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? Cfy30

# Hilbert Transform Phase

Started by ●April 16, 2010

Reply by ●April 16, 20102010-04-16

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 -- "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. ���������������������������������������������������������������������

Reply by ●April 16, 20102010-04-16

On Apr 16, 4:43�pm, "cfy30" <cfy30@n_o_s_p_a_m.yahoo.com> 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? > > Cfy30What wire are you using for your Hilbert Transformer? Beware of Eddie Current. Hardy

Reply by ●April 16, 20102010-04-16

>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?"Hilbert transform" is not an algorithm. Its a concept. Any algorithm you might use to realise a Hilbert transform is an approximation. It may work well over the central part of the band, but don't expect it work work well near the ends. 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. Steve

Reply by ●April 16, 20102010-04-16

..still don't understand. What I am observing is the phase change linearly with frequency! It is very clear when I plot(w, unwrap(angle(h))*180/pi); Cfy30>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 >-- >"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. >��������������������������������������������������������������������� >

Reply by ●April 16, 20102010-04-16

What is that suppose to mean? Eddie current, not Eddy current?>On Apr 16, 4:43=A0pm, "cfy30" <cfy30@n_o_s_p_a_m.yahoo.com> wrote: >> Hi all, >> >> I am a newbie to Hilbert transform. I found the follow definition from >> textbook >> >> H(f) =3D -j or -90degree, f>0 >> H(f) =3D =A0j or +90degree, f<0 >> H(f) =3D =A00, f=3D0 >> >> 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 =3D firpm(10,[.1 .9],[1 1],'Hilbert') >> w =3D linspace(-pi, pi, 2^12); >> h =3D freqz(b, 1, w); >> plot(w, angle(h)*180/pi); >> >> What do I miss? >> >> Cfy30 > >What wire are you using for your Hilbert Transformer? Beware of Eddie >Current. > > >Hardy >

Reply by ●April 16, 20102010-04-16

On 16 Apr., 06:43, "cfy30" <cfy30@n_o_s_p_a_m.yahoo.com> 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?You have to compensate for the delay of the FIR filter from the phase response. In your case, it is 5 samples, so try plot(w, angle(h.*exp(5i*w))/pi*180); Regards, Andor

Reply by ●April 16, 20102010-04-16

Thanks Andor! You are right! Cfy30>On 16 Apr., 06:43, "cfy30" <cfy30@n_o_s_p_a_m.yahoo.com> wrote: >> Hi all, >> >> I am a newbie to Hilbert transform. I found the follow definition from >> textbook >> >> H(f) =3D -j or -90degree, f>0 >> H(f) =3D =A0j or +90degree, f<0 >> H(f) =3D =A00, f=3D0 >> >> 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 =3D firpm(10,[.1 .9],[1 1],'Hilbert') >> w =3D linspace(-pi, pi, 2^12); >> h =3D freqz(b, 1, w); >> plot(w, angle(h)*180/pi); >> >> What do I miss? > >You have to compensate for the delay of the FIR filter from the phase >response. In your case, it is 5 samples, so try > >plot(w, angle(h.*exp(5i*w))/pi*180); > >Regards, >Andor >

Reply by ●April 17, 20102010-04-17

On 4/16/2010 1:51 AM, cfy30 wrote:> ..still don't understand. What I am observing is the phase change linearly > with frequency! It is very clear when I plot(w, unwrap(angle(h))*180/pi);What is the lower frequency of your observation? At the sample rate you use, how many samples represent 90 degrees at .1 Hz? 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. ���������������������������������������������������������������������

Reply by ●April 21, 20102010-04-21

In article <Xs6dnbGFup7rZ1rWnZ2dnUVZ_t-dnZ2d@giganews.com>, steveu@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 three-tap filter whose impulse response is -1, 0, +1.) The problem, as other posters have commented, is that for a given amplitude passband bandwidth, low-order FIR filters have larger amounts of amplitude ripple in the filter passband than higher-order 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.