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Table 6-2 n x n 0 2 1 4 2 11 3 5 4 7 5 6 6 9 7 10 8 3 9 6 10 8 11 4 12 1 13 3 14 5 15 12
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The time averages of x t and x2 t may be approximated from x n . Xavg 2 4 11 5 7 6 9 10 3 6 8 4 1 3 5 12 =16 6
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2 Xeff 22 42 112 52 72 62 92 102 33 62 82 42 12 32 52 122 =16 46 Xeff 6:78
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EXAMPLE 6.26 A binary signal v t 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 e ective 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 v t can be approximated as vavg 0:5 5000 0:5 5000 =10,000 0 The e ective value of v t is
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2 Veff 0:5 2 5000 0:5 2 5000 =10,000 0:5 2
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Veff 0:5 V
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The value of Veff is exact and independent of the number of intervals.
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6.1 Find the maximum and minimum values of v 1 2 sin !t  , given ! 1000 rad/s and  3 rad. Determine if the function v is periodic, and nd its frequency f and period T. Specify the phase angle in degrees.
Vmax 1 2 3 The function v is periodic. Thus, Vmin 1 2 1
To nd the frequency and period, we note that ! 2f 1000 rad/s. and T 1=f 2=1000 0:00628 s 6:28 ms
f 1000=2 159:15 Hz
Phase angle 3 rad 1808 3= 171:98
In a microwave range measurement system the electromagnetic signal v1 A sin 2ft, with f 100 MHz, is transmitted and its echo v2 t from the target is recorded. The range is computed from , the time delay between the signal and its echo. (a) Write an expression for v2 t and compute its phase angle for time delays 1 515 ns and 2 555 ns. (b) Can the distance be computed unambiguously from the phase angle in v2 t If not, determine the additional needed information.
(a) Let v2 t B sin 2f t  B sin 2ft  . For f 100 MHz 108 Hz,  2f  2 108  2k  where 0 <  < 2. For 1 515 10 9 , 1 2108 515 10 9 103 51 2 1 or k1 51 and 1 . For 2 555 10 9 , 2 2108 555 10 9 111 55 2 2 or k2 55 and 2 .
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WAVEFORMS AND SIGNALS
(b) Since phase angles 1 and 2 are equal, the time delays 1 and 2 may not be distinguished from each other based on the corresponding phase angles 1 and 2 . For unambiguous determination of the distance, k and  are both needed.
Show that if periods T1 and T2 of two periodic functions v1 t and v2 t have a common multiple, the sum of the two functions, v t v1 t v2 t , is periodic with a period equal to the smallest common multiple of T1 and T2 . In such case show that Vavg V1;avg V2;avg .
If two integers n1 and n2 can be found such that T n1 T1 n2 T2 , then v1 t v1 t n1 T1 and v2 t v2 t n2 T2 . Consequently, v t T v1 t T v2 t T v1 t v2 t v t and v t is periodic with period T. The average is 1 T 1 T 1 T v1 t v2 t dt v1 t dt v t dt V1;avg V2;avg Vavg T 0 T 0 T 0 2
Show that the average of cos2 !t  is 1/2.
Using the identity cos2 !t  1 1 cos 2 !t  , the notation h f i Favg , and the result of 2 Problem 6.3, we have h1 cos 2 !t  i h1i hcos 2 !t  i But hcos 2 !t  i 0. Therefore, hcos2 !t  i 1=2.
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