MATLAB: Difference between lines using linespace

linespace

Hi there, I need to know what is the difference between these: f=linspace(-fs/2,fs/2,samp_Tr); f = -fs/2 + (0:samp_Tr-1)*fs/samp_Tr; actually it gives me in my system different solution

thanks in advance

Best Answer

The two are very similar. The difference is that linspace taked some extra care to reduce the effect of accumulated round-off error. For example, because 1/3 cannot be exactly represented in binary, if you add the closest binary value to 1/3 together three times, you will get a value slightly less than 1. Imagine that you did that 300 times in a row; the result would have to end up 100 times that small difference less than 100. That would happen with the : operator. linspace works with that by dividing the range into two pieces and allowing accumulated error from the beginning to the middle and accumulated error from the end towards the middle, so right in the middle the gap might be a slightly different size.

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MATLAB: What are the frequencies when N in fft(x,N) is odd

In general, you can do it as

f = (0:N-1)*fs/N

If you want to map it to negative frequencies

if N is even

f(N/2+1:end) = f(N/2+1:end)-fs % sample -fs/2

f(N/2+2:end) = f(N/2+2:end)-fs  % sample fs/2

if N is odd

f((N+1)/2+1:end) = f((N+1)/2+1:end)-fs

HTH

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: What are the frequencies when N in fft(x,N) is odd

MATLAB: Amplitude Error in amplitude Spectrum of FFT

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