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# ch05 - Chapter 5 Random Signals and Noise 5.1 Problem...

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Chapter 5 Random Signals and Noise 5.1 Problem Solutions Problem 5.1 The various sample functions are as follows. Sample functions for case (a) are horizontal lines at levels A, 0 , A , each case of which occurs equally often (with probability 1/3). Sample functions for case (b) are horizontal lines at levels 5 A, 3 A, A, A, 3 A, 5 A , each case of which occurs equally often (with probability 1/6). Sample functions for case (c) are horizontal lines at levels 4 A, 2 A, 2 A, 4 A, or oblique straight lines of slope A or A , each case of which occurs equally often (with probability 1/6). Problem 5.2 a. For case (a) of problem 5.1, since the sample functions are constant with time and are less than or equal 2A, the probability is one. For case (b) of problem 5.1, since the sample functions are constant with time and 4 out of 6 are less than or equal to 2A, the probability is 2/3. For case (c) of problem 5.1, the probability is again 2/3 because 4 out of 6 of the sample functions will be less than 2 A at t = 4 . b. The probabilities are now 2/3, 1/2, and 1/2, respectively. c. The probabilities are 1, 2/3, and 5/6, respectively. Problem 5.3 a. The sketches would consist of squarewaves of random delay with respect to t = 0 . 1

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2 CHAPTER 5. RANDOM SIGNALS AND NOISE b. X ( t ) takes on only two values A and A , and these are equally likely. Thus f X ( x ) = 1 2 δ ( x A ) + 1 2 δ ( x + A ) Problem 5.4 a. The sample functions at the integrator output are of the form Y ( i ) ( t ) = Z t A cos ( ω 0 λ ) d λ = A ω 0 sin ω 0 t where A is Gaussian. b. Y ( i ) ( t 0 ) is Gaussian with mean zero and standard deviation σ Y = σ a | sin ( ω 0 t 0 ) | ω 0 c. Not stationary and not ergodic. Problem 5.5 a. By inspection of the sample functions, the mean is zero. Considering the time average correlation function, de fi ned as R ( λ ) = lim T →∞ 1 2 T Z T T x ( t ) x ( t + λ ) dt = 1 T 0 Z T 0 x ( t ) x ( t + λ ) dt where the last expression follows because of the periodicity of x ( t ) , it follows from a sketch of the integrand that R ( τ ) = A 2 (1 4 λ /T 0 ) , 0 λ T 0 / 2 Since R ( τ ) is even and periodic, this de fi nes the autocorrelation function for all λ . b. Yes, it is wide sense stationary. The random delay, λ , being over a full period, means that no untypical sample functions occur.
5.1. PROBLEM SOLUTIONS 3 Problem 5.6 a. The time average mean is zero. The statistical average mean is E [ X ( t )] = Z π / 2 π / 4 A cos (2 π f 0 t + θ ) d θ π / 4 = 4 A π sin (2 π f 0 t + θ ) ¯ ¯ ¯ ¯ π / 2 π / 4 = 4 A π [sin (2 π f 0 t + π / 2) sin (2 π f 0 t + π / 4)] = 4 A π ·μ 1 1 2 cos (2 π f 0 t ) 1 2 sin (2 π f 0 t ) ¸ The time average variance is A 2 / 2 . The statistical average second moment is E £ X 2 ( t ) ¤ = Z π / 2 π / 4 A 2 cos 2 (2 π f 0 t + θ ) d θ π / 4 = 2 A 2 π " Z π / 2 π / 4 d θ + Z π / 2 π / 4 cos (4 π f 0 t + 2 θ ) d θ # = 2 A 2 π " π 4 + 1 2 sin (4 π f 0 t + 2 θ ) ¯ ¯ ¯ ¯ π / 2 π / 4 # = A 2 2 + A 2 π h sin (4 π f 0 t + π ) sin ³ 4 π f 0 t + π 2 ´i = A 2 2 A 2 π [sin (4 π f 0 t ) + cos (4 π f 0 t )] The statistical average variance is this expression minus E 2 [ X ( t )] .

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