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 µ 1. Consider a random process X ´t defined by µ ´ π µ X ´t sin 2 fct ; ℄ in which the frequency fc is a random variable uniformly distributed over the range 0 W . µ Show that X ´t is nonstationary. Hint: Examine specific sample functions of the random µ = ; = 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 µ X ´t 1 Z W 1 ´ ℄ ´ π µ ´ π ℄ E X t sin 2 ft df 1 cos 2 Wt W 0 W ´ ℄ ´ µ 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 µ ´ π µ · ´ π µ Z ´t X cos 2 t Y sin 2 t µ ´ µ (a) Determine the joint probability density function of the random variables Z ´t1 and Z t2 µ obtained by observing Z ´t at times t1 and t2 respectively. µ SOLUTION: Since every weighted sum of the samples of the Gaussian process Z ´t µ ´ µ 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. ´ ℄ ´ π µ ℄ · ´ π µ ℄ E Z t1 cos 2 t1 E X sin 2 t1 E Y ℄ ℄ ´ ℄ ´ ℄ Since E X E Y 0,E Z t1 0. Similarly, E Z t2 0. ´ µ ´ ℄ ´ µ ´ ℄ CovZ t1 Z t2 E Z t1 Z t2 ´ π µ · ´ π ℄ ´ π µ · ´ π ℄ E X cos 2 t1 Y sin 2 t1 X cos 2 t2 Y sin 2 t2 2 π µ ´ π µ ℄ · ´ π µ ´ π µ · ´ π µ ´ π ℄ ℄ cos´2 t1 cos 2 t2 E X cos 2 t1 sin 2 t2 sin 2 t1 cos 2 t2 E XY 2 ´ π µ ´ π µ ℄ · sin 2 t1 sin 2 t2 E Y 2 2 ℄ ℄ ℄ ℄: ℄ Noting that, E X 1, E Y 1 and E XY E X E Y 0 (since X and Y are independent), we obtain, ´ µ ´ ℄ π´ ℄ CovZ t1 Z t2 cos 2 t1 t2 1 2 2 σ ´ ℄ ´ µ ´ ℄ E Z t1 1. This result is obtained by putting t1 t2 in Cov Z t1 Z t2 . µ Z ´t1 σ2 2 ´ ℄ Similarly, E Z t2 1 µ Z ´t2 Correlation coefficient is given by ´ µ ´ ℄ CovZ t1 Z t2 π´ ℄ ρ cos 2 t t σ σ2 1 2 ´ µ µ Z t1 Z ´t2 Hence the joint PDF ; µ : ´ ; ℄ ´ µ; ´ µ fZ ´t1 Z t2 z1 z2 C exp Q z1 z2 where, 1 1 Ô C π´ ℄ 2 π ´ π´ µµµ 2π ´1 cos 2 t1 t2 2 sin 2 t1 t2 1 2 2 ; µ π´ ℄ · ℄ Q´z1 z2 2 z1 2cos 2 t1 t2 z1z2 z2 π´ ℄ sin 2 t1 t2 µ (b) Is the process Z ´t stationary? Why? ´ ℄ ´ µ ´ µ SOLUTION: We find that E Z t 0 and covariance of Z t1 and Z t2 depends only ´ µ 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. µ 3. The square wave x´t of FIGURE 1 of constant amplitude A, period T0, and delay td, repre- µ Figure 1: Square wave for x´t µ sents the sample function of a random process X ´t . The delay is random, described by the probability density function 8 < 1 T0 T0 2 td 2 µ fTD ´td T0 : 0 otherwise 2 µ (a) Determine the probability density function of the random variable X ´tk obtained by µ observing the random process X ´t at time tk. µ 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 1 1 µ ´ µ · ´ µ ´ δ δ µ f ´ x x x A X t 2 2 µ (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 Ê ∞ ´ ℄ µ ´ µ E X t ∞ xf ´ x dx X t 1 1 : · : 0 2 A 2 A 2 Autocorrelation: Let’s denote the square wave with random delay time tD, period T0 ´ µ and amplitude A as A:SqT0 t tD . Then, the autocorrelation can be written as, : ´ µ: : ´ · τ℄ τµ RX ´ E A SqT0 t tD A SqT0 t tD 2 Ê ∞ ´ µ: ´ · τ℄ ´ µ A ∞ Sq t t Sq t t f t dt T0 D T0 D TD D D T Ê 0 2 2 1 ´ µ: ´ · τ℄ A Sq t t Sq t t dt T0 T0 D T0 D T0 D 2 2 τj A j T0 ´ µ; jτj 1 2 2 T0 2 µ Since the square wave is periodic with period T0, RX ´t must also be periodic with period T0. µ (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 A ´ µ > < x t 2 and time-autocorrelation is, Z T0 =2 2 τj 1 A j T0 ´ · τµ ´ µ > ´ · τµ ´ µ ´ µ; jτj < x t x t x t x t 1 2 = T0 T0 2 2 T0 2 Again, the autocorrelation is periodic with period T0. µ (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 µ 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 stationar- µ ity, it follows that X ´t must be wide-sense stationary. µ 4. Consider two linear filters connected in cascade as in FIGURE 2. Let X ´t be a stationary τµ process with autocorrelation function RX ´ . The random process appearing at the first filter µ ´ µ output is V ´t and second filter output is Y t . 3 Figure 2: Cascade of linear filters µ (a) Find the autocorrelation function of Y ´t SOLUTION: The cascade connection of two filters is equivalent to a filter with im- pulse response Z ∞ µ ´ µ ´ µ h´t h1 u h2 t u du ∞ µ The autocorrelation function of Y ´t is given by, Z Z ∞ ∞ τµ ´τ µ ´τ µ ´τ τ · τ µ τ τ RY ´ h 1 h 2 RX 1 2 d 1 d 2 ∞ ∞ τµ ´ µ ´ µ (b) Find the cross-correlation function RVY ´ of V t and Y t . µ ´ µ SOLUTION: The cross correlation function of V ´t and Y t is, τµ ´ · τµ ´ ℄ RVY ´ E V t Y t µ ´ µ V ´t and Y t are related as follows, Z ∞ µ ´λµ ´ λµ λ Y ´t V h2 t d ∞ Therefore, Ê ∞ τµ ´ · τµ ´λµ ´ λµ λ℄ R ´ E V t ∞ V h t d VY 2 Ê ∞ ´ λµ ´ · τµ ´λ℄ λ ∞ h t E V t V d 2 Ê ∞ ´ λµ ´ · τ λµ λ ∞ h t R t d 2 V µ 5. A random telegraph signal X ´t , characterized by the autocorrelation function τµ ´ jτjµ RX ´ exp 2v 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, Ê ∞ µ ´τµ ´ π µ S ´ f ∞ R exp j2 ft dt X X Ê Ê 0 ∞ ´ µ ´ π µ · ´ µ ´ π µ ∞ exp 2vt exp j2 ft dt exp 2vt exp j2 ft dt 0 1 1 · π µ ´ · π µ 2´v j f 2 v j f v 2 π2 2 v · f 4 Figure 3: RC Filter The transfer function of the filter is 1 µ H ´ f π 1 · j2 f RC Therefore, PSD of the filter output is, 2 v µ j ´ µj ´ µ SY ´ f H f SX f 2 2 2 · π µ · π ℄ ´v f 1 j2 f RC µ To find the autocorrelation function of the filter output, we first expand SY ´ f in partial fractions as follows, v 1 1 µ · ℄ SY ´ f 2 2 2 2 2 2 2 2 2 ´ = µ · π · π 1 4R C v 1 2RC f v f Recognizing that, 1 g ´ jτjµ IFT f exp 2v 2 π2 2 v · f = µ ´1 2RC g ´ j j= µ IFT f 2 2 2 exp t RC = µ · π ´1 2RC f τµ f ´ µg where IFT stands for Inverse Fourier Transform, we get RY ´ IFT SY f ´ jτjµ τj j v exp 2v τµ ´ ℄ RY ´ 2 2 2 2RC exp 1 4R C v v RC µ ´ µ 6. A stationary Gaussian process X ´t has zero mean and power spectral density SX f . Deter- mine the probability density function of a random variable obtained by observing the process µ X ´t at some time tk. 2 µ σ ´ µ SOLUTION: Let µ´x be the mean and x be the variance of the random variable Xk obtained by observing the random process at time tk.