Consider two finite-length sequences, x[n] = {0 n 3 and h[n] = delta[n - 1] - de
ID: 2315140 • Letter: C
Question
Consider two finite-length sequences, x[n] = {0 n 3 and h[n] = delta[n - 1] - delta[n - 2]. (a) Determine and plot the discrete convolution y[n] = x[n] * h[n] for these two sequences. (b) Note that the length of n[n] is 4 and the length of h[n] is 3 samples. Therefore, each sequence can be represented by an N-point DFT with N Greaterthan 4. Zero padding would be used to extend the lengths when N > 4. Determine expressions for the N-point DFTs X[k] and H[k]. Do not obtain numerical answers-instead express your answers in terms of e^-j(2pi/N)^k. (c) Now form the product Y[k] = X[k]H[k] again expressed in terms of e^-j(2pi k/N)n. (d) From the result in part (c), determine the IDFT of Y[k] when N = 6. Compare your answer to the result of part (a). (e) Repeat when N = 4. In this case, the complex exponentials must be changed so that the exponents are less than 2pi (i.e., = e^-j(2pi/N)(N+k) = e^-j(2pi/N)^k if 0 lessthanorequalto k lessthanorequalto N - 1). (f) In general, Consider x[n] to have length L and h[n] to have length M. How should N bechosen so that the MATLAB statement y = ifft(fft(x,N). *fft(h,N)) produces the same result as the MATLAB statement y = conv (x, h)?Explanation / Answer
clear all;
close all;
x=input('enter the sequence');
subplot(3,1,1);
stem(x);
xlabel('time');
ylabel('magnitude');
title('input sequence');
N=length(x);
y=(zeros(1,N));
x=bitrevorder(x);
g=1;
for m=1:1:(log(N)/log(2))
for n=1:1:N/2^m
for p=1:1:2^(m-1)
a=x(g);
b=x(g+2^(m-1));
y(g)=a+b+exp(-i*2*pi*(p-1)/2^m);
y(g+2^(m-1))=a-b*exp(-i*pi*(p-1)/2^m);
g=g+1;
end
g=2*m*n+1;
end
g=1;
for t=1:1:N
x(t)=y(t);
end
end
subplot(3,1,2);
stem(abs(x));
xlabel('time');
ylabel('magnitude');
title('magnitude of DIT-FFT sequence');
subplot(3,1,3);
stem(angle(x));
xlabel('angle');
ylabel('amplitude');
title('angle of DIT-FFT');
gtext('kluniversity');
disp(abs(x))