Question

Objective Conduct DTFT, DTFS on a periodic discrete signal. Task: Consider the system with impulse response Tth sin 8 h(n) S(n) Tn (1) Find the Fourier-series representation for the output y(n) when the input x(n) is the periodic extension of the sequence 3/2, -1,0, -3/2, 1,0 Plot the x(n), h(n), y(n) and Fourier coefficient bk using Matlab or handwriting (Example 7.2.6 irse material) in cour (2) Find the output y(n) of the system with the input 1 Tn Tn x(п) sin 5 COS 9 and plot H), Y, [Y(), x(n), y(n) using Matlab or handwriting (Example 7.4.6 in course material) 13

Answer 1

close all,
clear all,
clc,

Fs=1000;
n=0:(1/Fs):1;

X = [3/2,-1,0];

h=dirac(n);
idx = h == Inf;
h(idx) = 1; % set Inf to finite value

Hn=h - (sin(n*pi)/(n*pi));
for r=1:length(n)
temp = rem(r,3);
if(temp==0), temp=3; end
Xn(r) = X(temp);
end
subplot(2,2,1); plot(Xn); title('Repeating Signal, X = [3/2,-1,0]');
subplot(2,2,2); plot(Hn); title('Impulse Response, Hn=Delta(n) - (sin(n*pi)/(n*pi))');

Yn = convn(Xn,Hn);
subplot(2,2,3); plot(Yn); title('Output Yn = Conv(Xn,Hn)');

L=length(Yn);
NFFT = 2^nextpow2(L); % Next power of 2 from length of y
Y = fft(Yn,NFFT)/L;
f = Fs/2*linspace(0,1,NFFT/2);
% Plot single-sided amplitude spectrum.
No_of_Points=20;
subplot(2,2,4); plot(f(1:No_of_Points),2*abs(Y(1:No_of_Points)));
%subplot(2,2,4); plot(f,2*abs(Y(1:NFFT/2)));
str = strcat('FFT of Yn at N = ',num2str(NFFT),' Points, Showing only 1 to ',num2str(No_of_Points),' Points for good viewing');
title(str);
xlabel('Frequency (Hz)'); ylabel('|Y(f)|');
%subplot(3,1,3); plot(Xn);

figure,
Xn = cos((n*pi)/5) + sin(((n*pi)/9)+0.5);

h=dirac(n);
idx = h == Inf;
h(idx) = 1; % set Inf to finite value

Hn=h - (sin(n*pi)/(n*pi));
subplot(2,2,1); plot(Xn); title('Xn = cos((n*pi)/5) + sin(((n*pi)/9)+0.5)');
subplot(2,2,2); plot(Hn); title('Impulse Response, Hn=Delta(n) - (sin(n*pi)/(n*pi))');

Yn = convn(Xn,Hn);
subplot(2,2,3); plot(Yn); title('Output Yn = Conv(Xn,Hn)');

L=length(Yn);
NFFT = 2^nextpow2(L); % Next power of 2 from length of y
Y = fft(Yn,NFFT)/L;
f = Fs/2*linspace(0,1,NFFT/2);
% Plot single-sided amplitude spectrum.
No_of_Points=20;
subplot(2,2,4); plot(f(1:No_of_Points),2*abs(Y(1:No_of_Points)));
%subplot(2,2,4); plot(f,2*abs(Y(1:NFFT/2)));
str = strcat('FFT of Yn at N = ',num2str(NFFT),' Points, Showing only 1 to ',num2str(No_of_Points),' Points for good viewing');
title(str);
xlabel('Frequency (Hz)'); ylabel('|Y(f)|');

Impulse Response, Hn-Delta(n) (sin(n*pi)/(n"pi) Xn cos((n pi)5) + sin(n"p9)+05) 1.8 0.8 1.58 0.6 1,56 0.4 154 0.2 1.52 1.5 48 02 1.46 -04 1000 1200 800 FOC 200 400 1200 1000 80C 600 400 200 FFT of Yn at N 2048 Points, Showing only 1 to20 Points for good viewing Output Yn Convin,Hn) E00 100 400 -100 300 200 200 300 100 400 0 500 1500 2000 2500 1000 F00 Frequency (Hz)

Impulse Response, Hn-Delta(n) (sin(n*pi)/(n"pi) Repeating Signal, X- 13/2-101 1.5 0.8 0.6 0.4 0.2 0 -0.5 02 -04 1000 1200 800 FOC 200 400 1200 1000 800 600 200 400 FFT of Yn at N=2048 Points, Showing only 1 to20 Points for good viewing Output Yn Convin,Hn) 10 50 -10 40 -20 30 30 20 -40 10 50 0 1500 -60 1000 2500 2000 F00 Frequency (Hz)