On Thu, 3 Mar 2005 18:36:45 -0500, "Clay S. Turner"
<Physics@Bellsouth.net> wrote:
>
>"Richard Webb" <webby@REMOVEMErichardwebb-online.co.uk> wrote in message
>news:qoCdnYrml_iI5brfRVnytg@pipex.net...
>> I've had an interest in the Hilbert transform since I had to implement
>> one in a project a few years ago (with a bit of help from Rick!). I
>> haven't
>> looked at them since and your paper was a great refresher, it even taught
>> me
>> a few things I either forgor or never knew!! For some reason a lot of text
>> books don't make the Hilbert transform easy to understand, usually because
>> of their use of language getting in the way of conveying the information.
>> I
>> found your paper pitched at the right level!
>>
>> Rich.
>>
>>
>
>Thanks Rich,
> That's a great compliment. Writing is not my forte' - I'll leave that to
>Rick Lyons. But I do try to cut through all of the stuff and distill out
>what is important. I'm glad that this paper was at the right level for you.
>Hopefully others will think the same.
>
>Again Thanks,
>Clay
Ha ha Clay. You tickle me.
Your writing has four important characteristics.
It's gentle in approach, deliberate in delivery,
thorough in content, and ultimately readable.
I just glanced briefly at your material so far,
but I found a silly little typo.
The sentence after Equation 24 has the words:
"the direction that taken by".
In any case, Good goin' Clay!
[-Rick-]
Reply by Clay S. Turner●March 4, 20052005-03-04
"Martin Eisenberg" <martin.eisenberg@udo.edu> wrote in message
news:1109901638.416547@ostenberg.wh.uni-dortmund.de...
> Hi Clay,
>
>
Hello Martin,
Thanks for the thorough read on my paper. I'll work on fixing the manuscript
this weekend.
Clay
Reply by robert bristow-johnson●March 4, 20052005-03-04
in article 1109882654.091294.118040@g14g2000cwa.googlegroups.com, Clay at
physics@bellsouth.net wrote on 03/03/2005 15:44:
> robert bristow-johnson wrote:
>
>>
>> hey Clay, would you like a section on how, for minimum-phase filters, that
>> the phase response (in radians) is the negative of the Hilbert transform of
>> the natural log of the magnitude response? (and maybe, this one is easier,
>> how causal impulse responses correspond to a frequency response where the
>> imaginary part and real part are a hilbert pair?)
>
> These would be neat additions. I'll look into figuring these out or do
> you already have such proofs handy?
since your paper is pretty much exclusively continuous-time, there are two
ways i know to prove it.
one involves residue theory of complex functions. the complex log of the
frequency response is the log magnitude in the real part and the phase in
radians in the imaginary part and we want to show that, for a minimum-phase
filter, they are a Hilbert pair. looking at the whole Laplace H(s), in
which evaluating it on the jw axis is the frequency response H(jw). (BTW,
Clay, i would change *all* H(w) and X(w) and such to H(jw) and X(jw) to make
the notation compatible with Laplace Transform.) so if you look at the
log(H(s)) function, you want to relate the real and imag parts of the
complex function going up the jw axis. if all of the singularities are in
the left half plane, there is a theorem that says the real and imag parts of
an analytic function evaluated along the jw axis are Hilbert pairs. if
there are any zeros of H(s) in the right half plane, then the log(H(s)) is
not analytic in the entire right half plane (because the log(0) is a
singularity).
the crude but doable way to show it without residue theory is to show that
for q < 0,
Hilbert{ log|jw - q| } = -arg{ jw - q }
or
Hilbert{ 0.5*log(w^2 + q^2) } = -arctan(w/(-q))
which i've never been able to do directly.
but you *can* show that their derivatives are Hilbert pairs:
Hilbert{ (d/dw)0.5*log(w^2 + q^2) } = -(d/dw)arctan(w/(-q))
or
Hilbert{ w/(w^2 + q^2) } = q/(w^2 + q^2)
that's doable. (you have to do partial fraction expansion.) and you can
argue that because the Hilbert Transformer is an LTI system and that
convolution is commutative, that because
Hilbert{ w/(w^2 + q^2) } = q/(w^2 + q^2)
is true, then
Hilbert{ 0.5*log(w^2 + q^2) } = -arctan(w/(-q))
or
Hilbert{ log|jw - q| } = -arg{ jw - q }
is true, and therefore
Hilbert{ log|jw - q1| + log|jw - q2| - log|jw - p1| - log|jw - p2| }
= -( arg{jw - q1} + arg{jw - q2} - arg{jw - p1} - arg{jw - p2} )
is true for as many zeros, q1, q2 ..., and poles, p1, p2, ... you can get.
it's a bitch, i guess.
--
r b-j rbj@audioimagination.com
"Imagination is more important than knowledge."
Reply by Martin Eisenberg●March 3, 20052005-03-03
Hi Clay,
thanks for that! Here are some cosmetic things that occurred to me
during a quick read. But the writeup is very nice as is -- I'll
definitely keep it.
- In the paragraph below the middle of page 2, it should say:
"Likewise for w0 = w < 0..."; i.e., the sign on w0 is extraneous (or
the relator goes in the wrong direction).
- Maybe I'm being thick, but the last identity in (9) is unclear to
me. I found a manipulation that yields what you state, but I can't
formally justify it. Some hint would be nice there.
- I find it unfortunate that the explanatory comment on (17) ended up
on the next page. I know it's silly but PDF page gaps are bumps in my
reading flow, and I can imagine others are similarly conditioned by
meatspace books. Maybe you can put it in the preceding line.
- For (39) to hold, psi(t) doesn't have to be analytic, does it?
- Including two variables in Euler's identity at the bottom of page 9
doesn't add anything except characters, I think.
- The first line on page 11 lacks a closing parenthesis after
"singularity".
- There's an opening bracket missing in (54).
--
Quidquid latine dictum sit, altum viditur.
Reply by Clay S. Turner●March 3, 20052005-03-03
"Richard Webb" <webby@REMOVEMErichardwebb-online.co.uk> wrote in message
news:qoCdnYrml_iI5brfRVnytg@pipex.net...
> I've had an interest in the Hilbert transform since I had to implement
> one in a project a few years ago (with a bit of help from Rick!). I
> haven't
> looked at them since and your paper was a great refresher, it even taught
> me
> a few things I either forgor or never knew!! For some reason a lot of text
> books don't make the Hilbert transform easy to understand, usually because
> of their use of language getting in the way of conveying the information.
> I
> found your paper pitched at the right level!
>
> Rich.
>
>
Thanks Rich,
That's a great compliment. Writing is not my forte' - I'll leave that to
Rick Lyons. But I do try to cut through all of the stuff and distill out
what is important. I'm glad that this paper was at the right level for you.
Hopefully others will think the same.
Again Thanks,
Clay
Reply by Richard Webb●March 3, 20052005-03-03
"Clay S. Turner" <Physics@Bellsouth.net> wrote in message
news:a5sVd.42268$Rl5.34755@bignews4.bellsouth.net...
> Hello Guys,
>
> Since questions often arise about Hilbert transforms, I thought I'd put
> together a paper covering the topic. Basically I'm looking for feed back
on
> my paper. I know there are some formatting issues - but I haven't plans to
> publish it anywhere other than the web. Basically I'm trying to tie
together
> the DSP aspects of Hilbert transforms. So if there is anything you would
> like to see added, corrected, changed or what have you, please let me know
> what you think about it. And if the paper just plain sucks - well let me
> know that too ;-)
>
> http://personal.atl.bellsouth.net/p/h/physics/hilberttransforms.pdf
>
> Thanks,
> Clay
>
>
> --
> Clay S. Turner
> Vice President - Wireless Systems Engineering, Inc.
> Satellite Beach, FL
> www.WSE.biz
>
>
Thanks Clay,
I've had an interest in the Hilbert transform since I had to implement
one in a project a few years ago (with a bit of help from Rick!). I haven't
looked at them since and your paper was a great refresher, it even taught me
a few things I either forgor or never knew!! For some reason a lot of text
books don't make the Hilbert transform easy to understand, usually because
of their use of language getting in the way of conveying the information. I
found your paper pitched at the right level!
Rich.
Reply by Clay●March 3, 20052005-03-03
robert bristow-johnson wrote:
>
> hey Clay, would you like a section on how, for minimum-phase filters,
that
> the phase response (in radians) is the negative of the Hilbert
transform of
> the natural log of the magnitude response? (and maybe, this one is
easier,
> how causal impulse responses correspond to a frequency response where
the
> imaginary part and real part are a hilbert pair?)
>
> --
>
> r b-j rbj@audioimagination.com
>
> "Imagination is more important than knowledge."
Hello Robert,
These would be neat additions. I'll look into figuring these out or do
you already have such proofs handy?
Clay
Reply by robert bristow-johnson●March 3, 20052005-03-03
in article a5sVd.42268$Rl5.34755@bignews4.bellsouth.net, Clay S. Turner at
Physics@Bellsouth.net wrote on 03/02/2005 18:37:
> Since questions often arise about Hilbert transforms, I thought I'd put
> together a paper covering the topic. Basically I'm looking for feed back on
> my paper. I know there are some formatting issues - but I haven't plans to
> publish it anywhere other than the web. Basically I'm trying to tie together
> the DSP aspects of Hilbert transforms. So if there is anything you would
> like to see added, corrected, changed or what have you, please let me know
> what you think about it. And if the paper just plain sucks - well let me
> know that too ;-)
>
> http://personal.atl.bellsouth.net/p/h/physics/hilberttransforms.pdf
hey Clay, would you like a section on how, for minimum-phase filters, that
the phase response (in radians) is the negative of the Hilbert transform of
the natural log of the magnitude response? (and maybe, this one is easier,
how causal impulse responses correspond to a frequency response where the
imaginary part and real part are a hilbert pair?)
--
r b-j rbj@audioimagination.com
"Imagination is more important than knowledge."
Reply by Clay●March 3, 20052005-03-03
Olw wrote:
> Clay S. Turner wrote:
>> >
> Last page. The integral trick. 1/x = int_0^\infty (e^{-st}ds.
> This is 1/t, not 1/x.
>
> Olw
Thanks Olw,
I just fixed it in the manuscript.
Clay
Reply by Olw●March 3, 20052005-03-03
Clay S. Turner wrote:
> Hello Guys,
>
> Since questions often arise about Hilbert transforms, I thought I'd put
> together a paper covering the topic. Basically I'm looking for feed back on
> my paper. I know there are some formatting issues - but I haven't plans to
> publish it anywhere other than the web. Basically I'm trying to tie together
> the DSP aspects of Hilbert transforms. So if there is anything you would
> like to see added, corrected, changed or what have you, please let me know
> what you think about it. And if the paper just plain sucks - well let me
> know that too ;-)
>
> http://personal.atl.bellsouth.net/p/h/physics/hilberttransforms.pdf
>
> Thanks,
> Clay
>
>
Last page. The integral trick. 1/x = int_0^\infty (e^{-st}ds.
This is 1/t, not 1/x.
Olw