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Problem Solutions – Chapter 10 Problem 10.2.1 Solution In Example 10.3, the daily noontime temperature at Newark Airport is a discrete time, con- tinuous value random process. However, if the temperature is recorded only in units of one degree, then the process was would be discrete value. In Example 10.4, the the number of active telephone calls is discrete time and discrete value. The dice rolling experiment of Example 10.5 yields a discrete time, discrete value random process. The QPSK system of Example 10.6 is a continuous time and continuous value random pro- cess. Problem 10.2.2 Solution The sample space of the underlying experiment is S ={ s 0 , s 1 , s 2 , s 3 } . The four elements in the sample space are equally likely. The ensemble of sample functions is { x ( t , s i ) | i = 0 , 1 , 2 , 3 } where x ( t , s i ) = cos ( 2 π f 0 t + π/ 4 + i 2 )( 0 t T ) (1) For f 0 = 5 / T , this ensemble is shown below. 0 0.2T 0.4T 0.6T 0.8T T -1 -0.5 0 0.5 1 x(t,s 0 ) 0 0.2T 0.4T 0.6T 0.8T T -1 -0.5 0 0.5 1 1 0 0.2T 0.4T 0.6T 0.8T T -1 -0.5 0 0.5 1 2 0 0.2T 0.4T 0.6T 0.8T T -1 -0.5 0 0.5 1 3 t 351

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Problem 10.2.3 Solution The eight possible waveforms correspond to the the bit sequences { ( 0 , 0 , 0 ), ( 1 , 0 , 0 ), ( 1 , 1 , 0 ),. ..,( 1 , 1 , 1 ) } (1) The corresponding eight waveforms are: 0 T 2T 3T -1 0 1 0 T 2T 3T -1 0 1 0 T 2T 3T -1 0 1 0 T 2T 3T -1 0 1 0 T 2T 3T -1 0 1 0 T 2T 3T -1 0 1 0 T 2T 3T -1 0 1 0 T 2T 3T -1 0 1 Problem 10.2.4 Solution The statement is false . As a counterexample, consider the rectified cosine waveform X ( t ) = R | cos 2 π ft | of Example 10.9. When t = π/ 2, then cos 2 π = 0 so that X (π/ 2 ) = 0. Hence X (π/ 2 ) has PDF f X (π/ 2 ) ( x ) = δ( x ) (1) That is, X (π/ 2 ) is a discrete random variable. Problem 10.3.1 Solution In this problem, we start from first principles. What makes this problem fairly straightforward is that the ramp is defined for all time. That is, the ramp doesn’t start at time t = W . P [ X ( t ) x ] = P [ t W x ] = P [ W t x ] (1) Since W 0, if x t then P [ W t x ]= 1. When x < t , P [ W t x ] = Z t x f W (w) d w = e ( t x ) (2) Combining these facts, we have F X ( t ) ( x ) = P [ W t x ] = ± e ( t x ) x < t 1 t x (3) 352
We note that the CDF contain no discontinuities. Taking the derivative of the CDF F X ( t ) ( x ) with respect to x , we obtain the PDF f X ( t ) ( x ) = ± e x t x < t 0 otherwise (4) Problem 10.3.2 Solution (a) Each resistor has frequency W in Hertz with uniform PDF f R ( r ) = ± 0 . 025 9980 r 1020 0 otherwise (1) The probability that a test yields a one part in 10 4 oscillator is p = P [9999 W 10001] = Z 10001 9999 ( 0 . 025 ) dr = 0 . 05 (2) (b) To find the PMF of T 1 , we view each oscillator test as an independent trial. A success occurs on a trial with probability p if we find a one part in 10 4 oscillator. The first one part in 10 4 oscillator is found at time T 1 = t if we observe failures on trials 1 ,..., t 1 followed by a success on trial t . Hence, just as in Example 2.11, T 1 has the geometric PMF P T 1 ( t ) = ± ( 1 p ) t 1 pt = 1 , 2 ,... 9 otherwise (3) A geometric random variable with success probability p has mean 1 / p . This is derived in Theorem 2.5. The expected time to find the first good oscillator is E [ T 1 ]= 1 / p = 20 minutes.

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í™•ë¥ _ë°_ëžœë¤ë³€ìˆ˜_ì†”&e

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