robert bristow-johnson wrote: ...> 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.Yes. I wrote earlier that the magnitude and phase responses are Hilbert transform pairs. Wrong is wrong, but let me add in mitigation that I had Bode plots in mind. Those are phase, and log magnitude. Ouch! 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 20, 20042004-02-20
Reply by ●February 20, 20042004-02-20
This discussion has lead me to wonder what kind of
characteristics a "maximum phase" filter, with all its zeros
outside the unit circle, would be. Probably obvious to the
inhabitants of the complex plane but I'm only slowly taking
up residence there.
Bob
--
"Things should be described as simply as possible, but no
simpler."
A. Einstein
Reply by ●February 20, 20042004-02-20
"Jerry Avins" <jya@ieee.org> wrote in message news:4032f004$0$3067$61fed72c@news.rcn.com...> Since the minimum-phase frequency and phase responses are a Hilbert > transform pair, it seems to me that greatly affecting one can't affect > the other only a little. Of course, the back of my envelope is rather > smudged tonight. > > JerryNow that you mention it, Jerry, I can see that the phase response of a causal filter is the Hilbert transform of the log-magnitude response. But I also know that there are multiple causal fitlers with the same magnitude response, but different phase responses -- the minimum phase filter and all the non-minimum phase ones. But the magnitude response can have only *one* Hilbert tranform. Where is the loophole that lets these multiple filters exist?
Reply by ●February 21, 20042004-02-21
Matt Timmermans wrote:> "Jerry Avins" <jya@ieee.org> wrote in message > news:4032f004$0$3067$61fed72c@news.rcn.com... > >>Since the minimum-phase frequency and phase responses are a Hilbert >>transform pair, it seems to me that greatly affecting one can't affect >>the other only a little. Of course, the back of my envelope is rather >>smudged tonight. >> >>Jerry > > > Now that you mention it, Jerry, I can see that the phase response of a > causal filter is the Hilbert transform of the log-magnitude response. But I > also know that there are multiple causal fitlers with the same magnitude > response, but different phase responses -- the minimum phase filter and all > the non-minimum phase ones. But the magnitude response can have only *one* > Hilbert tranform. > > Where is the loophole that lets these multiple filters exist?I don't see a loophole. The Hilbert transform of the log magnitude response of any filter is the minimum-phase response, and vice versa. I don't know offhand what the Hilbert transform of a non mimimum-phase response represents. I didn't mean to leave the impression that I had any idea at all. Maybe Guillemin sheds some light on it, but that's an uphill drag for me. Procedure for determining the minimum-phase response for any filter you like: Take the log of the magnitude response you want, and the Hilbert transform of that. Now you have the phase response. Simple: just go build it! It really is simple with lumped constants and continuous signals. Plot the log magnitude vs. log frequency. That makes the approximations to the curve between poles or zeros asymptotically straight and easy to draw. The slope of the curve turns up at a pole and down at a zero by 20 dB/decade. Identify these critical points, and you have h(f) in factored form. (Well, almost.) The phase due to a single break is .1 radian off the asymptote a decade away from the break on either side, 45 degrees at the break and asymptotically 90 degrees on one side or the other, whichever makes sense. So the amplitude response gives the phase response and vice versa long before a student of circuit design ever needs to know about Herr Hilbert. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●February 21, 20042004-02-21
"Jerry Avins" <jya@ieee.org> wrote in message news:4036e811$0$3072$61fed72c@news.rcn.com...> I don't see a loophole. The Hilbert transform of the log magnitude > response of any filter is the minimum-phase response, and vice versa. > I don't know offhand what the Hilbert transform of a non mimimum-phase > response represents. I didn't mean to leave the impression that I had > any idea at all. Maybe Guillemin sheds some light on it, but that's an > uphill drag for me.The reason I want a loophole is that it seemas to that the Hilbert transform of the log magnitude should give you the *only* causal response, like so: 1 - the impulse response is zero for t<=0, 2 - so the complex frequency response is analytic; 3 - so the complex log of the frequency response is analytic; 4 - and that is just log_magnitude + i*phase, so the phase is the Hilbert transform of the log-magnitude. Since there are non-minimum phase filters, also causal, I must be missing something.> It really is simple with lumped constants and continuous signals. Plot > the log magnitude vs. log frequency. That makes the approximations to > the curve between poles or zeros asymptotically straight and easy to > draw. The slope of the curve turns up at a pole and down at a zero by 20 > dB/decade. Identify these critical points, and you have h(f) in factored > form. (Well, almost.) The phase due to a single break is .1 radian off > the asymptote a decade away from the break on either side, 45 degrees at > the break and asymptotically 90 degrees on one side or the other, > whichever makes sense. So the amplitude response gives the phase > response and vice versa long before a student of circuit design ever > needs to know about Herr Hilbert.That is pretty cool.
Reply by ●February 21, 20042004-02-21
"Matt Timmermans" <mt0000@sympatico.nospam-remove.ca> wrote in message news:osKZb.17912$Cd6.941881@news20.bellglobal.com...> > "Jerry Avins" <jya@ieee.org> wrote in message > news:4036e811$0$3072$61fed72c@news.rcn.com... > > I don't see a loophole. The Hilbert transform of the log magnitude > > response of any filter is the minimum-phase response, and vice versa. > > I don't know offhand what the Hilbert transform of a non mimimum-phase > > response represents. I didn't mean to leave the impression that I had > > any idea at all. Maybe Guillemin sheds some light on it, but that's an > > uphill drag for me. > > The reason I want a loophole is that it seemas to that the Hilberttransform> of the log magnitude should give you the *only* causal response,Matt, There are many filters with the same magnitude response that are causal. So starting with the log magnitude doesn't map to "the only" causal response because there are many. A mapping like that might get a particular member of that set - like minimum phase, etc. Example: Design a FIR filter using Parks-McClellan. It will by symmetric and not minimum phase. Now, move all the zeros outside the unit circle into the unit circle: same angle, reciprocal magnitude. The magnitude response of this filter will be the same as originally and it will now be minimum phase. Fred
Reply by ●February 21, 20042004-02-21
"Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message news:xfydnTfZa8-3H6rdRVn-hQ@centurytel.net...> Matt, > > There are many filters with the same magnitude response that are causal.So> starting with the log magnitude doesn't map to "the only" causal response > because there are many. A mapping like that might get a particular member > of that set - like minimum phase, etc.Yes -- that is my problem, because the mapping seems inevitable to me. If the filter is causal, how can its frequency response not be analytic? Or if the frequency response is analytic, how can the complex log response not be analytic? Or if the complex log response is analyitic, how can the phase response not be the Hilbert transform of the log magnitude response? Or if it is, how can there be more than one phase response for a given log magnitude response? I know there is an answer to one of these questions that allows all the non-minimal phase filters, but I don't know what it is.
Reply by ●February 21, 20042004-02-21
Matt Timmermans wrote:> "Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message > news:xfydnTfZa8-3H6rdRVn-hQ@centurytel.net... > >>Matt, >> >>There are many filters with the same magnitude response that are causal. > > So > >>starting with the log magnitude doesn't map to "the only" causal response >>because there are many. A mapping like that might get a particular member >>of that set - like minimum phase, etc. > > > Yes -- that is my problem, because the mapping seems inevitable to me. If > the filter is causal, how can its frequency response not be analytic? Or if > the frequency response is analytic, how can the complex log response not be > analytic? Or if the complex log response is analyitic, how can the phase > response not be the Hilbert transform of the log magnitude response? Or if > it is, how can there be more than one phase response for a given log > magnitude response? > > I know there is an answer to one of these questions that allows all the > non-minimal phase filters, but I don't know what it is.It's the log that forces minimum phase. O&S (1st ed.), Chap 7. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●February 21, 20042004-02-21
"Jerry Avins" <jya@ieee.org> wrote in message news:4037c767$0$3102$61fed72c@news.rcn.com...> > It's the log that forces minimum phase. O&S (1st ed.), Chap 7. >Thanks, Jerry! I don't have an O&S around, but that was enough for me to figure it out. If you have g(x) => e^f(x), then f(x) analytic => g(x) analytic, but not vice versa. For causal, but non-minimum phase filters, the phase response increases without bound (discrete case), or has a step component (continuous case), that prevents you from even evaluating the Hilbert transform to compare it to log-magnitude.
Reply by ●February 22, 20042004-02-22
"Bob Cain" <arcane@arcanemethods.com> wrote in message news:c105ag0t4p@enews1.newsguy.com...> > Now that's an interesting question that I would like the answer to aswell.> > In particular, knowing that a filter is minimum phase for a particlar > > magnitude response doesn't convince me that there isn't a *nearlyidentical*> > magnitude response with a minimum phase filter that is *much* more > > concentrated towards t=0. I would be very happy to have a conclusive > > demonstration of that. > > If you get further insight on that, please share it. >Done: What, exactly, a minimum phase filter minimizes is the average group delay (discrete case) or integral of the group delay (continuous) over the entire spectrum -- they're zero for minimum phase filters. Any minimum phase filter without zeros right on the frequency axis can be reconstructed from its log-magnitude spectrum, by using the Hilbert transform to derive its phase response. Exponentiate log-magnitude + j*phase to get the frequency response, and transform that back into the time domain to get the impulse response. The impulse response will be causal, because log-magnitude + j*phase is an analytic signal, and exponentiation preserves this property. From the above, we can deduce that, for minimum phase filters, the group delay is the Hilbert transform of the derivative of the -log-magnitude response, so I'm now quite confident that tiny changes in the log-magnitude response won't move large portions of the energy significantly forward or backward in time in the minimum phase response.
