## The approximate Difference between a DFT and an FFT

Both the Discreet Fourier Transform and the Fast Fourier Transform produce complex-numbered coefficients, the non-zero amplitudes of which will represent frequency components in the signal. They both produce a more accurate measure of this property of the signal, than the Discreet Cosine Transforms do.

Without getting into rigorous Math,

If we have a 1024-sample interval in the time-domain, then the DFT of that simply computes the coefficients from 0 through to 1023, half-cycles. A frequency component present at one coefficient, let us say an even-numbered coefficient, will also have a non-zero effect on the adjacent, odd-numbered coefficients, which can therefore not be separated fully, by a Fourier Transform that defines both sets. A DFT will generally compute them all.

An FFT has as a premise, a specific number of coefficients per octave. That number could be (1), but seldom actually is. In general, an FFT will at first compute (2 * n) coefficients over the full sampling interval, will then fold the interval, and will then compute another (n) coefficients, and will fold the interval again, until the highest-frequency coefficient approaches 1/2 the number of time-domain samples in the last computed interval.

This will cause the higher-octave coefficients to be more spread out and less numerous, but because they are also being computed for successively shorter sampling intervals, they also become less selective, so that all the signal energy is eventually accounted for.

Also, with an FFT, it is usually the coefficients which correspond to the even-numbered ones in the DFT which are computed, again because one frequency component from the signal does not need to be accounted for twice. Thus, whole-numbers of cycles per sampling interval are usually computed.

For example, if we start with a 1024-sample interval in the time-domain, we may decide that we want to achieve (n = 4) coefficients per octave. We therefore compute 8 over the full interval, including (F = 0) but excluding (F = 8). Then we fold the interval down to 512 samples, and compute the coefficients from (F = 4) through (F = 7).

A frequency component that completes the 1024-sample interval 8 times, will complete the 512-sample interval 4 times, so that the second set of coefficients continues where the first left off. And then again, for a twice-folded interval of 256 samples, we compute from (F = 4) through (F = 7)…

After we have folded our original sampling interval 6 times, we are left with a 16-sample interval, which forms the end of our series, because (F = 8) would fit in exactly, into 16 samples. And, true to form, we omit the last coefficient, as we did with the DFT.

210  =  1024

10 – 6 = 4

24  =  16

So we would end up with

(1 * 8) + (6 * 4) =  32  Coefficients .

And this type of symmetry seemed relevant in this earlier posting.

Dirk