Regarding "A weighted sum of some DFT outputs"
- 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) ? If yes, do you agree that no SINGLE C-array element represents, say 1.5 * Fs / 8 frequency ? 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 ? If yes, do you agree that had I wanted to change only gain of, say, (2 * Fs / 8), i.e amplitude of (C[2], C[6]) I wouldn't have changed the amplitude of, say, (1 * Fs / 8) spectral component or any other spectral component for that matter ? If yes, that's what I meant saying that central frequencies should be a multiple of frequency resolution ((Fs / 8) in this example). If central frequency is not a multiple of frequency resolution, its amplitude can NOT be changed without affecting the amplitude of other spectral components. On Mon, 2 Jan 2006 23:53:33 +0100 fons adriaensen <[EMAIL PROTECTED]> wrote: > On Mon, Jan 02, 2006 at 11:37:54PM +0200, Sergei Steshenko wrote: > > > So, how are going to implement a central frequency which is > > not a multiple of (Fs / N) in a DFT equalizer ? > > > > That is, what data resulting from direct DFT represents such frequencies ? > > A weighted sum of some DFT outputs, instead of just one of them. > > Look at what you have now. You do a forward FFT and get N complex > values A [i]. You multiply each of these A [i] by some factor, and > then do an IFFT. > > The second step can be regarded as the multiplication of the vector > A [i] by a diagonal matrix. Using central frequencies in between > the 'integer' bins would just mean this matrix is no longer diagonal. > In practice it would become a band matrix and still be quite sparse, > and the product can be done efficiently. > > Once you have the matrix instead of the term by term product, the > windowing can also be absorbed into it. > > -- > 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
