MATLAB: A doubt in ZERO PADDING IMPLEMENTATION

stftzero padding

actually i am applying short time fourier transform and i am implementing zero's to the end of the signal to match with the window length but i can see some artifacts in my plot with less power how can i avoid it can i avoid it by applying zeros between the signal like ex: 102030….. how can i do it…..

Best Answer

Not sure I follow why you're trying to do this, but here is code for interleaving zeros in your array:

m = rand(1,10); % Sample data.
interleaved = zeros(1, 2*length(m));
interleaved(1:2:end) = m

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MATLAB: How I can analyze this magnitude of 1D signal

If the large value shown in the figure corresponds to 0 frequency, then that simply means the signal has a nonzero DC value. The 0 frequency component in the DFT is simply the sum of all elements in the input signal, or N times the mean value of the signal, where N is the number of elements in the signal.

Quite often, the zero frequency component is not of interest when doing a frequency analysis. To remove that, simply remove the mean from the signal.

For example, compare:

Fs = 1000;
t = 0:1/Fs:1-1/Fs;
  x = 20+cos(2*pi*100*t)+randn(size(t));
  xdft = fft(x);
  xdft = xdft(1:length(x)/2+1);
  plot(abs(xdft))
  % remove mean
  xnew = x-mean(x);
  xdftnew = fft(xnew);
  xdftnew = xdftnew(1:length(x)/2+1);
  plot(abs(xdftnew))

MATLAB: Amplitude Error in amplitude Spectrum of FFT

I suspect that your frequency of interest is simply not falling on a DFT bin directly. Why do you think you need to use a power of two for the DFT?

For example, if I have data sampled at 1 kHz and I have 1000 points, then frequency increment in DFT bins is 1 Hz (the bins are 1 Hz apart) so a frequency of 100 Hz for example will fall directly on a bin.

   Fs = 1000;
   t = 0:1/Fs:1-1/Fs;
   x = 1.5*cos(2*pi*100*t);
   xdft = fft(x);
   freq = 0:Fs/length(x):Fs/2;
   xdft = xdft(1:length(x)/2+1)./length(x);
   xdft(2:end) = 2*xdft(2:end);
   plot(freq,abs(xdft))

See that the magnitude is estimated exactly. Now watch what happens when I pad to 1024:

   xdft = fft(x,1024);
   freq = 0:Fs/length(xdft):Fs/2;
   xdft = xdft(1:length(xdft)/2+1)./length(x);
   xdft(2:end) = 2*xdft(2:end);
   plot(freq,abs(xdft))

by padding to 1024, I have now made the frequency of interest fall between DFT bins and that makes my amplitude estimate inaccurate.

See https://www.mathworks.com/help/signal/ug/amplitude-estimation-and-zero-padding.html

in the Signal Processing Toolbox documentation

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MATLAB: How I can analyze this magnitude of 1D signal

MATLAB: Amplitude Error in amplitude Spectrum of FFT

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