Re: [Alsa-user] multichannel multiband FFT-based equalizer - README

[fons a. <fon...@sk...>]

On Tue, Jan 03, 2006 at 01:22:56AM +0200, Sergei Steshenko wrote:

> - do you agree that if, say, I have an 8 point FFTW, the following
> frequencies are represented in the FFTW output array C (the result of time ->
> frequency conversion, i.e. direct FFT):
> 
> C[0]       <=> DC (only real part)
> C[1], C[7] <=> 1 * Fs / 8;
> C[2], C[6] <=> 2 * Fs / 8;
> C[3], C[5] <=> 3 * Fs / 8;
> C[4]       <=> 4 * Fs / 8; Nyquist frequency (only imaginary part)

Yes, except that it's the cosine (real) part of Fs/2, that is in C[4].

> If yes, do you agree that no SINGLE C-array element represents, say
> 1.5 * Fs / 8 frequency ?

Yes. 
 
> If yes, do you agree that changing simultaneously gain of
> C[1], C[7] and C[2], C[6] pairs. i.e of the pairs that represent
> (1 * Fs / 8)  and (2 * Fs / 8) pairs I will not only change gain
> of (1.5 * Fs / 8) frequency, but also of the whole
> 1 * Fs / 8) .. (2 * Fs / 8) frequency range ?

Yes. This is no different from changing only one value - it represents
more than just the exact central frequency (you'd have a very bad equaliser
otherwise !). The minimum bandwidth you can make is Fs/N, and a bit more
with windowing. You may have some difficulty in believing that a 'non-integer'
band could have the same bandwidth as an 'integer' one, but it *is* possible,
and even quite straightforward. 

Look at it like this: there is no essential difference between the time
and the frequency domains, they are 'duals'. Just as you can delay a
sampled signal by half (or any fraction of) a sample by interpolation
and without impairing bandwidth (i.e. resolution in the time domain),
you can interpolate in the frequency domain without impairing resolution.
The output of a DFT is just 'samples in the frequency domain', like the
input is 'samples in the time domain'.

Or one more variation: you say "no SINGLE C-array element represents,
say 1.5 * Fs / 8 frequency", and that is correct. In the same way,
in a sampled signal, no single sample represents the value of the
original analog signal halfway between samples i and i+1. It is
represented by all surrounding samples, with sin(x)/x weighting.
And it can be reconstructed. The same is true in the frequency domain.

-- 
FA