abariska@student.ethz.ch wrote: ...> But the delay in b[n] is fractional, 1/7 sample, and the resulting > windowed and sampled sinc does not have linear phase response. It is > interesting that with infinite many samples, strict symmetry of the > coeffcients is not needed anymore for linear phase.Silly me! I took it to be 7 samples. 1/7 is so bizarre, I filtered it out. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������

# questions raised by reading and thinking with possibly missing background

Started by ●December 10, 2005

Reply by ●December 15, 20052005-12-15

Reply by ●December 15, 20052005-12-15

On 14 Dec 2005 20:44:03 -0800, "robert bristow-johnson" <rbj@audioimagination.com> wrote:>but it's not so much the ripples ramping up to the main lobe of the >impulse response, Al. it's a consequence of using the Parks-McClellan >algorithm to design an "equi-ripple" filter. those ripples in the >pass-band are like multiplying by a cosine which has the effect of >translation in the other domain.Harris describes an analogous phenomenon in his discussion of the Dolph-Chebyshev window; "On the Use of Windows for Harmonic Analysis with the Discrete Fourier Transform", Proceedings of the IEEE, Vol. 66, No. 1, January 1978. This window is characterized by equal-amplitude sidelobes (Harris describes them as "almost sinusoidal!") that result in displaced impulses when transformed into the other domain. Greg

Reply by ●December 15, 20052005-12-15

robert bristow-johnson wrote:> abariska@student.ethz.ch wrote:> > You claim that the infinitely long sequence > > > > b[n] = sinc(n + 1/7), > > > > for all n, is linear-phase. That could be true. Do you have an idea for > > a proof? > > it appears to me that you already have it proved. at least for > frequencies between -Nyquist to +Nyquist. > > since the acausal continuous-time impulse response > > h(t) = sinc(t - d) > > has frequency response as > > H(f) = exp(-j*2*pi*f*d) for |f| < 1/2 , zero otherwize, > > then H(f) is clearly linear phase for any constant delay, d. and since > H(f) is 0 for |f| >= 1/2, h(t) can be sampled at a sampling rate,1/T, > of 1 and have no aliasing: > > h[n] = h(n*T) = sinc(n - d) > > according to the sampling theorem the frequency response of the sampled > function is the same as the frequency response of the continuous-time > function getting sampled for frequencies with magnitude less than > Nyquist (in this case, Nyquist is 1/2) and that frequency response is > periodically extended for frequencies above Nyquist.Sounds good to me. The truncated sinc in Randy's example was not bandlimited, and the aliasing when sampling messed up the phase linearity. I think Randy's proof outline was somewhere on along your lines as well. What do you think of that theorem that I posted earlier from this paper: http://0xdc.com/paper.pdf (at the bottom of the first page). I have a feeling that anti-symmetric linear-phase sequences are not covered in the statement of the proof. This would make the "iff" claim false.> > does that do it for you, Abariska?In case you hadn't noticed, "abariska" stands for Andor Bariska. It's my old and inactive student account (go spam that) :-). Regards, Andor