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> When I was in third year of Uni, we did a course on complex maths that
> really threw me. It was all about analytic functions, the
> Cauchy-Riemann equations, etc. and involved all sorts of integrations
> of curves in the complex plane. Now I'd always been good at maths, but
> what with all the other stuff going on, and the fact that this stuff
> was quite difficult, I never really conquered the subject, which I
> consider a serious hole in my knowledge. This is what Wolfram says
> about analytic functions:

[- snip -] 

> I hadn't thought of most of this stuff in ages until last night
> something clicked in my head. I have been working for a few months
> with Single-Sideband Signals generated by Hilbert Transforms. These
> are of course analytic signals, since they contain only positive
> frequencies. My question is this: Are analytic signals related in
> anyway to analytic functions? Does an analytic function necessarily
> produce a single sideband signal? Is there no connection at all?

You have spotted the very essence of DSP. Have you ever wondered 
why you use those _rational_ transfer functions in z domain, and 
not arbitrary experssions of z, like Log(z) or exp(Z)? It's 
because rational complex-valued functions are analytic.

Have you ever wondered why you use the tabulated inverse Laplace 
or inverse Fourier transforms instead of computing an integral?
It's becuase the Cauchy residual theorem applies with analystic 
functions.

To my knowledge, there is only one book that shows these things 
in all their gory mathemathical details:

Oppenheim & Schafer: "Digital Signal Processing"
    Prentice Hall, 1975.

(The Rabiner & Gold book from about the same time could very well 
be close, but I have never actually seen it, so I don't know.)

Oppenheim & Schafer have written lots of books, so make sure you get 
the one from 1975. Their discussion of the Hilbert transform is 
basically a full semester course on complex functions, much like the 
course you took at university, condensed to 10-pages.

DSP is applied maths. You have spotted the essence of it.

Rune

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Rune Allnor wrote:
...
> 
> You have spotted the very essence of DSP. Have you ever wondered 
> why you use those _rational_ transfer functions in z domain, and 
> not arbitrary experssions of z, like Log(z) or exp(Z)? It's 
> because rational complex-valued functions are analytic.

They are not, Rune. Rational functions are meromorphic, which basically 
means they are analytic on the complex numbers C minus the set of their 
poles.

If you include a complex infinity into the set of the complex numbers, 
which is a one-point compactification of C, sometimes denoted as C-bar, 
then one can say that the meromorphic fucntions are analytic on C-bar. A 
nice model of C-bar is the complex number sphere (where 0 is the 
south-pole and complex infinity is the north-pole and the unit circle is 
the equator).

The meromorphic functions constitute a field which naturally embedds the 
complex numbers, similar to the way the complex number field embedds the 
real numbers.


> Have you ever wondered why you use the tabulated inverse Laplace 
> or inverse Fourier transforms instead of computing an integral?

To avoid having to teach complex integration?

:-)

Regards,
Andor

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Andor Bariska <a...@nospam.net> writes:

> Rune Allnor wrote:
> ...
> > You have spotted the very essence of DSP. Have you ever wondered why
> > you use those _rational_ transfer functions in z domain, and not
> > arbitrary experssions of z, like Log(z) or exp(Z)? It's because
> > rational complex-valued functions are analytic.
> 
> 
> They are not, Rune. Rational functions are meromorphic, which
> basically means they are analytic on the complex numbers C minus the
> set of their poles.

Andor,

It's good that you point out this fact for the OP, but don't you
think Rune probably already knows this? He probably just posted in
a hurry and/or without feeling it necessary to spell out all the 
details. As you imply, "analytic" is meaningless without a domain.

Note also that in my old Complex Variables class, we used to call a
function that was analytic over all C "entire." (That class was based
on the Churchill and Brown text.)

> The meromorphic functions constitute a field which naturally embedds
> the complex numbers, similar to the way the complex number field
> embedds the real numbers.

I believe you've got some detail wrong here because if what you say
is true, the Fundamental Theorem of Algebra is wrong. There are no
extension fields (splitting fields?) of the complex. 
-- 
Randy Yates
Sony Ericsson Mobile Communications
Research Triangle Park, NC, USA
r...@sonyericsson.com, 919-472-1124

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Randy Yates wrote:

> Andor,
> 
> It's good that you point out this fact for the OP, but don't you
> think Rune probably already knows this? 

If he does, then he forgot to mention it. I thought I'd jump in and jog 
everybody's memories.

 > He probably just posted in
> a hurry and/or without feeling it necessary to spell out all the 
> details. As you imply, "analytic" is meaningless without a domain.

Indeed.


>>The meromorphic functions constitute a field which naturally embedds
>>the complex numbers, similar to the way the complex number field
>>embedds the real numbers.
> 
> 
> I believe you've got some detail wrong here because if what you say
> is true, the Fundamental Theorem of Algebra is wrong. There are no
> extension fields (splitting fields?) of the complex.

I meant it in the following sense:

Def: A subset G of a set F is called a subfield, if (F,+,*) and (G,+,*) 
are both fields.

Examples:
- Q is a subfield of R
- R is a subfield of C
- C is a subfield of M

(M is the set of meromorphic functions)

Notice my choice of "embed" rather than "extend".

Also, I believe the Fundamental Theorem of Algebra was concerned with 
the number of zeroes of a polynomial? The fact that there are no 
commutative division algebras with order larger than 2 is known to me as 
  Theorem of Frobenius (well, at least it is a direct corollary thereof).

Perhaps you got confused by the fact that M is not, unlike R and C, an 
algebra.

Regards,
Andor

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Andor Bariska wrote:
> Rune Allnor wrote:
> ...
> >
> > You have spotted the very essence of DSP. Have you ever wondered
> > why you use those _rational_ transfer functions in z domain, and
> > not arbitrary experssions of z, like Log(z) or exp(Z)? It's
> > because rational complex-valued functions are analytic.
>
> They are not, Rune. Rational functions are meromorphic, which
basically
> means they are analytic on the complex numbers C minus the set of
their
> poles.

If I ever heard the term "meromorphic" in use before, I've forgotten
about it. The "analytic everywhere except at the poles and branch
cuts" goes without saying, particularly since the OP both works in
DSP and has the complex maths course.

What I forgot to mention was that the Exp(z) function probably is
analytic (almost) everywhere, but has a Laurent series representation
with infinitely many terms. Which is a bit awakward. And the Log(z)
function has branch cuts as well. Which is even more awkward.
Interestingly, the branch cuts of the Log(z) function appear to be
the main obstacle for getting the complex cepstra to work.

> If you include a complex infinity into the set of the complex
numbers,
> which is a one-point compactification of C, sometimes denoted as
C-bar,
> then one can say that the meromorphic fucntions are analytic on
C-bar. A
> nice model of C-bar is the complex number sphere (where 0 is the
> south-pole and complex infinity is the north-pole and the unit circle
is
> the equator).
>
> The meromorphic functions constitute a field which naturally embedds
the
> complex numbers, similar to the way the complex number field embedds
the
> real numbers.

Interesting. I didn't know that.

> > Have you ever wondered why you use the tabulated inverse Laplace
> > or inverse Fourier transforms instead of computing an integral?
>
> To avoid having to teach complex integration?

Yes. My main reaction last year when I saw the Oppenheim & Schafer
book from 1975 for the first time, was that that book had little to
do with DSP as we know it today. Their book was more about applied
maths. If one had to compute the IFT in the "algebraically correct"
way, no electrical engineer would ever qualify for the DSP intro
class, let alone get any work done.

Those who have a copy available, check out section 2.2 in O&S, 1975.

What saves the day is Cauchy's residue theorem that ensures that
given a H(z) on the form of a ratio of two finite polynomials,
all we need is to know the zeros and poles and then we can set up
a table of formulas that takes from z domain to time domain.

All of that saves ridiculous amounts of work, both in teaching DSP
and in working with DSP, and it works only because it is based
on the theory of analytic functions.

Rune

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The Google guys still seem to be experimenting with an optimal user
interface. Right now, they don't give me the original message when
replying, so please excuse the odd quote:

Rune wrote:
"
What I forgot to mention was that the Exp(z) function probably is
analytic (almost) everywhere, but has a Laurent series representation
with infinitely many terms. Which is a bit awakward.
"
The Laurent series expansion of exp(z) stops dead at 0. Exp(z) is about
as analytic as they come. Did you mean exp(1/z)? This has a "major"
singularity (what is the correct term in English?) at zero, meaning
exactly that the Laurent series expansion around zero has infinitely
many (negative power) terms.

"
And the Log(z) function has branch cuts as well. Which is even more
awkward.
Interestingly, the branch cuts of the Log(z) function appear to be
the main obstacle for getting the complex cepstra to work.
"

This was something I never quite understood. Is the complex cepstrum of
s(t) not defined as  FT^(-1) ( log( Abs( FT( s(t) ) ) ) )? What kind of
information can one possibly gather from that? What do you mean by
"getting the complex cepstra to work"?

Regards,
Andor

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Andor wrote:
> The Google guys still seem to be experimenting with an optimal user
> interface. Right now, they don't give me the original message when
> replying, so please excuse the odd quote:
>
> Rune wrote:
> "
> What I forgot to mention was that the Exp(z) function probably is
> analytic (almost) everywhere, but has a Laurent series representation
> with infinitely many terms. Which is a bit awakward.
> "
> The Laurent series expansion of exp(z) stops dead at 0. Exp(z) is
about
> as analytic as they come. Did you mean exp(1/z)? This has a "major"
> singularity (what is the correct term in English?) at zero, meaning
> exactly that the Laurent series expansion around zero has infinitely
> many (negative power) terms.

I am on thin ice here, and I may very well have messed up the very
precise terminology. I took a basic course on maths where complex
numbers were taught, but I have forgotten most of it. At least the
subtle details in the levels slightly above basic, that we discuss
here.

> "
> And the Log(z) function has branch cuts as well. Which is even more
> awkward.
> Interestingly, the branch cuts of the Log(z) function appear to be
> the main obstacle for getting the complex cepstra to work.
> "
>
> This was something I never quite understood. Is the complex cepstrum
of
> s(t) not defined as  FT^(-1) ( log( Abs( FT( s(t) ) ) ) )? What kind
of
> information can one possibly gather from that? What do you mean by
> "getting the complex cepstra to work"?

What you cite is the *real* cepstrum. The complex cepstrum is
defined as

S(t) = IFT{Log{FT{x(t)}}}                                  [1]

where X(w) = FT{x(t)} is complex-valued and Log is the complex
logarithm,

Log(z) = Log{|z|*exp(j*phi)} = log(|z|) +j(phi + n*2*pi).  [2]

The requirement is that the complex cepstrum is real-valued and
causal (the term "complex" refers to the complex Log function
being used) so the real and imaginary parts in [2] must be Hilbert
transform pairs. Which in turn means that the real and imaginary
parts must be continuous on the unit circle in complex z domain.

The real part, log(|z|), is no problem except for z=0. The imaginary
part is a problem, since the periodicity in phi introduces a
discontinuity, a branch cut. Basically, it's very awkward to unwrap
the phase of the spectrum in a way that is both continuous on the
unit circle, satisfies the Hilbert tranform relation to the
log(|z|) function, and also results in a causal, real-valued
cepstrum S(t). Check out chapters 7 and 10 in Oppenheim and
Schafer's 1975 book for details of the statement of the problem.

Rune

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I wrote:

"
This was something I never quite understood. Is the complex cepstrum of
s(t) not defined as FT^(-1) ( log( Abs( FT( s(t) ) ) ) )? What kind of
information can one possibly gather from that?
"

I did a quick search, and the first link churned out by google was
this:

http://www.lsv.uni-saarland.de/teaching/pattern_and_speech_recognition/ws0405/Pattern_and_Speech_Recognition_Ch4.pdf

It seems that the cepstrum (at least in that article) is used to find
formants for speech. This does make sense, as the regular spacing of
the formant harmonics in the log magnitude of the spectrum will show up
as a peak if you take the fourier transform again. Quite clever.
Any other uses?

Regards,
Andor

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Andor wrote:
> I wrote:
> 
> "
> This was something I never quite understood. Is the complex cepstrum of
> s(t) not defined as FT^(-1) ( log( Abs( FT( s(t) ) ) ) )? What kind of
> information can one possibly gather from that?
> "
> 
> I did a quick search, and the first link churned out by google was
> this:
> 
>
http://www.lsv.uni-saarland.de/teaching/pattern_and_speech_recognition/ws0405/Pattern_and_Speech_Recognition_Ch4.pdf
> 
> It seems that the cepstrum (at least in that article) is used to find
> formants for speech. This does make sense, as the regular spacing of
> the formant harmonics in the log magnitude of the spectrum will show up
> as a peak if you take the fourier transform again. Quite clever.
> Any other uses?
> 
> Regards,
> Andor
> 

Yup, a couple:
1)  convolution in time = Multiplication in frequency = Addition in 
Ceptstrum (quefrency ?). It has been used for deconvolution.

2) It can also be used to form the minimum phase decomposition of a 
frequency response.

I believe one of the other problems is that to get back you must take 
the exponential - thus any error can be hugely magnified.

Cheers,
David

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David Kirkland wrote:
...
> Yup, a couple:
> 1)  convolution in time = Multiplication in frequency = Addition in 
> Ceptstrum (quefrency ?).

David,

do you know who invented these ridiculous terms? Cepstrum, Quefrency, 
... they produce knots in my tongue!

Regards,
Andor

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