FFT the whole input vs FFT segment

Question

suppose I have a 4 seconds signal captured at $f_s$ = 1024Hz, which gives 4,096 data array. What is the different between these 2 FFT methods (MATLAB code)

[1] FFT the entire input with 4096-points FFT

N = length(input);
Y = fft(input)/N;

[2] Divide the input into 8 segments (4096/8=512 data point each), and feed each segment to 512-points FFT, and then average the output:

N = length(input);
Y = zeros(1,512);
for i = 1:8
  segment = input(512*(i-0)+1 : 512*i);
  Y = Y + abs(fft(segment,512))/512/8;
end

Question #1: Are these 2 methods equivalent?

Question #2: Is the frequency resolution for both methods equal to $\frac{f_s}{N}$ = 0.25Hz per bin?

Answer 1

Question #1: Are these 2 methods equivalent?

No, method #1 is taking an FFT of the entire sequence resulting in frequency bin widths of fs/N = 1024/4096 = 0.25 Hz. Method #2 is summing shorter FFTs at a frequency resolution (bin width) of 1024/512 = 2.0 Hz. For #2, if you divide by the number of FFTs after you sum them you are then averaging them. This is often times used to reduce the variance of the result.

Question #2: Is the frequency resolution for both methods equal to 0.25Hz per bin?

No, they are 0.25 & 2.0 Hz. The resolution, or bin width, of the FFT is fs/N.

Answer 2

You are absolutely correct, the frequency resolution $f_R$ (or bin width) should be calculated based on the length of FFT, $N_{FFT}$ (i.e. $f_R=\frac{f_s}{N_{FFT}}$), instead of the length of the original captured signal $N$.

I believe most journals/books out there simply assume ($N=N_{FFT}$) in their sample calculation, and use the notation of $f_R=\frac{f_s}{N}$ throughout their books/papers. But when $N{\neq}N_{FFT}$ as in method #2, then the $f_R=\frac{f_s}{N}$ could to lead confusion among newbies (like myself). So a safer way is to use $f_R=\frac{f_s}{N_{FFT}}$ instead.

To prove this, below are the output of 5Hz + 30Hz sinusoids ($f_s$=1024Hz, duration=4s, 4096 data points), and it's output with the FFT-512, and FFT-4096 respectively. You can see the frequency resolution (or bin width) of the FFT-512 output (3rd plot) is wider than FFT-4096 (2nd plot), since $\frac{f_s}{512}>\frac{f_s}{4096}$.

Answer 3

The frequency resolution of a DFT operation is solely dependent on the #of samples ($N_s$)only. Taking a larger than $N_s$ length DFT implicitly makes use of zero-padding in matlab. This does not increase the frequency resolution (though it may appear so when you see the plots) and rather makes use of spectrum obtained without zero-padding to interpolate to the increased DFT length sample values. This operation helps in smooting the appearance of the spectrum and resolving potential ambiguities between the peaks.

In your case you are dividing the signal into frames and DFT of each frame will have thus a lower fundamental resolution (dependent on the number of sample you use 512 now). You are seeing good resolution (though the second case has poor resolution than first) because the sampling fvequency is much higher than the signal content frequencies hence you have captured enough number of cycles of the constituent components in both cases.

The approach of dividing the signal into frames and doing then averaging the resulting spectrum^2 (energy in the frequency domain) is used in power spectral density computation of stochastic (deterministic (sine waves)+noise) signals where number of available samples is always a limitation, hence it is appropriate to deduce interpretation from an average PSD than a single PSD estimate.