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From: Clay on 23 Apr 2010 11:17 On Apr 22, 6:27 pm, Dirk Bell <bellda2...(a)cox.net> wrote: > On Apr 22, 5:42 pm, Jerry Avins <j...(a)ieee.org> wrote: > > > > > > > On 4/22/2010 2:26 PM, Dirk Bell wrote: > > > > On Apr 16, 1:26 am, Jerry Avins<j...(a)ieee.org> wrote: > > >> On 4/16/2010 12:43 AM, cfy30 wrote: > > > >>> Hi all, > > > >>> I am a newbie to Hilbert transform. I found the follow definition from > > >>> textbook > > > >>> H(f) = j or 90degree, f>0 > > >>> H(f) = j or +90degree, f<0 > > >>> H(f) = 0, f=0 > > > >>> But when I plot the phase out in Matlab, I saw that the phase is not > > >>> constant at 90 or +90degree across frequency. The code I have is as > > >>> follow > > > >>> b = firpm(10,[.1 .9],[1 1],'Hilbert') > > >>> w = linspace(pi, pi, 2^12); > > >>> h = freqz(b, 1, w); > > >>> plot(w, angle(h)*180/pi); > > > >>> What do I miss? > > > >> The phase is 90 degrees only over a limited band. All bets are off at > > >> Fs/2, and you can't include enough taps to get 90 degrees of delay at > > >> DC. What's more, there will be some amplitude ripple in the passband.. > > > Jerry, > > > > The phase is not really 90 degrees only over a "limited" band. Try a > > > 100K point FFT on the OP's filter (remove delay) and the only 2 values > > > that are not +pi/2 with error magnitude<4*10^13 radians are at 0 > > > and pi radians where they are zero. The phase is essentially correct > > > over the whole band. The amplitude is another story. > > > Dirk, > > > Of course you're right. Antisymmetry assures that the phase be perfect > > even with a balanced differentiator, [1, 0, +1]. I had in mind the > > implementation with parallel filters whose difference approximates 90 > > degrees over the bans of interest. Bad form! > > > Jerry > >  > > "It does me no injury for my neighbor to say there are 20 gods, or no > > God. It neither picks my pocket nor breaks my leg." > > Thomas Jefferson to the Virginia House of Delegates in 1776. Hide quoted text  > > >  Show quoted text  > > Hi Jerry, > > I took my post down because I looked at later posts and thought it was > basically covered. However antisymmetry by itself is not enough to > guarantee perfect phase. For example [1 0 1 0 1 > 0 1] does not work. > > I was curious about another comment somone here made. What is the > best way to design a 45 degree phase shifter? > > Dirk Hide quoted text  > >  Show quoted text  I don't know if it is the best way, but it is a simple way. See: http://www.claysturner.com/dsp/ASG.pdf Clay
From: Dirk Bell on 23 Apr 2010 11:41 On Apr 23, 11:17 am, Clay <c...(a)claysturner.com> wrote: > On Apr 22, 6:27 pm, Dirk Bell <bellda2...(a)cox.net> wrote: > > > > > > > On Apr 22, 5:42 pm, Jerry Avins <j...(a)ieee.org> wrote: > > > > On 4/22/2010 2:26 PM, Dirk Bell wrote: > > > > > On Apr 16, 1:26 am, Jerry Avins<j...(a)ieee.org> wrote: > > > >> On 4/16/2010 12:43 AM, cfy30 wrote: > > > > >>> Hi all, > > > > >>> I am a newbie to Hilbert transform. I found the follow definition from > > > >>> textbook > > > > >>> H(f) = j or 90degree, f>0 > > > >>> H(f) = j or +90degree, f<0 > > > >>> H(f) = 0, f=0 > > > > >>> But when I plot the phase out in Matlab, I saw that the phase is not > > > >>> constant at 90 or +90degree across frequency. The code I have is as > > > >>> follow > > > > >>> b = firpm(10,[.1 .9],[1 1],'Hilbert') > > > >>> w = linspace(pi, pi, 2^12); > > > >>> h = freqz(b, 1, w); > > > >>> plot(w, angle(h)*180/pi); > > > > >>> What do I miss? > > > > >> The phase is 90 degrees only over a limited band. All bets are off at > > > >> Fs/2, and you can't include enough taps to get 90 degrees of delay at > > > >> DC. What's more, there will be some amplitude ripple in the passband. > > > > Jerry, > > > > > The phase is not really 90 degrees only over a "limited" band. Try a > > > > 100K point FFT on the OP's filter (remove delay) and the only 2 values > > > > that are not +pi/2 with error magnitude<4*10^13 radians are at 0 > > > > and pi radians where they are zero. The phase is essentially correct > > > > over the whole band. The amplitude is another story. > > > > Dirk, > > > > Of course you're right. Antisymmetry assures that the phase be perfect > > > even with a balanced differentiator, [1, 0, +1]. I had in mind the > > > implementation with parallel filters whose difference approximates 90 > > > degrees over the bans of interest. Bad form! > > > > Jerry > > >  > > > "It does me no injury for my neighbor to say there are 20 gods, or no > > > God. It neither picks my pocket nor breaks my leg." > > > Thomas Jefferson to the Virginia House of Delegates in 1776. Hide quoted text  > > > >  Show quoted text  > > > Hi Jerry, > > > I took my post down because I looked at later posts and thought it was > > basically covered. However antisymmetry by itself is not enough to > > guarantee perfect phase. For example [1 0 1 0 1 > > 0 1] does not work. > > > I was curious about another comment somone here made. What is the > > best way to design a 45 degree phase shifter? > > > Dirk Hide quoted text  > > >  Show quoted text  > > I don't know if it is the best way, but it is a simple way. See: > > http://www.claysturner.com/dsp/ASG.pdf > > Clay Hide quoted text  > >  Show quoted text  Thanks Clay. Dirk
From: John Monro on 22 Apr 2010 03:49 Robert Orban wrote: > In article <Xs6dnbGFup7rZ1rWnZ2dnUVZ_tdnZ2d(a)giganews.com>, > steveu(a)n_o_s_p_a_m.coppice.org says... > >> Just a few terms for an FIR implementation of a Hilbert transform can give >> you pretty close to 90 degrees over a large part of the band. Don't expect >> a perfect brick wall transition from + to  90 at DC, though. Its the >> amplitude response that is the greater problem. It takes a lot of terms to >> get that close to flat at low and high frequencies. > > IIRC, as long as the impulse response of the FIR is stictly antimetric > around the center tap, you will have an exact 90 degree phase shift (+ a > fixed delay) at all frequencies regardless of the number of taps. (A > trivial example is a threetap filter whose impulse response is 1, 0, +1.) > The problem, as other posters have commented, is that for a given amplitude > passband bandwidth, loworder FIR filters have larger amounts of amplitude > ripple in the filter passband than higherorder filters and no filter with > a finite number of taps can have an ampltude bandwidth extending from DC to > fs/2. > > It is possible to transform the filter structure such that the amplitude > response is flat but the phase shift error varies over the passband. What > is not possible with a finite number of taps is to obtain a flat passband > from 0 to fs/2 Hz and a 90 degree phase shift (+ fixed delay) > simultaneously. > Usually the problem is not that you can't tolerate a little gain variaton across the passband but that you need to match closely the frequency response of the inphase channel to the frequency response of the quadrature channel. A common solution is to produce an inphase signal that has the same ripple as the quadrature signal. This is done by applying the original signal to a HP FIR filter that is a symmetrical version of the Hilbert transformer. Regards, John
From: Clay on 23 Apr 2010 12:02 On Apr 23, 11:41 am, Dirk Bell <bellda2...(a)cox.net> wrote: > On Apr 23, 11:17 am, Clay <c...(a)claysturner.com> wrote: > > > > > > > On Apr 22, 6:27 pm, Dirk Bell <bellda2...(a)cox.net> wrote: > > > > On Apr 22, 5:42 pm, Jerry Avins <j...(a)ieee.org> wrote: > > > > > On 4/22/2010 2:26 PM, Dirk Bell wrote: > > > > > > On Apr 16, 1:26 am, Jerry Avins<j...(a)ieee.org> wrote: > > > > >> On 4/16/2010 12:43 AM, cfy30 wrote: > > > > > >>> Hi all, > > > > > >>> I am a newbie to Hilbert transform. I found the follow definition from > > > > >>> textbook > > > > > >>> H(f) = j or 90degree, f>0 > > > > >>> H(f) = j or +90degree, f<0 > > > > >>> H(f) = 0, f=0 > > > > > >>> But when I plot the phase out in Matlab, I saw that the phase is not > > > > >>> constant at 90 or +90degree across frequency. The code I have is as > > > > >>> follow > > > > > >>> b = firpm(10,[.1 .9],[1 1],'Hilbert') > > > > >>> w = linspace(pi, pi, 2^12); > > > > >>> h = freqz(b, 1, w); > > > > >>> plot(w, angle(h)*180/pi); > > > > > >>> What do I miss? > > > > > >> The phase is 90 degrees only over a limited band. All bets are off at > > > > >> Fs/2, and you can't include enough taps to get 90 degrees of delay at > > > > >> DC. What's more, there will be some amplitude ripple in the passband. > > > > > Jerry, > > > > > > The phase is not really 90 degrees only over a "limited" band. Try a > > > > > 100K point FFT on the OP's filter (remove delay) and the only 2 values > > > > > that are not +pi/2 with error magnitude<4*10^13 radians are at 0 > > > > > and pi radians where they are zero. The phase is essentially correct > > > > > over the whole band. The amplitude is another story. > > > > > Dirk, > > > > > Of course you're right. Antisymmetry assures that the phase be perfect > > > > even with a balanced differentiator, [1, 0, +1]. I had in mind the > > > > implementation with parallel filters whose difference approximates 90 > > > > degrees over the bans of interest. Bad form! > > > > > Jerry > > > >  > > > > "It does me no injury for my neighbor to say there are 20 gods, or no > > > > God. It neither picks my pocket nor breaks my leg." > > > > Thomas Jefferson to the Virginia House of Delegates in 1776. Hide quoted text  > > > > >  Show quoted text  > > > > Hi Jerry, > > > > I took my post down because I looked at later posts and thought it was > > > basically covered. However antisymmetry by itself is not enough to > > > guarantee perfect phase. For example [1 0 1 0 1 > > > 0 1] does not work. > > > > I was curious about another comment somone here made. What is the > > > best way to design a 45 degree phase shifter? > > > > Dirk Hide quoted text  > > > >  Show quoted text  > > > I don't know if it is the best way, but it is a simple way. See: > > >http://www.claysturner.com/dsp/ASG.pdf > > > Clay Hide quoted text  > > >  Show quoted text  > > Thanks Clay. > > Dirk Hide quoted text  > >  Show quoted text  You are welcome.
From: steveu on 22 Apr 2010 04:20 >In article <Xs6dnbGFup7rZ1rWnZ2dnUVZ_tdnZ2d(a)giganews.com>, >steveu(a)n_o_s_p_a_m.coppice.org says... > >> >>Just a few terms for an FIR implementation of a Hilbert transform can give >>you pretty close to 90 degrees over a large part of the band. Don't expect >>a perfect brick wall transition from + to  90 at DC, though. Its the >>amplitude response that is the greater problem. It takes a lot of terms to >>get that close to flat at low and high frequencies. > >IIRC, as long as the impulse response of the FIR is stictly antimetric >around the center tap, you will have an exact 90 degree phase shift (+ a >fixed delay) at all frequencies regardless of the number of taps. (A >trivial example is a threetap filter whose impulse response is 1, 0, +1.) >The problem, as other posters have commented, is that for a given amplitude >passband bandwidth, loworder FIR filters have larger amounts of amplitude >ripple in the filter passband than higherorder filters and no filter with >a finite number of taps can have an ampltude bandwidth extending from DC to >fs/2. > >It is possible to transform the filter structure such that the amplitude >response is flat but the phase shift error varies over the passband. What >is not possible with a finite number of taps is to obtain a flat passband >from 0 to fs/2 Hz and a 90 degree phase shift (+ fixed delay) >simultaneously. For some bizarre amplitude versus frequency response curves you can indeed reduce the transform to something trivial, like the simple subtraction of two terms you describe, and have an accurate 90 degree shift. I supposed there are applications, where only the phase is important, where this would be fine. More interestingly for most purposes, a few terms can give you a flattish frequency response over the central part of the band, with exactly 90 degrees shift over a very large part of the band, but I don't think the converse is true  i.e. I don't think you can use a small number of terms to achieve a roughly 90 degree shift over the central area of the band with a dead flat response over a substantial part of the band. Steve
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