Period_13_Exercise

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University of Wisconsin, Madison *

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431

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Electrical Engineering

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May 5, 2024

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pdf

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ECE 431 Period 13 z-Transform and LTI Systems ######NAME##### Problem 1 Answer (use Equation Editor to replace giving your answer) Problem 2a Answe r (use Equation Editor to replace symbols with your answer) Set up workspace for Problem 2b and later clear close all clc Problem 2b Answer (give your written description comparing 2a with 2b) The graphical result agree with the calculations from 2a. b = [12 1]; a = [6 1 -1]; imp = [1 zeros(1,99)]; % Look up filter() command to have it operate on imp h = filter(b,a,imp); figure stem([0:10],h(1:11)) xlabel( 'Time (n)' ) ylabel( 'Impulse Response' ) title( 'Prob 2 System Impulse Response' ) 1
Problem 2c Answer ___2.153____ x = 4*cos(3*pi*[0:499]/64); y = filter(b,a,x); figure stem(0:499,y(1:500)) title( 'Output of system described by difference equation' ) xlabel( 'Time (n)' ) ylabel( 'Output' ) 2
Problem 2d Answer (your written answer if the system is stable or not and why) The system is stable, because all the poles are inside the unit circle, and the ROC contains the unit circle. figure % Look up zplane command to have it operate on coefficients b, a zplane(b,a) %% Look up zplane command to have it operate on coefficients b, a 3
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Problem 3a b = [0 0 0 1 0 0 0 1]; a = [1 -1/3 -1/4 1/12 21/20 -7/20 -21/80 7/80]; x = cos(pi*[0:499]/128); ya = filter(b,a,x); figure stem(0:499,ya(1:500)) title( 'Prob 3 system output' ) xlabel( 'Time (n)' ) ylabel( 'Output' ) 4
Problem 3b imp = [1 zeros(1,499)]; h = filter(b,a,imp); figure stem(0:499,h) title( 'Prob 3 System Impulse Response' ) xlabel( 'Time (n)' ) ylabel( 'Output' ) 5
Problem 3c Answer (your written explanation whether the system is stable or not and why) The system is unstable, because there are poles existing outside the unit circle. figure zplane(b,a) title( 'Prob 3 Poles and Zeros' ) 6
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Problem 3d Answer (give index n for the first non-zero samples of the response in parts a) and b) and explanation why) At n=3, the output and the impulse response are first nonzero, because the first entry of the numerator of the transfer function resembles n=3. figure stem(0:9,ya(1:10)) title( 'Output from sinusoid input' ) xlabel( 'Time (n)' ) ylabel( 'Output' ) 7
figure stem(0:9,h(1:10)) title( 'Impulse Response' ) xlabel( 'Time (n)' ) ylabel( 'Output' ) 8
9
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