# A4 - Signals and Systems Issued 16 September 2015 Fall 2015...

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Signals and Systems Issued: 16 September 2015 Fall 2015 Due: 23 September 2015 Homework Assignment #4 1. (a) For the given simulation diagram, find the operational transfer function relating y ( k ) and x ( k ); (b) Write a computer program to simulate the system and plot its output. Make your program correspond to the given simulation diagram; that is, program coupled difference equations in variables that correspond to the outputs of the delay blocks. (c) Run the program to determine the zero-input response of y 1 ( k ), y 2 ( k ), y 3 ( k ), and y ( k ) if the initial conditions are y 1 (0) = y 2 (0) = 1, y 3 (0) = 1. (d) Run the program to determine the zero-state response of y 1 ( k ), y 2 ( k ), y 3 ( k ), and y ( k ) if the input x ( k ) is the unit step sequence. 2. Find the complete closed-form solution of the following difference equations: (a) (E 0.5){ y ( k )} = 0, y (0) = 7. (b) (E 1){ y ( k )} = 3 (0.5) k , y (0) = 0. (c) (E 2 + 3E + 2){ y ( k )} = 0, y (0) = 1, y (1) = 0. (d) (E 2 + 1){ y ( k )} = 3 2 k , y (0) = y (1) = 0. (e) (E 2 1){ y ( k )} = 0.5, y (0) = 1, y (1) = 2. (f) (E 2 E + 2){ y ( k )} = k 2 k , y (0) = y (1) = 1. (g) (E 3 E 2 E + 1){ y ( k )} = 2, y (0) = 0, y (1) = 1, y (2) = 2. 3. Are the following systems BIBO stable? Justify your answer. (a) (E 2 + 3E + 2) y ( k ) = x ( k ) (b) (10E 2 + 3E + 2) y ( k ) = (E 2){ x ( k )} (c) (E 2 + 1.2E + 0.2) y ( k ) = x ( k ) (d) (E 3 + 3E 2 + 3E + 1) y ( k ) = (E 2 0.1E){ x ( k )} x ( k ) y ( k ) 0.45 E 1 1.05 E 1 1.4 E 1 2 y 1 ( k ) y 2 ( k ) y 3 ( k )

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Signals and Systems Fall 2015 Solution to Homework Assignment #4 1-a) x ( k ) y ( k ) 0.45 E 1 1.05 E 1 1.4 E 1 2 y 1 ( k ) y 2 ( k ) y 3 ( k ) (a) From the diagram, we have: 2 1 ( 1) ( ) y k y k + = 2 1 { ( )} ( ) E y k y k = (1) 1 1 2 ( 1) 1.4 ( ) 0.45 ( ) ( ) y k y k y k x k + = + 1 2 ( 1.4){ ( )} 0.45 ( ) ( ) E y k y k x k = − + (2) Substituting (1) into (2) yields 2 2 ( 1.4)( ){ ( )} 0.45 ( ) ( ) E E y k y k x k = − + 2 2 ( 1.4 0.45){ ( )} ( ) E E y k x k + = (3) 3 2 ( ) 2 ( ) ( ) y k y k y k = + (4) 3 3 ( 1) 1.05 ( ) y k y k + = 3 ( 1.05){ ( )} 0 E y k = (5) 3 ( ) (1.05) k y k m = × . Note that m depends on the initial condition. Observe that 3 ( ) y k does not depend on the input ( ) x k . Furthermore, we have 2 3 ( ) ( ) 2 ( ) y k y k y k = + . As such, the operational transfer function can be obtained by using just the component 2 ( ) y k in the output. In particular, substituting (4) into (3) yields 2 ( 1.4 0.45){ ( )} ( ) E E y k x k + = Finally, we obtain the operational transfer function of the system as: 2 1 ( ) 1.4 0.45 H E E E = +
1-b) MATLAB Program for the system in the given diagram: From equations (1), (2) and (4) we know that: ( ) ( ) ( ) ( ) 1 1 2 1 1.4 0.45 y k x k y k y k + = + ( ) ( ) 2 1 1 y k y k + = ( ) ( ) 3 3 1 1.05 y k y k + = ( ) ( ) ( ) 2 3 1 1 2 1 y k y k y k + = + + + 1-c) Zero input response: %% Assignment 4, problem 1-c: zero input N = 40; % Simulation end time y1 = zeros(1,1+N); % vector size definition decrease the simulation y2 = zeros(1,1+N); % time since total number of simulation instant

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