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

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Unformatted text preview: THE STATE UNIVERSITY OF NEW JERSEY RUTGERS College of Engineering Department of Electrical and Computer Engineering 332:322 Principles of Communications Systems Problem Set 9 Haykin: 1.11.10 Spring 2004 1. Consider a random process X t defined by 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 1 W 1 sin 2 f t d f EX t 1 cos 2W t W 0 W 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 (a) Determine the joint probability density function of the random variables Z t 1 and Z t2 obtained by observing Z t at times t1 and t2 respectively. SOLUTION: Since every weighted sum of the samples of the Gaussian process Z t is Gaussian, Z t1 , Z t2 are jointly Gaussian random variables. Hence we need to find mean, variance and correlation co-efficient to evaluate the joint Gaussian PDF. 1 Cov Z t1 Z t2 cos 2 t1 # % $ # Noting that, E X 2 1, E Y 2 independent), we obtain, 1 and E XY EX EY t2 0 (since X and Y are "! Cov Z t1 Z t2 E Z t 1 Z t2 E X cos 2t1 Y sin 2t1 X cos 2t2 Y sin 2t2 cos 2t1 cos 2t2 E X 2 cos 2t1 sin 2t2 sin 2t1 cos 2t2 E XY sin 2t1 sin 2t2 E Y 2 Since E X EY 0, E Z t1 0. Similarly, E Z t2 E Z t1 cos 2t1 E X sin 2t1 E Y 0. Z t X cos 2t Y sin 2t X t sin 2 fct Correlation coefficient is given by Hence the joint PDF t1 Z t2 where, (b) Is the process Z t stationary? Why? SOLUTION: We find that E Z t 0 and covariance of Z t1 and Z t2 depends only on the time difference t1 t2 . The process Z t is hence wide sense stationary. Since it is Gaussian, it is also strict sense stationary. 3. The square wave x t of FIGURE 1 of constant amplitude A, period T0 , and delay td , repre- sents the sample function of a random process X t . The delay is random, described by the probability density function 2 fTD td 1 T0 td 2 T0 0 otherwise Figure 1: Square wave for x t T0 2 Q z 1 z2 t2 z 1 z 2 2 cos2 2 t1 t2 1 z2 2 cos 2 t1 1 2 sin 2 t1 t2 1 C 1 1 2 sin 2 t1 $ fZ z1 z2 C exp Q z 1 z2 2 cos 2 t1 Cov Z t1 Z t2 Z t1 2 t Z t2 Similarly, 2 Z E Z 2 t2 t1 1 t2 t2 z2 2 2 Z E Z 2 t1 1. This result is obtained by putting t1 t2 in Cov Z t1 Z t2 . (a) Determine the probability density function of the random variable X t k obtained by observing the random process X t at time tk . SOLUTION: X t is a square wave, and it takes on the two values 0 or A with equal probability. Hence the PDF can be given as t (b) Determine the mean and autocorrelation function of X t using ensemble-averaging SOLUTION: Using our definition of ensemble average for the mean of a stochastic process, we find EX t x fX t x dx 1 02 A1 2 A 2 Autocorrelation: Let's denote the square wave with random delay time t D , period T0 and amplitude A as A SqT0 t tD . Then, the autocorrelation can be written as, Since the square wave is periodic with period T0 , RX t must also be periodic with period T0 . (c) Determine the mean and autocorrelation function of X t using time-averaging. SOLUTION: On a time-averaging basis we note by inspection that the mean is and time-autocorrelation is, T0 2 Again, the autocorrelation is periodic with period T0 . (d) Establish whether or not X t is stationary. In what sense is it ergodic? SOLUTION: We note that the ensemble-averaging and time-averaging yield the same set of results for the mean and autocorrelation functions. Therefore, X t is ergodic in the mean and autocorrelation function. Since ergodicity implies wide-sense stationarity, it follows that X t must be wide-sense stationary. 4. Consider two linear filters connected in cascade as in FIGURE 2. Let X t be a stationary process with autocorrelation function RX . The random process appearing at the first filter output is V t and second filter output is Y t . 3 xt xt 2 xt xt 1 T0 T0 2 xt A 2 A2 1 2 A2 2 1 2 T0 T0 2 A2 SqT0 t tD SqT0 t T0 2 T0 2 tD 1 T0 dtD T0 $ $ $ $ $ RX E A SqT0 t tD A SqT0 t tD A2 SqT0 t tD SqT0 t tD fTD tD dtD $ $ fX x 1 x 2 1 x 2 A $ T0 2 Figure 2: Cascade of linear filters (a) Find the autocorrelation function of Y t SOLUTION: The cascade connection of two filters is equivalent to a filter with impulse response The autocorrelation function of Y t is given by, (b) Find the cross-correlation function RVY of V t and Y t . SOLUTION: The cross correlation function of V t and Y t is, V t and Y t are related as follows, 5. A random telegraph signal X t , characterized by the autocorrelation function where v is a constant, is applied to a low-pass RC filter of FIGURE 3. Determine the power spectral density and autocorrelation function of the random process at the filter output. SOLUTION: The power spectral density of the random telegraph wave is given as, SX f j2 f t dt RX exp 0 j2 f t exp 2vt exp 1 1 2 v j f 2 v j f v v2 2 f 2 4 dt 2vt exp 0 exp RX exp 2v RVY E V t V h2 t d h2 t E V t V d h2 t RV t d Therefore, Y t V h2 t d RVY EV t Y t RY h 1 h 2 RX 1 ht h1 u h 2 t u du 2 d1 d2 j2 f t dt Figure 3: RC Filter The transfer function of the filter is Therefore, PSD of the filter output is, To find the autocorrelation function of the filter output, we first expand SY f in partial fractions as follows, 1 v2 Recognizing that, 6. A stationary Gaussian process X t has zero mean and power spectral density S X f . Determine the probability density function of a random variable obtained by observing the process X t at some time tk . SOLUTION: Let x be the mean and 2 x be the variance of the random variable Xk obtained by observing the random process at time tk . Then, 5 2 x 2 E Xk x 0 2 x 2 E Xk 1 v 2RC exp RY v 4R2C 2 v2 exp 2v RC IFT SY f where IFT stands for Inverse Fourier Transform, we get RY IFT exp 1 2RC 1 2RC 2 2 f 2 v2 t RC IFT exp 1 2 f 2 2v SY f v 1 2C 2 v2 1 2RC 2 2 f 2 4R 1 2 f 2 ! SY f H f 2 SX f H f 1 1 j2 f RC v2 2 f 2 v 1 j2 f RC The PDF of Gaussian random variable Xk is given by 22 x 7. A stationary Gaussian process X t with mean x and variance 2 is passed through two X linear filters with impulse responses h1 t and h2 t , yielding processes Y t and Z t , as shown in FIGURE 4 Figure 4: Parallel systems. (a) Determine the joint probability density function of the random variables Y t 1 and Z t2 . SOLUTION: Since X t is a Gaussian random process, the random variables Y t 1 and Z t2 are jointly Gaussian. Hence to find the joint PDF, we need to find variance cov Y t1 Z t2 2 Y1 , 2 2 , mean Y1 , Z2 and correlation coefficient Z Y Z 1 2 6 where H2 0 u du and x is the mean of the stationary random process X t . h2 Z2 H2 0 x Z t2 X t2 u h2 u du where H1 0 Similarly, d and x is the mean of the stationary random process X t . h1 Y1 H1 0 x Y t1 X t1 h1 d fXk x exp 1 x2 22 x 2 x 2 E Xk SX f d f We note that The covariance of Y t1 and Z t2 is where CX is the autocovariance function of X t . h2 h2 u d du Finally, the joint Gaussian PDF can be written as, t1 Z t2 where, (b) What conditions are necessary and sufficient to ensure that Y t1 and Z t2 are statistically independent? SOLUTION: The random variables Y t1 and Z t2 are uncorrelated if and only if their covariance is zero. Since Y t and Z t are jointly Gaussian processes, it follows that Y t1 and Z t2 are statistically independent if Cov Y t1 Z t2 0. Therefore the necessary and sufficient condition for Y t1 and Z t2 to be statistically independent is that CX t1 t2 u h1 h2 u d du 8. A stationary Gaussian process X t with zero mean and power spectral density S X f is applied to a linear filter whose impulse response h t is shown in FIGURE 5. A sample Y is Figure 5: h t for problem 8 taken of the random process at the filter output at time T . 7 21 2 Y1 Y1 Z2 Z2 Qyz 2 2 1 y Y1 y k 1 2Y1 Z2 1 2 Y1 z Z2 z Z2 2 fY yz k exp Q y z 2 2 Z h1 h1 u d du E Z 2 Y1 E Y t1 Y1 2 CX u t 2 Z2 2 CX u Cov Y t1 Z t2 E Y t1 Y1 Z t2 Z2 E X t1 x X t2 x h1 h2 u d du x X t2 x h1 h2 u d du E X t1 CX t1 t2 u h1 h2 u d du (a) Determine the mean and variance of Y SOLUTION: The filter output is 0 The mean of Y is therefore 0 Variance of Y (b) What is the probability density function of Y ? SOLUTION: Since the filter output is Gaussian, it follows that Y is also Gaussian. Hence the PDF of Y is 1 y2 exp fY y 2 2Y 2 2Y 8 2 Y SX f sinc2 f T d f Therefore, H f 2 Y E Y2 E Y 2 RY 0 SY f d f 2 df SX f H f j2 f t dt h t exp 1 T j2 f t dt T 0 exp sinc f T exp j f T EY E 1 T T 0 1 T T 0 E X u du X u du Y X u du 1 T T Put T u. Then the sample value of Y t at t T equals Y t h X 1 T T 0 X T t d d ...
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This homework help was uploaded on 04/03/2008 for the course ECE ECE332 taught by Professor Rose during the Spring '08 term at Rutgers.

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