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From: cfy30 on 16 Apr 2010 01:55
What is that suppose to mean? Eddie current, not Eddy current?


>On Apr 16, 4:43=A0pm, "cfy30" <cfy30(a)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
>
From: Andor on 16 Apr 2010 02:33
On 16 Apr., 06:43, "cfy30" <cfy30(a)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
From: cfy30 on 16 Apr 2010 13:51
Thanks Andor! You are right!


Cfy30

>On 16 Apr., 06:43, "cfy30" <cfy30(a)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
>
From: Jerry Avins on 16 Apr 2010 23:01
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.
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From: Clay on 22 Apr 2010 10:34
On Apr 22, 3:49 am, John Monro <johnmo...(a)optusnet.com.au> wrote:
> Robert Orban wrote:
> > In article <Xs6dnbGFup7rZ1rWnZ2dnUVZ_t-dn...(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- Hide quoted text -
>
> - Show quoted text -

Or you can simply design a 45 degree phase shift filter. Then simply
pass your original signal through the filter and a copy of the
original signal through a time reversed version of the filter. The
difference between the two is 90 degrees and the amplitude responses
are identical.

Clay
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