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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_t-dnZ2d(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 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.
>
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 in-phase channel to
the frequency response of the quadrature channel.

A common solution is to produce an in-phase 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_t-dnZ2d(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 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.

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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