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PS5solns

# PS5solns - Alternatively we can do this in the f domain...

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Alternatively, we can do this in the f domain, x 1 ( t ) = Sinc ( t ) = X 1 ( j 2 πf ) = I [ - 1 2 , 1 2 ] ( f ). x 2 ( t ) = sin(2 π ( t - 1)) π ( t - 1) = 2 Sinc (2( t - 1)) = X 2 ( j 2 πf ) = I [ - 1 2 , 1 2 ] ( f ) e - j 2 πf . x ( t ) = x 1 ( t ) x 2 ( t ) = X ( j 2 πf ) = X 1 ( j 2 πf ) * X 2 ( j 2 πf ). Peforming the convolution as we did in the ω domain, we get X ( j 2 πf ) = 1 j 2 π (1 + e - j 2 πf ) - 3 2 < f < - 1 2 - 1 j 2 π (1 + e - j 2 πf ) 1 2 < f < 3 2 0 elsewhere Problem 2 4.4 ( a ) x 1 ( t ) = 1 2 π R -∞ [2 πδ ( ω ) + πδ ( ω - 4 π ) + πδ ( ω + 4 π )] e jωt = 1 + 1 2 e j 4 πt + 1 2 e - j 4 πt = 1 + cos(4 πt )

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Problem 6 ts = 0.1; fs = 0.05; % For b), ts = 0.05, fs = 0.02 N = 256; % 1/(ts*fs) = 200; % For b), N = 1024, 1/(ts*fs) = 1000 x = zeros(1,N); m = (1-0)/ts; % number of non-zero samples x(1:m) = 1; % now we have x as a function of the sample number y = fft(x)*ts; % the function fft() implements H[m] fs1 = 1/(ts*N); % actual frequency granularity < 0.05 n = 1:N; % The analysis frequencies are multiples of fs1 figure, subplot(2,1,1);plot(n*fs1, abs(y)); grid on title(’Magnitude of H ( j2\pif ) [ts = 0.1, fs = 0.05]’,’Fontsize’, 12); xlabel(’Frequency f (Hz)’);ylabel(’| H ( j2\pif ) |’);
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PS5solns - Alternatively we can do this in the f domain...

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