period, we note that ! ¼ 2 f ¼ 1000 rad/s.
Thus,
f ¼ 1000=2 ¼ 159:15 Hz and T ¼ 1=f ¼ 2 =1000 ¼ 0:00628 s ¼ 6:28 ms
Phase angle ¼ 3 rad ¼ 1808 3= ¼ 171:98

6.2 In a microwave range measurement system the electromagnetic signal v 1 ¼ A sin 2 ft, with
f ¼ 100 MHz, is transmitted and its echo v 2 ðtÞ from the target is recorded. The range is com-
puted from , the time delay between the signal and its echo. (a) Write an expression for v 2 ðtÞ
and compute its phase angle for time delays 1 ¼ 515 ns and 2 ¼ 555 ns. (b) Can the distance
be computed unambiguously from the phase angle in v ðtÞ? If not, determine the additional
2
needed information.
(a) Let v 2 ðtÞ¼ B sin 2 f ðt Þ¼ B sinð2 ft Þ.
8
8
For f ¼ 100 MHz ¼ 10 Hz, ¼ 2 f ¼ 2 10 ¼ 2 k þ where 0 < < 2 .
9
8
For 1 ¼ 515 10 , 1 ¼ 2 10 515 10 9 ¼ 103 ¼ 51 2 þ 1 or k 1 ¼ 51 and 1 ¼ .
9
8
For 2 ¼ 555 10 , 2 ¼ 2 10 555 10 9 ¼ 111 ¼ 55 2 þ 2 or k 2 ¼ 55 and 2 ¼ .