How would you compare the SNR gain from a longer FFT (more samples) to overlapping FFTs? Given the choice, which one would you pick and why?

Update: Suppose you have 2048 samples of a signal. Assuming the sampling rate divided by the fundamental frequency is an integer, compare/constrast the SNR gain/noise variance reduction of the following:

  1. A single 2048 point FFT.
  2. 16, 128 point FFTs added coherently.
  3. 16, 128 point FFTs added incoherently.
  4. 24, 128 point FFTs (50% overlap), added incoherently.
  • $\begingroup$ Don't the answers to this question answer your question? $\endgroup$ – Matt L. Sep 12 '14 at 10:25
  • $\begingroup$ No, my question is comparing the processing gain to overlap and add. $\endgroup$ – random_dsp_guy Sep 12 '14 at 15:39
  • $\begingroup$ 'overlap and add' refers to fast convolution. Do you mean Welch's method for power spectrum estimation? $\endgroup$ – Matt L. Sep 12 '14 at 15:41

That depends on your application, signal and noise.
Increasing the FFT length means better frequency resolution.
It can be shown that the FFT of a zero-mean white noise is also white noise. Now, given that observation, this is the motivation to performing multiple shorter FFTs since the noise averages and its variance decreases by a factor of the number of FFTs.
For a stationary signal, e.g. a sine wave, the FFT should not (theoretically) change over time. Thus, averaging would have no effect. So in this case, since the noise decreases, you should expect a gain in SNR by a factor of the number of FFTs.

However, one should also take into account the scalloping loss (or spectral leakage) which can reduce the SNR significantly, and choose the FFT length correctly, according to the application.

Personally I always use overlapping if possible.

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.