pcshw9_soln

# pcshw9_soln - THE STATE UNIVERSITY OF NEW JERSEY RUTGERS...

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THE STATE UNIVERSITY OF NEW JERSEY RUTGERS College of Engineering Department of Electrical and Computer Engineering 332:322 Principles of Communications Systems Spring 2004 Problem Set 9 Haykin: 1.1–1.10 1. Consider a random process X t defined by X t sin 2 π f c t in which the frequency f c is a random variable uniformly distributed over the range 0 W . Show that X t is nonstationary. Hint: Examine specific sample functions of the random process X t for the frequency f W 2 W 4 and W say. SOLUTION: An easy way to solve this problem is to find the mean of the random process X t E X t 1 W W 0 sin 2 π ft d f 1 W 1 cos 2 π Wt Clearly E X t is a function of time and hence the process X t is not stationary. 2. Let X and Y be statistically independent Gaussian-distributed random variables each with zero mean and unit variance. Define the Gaussian process Z t X cos 2 π t Y sin 2 π t (a) Determine the joint probability density function of the random variables Z t 1 and Z t 2 obtained by observing Z t at times t 1 and t 2 respectively. SOLUTION: Since every weighted sum of the samples of the Gaussian process Z t is Gaussian, Z t 1 , Z t 2 are jointly Gaussian random variables. Hence we need to find mean, variance and correlation co-efficient to evaluate the joint Gaussian PDF. E Z t 1 cos 2 π t 1 E X sin 2 π t 1 E Y Since E X E Y 0 , E Z t 1 0 . Similarly, E Z t 2 0 . Cov Z t 1 Z t 2 E Z t 1 Z t 2 E X cos 2 π t 1 Y sin 2 π t 1 X cos 2 π t 2 Y sin 2 π t 2 cos 2 π t 1 cos 2 π t 2 E X 2 cos 2 π t 1 sin 2 π t 2 sin 2 π t 1 cos 2 π t 2 E XY sin 2 π t 1 sin 2 π t 2 E Y 2 Noting that, E X 2 1 , E Y 2 1 and E XY E X E Y 0 (since X and Y are independent), we obtain, Cov Z t 1 Z t 2 cos 2 π t 1 t 2 1

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σ 2 Z t 1 E Z 2 t 1 1 . This result is obtained by putting t 1 t 2 in Cov Z t 1 Z t 2 . Similarly, σ 2 Z t 2 E Z 2 t 2 1 Correlation coefficient is given by ρ Cov Z t 1 Z t 2 σ Z t 1 σ 2 Z t 2 cos 2 π t 1 t 2 Hence the joint PDF f Z t 1 Z t 2 z 1 z 2 C exp Q z 1 z 2 where, C 1 2 π 1 cos 2 2 π t 1 t 2 1 2 π sin 2 π t 1 t 2 Q z 1 z 2 1 sin 2 2 π t 1 t 2 z 2 1 2cos 2 π t 1 t 2 z 1 z 2 z 2 2 (b) Is the process Z t stationary? Why? SOLUTION: We find that E Z t 0 and covariance of Z t 1 and Z t 2 depends only on the time difference t 1 t 2 . The process Z t is hence wide sense stationary. Since it is Gaussian, it is also strict sense stationary.
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