CAUTION! was "What is the advantage on high-sampling rate ?"

rhn@mauve.rahul.net (Ronald H. Nicholson Jr.) writes:

> In article <xxpu0z1mgfo.fsf@usrts005.corpusers.net>,
> Randy Yates  <randy.yates@sonyericsson.com> wrote:
> >The DFT, BY DEFINITION, produces a *FINITE* set of values as output.
> >These are, BY DEFINITION, frequency-domain output samples. 
> 
> Only if you make certain assumptions about the input.  You've gone
> in a complete circle.

Yeah, and THAT'S only true if you make assumptions about the
meaning of the word "Only" and the word "if" and the word "you"
and the word "make" and the word "certain" and the word ...

This is black and white, or at least as black and white as anything
can be. We could all be the dream of some mythical sentient being
for all we *really* know.
-- 
Randy Yates
Sony Ericsson Mobile Communications
Research Triangle Park, NC, USA
randy.yates@sonyericsson.com, 919-472-1124

Randy Yates wrote:

> 
> You're wrong. I don't mean to be disrespectul, and Lord knows
> I not tactful, but you are flat-out wrong. Here's why.
> 

'Fraid I have the same opinion of your position.  There is 
nothing whatsoever in the DFT definition that speaks to 
infinity or anything outside the interval including the 
sinusoidal intervals that are used as the basis.  The only 
time considered in the definition is that within the 
interval being transformed and no more need be to qualify 
those sinusoidal sections as a complete and orthogonal basis 
for band limited functions defined within that interval.


Bob
-- 

"Things should be described as simply as possible, but no 
simpler."

                                              A. Einstein

Ronald H. Nicholson Jr. wrote:

> 
> Only if you make certain assumptions about the input.  You've gone
> in a complete circle.

Exactly.

> 
> A DFT is only an operator which munges one finite set of bits into
> another.  

Period.  End of story.


Bob
-- 

"Things should be described as simply as possible, but no 
simpler."

                                              A. Einstein

Bob Cain <arcane@arcanemethods.com> writes:

> Randy Yates wrote:
>
>> You're wrong. I don't mean to be disrespectul, and Lord knows
>> I not tactful, but you are flat-out wrong. Here's why.
>>
>
> 'Fraid I have the same opinion of your position.  There is nothing
> whatsoever in the DFT definition that speaks to infinity 

As long as you agree that we have a set of discrete values in frequency
spaced evenly apart (i.e., of the form \sum_{k=-\infty}^{+\infty}
a[k]delta[f-kF]" it doesn't have to have anything to do with the DFT
definition. If you have a set of impulses in frequency, then the
corresponding time-domain function is necessarily periodic. Those
impulses can come from a DFT, a Fourier series analysis, or your
mother's kitchen.

  periodic time <==> discrete frequency

Live it, love it.
-- 
%  Randy Yates                  % "How's life on earth? 
%% Fuquay-Varina, NC            %  ... What is it worth?" 
%%% 919-577-9882                % 'Mission (A World Record)', 
%%%% <yates@ieee.org>           % *A New World Record*, ELO
http://home.earthlink.net/~yatescr

Ronald H. Nicholson Jr. wrote:
> Randy Yates  <randy.yates@sonyericsson.com> wrote:
> >The DFT, BY DEFINITION, produces a *FINITE* set of values as output.
> >These are, BY DEFINITION, frequency-domain output samples. 
> 
> Only if you make certain assumptions about the input.  You've gone
> in a complete circle.

Haha, pun intended?

 > A DFT is only an operator which munges one finite set of bits into
> another.  

That's what I would say also. The DFT is a unitary linear map on a
hilbert space:
DFT: CC^n -> CC^n

(CC: complex nubmers, n dimension of hilbert space)

As such it transforms a set of o.n vectors into another set of o.n
vectors while keeping the inner product invariant. It has some nice
properties: one of them is that is diagonalizes the shift operator -
linear algebra version of the FT shifting theorms. There are
absolutely no assumptions made on what basis you are acting on, or
what physical meaning is attached to that basis.

Regards,
Andor

Randy Yates wrote:

> Bob Cain <arcane@arcanemethods.com> writes:
> 
> 
>>Randy Yates wrote:
>>
>>
>>>You're wrong. I don't mean to be disrespectul, and Lord knows
>>>I not tactful, but you are flat-out wrong. Here's why.
>>>
>>
>>'Fraid I have the same opinion of your position.  There is nothing
>>whatsoever in the DFT definition that speaks to infinity 
> 
> 
> As long as you agree that we have a set of discrete values in frequency
> spaced evenly apart (i.e., of the form \sum_{k=-\infty}^{+\infty}
> a[k]delta[f-kF]" it doesn't have to have anything to do with the DFT
> definition. 

And if I don't agree with that?  All I see after a DFT is 
the weights to be applied to set of harmonic functions 
(sinusoids and cosinusoids) that are defined only in the 
same interval as that of the set of time domain points. 
That set is comprised of all those harmonics which have an 
integral number of cycles within the interval with 
frequencies up to the Nyquist limit.

Nothing in the DFT definition is assumed about their value 
outside that interval.  Infinity only appears if you define 
those sinusoids and cosinusoids to extend to +/- infinity in 
which case you will obviously will the periodicity you can't 
shake but there is nothing in the DFT definition which makes 
that extension to the infinities explicit.  That falls out 
of your assumption about the basis set.

In the use that I almost exclusively have for the DFT, the 
assumption that the basis functions are zero outside that 
interval matches intuition and the physical systems I work 
with much better than extension of them as harmonic 
functions.  The problems you attack with it may be different 
so your assumptions may yield a better match.


Bob
-- 

"Things should be described as simply as possible, but no 
simpler."

                                              A. Einstein

an2or@mailcircuit.com (Andor) writes:

> Ronald H. Nicholson Jr. wrote:
>> Randy Yates  <randy.yates@sonyericsson.com> wrote:
>> >The DFT, BY DEFINITION, produces a *FINITE* set of values as output.
>> >These are, BY DEFINITION, frequency-domain output samples. 
>> 
>> Only if you make certain assumptions about the input.  You've gone
>> in a complete circle.
>
> Haha, pun intended?
>
>  > A DFT is only an operator which munges one finite set of bits into
>> another.  
>
> That's what I would say also. The DFT is a unitary linear map on a
> hilbert space:
> DFT: CC^n -> CC^n
>
> (CC: complex nubmers, n dimension of hilbert space)
>
> As such it transforms a set of o.n vectors into another set of o.n
> vectors while keeping the inner product invariant. It has some nice
> properties: one of them is that is diagonalizes the shift operator -
> linear algebra version of the FT shifting theorms. There are
> absolutely no assumptions made on what basis you are acting on, or
> what physical meaning is attached to that basis.

Tell you what - you go ahead and constrain yourselves to thinking of
the FFT output as nothing more than the result of a unitary linear
operator operating on a set of vectors in a Hilbert space.

I much prefer to use it as a tool to operate on real-world data and
real-world problems to accomplish real-world goals, and thus to have
real-world significance.

I just as capable of abstracting the math into meaninglessness as
the rest of you - I choose instead to use it as a tool.
-- 
%  Randy Yates                  % "And all that I can do
%% Fuquay-Varina, NC            %  is say I'm sorry, 
%%% 919-577-9882                %  that's the way it goes..."
%%%% <yates@ieee.org>           % Getting To The Point', *Balance of Power*, ELO
http://home.earthlink.net/~yatescr

Bob Cain <arcane@arcanemethods.com> writes:

> Randy Yates wrote:
>
>> Bob Cain <arcane@arcanemethods.com> writes:
>>
>>>Randy Yates wrote:
>>>
>>>
>>>>You're wrong. I don't mean to be disrespectul, and Lord knows
>>>>I not tactful, but you are flat-out wrong. Here's why.
>>>>
>>>
>>>'Fraid I have the same opinion of your position.  There is nothing
>>> whatsoever in the DFT definition that speaks to infinity
>> As long as you agree that we have a set of discrete values in
>> frequency
>> spaced evenly apart (i.e., of the form \sum_{k=-\infty}^{+\infty}
>> a[k]delta[f-kF]" it doesn't have to have anything to do with the DFT
>> definition.
>
> And if I don't agree with that?  All I see after a DFT is the weights
> to be applied to set of harmonic functions (sinusoids and cosinusoids)
> that are defined only in the same interval as that of the set of time
> domain points. That set is comprised of all those harmonics which have
> an integral number of cycles within the interval with frequencies up
> to the Nyquist limit.

I see this viewpoint, and agree with its validity. 

> Nothing in the DFT definition is assumed about their value outside
> that interval.  Infinity only appears if you define those sinusoids
> and cosinusoids to extend to +/- infinity in which case you will
> obviously will the periodicity you can't shake but there is nothing in
> the DFT definition which makes that extension to the infinities
> explicit.  That falls out of your assumption about the basis set.
>
> In the use that I almost exclusively have for the DFT, the assumption
> that the basis functions are zero outside that interval matches
> intuition and the physical systems I work with much better than
> extension of them as harmonic functions.  The problems you attack with
> it may be different so your assumptions may yield a better match.

I guess it depends on the assumptions you make about the meaning of
the DFT output, which is equivalent to making assumptions about the
basis set. As I stated, IF you consider the output to be a spectrum
in the normal Fourier sense, then you get the result I put forth. If
you interpret the output differntly, you can certainly get a different
result. 
-- 
%  Randy Yates                  % "So now it's getting late,
%% Fuquay-Varina, NC            %    and those who hesitate
%%% 919-577-9882                %    got no one..."
%%%% <yates@ieee.org>           % 'Waterfall', *Face The Music*, ELO
http://home.earthlink.net/~yatescr

Randy Yates wrote:

> 
> I guess it depends on the assumptions you make about the meaning of
> the DFT output, which is equivalent to making assumptions about the
> basis set. 

I would only modify that to say that it depends on the 
assumptions you make about the meaning of the DFT _input_, 
specifically the basis set.

In the problems I work on, FIR functions and their 
application by convolution to audio recordings, viewing the 
basis set as harmonics extended to infinity will make 
repetitive something that actually 'happens' once.

Thus I fail to see the usefulness of the infinite, 
...periodicperiodic... assumption when working with filters. 
  There is a real difference in the result of convolving a 
single IR with a signal and convolving an infinite train of 
them.

> As I stated, IF you consider the output to be a spectrum
> in the normal Fourier sense, then you get the result I put forth. If
> you interpret the output differntly, you can certainly get a different
> result. 

In the end we don't disagree at all, really, and are both 
correct provisionally.  I love arguments that end that way.

I doubt that Robert is done with me, however.  :-)


Bob
-- 

"Things should be described as simply as possible, but no 
simpler."

                                              A. Einstein

In article <4qr0fe5c.fsf@ieee.org>, Randy Yates  <yates@ieee.org> wrote:
>Tell you what - you go ahead and constrain yourselves to thinking of
>the FFT output as nothing more than the result of a unitary linear
>operator operating on a set of vectors in a Hilbert space.
>
>I much prefer to use it as a tool to operate on real-world data and
>real-world problems to accomplish real-world goals, and thus to have
>real-world significance.

I do to.  But all the real-world data I deal with doesn't have infinite
extent (e.g. is much more recent phenomena than stuff periodic since
the Big-Bang), and doesn't have an infinite spectra (I doubt my audio
system produces any X-ray radiation above background, maybe I need bigger
quantum-plated speaker cables? :^).  I'm not sure what real-world goals
would be accomplished by assuming such.


IMHO. YMMV.
-- 
Ron Nicholson   rhn AT nicholson DOT com   http://www.nicholson.com/rhn/ 
#include <canonical.disclaimer>        // only my own opinions, etc.