"robert bristow-johnson" <rbj@surfglobal.net> wrote in message news:BC6313C3.8E9C%rbj@surfglobal.net...> it may be true, but doesn't appear obvious to me. if Re{f(w)} andIm{f(w)}> are a Hilbert pair, why is Re{exp(f(w))} and Im{exp(f(w))}?Because pointwise sums and products preserve that property, and are sufficient to construct exp(f(w)) from f(w), using the Taylor series for exp(x).> i'm not sure that you aren't tossing about the term "analytic" here since > there are two different usages of it, *both* possibly legitimately > referenced when one is discussing minimum phase and Hilbert Transform.I was careful to tell you which I meant -- "has one-sided Fourier transform".> [...] any phase response different from that would have to that of a > non-minimum phase filter and, as the name implies, it would have to > be *more* than the min-phase.Either that or non-causal, yes.> if you set > up the limits of the integrals well, you can compute the HT of a lot of > functions despite the problems of integrating 1/t. [...] that will workfor> an awful lot of odd symmetry functions (which the phase > guys are) because whatever nasty thing that happens around +a will get > virtually cancelled by whatever nasty thing that happens around -a as > a->inf.No, that does not work for odd symmetry functions. The HT kernel has odd symmetry too, so the +a and -a sides add constructively.
FIR roots and frequency response
Started by ●February 13, 2004
Reply by ●February 26, 20042004-02-26
Reply by ●February 26, 20042004-02-26
Ronald H. Nicholson Jr. wrote:> In article <403a23f0$0$3095$61fed72c@news.rcn.com>, > Jerry Avins <jya@ieee.org> wrote: > >>Bob Cain wrote: >> >>>Matt Timmermans wrote: >>> >>>>You actually have to put the poles outside the unit circle, as well as >>>>the zeros, to get that effect. I believe the term "maximum phase", >>>>though it isn't used much means with poles inside and zeros outside. >>> >>>Curious how to do that if they are all at the origin. >> >>They would all go to infinity, the only question being the angle. But it >>doesn't happen. I meant an FIR structure; how else could the tap >>coefficients be swapped end for end?, so the poles stay at the origin. > > > The poles do go to infinity if you truly swap the "entire" FIR structure > end for end. That's because you'd have to swap the entire positive real > axis to cover all possible FIR stuctures. All the FIR coeff's end up > at positive infinity (in reverse order), so the impulse response in any > finite interval around the origin becomes zero, corresponding to the > denominator of the transfer function being infinite, corresponding to > a pole at infinity. > > If you reflect the FIR structure around the origin you end up with no > poles, but also a what's usually called a non-causal filter. > > If instead you reflect the FIR coeffs around some finite tap number > (even an infinite extent of coeffs), a number of poles corresponding to > the delay of that reflection point stay put or get added at zero. > > > (Disclaimer: This post is probably due to line noise, no doubt caused > by too large an evening glass of red wine placed on the modem cable.) > IMHO. YMMV.You lost me. I hold this truth to be, if not self evident, at least well established: every transversal (tapped delay line) filter has as many poles as zeros, and they are all at the origin on the z plane. Assuming that's correct, how can scrambling the values of the tap coefficients in any way put poles at infinity? Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●February 26, 20042004-02-26
In article <403e0f1d$0$3094$61fed72c@news.rcn.com>, Jerry Avins <jya@ieee.org> wrote:>You lost me. I hold this truth to be, if not self evident, at least well >established: every transversal (tapped delay line) filter has as many >poles as zeros, and they are all at the origin on the z plane.Not true, IMHO, unless you add more conditions. Dividing by the transfer function by z (or more canonically multiplying by z^-1) is equivalent to adding a one sample delay. You can have less poles than zeros in a non-causal FIR filter (less by the number of samples by which the output precedes its first input). This is equivalent to shifting all the tap coeffs to the left. You can have more poles than zeros simply by adding a delay line to the input or output. This is equivalent to shifting all the tap coeffs to the right.>Assuming >that's correct, how can scrambling the values of the tap coefficients >in any way put poles at infinity?How else could you get a transfer function (with a non-zero impulse response) to go to zero over any finite domain, which would happen if all the usual coeffs got swapped end for end with those at +infinity? (don't forget. this would not have been posted except for line noise caused by a glass of red wine pinching the modem cable. :) IMHO. YMMV. -- Ron Nicholson rhn AT nicholson DOT com http://www.nicholson.com/rhn/ #include <canonical.disclaimer> // only my own opinions, etc.
Reply by ●February 26, 20042004-02-26
Ronald H. Nicholson Jr. wrote:> In article <403e0f1d$0$3094$61fed72c@news.rcn.com>, > Jerry Avins <jya@ieee.org> wrote: > >>You lost me. I hold this truth to be, if not self evident, at least well >>established: every transversal (tapped delay line) filter has as many >>poles as zeros, and they are all at the origin on the z plane. > > > Not true, IMHO, unless you add more conditions. Dividing by the transfer > function by z (or more canonically multiplying by z^-1) is equivalent > to adding a one sample delay. You can have less poles than zeros in a > non-causal FIR filter (less by the number of samples by which the output > precedes its first input). This is equivalent to shifting all the tap > coeffs to the left. You can have more poles than zeros simply by adding > a delay line to the input or output. This is equivalent to shifting all > the tap coeffs to the right. > > >>Assuming >>that's correct, how can scrambling the values of the tap coefficients >>in any way put poles at infinity? > > > How else could you get a transfer function (with a non-zero impulse > response) to go to zero over any finite domain, which would happen > if all the usual coeffs got swapped end for end with those at > +infinity? > > > (don't forget. this would not have been posted except for line > noise caused by a glass of red wine pinching the modem cable. :) > IMHO. YMMV.I'm still lost. Why does swapping the tap coefficients of a delay line end for end, or shuffling them in any other way, make the transfer function go to zero over over a finite domain? Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
