Analytic Functions and Single Side Band Signals

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:

"A complex function is said to be analytic on a region R if it is
complex differentiable at every point in R. The terms holomorphic
function, differentiable function, complex differentiable function,
and regular function are sometimes used interchangeably with "analytic
function" (Krantz 1999, p. 16). Many mathematicians prefer the term
"holomorphic function" (or "holomorphic map") to "analytic function"
(Krantz 1999, p. 16), while "analytic" appears to be in widespread use
among physicists, engineers, and in some older texts (e.g., Morse and
Feshbach 1953, pp. 356-374; Knopp 1996, pp.
 83-111; Whittaker and Watson 1990, p. 83)."

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?

Cheers
Porterboy

porterboy76@yahoo.com (porterboy) writes:
> [...]
> Does an analytic function necessarily
> produce a single sideband signal? 

Definitely not! F(z) = cos(x), where z = x + i*y, 
is analytic over all C but it is not an analytic
signal.
-- 
Randy Yates
Sony Ericsson Mobile Communications
Research Triangle Park, NC, USA
randy.yates@sonyericsson.com, 919-472-1124

in article BCYvd.640$Z27.454@bignews5.bellsouth.net, Clay S. Turner at
Physics@Bellsouth.net wrote on 12/15/2004 10:23:

> 
> "porterboy" <porterboy76@yahoo.com> wrote in message
> news:c4b57fd0.0412150037.5615c7c0@posting.google.com...
> 
> 
>> 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?
>> 
> 
> Hello Porterboy,
> 
> Yes analytic functions and analytic signals are related.
> 
> I.e.,
> 
> An analytic function may be written as
> 
> f(t,tau) + j*g(t,tau) where it obeys the C-R relations.
> 
> If you let tau==0 in an analytic function you get an analytic signal. And in
> this case g(t,0) is the Hilbert transform of f(t,0). And using this you can
> easily show that an analytic signal has a one sided transform. (Use the
> sigum function in the fourier domain) One of the simplest analytic signals
> is cos(t)+j*sin(t).

okay, Clay, now i'm confused.  maybe it's semantic.  first of all, what are
f() and g()?  are they complex functions where the complex argument is
z = t + j*tau ?  if so, by "C-R", do you mean that df/dtau = dg/dt ?  when i
crack a complex variables book, that what i think they mean by "analytic
function".  checking wikipedia, it looks like "holomorphic" is the new word
for the same thing, but my Levison and Redheffer book doesn't say
"holomorpic".  i can be 25 - 30 years behind the times.

now, assuming i got the semantics right, *why* should that mean that

    g(t,0) = Hilbert{ f(t,0) } ?

i don't see the two as related.

Of course the last thing you said is fundamentally true.  if

    g(t) = Hilbert{ f(t) } ,

then 

    a(t) = f(t) + j*g(t)

is one-sided.

i've always thought that this term "analytic" has been the victim of
semantics between two related fields: complex variables and
communications/signal_processing .  i just accepted the semantic difference
in the same way that i accept the word "sample" means different things
within the audio DSP community, depending on who you're talking to.

so, Clay, please clear this up for me.

thnx.

-- 

r b-j                  rbj@audioimagination.com

"Imagination is more important than knowledge."

Randy Yates wrote:
> [snip]
> 
> Definitely not! F(z) = cos(x), where z = x + i*y, 
> is analytic over all C but it is not an analytic
> signal.

I think I'm missing something. Then again I tended to have to retake my 
college math courses ~40 years ago ;{

I can see writing something of the *form*
[ That's not to say I would know what to do with it. ]
F(z) = cos(z)
z = x + i*y

But I don't understand writing
F(z) = cos(x)
z = x + i*y

I think I would write something like
F(x,y) = cos( x + i*y )

Thanks

Hello Robert,
Comments below:

"robert bristow-johnson" <rbj@audioimagination.com> wrote in message 
news:BDE5D304.31EB%rbj@audioimagination.com...
> in article BCYvd.640$Z27.454@bignews5.bellsouth.net, Clay S. Turner at
> Physics@Bellsouth.net wrote on 12/15/2004 10:23:
>
>>
>> "porterboy" <porterboy76@yahoo.com> wrote in message
>> news:c4b57fd0.0412150037.5615c7c0@posting.google.com...
>>
>>
>>> 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?
>>>
>>
>> Hello Porterboy,
>>
>> Yes analytic functions and analytic signals are related.
>>
>> I.e.,
>>
>> An analytic function may be written as
>>
>> f(t,tau) + j*g(t,tau) where it obeys the C-R relations.
>>
>> If you let tau==0 in an analytic function you get an analytic signal. And 
>> in
>> this case g(t,0) is the Hilbert transform of f(t,0). And using this you 
>> can
>> easily show that an analytic signal has a one sided transform. (Use the
>> sigum function in the fourier domain) One of the simplest analytic 
>> signals
>> is cos(t)+j*sin(t).
>
> okay, Clay, now i'm confused.  maybe it's semantic.  first of all, what 
> are
> f() and g()?  are they complex functions where the complex argument is
> z = t + j*tau ?  if so, by "C-R", do you mean that df/dtau = dg/dt ?  when 
> i
> crack a complex variables book, that what i think they mean by "analytic
> function".  checking wikipedia, it looks like "holomorphic" is the new 
> word
> for the same thing, but my Levison and Redheffer book doesn't say
> "holomorpic".  i can be 25 - 30 years behind the times.

f(t,tau) and g(t,tau) are purely real functions. So the complex function is 
built by

psi(t,tau) = f(t,tau) + j*g(t,tau)

The CR relations are:

df/dt = dg/d tau

and

df/d tau = - dg/dt

where the derivatives are partial derivatives.


>
> now, assuming i got the semantics right, *why* should that mean that
>
>    g(t,0) = Hilbert{ f(t,0) } ?
>
> i don't see the two as related.

This is not immediately obvious but it may be derived from a neat 
application of the Cauchy integral formula that says

f(z_0) = (1/(2pi*j)) * path integral f(z)/(z-z_0) dz

where the path is closed and the singular point is inside of the path. Also 
f(z) needs to be analytic (obeys the C-R relations) both on and inside of 
the path. You probably recall this when integrating Laurent series.

So now pick a path that has a straight segment along the real axis going 
from -R to R and close the path with a semicircle of radius R. However the 
straight part also has a little semicircle to hop around the singular point 
z_0.  (also the singular point is chosen to be on the real axis) When R is 
allowed to grow towards infinity, it is helpful to think of the three parts 
of the path and their contribution to the overall integral. The outer 
semicircle will force its integral's contribution to go to zero. You may 
wish to look up Jordan's lemma. So in the limit of large R

There is also a correlary to Cauchy's theorem that concerns using paths that 
don't completely enclose the singular point. Basically you just use the 
portion of a circle that a path uses around the point to proportion Cauchy' 
standard result. For example if a semicircle is used, then instead of the 
2pi*j, you just use pi*j.

So using the above path in a way analogous to the Cauchy integral formula we 
find (I know many details skipped)

psi(t,0) = (1/(2pi*j)) * principal value integral psi(s)/(s-t) ds

The term "principal value" refers to the nature of the limiting process so 
exact cancelation may occur. This type of integral just shows up in Hilbert 
transforms.

So from here with a little work one can show that

f(t,0) = (1/pi) principal value integral g(s,0)/(s-t) ds

And apart from the ,0 in the f(t,0) and g(s,0) terms - this is the 
definition for a Hilbert transform.

>
> Of course the last thing you said is fundamentally true.  if
>
>    g(t) = Hilbert{ f(t) } ,
>
> then
>
>    a(t) = f(t) + j*g(t)
>
> is one-sided.
>
> i've always thought that this term "analytic" has been the victim of
> semantics between two related fields: complex variables and
> communications/signal_processing .  i just accepted the semantic 
> difference
> in the same way that i accept the word "sample" means different things
> within the audio DSP community, depending on who you're talking to.
>
> so, Clay, please clear this up for me.

Having come from a Math background, I learned to associate the term 
"analytic function" as obeying the C-R relations, with the usage of saying a 
function is "analytic" also meaning the same thing in the context we are 
talking about a complex function. Dennis Gabor in 1946 extends the analytic 
concept of evaluating the analytic function along just one axis to create a 
"analytic signal." Even though I showed the mapping along the t axis, one 
may also use the tau axis. So I'm just careful to use the term "analytic 
signal" when referring to functions with one-sided Fourier transforms. Using 
"analytic" by itself is where one may get into trouble without ensuring 
proper clarification.

I hope this helps some. If I get a chance, I may work out all of the gory 
details on this and post it on my site. It has been a long time since I 
trudged through this. However the result is quite neat and has practical 
applications as we all know.

Clay


>
> thnx.
>
> -- 
>
> r b-j                  rbj@audioimagination.com
>
> "Imagination is more important than knowledge."
>
> 

Richard Owlett <rowlett@atlascomm.net> writes:

> Randy Yates wrote:
> > [snip]
> > Definitely not! F(z) = cos(x), where z = x + i*y, is analytic over
> > all C but it is not an analytic
> 
> > signal.
> 
> I think I'm missing something. Then again I tended to have to retake
> my college math courses ~40 years ago ;{
> 
> 
> I can see writing something of the *form*
> [ That's not to say I would know what to do with it. ]
> F(z) = cos(z)
> z = x + i*y
> 
> But I don't understand writing
> F(z) = cos(x)
> z = x + i*y
> 
> I think I would write something like
> F(x,y) = cos( x + i*y )

Hey Richard,

It does look odd, doesn't it? But I believe it's
correct.

Note that a complex function F is a mapping from the complex to the
complex, F: A \in C --> B \in C. That is, it maps the set of complex
numbers in its domain, A, to the set of complex numbers in its range,
B (remember domain and range from high school?). So in general a
complex function F(z) = f(x,y) + i*g(x,y), where z = x + i*y.

In my function, f(x,y) = cos(x) and g(x,y) = 0. In your function,
f(x,y) = cosh(y)*sin(x) and g(x,y) = sinh(y)*sin(x) - quite a 
different function. 

Does that help?
-- 
Randy Yates
Sony Ericsson Mobile Communications
Research Triangle Park, NC, USA
randy.yates@sonyericsson.com, 919-472-1124

Randy Yates <randy.yates@sonyericsson.com> writes:
> [...]
> In your function,
> f(x,y) = cosh(y)*sin(x) and g(x,y) = sinh(y)*sin(x) 

Typo: should be 

  f(x,y) = cosh(y)*cos(x) and g(x,y) = sinh(y)*sin(x) 
-- 
Randy Yates
Sony Ericsson Mobile Communications
Research Triangle Park, NC, USA
randy.yates@sonyericsson.com, 919-472-1124

Randy Yates wrote:

> Richard Owlett <rowlett@atlascomm.net> writes:
> 
> 
>>Randy Yates wrote:
>>
>>>[snip]
>>>Definitely not! F(z) = cos(x), where z = x + i*y, is analytic over
>>>all C but it is not an analytic
>>
>>>signal.
>>
>>I think I'm missing something. Then again I tended to have to retake
>>my college math courses ~40 years ago ;{
>>
>>
>>I can see writing something of the *form*
>>[ That's not to say I would know what to do with it. ]
>>F(z) = cos(z)
>>z = x + i*y
>>
>>But I don't understand writing
>>F(z) = cos(x)
>>z = x + i*y
>>
>>I think I would write something like
>>F(x,y) = cos( x + i*y )
> 
> 
> Hey Richard,
> 
> It does look odd, doesn't it? But I believe it's
> correct.
> 

I'll accept you as an expert witness -- see below ;)


> Note that a complex function F is a mapping from the complex to the
> complex, F: A \in C --> B \in C. That is, it maps the set of complex
> numbers in its domain, A, to the set of complex numbers in its range,
> B (remember domain and range from high school?). So in general a
> complex function F(z) = f(x,y) + i*g(x,y), where z = x + i*y.
> 
> In my function, f(x,y) = cos(x) and g(x,y) = 0. In your function,
> f(x,y) = cosh(y)*sin(x) and g(x,y) = sinh(y)*sin(x) - quite a 
> different function. 
> 
> Does that help?

In that it seems to prove my lack of comprehension of maths in general.
The next to last time I flunked out of an Ivy League BSEE program, the 
"Testing and Guidance Center" said my aptitude/interest profile 
resembled 'law' rather than 'engineering' students ;}

So my *INTERESTS* are in one area
    my *APTITUDE* in another

Thanks for the feedback.

[ PS once got a job interview solely on the basis that owner of company 
wanted to meet someone who had a resume (with verifiable references) as 
strange as mine ;]

"Richard Owlett" <rowlett@atlascomm.net> wrote in message 
news:10s0uu14q3r1uac@corp.supernews.com...
> Randy Yates wrote:
>> [snip]
>>
>> Definitely not! F(z) = cos(x), where z = x + i*y, is analytic over all C 
>> but it is not an analytic
>> signal.
>
> I think I'm missing something. Then again I tended to have to retake my 
> college math courses ~40 years ago ;{
>
> I can see writing something of the *form*
> [ That's not to say I would know what to do with it. ]
> F(z) = cos(z)
> z = x + i*y
>
> But I don't understand writing
> F(z) = cos(x)
> z = x + i*y

I believe Randy meant to say

F(z) = cos(z)

To show it is analytic (function)

so cos(z) = cos(x+iy)

= cos(x)*cos(iy) - sin(x)*sin(iy)

recall trig functions with imaginary arguments go into hyperspace.

= cos(x)*cosh(y)-i*sin(x)*sinh(y)

so when put into the  u(x,y) + j*v(x,y) form

u(x,y) = cos(x)*cosh(y)

and

v(x,y) = -sin(x)*sinh(y)

Now test to see if the Cauchy Riemann relations hold.

du/dx = -sin(x)*cosh(y)

and

dv/dy = -sin(x)*cosh(y)

so du/dx = dv/dy  so far so good (CR relation #1 holds)

now

du/dy = cos(x)*sinh(y)

and

dv/dx = -cos(x)*sinh(y)

thus

du/dy = - dv/dx and CR relation #2 holds

so cos(z) is an analytic function.

Clay






>
> I think I would write something like
> F(x,y) = cos( x + i*y )
>
> Thanks 

"Clay S. Turner" <Physics@Bellsouth.net> writes:

> "Richard Owlett" <rowlett@atlascomm.net> wrote in message 
> news:10s0uu14q3r1uac@corp.supernews.com...
> > Randy Yates wrote:
> >> [snip]
> >>
> >> Definitely not! F(z) = cos(x), where z = x + i*y, is analytic over all C 
> >> but it is not an analytic
> >> signal.
> >
> > I think I'm missing something. Then again I tended to have to retake my 
> > college math courses ~40 years ago ;{
> >
> > I can see writing something of the *form*
> > [ That's not to say I would know what to do with it. ]
> > F(z) = cos(z)
> > z = x + i*y
> >
> > But I don't understand writing
> > F(z) = cos(x)
> > z = x + i*y
> 
> I believe Randy meant to say
> 
> F(z) = cos(z)

Hi Clay,

No, I meant to say F(z) = cos(x). The problem is, I 
am flat-out wrong. That function is not analytic! I
made the same mistake I've made several times before:
differentiable at all points is not the same as 
analytic!

Have mercy on me - I was going from memory: my Churchill
and Brown text is at home. I was able to see my error 
from the Cauchy-Riemann conditions you gave in another
post of yours today, Clay.

To Porterboy, you have my sincere apologizes for giving
you bad information. 
-- 
Randy Yates
Sony Ericsson Mobile Communications
Research Triangle Park, NC, USA
randy.yates@sonyericsson.com, 919-472-1124