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 <[EMAIL PROTECTED]> 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 > ------------------------------------------------------- This SF.net email is sponsored by: Splunk Inc. Do you grep through log files for problems? Stop! Download the new AJAX search engine that makes searching your log files as easy as surfing the web. DOWNLOAD SPLUNK! http://ads.osdn.com/?ad_id=7637&alloc_id=16865&op=click _______________________________________________ Alsa-user mailing list Alsa-user@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/alsa-user
