Why and how does FFT introduce a processing gain ?

Hi all,

In general the quantisation noise, assuming a sinusoidal input, gives a SNR of 1.7+6*n. But in FFT, the total noise floor is indeed lower than this, by a factor log10(N/2), where N is the no. of points in FFT. Why and How does the transform introduce a processing gain ?

Plz help me out...

Re: processing gain in FFT ?

You cant achieve any input signal improoving doing the FFT. You only change the representation of signal from time domain to frequency domain, but FFT doesnt perform any filtering, amplifying or something else what may be called processing gain.
Please, show me if you found an ADC with 16 bits and 97.7 dB SNR for example according to your formula. And how to improve it to 100.1 using 512FFT.

Hi,

Indeed, each bin of the DFT is the result (at one specific time) of the output of the input sequence applied to a FIR filter.

The "impulse response" of the filter associated to the k-th bin is a complex exponential that has k cycles in the duration of the data. Its frequency response is maximum at the frequency of the bin and crosses by 0 at the frequencies of all the other bins.

If you apply a sinusoidal input that has exactly k cycles in the N data points, only the k and N-k bins have non-zero output. The power (or energy) of the signal can be seen at these outputs.
The noise, if assumed white, is filtered by all the filters of the bank and its power is distributed among all the bins.
So, for pure sinusoidal inputs, the SNR at its specific bin is improved. This allows to peform spectral analysis that reveals the presence of ver weak narrowband signals immersed in strong wideband noise.

Regards

Z

Thanks zorro.
One more doubt... When a quantized sinusoid is taken FFT, the quantisation noise is del square by 12, which will be distributed in the entire frequency range, so as the no. of points increases, noise floor goes down. But at the same time if the no. of points increases, the bin peak amplitude which is A*N/2 also would increase, so wont there be a 6dB improvement in the NOISE FLOOR when the no. of FFT bins is doubled ? In many texts, i read about the 3dB improvement in NOISE FLOOR with respect to the quantisation noise distribution over all the bins, but i didnt find any reference to the increase in the corresponding bin amplitude.... Plz clarify me....

Defining

we have

Assume x is a sinusoid in a specific bin. The amplitude of the corresponding Xk increases linearly with N, as you said; there are more signal samples that are coherenty added and the signal contribution increases the power by 6 dB if N increases by a factor 2.
But increasing N you add more noise samples too in Xk; nevertheless, assuming that the noise samples are not correlated, they add noncoherently and its contribution to the amplitude of Xk increases sqrt(2) times, i.e. power increases 2 times or 3 dB.
The conlusion is that SNR improves 3 dB by doubling N.
If in the equations above you take a different normalization applying the 1/N factor to the first equation (not to the second), then for the same signal its contribution to Xk would be independent of N, and the contribution of noise to all the Xi decreases 3 dB. Of course, the result is the same

Regards

Z