EE112 HW 4 (1)

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May 8, 2024

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Rohan Pradhan EE 112 Section 2 19 March 2024 Homework #4 1. Code: % Define the signal x3[n] n = 0 : 20 ; x3 = 7 * (heaviside(n - 1 ) - heaviside(n - 9 )); % Multiply x3[n] by 3 to get x[n] x = 3 * x3; % Plot the signal x[n] subplot( 3 , 1 , 1 ); stem(n, x); xlabel( 'n' ); ylabel( 'Amplitude' ); title( 'Signal x[n]' ); % Compute the frequency spectrum of x[n] [H, w] = freqz(x); % Plot the magnitude of the frequency spectrum subplot( 3 , 1 , 2 ); plot(w, abs (H)); xlabel( 'Frequency (radians/sample)' ); ylabel( 'Magnitude' ); title( 'Magnitude Spectrum' ); % Plot the phase of the frequency spectrum subplot( 3 , 1 , 3 ); plot(w, angle (H)); xlabel( 'Frequency (radians/sample)' ); ylabel( 'Phase (radians)' );
title( 'Phase Spectrum' ); Plots: There are phase jumps in the plot sue to the discontinuous portion of x[n] 2. a.
b. 3. a.
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b. c. 4. a. % Given parameters a = 0.95 ; w_hat = - pi : 0.01 : pi ; % Calculate |V(e^(jw_hat))|^2 V_mag_squared = 1 ./ abs ( 1 - a * exp ( -1 j * w_hat)).^ 2 ; % Plotting
plot(w_hat, V_mag_squared); xlabel( 'Frequency (\omega)' ); ylabel( '|V(e^{j\omega^})|^2' ); title( '|V(e^{j\omega^})|^2 for -\pi < \omega^ \leq \pi' ); grid on; b. The response is a low pass filter c. % Given parameters a = 0.95 ; w_hat = - pi : 0.01 : pi ; % Calculate W(e^(jw_hat)) W = zeros ( size (w_hat)); for n = 0 : 20 w_n = a * n * cos ( 0.3 * pi * n); W = W + w_n * exp ( -1 j * w_hat * n); end % Calculate |W(e^(jw_hat))|^2 W_mag_squared = abs (W).^ 2 ;
% Plotting plot(w_hat, W_mag_squared); xlabel( 'Frequency (\omega)' ); ylabel( '|W(e^{j\omega^})|^2' ); title( '|W(e^{j\omega^})|^2 for -\pi < \omega^ \leq \pi' ); grid on; d. The response is a bandwidth filter 5. a.
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b. w = linspace ( 0 , 2 * pi , 1000 ); % Frequency vector from 0 to 2*pi H = 1. /( 1 + 0.81 * exp ( -1 j * 2 *w)); % Frequency response % Plot magnitude response figure; plot(w, abs (H)); xlabel( 'Frequency (radians/sample)' ); ylabel( '|H(e^{j\omega})|' ); title( 'Magnitude Response' ); grid on; c. % Calculate the phase response phase_H = angle (H);
% Unwrap the phase unwrapped_phase_H = unwrap (phase_H); % Plot wrapped and unwrapped phase responses figure; subplot( 2 , 1 , 1 ); plot(w, phase_H); xlabel( 'Frequency (radians/sample)' ); ylabel( 'Wrapped Phase (radians)' ); title( 'Wrapped Phase Response' ); grid on; subplot( 2 , 1 , 2 ); plot(w, unwrapped_phase_H); xlabel( 'Frequency (radians/sample)' ); ylabel( 'Unwrapped Phase (radians)' ); title( 'Unwrapped Phase Response' ); grid on; sgtitle( 'Wrapped and Unwrapped Phase Responses' ); d.
n = 0 : 50 ; % Define the range of n x = 2 * cos ( 0.5 * pi *n + deg2rad( 60 )); % Input signal b = 1.81 ; %value of b % Initialize y with zeros y = zeros ( size (n)); % Calculate y[n] using the difference equation for i = 3 : length (n) y( i ) = b * x( i ) - 0.81 * y( i -2 ); end e. % Define the system parameters b = sqrt ( 1 + 0.81 ^ 2 ); % Define the input signal x[n] n = 0 : 99 ; x = 2 * cos ( 0.5 * pi * n + deg2rad( 60 )); % Compute the steady-state response using the filter function y_ss = filter(b, [ 1 , 0 , -0.81 ], x); % Plot the steady-state response stem(n, y_ss); xlabel( 'n' ); ylabel( 'y[n]' ); title( 'Steady-State Response y[n] to x[n]' ); grid on; % Analytically calculate the response y[n] to the input x[n] y = b * x - 0.81 * [ 0 , 0 , y_ss( 1 :end -2 )]; % Compare y and y_ss to verify the result
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