Paavo Jumppanen wrote:> rhn@mauve.rahul.net (Ronald H. Nicholson Jr.) wrote in message news:<c0jn3f$6dn$1@blue.rahul.net>......>>The location of both the poles and the zeros is important. Imagine a >>crank handle at the location of every pole or zero on the complex >>plane. ... >...> > This explanation is fine for IIR's but doesn't work for FIR's because > FIR's don't have any poles, only zeros. Do you have a similarly clear > explanation for FIR's? > > Regards, > > > Pavo Jumppanen > Author of Har-Bal Harmonic Balancer > http://www.har-bal.comHow does the absence of poles invalidate the explanation? Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
FIR roots and frequency response
Started by ●February 13, 2004
Reply by ●February 19, 20042004-02-19
Reply by ●February 19, 20042004-02-19
In article 4034d725$0$3105$61fed72c@news.rcn.com, Jerry Avins at jya@ieee.org wrote on 02/19/2004 10:32:> Paavo Jumppanen wrote: > >> rhn@mauve.rahul.net (Ronald H. Nicholson Jr.) wrote in message >> news:<c0jn3f$6dn$1@blue.rahul.net>... > > ... > >>> The location of both the poles and the zeros is important. Imagine a >>> crank handle at the location of every pole or zero on the complex >>> plane. ... >> > ... >> >> This explanation is fine for IIR's but doesn't work for FIR's because >> FIR's don't have any poles, only zeros. Do you have a similarly clear >> explanation for FIR's? >> > > How does the absence of poles invalidate the explanation?in addition, causal FIRs *do* have poles. as many as zeros. but they're all located at the origin. so they don't contribute much to the frequency response except for linear-phase delay. r b-j
Reply by ●February 19, 20042004-02-19
Jerry Avins wrote:> Paavo Jumppanen wrote: >>> >> This explanation is fine for IIR's but doesn't work for FIR's because >> FIR's don't have any poles, only zeros. Do you have a similarly clear >> explanation for FIR's? >> > > > How does the absence of poles invalidate the explanation? >Yeah, me too. Bob -- "Things should be described as simply as possible, but no simpler." A. Einstein
Reply by ●February 19, 20042004-02-19
Jerry Avins <jya@ieee.org> wrote in message news:<4034d725$0$3105$61fed72c@news.rcn.com>...> Paavo Jumppanen wrote: > > > rhn@mauve.rahul.net (Ronald H. Nicholson Jr.) wrote in message news:<c0jn3f$6dn$1@blue.rahul.net>... > > ... > > >>The location of both the poles and the zeros is important. Imagine a > >>crank handle at the location of every pole or zero on the complex > >>plane. ... > > > ... > > > > This explanation is fine for IIR's but doesn't work for FIR's because > > FIR's don't have any poles, only zeros. Do you have a similarly clear > > explanation for FIR's? > > > > Regards, > > > > > > Pavo Jumppanen > > Author of Har-Bal Harmonic Balancer > > http://www.har-bal.com > > How does the absence of poles invalidate the explanation? > > JerryFrom my understanding of what he said, minimum phase is minimum phase becauses the full cycles of phase shift introduced by each pole is cancelled by a complimentary phase shift introduced by the zeros that are inside the unit circle. If any lie outside they wont give a complimentary phase shift to cancel the poles, hence mixed or and or maximum phase. So if you now apply this argument to and FIR which has no poles what is the phase shift introduced by a zero going to cancel with? That is the problem I have with understanding this explanation as applied to an FIR. Please enlighten me. Regards, Paavo Jumppanen Author of HarBal Harmonic Balancer http://www.har-bal.com
Reply by ●February 19, 20042004-02-19
"Matt Timmermans" <mt0000@sympatico.nospam-remove.ca> writes:> "Peter J. Kootsookos" <p.kootsookos@remove.ieee.org> wrote in message > news:s68znbh3qdi.fsf@mango.itee.uq.edu.au... > > Another observation is that the overall energy of the *much* more > > bunched filter cannot be different from the minimum phase original (by > > Parseval's theorem). > > > > I suspect you could probably extend that to an argument showing that > > the "energy delay" (see Oppenheim and Schafer, 1989, prob 5.36) of > > each cannot be significantly different. > > Any argument along these lines, or Jerry's, to which you replied, would need > some subtle refinement to avoid arguing against the existence of the > non-minimum phase filters we can make just by flipping zeros outside the > unit circle, as well as good linear phase FIR approximations to minimum > phase responses.If you read the problem I referred to, this is exactly what it does.> And if we truncate a minimum phase IIR to make a FIR, trading the > poles for a bunch of zeros, do we really expect the result to be > minimum phase?Clearly not in the general case, but for the conditions you've put on the problem (minimal magnitude response differences), it's possible.> If not, then couldn't flipping all the outer zeros into the unit > circle significantly change the impulse response, making it more > concentrated towards 0, even though the frequency response hasn't > changed much?The frequency response will have changed significantly in energy because sum_{k=0}^{N-1} |h[k]|^2 < sum_{k=0}^{\infty} |h[k]|^2 or 0 < sum_{k=N}^{\infty} |h[k]|^2 which, by Parseval, would go against your condition that the two are not significantly different. Ciao, Peter K. -- Peter J. Kootsookos "I will ignore all ideas for new works [..], the invention of which has reached its limits and for whose improvement I see no further hope." - Julius Frontinus, c. AD 84
Reply by ●February 19, 20042004-02-19
Paavo Jumppanen wrote:> Jerry Avins <jya@ieee.org> wrote in message news:<4034d725$0$3105$61fed72c@news.rcn.com>... > >>Paavo Jumppanen wrote: >> >> >>>rhn@mauve.rahul.net (Ronald H. Nicholson Jr.) wrote in message news:<c0jn3f$6dn$1@blue.rahul.net>... >> >> ... >> >> >>>>The location of both the poles and the zeros is important. Imagine a >>>>crank handle at the location of every pole or zero on the complex >>>>plane. ... >>> >> ... >> >>>This explanation is fine for IIR's but doesn't work for FIR's because >>>FIR's don't have any poles, only zeros. Do you have a similarly clear >>>explanation for FIR's? >>> >>>Regards, >>> >>> >>>Pavo Jumppanen >>>Author of Har-Bal Harmonic Balancer >>>http://www.har-bal.com >> >>How does the absence of poles invalidate the explanation? >> >>Jerry > > > From my understanding of what he said, minimum phase is minimum phase > becauses the full cycles of phase shift introduced by each pole is > cancelled by a complimentary phase shift introduced by the zeros that > are inside the unit circle. If any lie outside they wont give a > complimentary phase shift to cancel the poles, hence mixed or and or > maximum phase. > > So if you now apply this argument to and FIR which has no poles what > is the phase shift introduced by a zero going to cancel with? That is > the problem I have with understanding this explanation as applied to > an FIR. > > Please enlighten me. > > Regards, > > > Paavo Jumppanen > Author of HarBal Harmonic Balancer > http://www.har-bal.comFirst off, it's not that you don't have poles, but they're all on top of one another, stuck where they don't get noticed. Second, suppose that the poles didn't exist. Since they make the delay for the inboard zeros to offset, without them, there's be less delay. So much less that without them, the delay would be negative, which proves that they exist even though they hide well enough to stay out of sight. Please excuse mind my flippant explanation. With a glass of good brandy in hand, it's about all I can manage. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●February 19, 20042004-02-19
robert bristow-johnson <rbj@surfglobal.net> wrote in message news:<BC5A5DF0.8B96%rbj@surfglobal.net>...> In article 4034d725$0$3105$61fed72c@news.rcn.com, Jerry Avins at > jya@ieee.org wrote on 02/19/2004 10:32: > > > Paavo Jumppanen wrote: > > > >> rhn@mauve.rahul.net (Ronald H. Nicholson Jr.) wrote in message > >> news:<c0jn3f$6dn$1@blue.rahul.net>... > > > > ... > > > >>> The location of both the poles and the zeros is important. Imagine a > >>> crank handle at the location of every pole or zero on the complex > >>> plane. ... > >> > ... > >> > >> This explanation is fine for IIR's but doesn't work for FIR's because > >> FIR's don't have any poles, only zeros. Do you have a similarly clear > >> explanation for FIR's? > >> > > > > How does the absence of poles invalidate the explanation? > > in addition, causal FIRs *do* have poles. as many as zeros. but they're > all located at the origin. so they don't contribute much to the frequency > response except for linear-phase delay. > > r b-jCan you explain this further, please. I'm struggling to see how to factor a polynomial is z that has no denominator (other than 1) into poles and zeros. What am I missing or am I mixing up my analog and digital domains? Regards, Paavo Jumppanen. Author of HarBal Harmonic Balancer http://www.har-bal.com
Reply by ●February 19, 20042004-02-19
"Paavo Jumppanen" <PaavoJumppanen@iname.com> wrote in message news:1a72ed76.0402191612.17172427@posting.google.com...> robert bristow-johnson <rbj@surfglobal.net> wrote in messagenews:<BC5A5DF0.8B96%rbj@surfglobal.net>...> > In article 4034d725$0$3105$61fed72c@news.rcn.com, Jerry Avins at > > jya@ieee.org wrote on 02/19/2004 10:32: > > > > > Paavo Jumppanen wrote: > > > > > >> rhn@mauve.rahul.net (Ronald H. Nicholson Jr.) wrote in message > > >> news:<c0jn3f$6dn$1@blue.rahul.net>... > > > > > > ... > > > > > >>> The location of both the poles and the zeros is important. Imaginea> > >>> crank handle at the location of every pole or zero on the complex > > >>> plane. ... > > >> > > ... > > >> > > >> This explanation is fine for IIR's but doesn't work for FIR's because > > >> FIR's don't have any poles, only zeros. Do you have a similarly clear > > >> explanation for FIR's? > > >> > > > > > > How does the absence of poles invalidate the explanation? > > > > in addition, causal FIRs *do* have poles. as many as zeros. butthey're> > all located at the origin. so they don't contribute much to thefrequency> > response except for linear-phase delay. > > > > r b-j > > Can you explain this further, please. I'm struggling to see how to > factor a polynomial is z that has no denominator (other than 1) into > poles and zeros. What am I missing or am I mixing up my analog and > digital domains? >H(z)= a0 + a1*z^-1 +a2*z^-2 + ..... +an*z^-N is the normal way a FIR filter polynomial is written where z^-1 represents a unit delay. Multiplying by z^N/Z^n yields: H(z) = [a0*z^N + a1*z^N-1 + ...... + an]/z^N is the same function in terms of a rational function in z. Thus it has N poles at zero in the z plane and the zeros in the z plane are the zeros in the z^-1 plane inverted I do believe. Fred
Reply by ●February 19, 20042004-02-19
"Peter J. Kootsookos" <p.kootsookos@remove.ieee.org> wrote in message news:s68n07e4ok0.fsf@mango.itee.uq.edu.au...> If you read the problem I referred to, this is exactly what it does.I'll do that.> > And if we truncate a minimum phase IIR to make a FIR, trading the > > poles for a bunch of zeros, do we really expect the result to be > > minimum phase? > > Clearly not in the general case, but for the conditions you've put on > the problem (minimal magnitude response differences), it's possible. > > > If not, then couldn't flipping all the outer zeros into the unit > > circle significantly change the impulse response, making it more > > concentrated towards 0, even though the frequency response hasn't > > changed much? > > The frequency response will have changed significantly in energy > because > > sum_{k=0}^{N-1} |h[k]|^2 < sum_{k=0}^{\infty} |h[k]|^2 > > or 0 < sum_{k=N}^{\infty} |h[k]|^2 > > which, by Parseval, would go against your condition that the two are > not significantly different.The idea is to lop off only a small portion of the energy, which you can do for any stable IIR.
Reply by ●February 20, 20042004-02-20
In article 1a72ed76.0402191612.17172427@posting.google.com, Paavo Jumppanen at PaavoJumppanen@iname.com wrote on 02/19/2004 19:12:> robert bristow-johnson <rbj@surfglobal.net> wrote in message > news:<BC5A5DF0.8B96%rbj@surfglobal.net>......>> in addition, causal FIRs *do* have poles. as many as zeros. but they're >> all located at the origin. so they don't contribute much to the frequency >> response except for linear-phase delay....> > Can you explain this further, please. I'm struggling to see how to > factor a polynomial is z that has no denominator (other than 1) into > poles and zeros.okay, Fred, and Jerry with a little bit of hand waving (spilled brandy), pointed out why there are poles.> What am I missing or am I mixing up my analog and > digital domains?i dunno if you are or not. but the reason that zeros inside the unit circle make for a minimum phase filter is the same for zeros in the left half-plane for analog. In article 1a72ed76.0402191407.3ea5e23@posting.google.com, Paavo Jumppanen at PaavoJumppanen@iname.com wrote on 02/19/2004 17:07:> From my understanding of what he said, minimum phase is minimum phase > because the full cycles of phase shift introduced by each pole is > cancelled by a complimentary phase shift introduced by the zeros that > are inside the unit circle. If any lie outside they wont give a > complimentary phase shift to cancel the poles, hence mixed or and or > maximum phase.unless the zero lies right on top of a pole, it won't completely cancel the phase shift due to the pole. the reason is simply, for geometric reason (draw it out), zeros inside the unit circle have less of an angle with exp(j*w) than do zeros outside the unit circle. first, start with a real zero: 0<q<1. it is clearly true that arg{e^jw - q} < arg{e^jw - 1/q} for 0 < w < pi (if it was analog and we were in the s-plane i could say categorically that the angle with the minimum phase zero was always less than 90 degrees and that the angle with the maximum phase zero was always greater than 90 degrees. that isn't entirely true in the z-plane regarding the min-phase zero, but it *is* true for the max-phase zero and the inequality above is still always true and should be obvious if you draw it.) now for a complex zero, q, the reflected zero is 1/conj{q} (or conj{1/q}) rather than just 1/q and both zeros lie on the same radial line (have the same angle). so, geometrically, it's the same thing as the real zero (and reflection) except that the picture is rotated a little. as i tried to hand wave in my explanation with Bob Cain, it's easiest to see this in the s-plane with a continuous-time system. then the harder part is to make a BiLinear Transform argument to say the same is true in the z-plane. now what's *really* a bitch is to show that for a minimum phase system (as defined as one with all the zeros inside the unit circle or in the left half s-plane), the natural log of the magnitude response and the phase response (measured in radians) are a Hilbert Transform pair. it's doable, but a bitch. <arf> r b-j
