Hello all, I am working on a project to equalize the response of a loudspeaker using dsp prefiltering. I want to use a logarithmically-swept sine excitation function to measure the linear impulse response and harmonic distortion as described in Farina's paper "Simultaneous measurement of impulse response and distortion with a swept-sine technique" (paper#22 at http://pcfarina.eng.unipr.it/ASK/home.htm). In the paper, he claims the inverse filter of the log-swept sine for deconvolving the speaker's output is simply the time-reversed input with an exponentially-decreasing envelope applied. This is to compensate for the fact that the power is more spread out in the high frequency range. All of this makes sense to me. But, when I convolve my excitation function (20Hz-20kHz) with its inverse (using a fast convolution routine in MatLab), I don't get a delta function. The convolution sum has some residual high frequency information (a peak near 20kHz when I convert to freq). I am wondering if this is problematic, or just an artifact of the convolution routine. Since the inverse filter has a much higher amplitude at large frequency, I imagine a convolution would produce the max response when the high-f sinusoids are lined up, and conversely the most negative at the convolution steps when they are 180deg out of phase. So I guess it makes sense that the conv result would be wavy near the place where the input function directly overlaps its inverse... Of course, when I use the inverse filter to deconvolve the speaker's output, I don't want this artifact to corrupt my measured impulse response. Is it possible for me to find an inverse filter that does a better job?

# inv filter deconvolution

Started by ●June 24, 2004

Reply by ●June 26, 20042004-06-26

> In the paper, he claims the inverse filter of the log-swept sine for > deconvolving the speaker's output is simply the time-reversed input > with an exponentially-decreasing envelope applied.The author also claims that envelope applied to the inverse filter has a dB slope of 6dB/octave. When I generate a log-swept sine, take the fft, and convert to dB scale, I measure the fall-off from low to high frquency as 3dB/octave. (??)

Reply by ●June 26, 20042004-06-26

Dan Shorb wrote:>> In the paper, he claims the inverse filter of the log-swept sine for >> deconvolving the speaker's output is simply the time-reversed input >> with an exponentially-decreasing envelope applied. > > The author also claims that envelope applied to the inverse filter has > a dB slope of 6dB/octave. When I generate a log-swept sine, take the > fft, and convert to dB scale, I measure the fall-off from low to high > frquency as 3dB/octave. (??)Hi Dan, I use the same sort of procedure for HTRFs for which I have to deconvolve the headset transfer (for me its all is done in the freq domain using complex division). I checked the procedure above by extracting the inverse chirp using prihc=real(fftshift(fft(dirac)./fft(chirp)) % (dirac & chirp I guess you know what they are) next I convolved conv(prihc,chirp) and I got my dirac pulse back. Slightly noisy though, but time anti-aliasing solves this, a little. It seems to work, but there's 1 catch. The chirp needs to run from dc to fs/2. You can imagine what happens with fft(dirac)./fft(chirp) when fft(chirp) becomes zero. You have to block out those bins in the complex division and set them to zero. Same problems occur when you obtain steep notches in during measurements My guess is that undefined values (zero?) for high freq bins causes the error. Suo