Problems in plotting a Sinc signal, applying a FFT with noise and then the IFFT.

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claudio
claudio on 12 Sep 2015
Commented: Walter Roberson on 20 Sep 2015
Hello everyone!
I am trying to plot the signal above, with no success. The code is:
f = 10e3
Ts = 1/(32*f)
n = -160:160
ruido = 2*randn(1,numel(n));
s = sinc(2*pi*f*n*Ts)
%s = sinc(f*n*Ts)
figure
plot (s)
title ('SINC')
xlabel('Tempo (s)')
ylabel ('Amplitude')
x = fft (s);
x = fftshift (s);
fc = [-numel(x)/2:numel(x)/2-1]./numel(x)
figure
plot(fc,fftshift(abs(x)));
title('Transformada de Fourier')
xlabel('Frequencia (Hz)')
ylabel ('Magnitude (dB)')
c = sinc(2*pi*f*n*Ts) + ruido;
figure
plot(c);
title('Sinc com ruido')
xlabel('Frequencia (Hz)')
ylabel ('Magnitude (dB)')
This should be something like this:
But I´m getting this one instead:
Any help is appreciated! Thank you.
  1 Comment
Walter Roberson
Walter Roberson on 20 Sep 2015
What is Ts ? You define it as 1/(32*f) and every time you use it you multiply it by f so the net result is to cancel out the f and Ts and leave just 1/32 . Why ? What does it stand for?

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Accepted Answer

Hamoon
Hamoon on 12 Sep 2015
You've got some errors in your code, for example, you defined x=fftshift(s) which is wrong. You also add a normally distributed random amount with standard deviation 2, which is larger than the maximum amplitude of sinc function. And also you passed sinc function value without noise to the fft function. I corrected these mistakes, and I tried not to change your code a lot so you can trace the changes, here is the code:
f = 10e3;
Ts = 1/(32*f);
n = -160:160;
noiseSTD = .01;
ruido = noiseSTD*randn(1,numel(n));
s = sinc(2*pi*f*n*Ts);
c = sinc(2*pi*f*n*Ts) + ruido;
%s = sinc(f*n*Ts)
figure
plot (s)
title ('SINC')
xlabel('Tempo (s)')
ylabel ('Amplitude')
x = fft (c);
x = fftshift (abs(x));
fc = (-numel(x)/2:numel(x)/2-1)./numel(x);
figure
plot(fc,x);
title('Transformada de Fourier')
xlabel('Frequencia (Hz)')
ylabel ('Magnitude (dB)')
figure
plot(c);
title('Sinc com ruido')
xlabel('Frequencia (Hz)')
ylabel ('Magnitude (dB)')
  2 Comments
Hamoon
Hamoon on 12 Sep 2015
Yes. thank you Image Analyst. But I just tried to fix obvious mistakes here. he still needs to change some parameters. The result without noise would be like this:

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More Answers (1)

claudio
claudio on 20 Sep 2015
Thank you so much, guys. Codes worked like a charm.

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