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

[Sergei S. <ste...@li...>]

Fons,

when I said "central frequencies" I meant the frequencies which
can be reproduced exactly using one direct DFT and one inverse
DFT buffer.

That is, I claim that for the 8 points DFT/FFT example I gave if
input signal contains only

0 * Fs / 8
1 * Fs / 8
2 * Fs / 8
3 * Fs / 8

frequencies, i.e if it is periodic band limited signal, it
can be reproduced with infinite precision (provided
calculations are infinitely precise) using only one 8-element
array.

Any other signal, even if it is band limited, can not be
reproduced precisely using only one 8-element
array because that signal is not a periodic function with
the (4 / Fs) period.

Look at this another way.

For the small 8 points FFT example one can build at maximum
4-band equalizer with

0 * Fs / 8
1 * Fs / 8
2 * Fs / 8
3 * Fs / 8

central frequencies

Any attempt to increase the number of central frequencies will cause
ambiguity, or, as you said,

"The minimum bandwidth you can make is Fs/N"
.

My choice of central frequencies means I'm using the strict subset
of the maximum number of them - 4 in this example.

I know about interpolation, resampling and stuff like that, but it is
not what a simple FFT-based equalizer is doing.


And yes, you are correct regarding

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

though it doesn't change the point.

My point was that with an IIR-based equalizer central frequency
can really be chosen to have whatever value in 0..Fs/2 range,
I mean, the bandpass IIRs can be built that their magnitude maximum
will be at really arbitrary value in 0..Fs/2 range, and their Q factor
can be made arbitrary high (I mean infinite precision computer
arithmetics) - it's not the case with DFT when N is limited.

In the IIR case moral equivalent of limited N is limited IIR filter order,
still, its resonance frequency can be whatever in 0..Fs/2 range.

In the case of DFT limited N means both limited Q and discrete
central frequencies in the terms of precise signal restoration.

Regards,
  Sergei.

P.S, I've just seen a very good reply from Bill Unruh:

http://article.gmane.org/gmane.linux.alsa.user/21688
.


On Tue, 3 Jan 2006 01:25:39 +0100
fons adriaensen <fon...@sk...> wrote:

> 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
>