ece_351_hwset1_more_sol_fall_20

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Electrical Engineering

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Oct 30, 2023

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1 ECE 351 Homework Set 1: More Solutions (60 – 65)

2 Problem 60 a. h[n] = (1/4) n u[n] S | h[n] | = 1 + (1/4) + (1/16) + ... = 1/(1-1/4) = 4/3 <   BIBO Stable b. y[n + 2] – 5.5 y[n + 1] – 3 y[n] = 2 x[n]  char. eq: E 2 – 5.5E – 3 = 0  g = 6, -.5  not asymptotically stable  not BIBO stable (see p. 314)

3 Problem 61 a. y[n] = x[n] * h[n] = x[n] * { d [n] + d [n – 1] + d [n – 2] + d [n – 3] }/4 = { x[n] + x[n – 1] + x[n – 2] + x[n - 3] }/4 b. y[n] = x[n] * h[n] = (-.9) n u[n] * { d [n] + d [n – 1] + d [n – 2] + d [n – 3] }/4 = (1/4) {(-.9) n u[n] + (-.9) n-1 u[n-1] + (-.9) n-2 u[n-2] + (-.9) n-3 u[n-3] } = (1/4) (-.9) n { u[n] + (-.9) -1 u[n-1] + (-.9) -2 u[n-2] + (-.9) -3 u[n-3] } = (1/4) (-.9) n { u[n] – 1.1111 u[n-1] + 1.2346 u[n-2] - 1.3717 u[n-3] }

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4 Problem 61, continued c. y[n] = x[n] * h[n] = {5, 5, 5, 5, 5, 5, 5} * { d [n] + d [n – 1] + d [n – 2] + d [n – 3] }/4 =(1/4) [ {5 5 5 5 5 5 5} 1 st copy {5 5 5 5 5 5 5} 2 nd copy {5 5 5 5 5 5 5} {5 5 5 5 5 5 5}] = (1/4) { 5 10 15 20 20 20 20 15 10 5} sum = { 5/4, 10/4, 15/4, 5, 5, 5, 5, 15/4, 10/4, 5/4}

5 Problem 61, continued d. MATLAB code >> h = (1/4) * [1 1 1 1]; >> y = conv(x_noisy, h); >> n1 = 0: .1 : (.1)*(length(y)-1); >> subplot(3, 1, 1) >> stem(n, x) >> subplot(3, 1, 2) >> stem(n, x_noisy) >> subplot(3, 1, 3) >> stem(n1, y) 0 5 10 15 20 -5 0 5 Clean Sinusoid 0 5 10 15 20 -5 0 5 Noisy Sinusoid 0 5 10 15 20 -5 0 5 Filtered Noisy Sinusoid FIR Filter

6 Problem 62 y[n + 2] – (1/4) y[n + 1] + (1/8) y[n] = 2 x[n + 1] – 3 x[n] H(r) = (2r – 3)/(r 2 – .25r + .125) Steady-state Outputs: a. dc gain: H(r=1) = -1/(7/8) = -8/7 b. x[n] = 2  y[n] = (-8/7)*2 = -16/7 (using the dc gain c. x[n] = .7 n , gain = H(.7) = -3.6364  y[n] = -3.6364 (.7) n d. x[n] = cos( p n/8), using MATLAB: complex gain H(e j p /8 ) = -.31 + 1.58j; magnitude gain: 1.61, angle: 1.76 rad = 100.92º  y[n] =1.61 cos(n p /8 + 100.92º)

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7 Problem 62e – MATLAB Code >> b = [2 -3]; a = [1 -1/4 1/8]; >> n = 0 : 30; >> x_parta = 2 * ones(1,31); >> x_partb = .7 .^n .* ones(1,31); >> x_partc = cos(n*pi/8); >> y_parta = filter(b, a, x_parta); >> y_partb = filter(b, a, x_partb); >> y_partc = filter(b, a, x_partc); >> subplot(3,1,1), stem(n, y_parta) >> subplot(3,1,2), stem(n, y_partb) >> subplot(3,1,3), stem(n, y_partc)

8 Problem 62e – MATLAB 0 5 10 15 20 25 30 -5 0 5 yparta 0 5 10 15 20 25 30 -2 0 2 ypartb 0 5 10 15 20 25 30 -2 0 2 ypartc

9 Problem 63 Filter syntax: y = filter(b, a, x) where b is the vector of coefficients in P(z) and a is the vector of coefficients in Q(z), readily obtained from either H(z) = P(z)/Q(z) or from the difference equation, and x is the input Conv syntax: conv(x, h) where x is the input and h is the inpulse response Thus, use filter if the problem gives you the transfer function or the difference equation; use conv if the problem gives you the impulse response.

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10 Problem 64 y[n] – 1.5 y[n-1] + .5 y[n-2] = x[n] Find Char. Modes: Adv. Form: y[n + 2] – 1.5 y[n+1] + .5 y[n] = 0 E 2 – 1.5 E + .5 = 0 g 2 – 1.5 g + .5 = 0 ( g -1) ( g -.5) = 0  g = 1, .5 Char. Modes: 1 n , .5 n (lead to resonance) Input closest to char. mode: d. (.6) n

Problem 65 • Given spectrum X(f): 11 -100 100 X(f) f -500 -300 -100 100 300 500 X(f) f Spectrum of signal sampled at Nyquist rate, 200 samples/sec. 8 Note replicas are spaced apart by f s = 200 Hz; amplitude of sampled spectrum: 8/T s = 8 * f s = 1600 1600 … …