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PS6Solns_rev

# PS6Solns_rev - mf s,m = 0 1,N-1 The range m = 0 to N 2...

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Problem 2 clear;clc; % PROBLEM 2 a) ts = 1/8; t = -4:ts:4; x = sinc(4*t).*sinc(2*t); figure, plot(t,x);grid on % PROBLEM 2 b) N = 1024; x1 = zeros(1,N); % zero padding x1(1:numel(t)) = x; X = abs(ts*fft(x1)); fs = 1/(N*ts); % frequency resolution figure, plot([-N/2+1:N/2]*fs, [X(N/2+2:N) X(1:N/2+1)]);grid on % PROBLEM 2 c) ts = 1/20; t = -4:ts:4; y = sinc(4*t).*sinc(2*t).*sin(12*pi*t); figure,plot(t,y);grid on % PROBLEM 2 d) y1 = zeros(1,N); % zero padding y1(1:numel(t)) = y; Y = abs(ts*fft(y1)); fs1 = 1/(N*ts); % frequency resolution figure,plot([(-N/2 + 1):N/2]*fs1, [Y(N/2 +2:N) Y(1:N/2+1)]);grid on To plot the Fourier transform in the range - f max < f < f max , the sampling interval in the time domain should not be greater than 1 2 * f max . Hence, for f max = 4, we choose t s = 1 2 * 4 = 1 8 s , and for f max = 10, we choose t s = 1 2 * 10 = 1 20 s . By choosing the number of samples in the DFT as N = 1024, and zero-padding accordingly in the time domain, we get a frequency resolution f s = 1 1024 * ts Hz . The DFT gives the sampled spectrum of the signal at the discrete frequencies

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Unformatted text preview: mf s ,m = 0 , 1 ,...,N-1. The range m = 0 to N 2 corresponds to the range of frequencies 0 ≤ f ≤ f max , and the range m = N 2 + 1 to N-1 corresponds to the range of frequencies-f max < f < 0. We see that the magnitude spectrum plots for X ( j 2 πf ) and Y ( j 2 πf ) match exactly with the Fourier transform magnitude plots in Problems 1( a ) and ( c ) respectively. 1-4-3-2-1 1 2 3 4-0.2 0.2 0.4 0.6 0.8 1 1.2 Time t (s) sinc(4t) sinc(2t) Plot of x(t) = sinc(4t) sinc(2t)-4-3-2-1 1 2 3 4 0.05 0.1 0.15 0.2 0.25 0.3 0.35 Frequency f (Hz) | X (j2 π f) | Plot of | X (j2 π f) | vs f-4-3-2-1 1 2 3 4-1-0.8-0.6-0.4-0.2 0.2 0.4 0.6 0.8 1 Time t (s) x(t) sin(12 π t) Plot of y(t) = sinc(4t) sinc(2t) sin(12 π t)-10-8-6-4-2 2 4 6 8 10 0.02 0.04 0.06 0.08 0.1 0.12 0.14 Frequency f (Hz) | Y (j2 π f) | Plot of | Y (j2 π f) | vs f...
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