Hi Mohamud,
I understand that you are trying to compute and plot the DFT of the array Cn that contains zeros and ones, and you are expecting to find 64 peaks in the DFT output. Based on the code you provided, it appears that the issue is with the way you are computing the DFT. Fortunately, MATLAB provides the functions "fft" and "ifft" to compute the discrete Fourier Transform and its inverse, respectively, so you should not have to compute it the long way.
The issue in your code is in how you are computing the 'Sf_n' term. You do not need to include (f/fs) in this computation process, and you do not need to vectorize this line since it is only computing a single value in the overall DFT. I have provided example code below that corrects this portion of your code; additionally, I have a commented-out line of code below this computation that would take care of this in one line of code.
N=1024;
fs=100*10^6;
T=1/fs;
vec=zeros(N,1);
positions=[512 128 192 256 320 384 448 512 576 640 704 768 832 896 960 1024];
vec(positions)=1;
Cn=vec;
one_sided_span = 200*10^6;
num_steps = N;
step_value = (2*one_sided_span)/num_steps;
f = -one_sided_span:step_value:one_sided_span-1;
Sf = zeros(1,length(f));
for k=0:N-1
Sf_n = 0;
for n=0:N-1
Sf_n = Sf_n + Cn(n+1)*exp((-1i*(2*pi*k))*(n/N));
end
Sf(k+1) = Sf_n;
end
% DFT computation in one line
%Sf = fft(Cn,N);
figure(1)
plot(Cn)
title('The values of Cn');
xlabel('NFFT-Points');
ylabel('Amplitude');
axis([0 1050 0 1.5]);
figure(2)
plot(f,20*log10(abs(Sf)));
title('Sf in log scale');
xlabel('Frequency(Hz)');
ylabel('Amplitude (dB)');
axis([-2*10^8 2*10^8 -10 40]);
The plot of the DFT shows 64 peaks over the specified frequency range, which corresponds to what you were expecting.
I hope this helps.
Matt
