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

robert bristow-johnson wrote:

> no.  to use the DFT in anything but the most trivial applications, you also
> need to know how to either conceptually do modulo N arithmetic on the
> indices of x[n] and X[k] or to conceptually periodically extend x[n] and
> X[k] beyond the original limits of [0 .. N-1].  i contend these are the same
> thing and that if you can't do them, you cannot make the DFT work in
> applications requiring shifting (multiplying X[k] by exp(-j*2*pi*n0*k/N)) or
> convolution (multiplying X[k[ by H[k]).

Nah, that's baloney. I've seen people implement overlap-add and overlap-save
FFT filters without any real grasp of what's going on at all. What you
conceptualize happening outside the interval has no effect on the way the
computer crunches the numbers. I don't disagree that to periodically extend
the data can be a useful way to view it, but would never be so arrogant as to
claim that its the only way you can look at it. All you've offered in proof of
your contention is that 'I can't do it so no one else can'. That's pretty
lame.

-jim


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Bob Cain <arcane@arcanemethods.com> wrote in message news:<c7n91s02trp@enews3.newsguy.com>...
> robert bristow-johnson wrote:
> 
> > i didn't understand why "double" length in particular.  
> 
> It comes out of my unstated assumption that the low level 
> chunks we convolve are of equal length.  Yeah, I know about 
> the double less one.
> 
> > i know there are a
> > few in the music-dsp community (not necessarily the list) who, when they are
> > doing fast convolution, always choose the FFT length to be twice the buffer
> > length (which is also the hop length), which allows them to have an FIR
> > length of the same size (1/2 of the FFT).  they wouldn't have to do it that
> > way.  given an FIR length and knowledge of the processing costs of the FFT
> > and iFFT, there is a more optimal selection of the FFT length and the audio
> > buffer size can be the FFT length less the FIR (plus 1).  i figgered you
> > knew that, but still wondered where "double" came from.
> 
> You're right, I didn't know that.  How does one determine 
> those optimal sizes?

look at 
http://groups.google.com/groups?selm=383650AD.13EE%40viconet.com

i think usually the optimal N (which is a power of 2) is usually a bit
larger than twice L (the FIR length).

r b-j

jim <"N0sp"@m.sjedging@mwt.net> wrote in message news:<409f9634_5@corp.newsgroups.com>...
> robert bristow-johnson wrote:
> 
> > no.  to use the DFT in anything but the most trivial applications, you also
> > need to know how to either conceptually do modulo N arithmetic on the
> > indices of x[n] and X[k] or to conceptually periodically extend x[n] and
> > X[k] beyond the original limits of [0 .. N-1].  i contend these are the same
> > thing and that if you can't do them, you cannot make the DFT work in
> > applications requiring shifting (multiplying X[k] by exp(-j*2*pi*n0*k/N)) or
> > convolution (multiplying X[k[ by H[k]).
> 
> Nah, that's baloney. I've seen people implement overlap-add and overlap-save
> FFT filters without any real grasp of what's going on at all. What you
> conceptualize happening outside the interval has no effect on the way the
> computer crunches the numbers. I don't disagree that to periodically extend
> the data can be a useful way to view it, but would never be so arrogant as to
> claim that its the only way you can look at it. All you've offered in proof of
> your contention is that 'I can't do it so no one else can'. That's pretty
> lame.

i guess i must be pretty lame.  what proof would you like, Jim?  that
multiplying X[k] by exp(-j*2*pi*n0*k/N)) results in circular shifting
of x[n] by n0 (or OTOH linear shifting of the periodically extended
x[n])?  or that multiplying X[k] by H[k] results in circular
convolution of x[n] with h[n] (or, alternatively linear convolution of
the periodically extended x[n] with h[n] or vise versa)?  this is
pretty textbooky stuff and i didn't expect to have to explicitly prove
that. my only contention is that to neglect it "invites trouble" and
that it is conveptually a lot simpler (and closer to the truth) to
just always keep in mind that while *we* or *you* may not be thinking
of your data as periodically extended, the DFT is always thinking of
you data as periodic.  i've gone over why in pretty explicit detail. 
would you care to provide a counter example?

r b-j

robert bristow-johnson wrote:

> i guess i must be pretty lame.  what proof would you like, Jim?  that
> multiplying X[k] by exp(-j*2*pi*n0*k/N)) results in circular shifting
> of x[n] by n0 (or OTOH linear shifting of the periodically extended
> x[n])?  or that multiplying X[k] by H[k] results in circular
> convolution of x[n] with h[n] (or, alternatively linear convolution of
> the periodically extended x[n] with h[n] or vise versa)?  this is
> pretty textbooky stuff and i didn't expect to have to explicitly prove
> that.

More baloney. How does that prove your assertion that the DFT will not work
unless one shares your perception of the patterns that you see?


> my only contention is that to neglect it "invites trouble" 

That would be good if that really was your only contention.

>and
> that it is conveptually a lot simpler (and closer to the truth) to
> just always keep in mind that while *we* or *you* may not be thinking
> of your data as periodically extended, the DFT is always thinking of
> you data as periodic.  i've gone over why in pretty explicit detail.
> would you care to provide a counter example?

A counter example? You mean like an example of the DFT thinking something
else? No, I guess I don't.

-jim


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glen herrmannsfeldt <gah@ugcs.caltech.edu> wrote in message news:<H6Ujc.7709$lz5.854606@attbi_s53>...
> If the function is not periodic, then you should not use a DFT.

Is, or is not, a DFT a perfectly functional, orthonormal
linear-algebraic transform, taken in and of itself, on its own limited
length of basis vectors.

In short, does A At == I

If it does, a DFT is a perfectly functional orthonormal
linear-algebraic transform, and there's nothing at all wrong with
using it all by its lonesome on any data.

> If you don't like the assumption, don't use the DFT.

See above.

> There are other transforms (and other sets of basis functions)
> that don't make that assumption.

Ditto.

Bob Cain <arcane@arcanemethods.com> wrote in message news:<c6u4pc0cea@enews4.newsguy.com>...
> You once challenged me with the periodicity (or wrap) of 
> convolution by frequency domain multiplication and that 
> stumped me.  I now wonder if that isn't a property of the 
> convolution theorem 

'Tis

Randy Yates <randy.yates@sonyericsson.com> wrote in message news:<xxpu0z1mgfo.fsf@usrts005.corpusers.net>...
> The DFT, BY DEFINITION, produces a *FINITE* set of values as output.

On a finite set of input values. There is no need for these input
samples to have any assumptions whatsoever.

> These are, BY DEFINITION, frequency-domain output samples. It is
> an absolute IRREFUTABLE property that discrete-time samples in the
> frequency REQUIRE a periodic time-domain function. Period. End of
> argument.

Then you must argue that  IFFT (FFT (x) ) != x

Do you so argue?

An FFT or DFT, taken by itself, on its limited-time data, with no
other considerations, constitutes a fully functional, precisely
orthonormal linear-algebraic transform.

That's a fact.

It's one that ALSO can be explained by the properties of the integral
transform that originally created it, the period-interpretation
Fourier Integral.  One interpretation does not nullify the other.

an2or@mailcircuit.com (Andor) wrote in message news:<ce45f9ed.0404302355.3f7a72f7@posting.google.com>...
> 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


Thank you, thank you, thank you!

an2or@mailcircuit.com (Andor) wrote in message news:<ce45f9ed.0405020005.6e601124@posting.google.com>...
> It is you who is constraining himself. The DFT (as well as the FT) is
> a transform with certain properties. If you think of it only in
> time-frequency domain you are missing plenty of applications to areas
> where time and frequency are not applicable terms.

Agreed, by abandoning the "classical" interpretation, it is easy to
see how to formulate an MDCT, OBT, MLT, etc, in purely
linear-algebraic fashions, something that is quite difficult in the
original Fourier domain.
 
> It is very important to be able to map mathematical theorems into real
> world applications. But by keeping the abstract core of these theorems
> in the back of your head you may be able to find new and innovative
> solutions to new real world problems.

Hear, hear.

You may also discover that they are useful for, and provide solutions
to, things that you would never imagine had you not investigated the
abstract side of things.

Meta-logic: The means by which one can often avoid a great lot of
regular logic, especially when it shows that "you can't".

mark_ovchain@yahoo.com (Mark Ovchain) writes:

> Randy Yates <randy.yates@sonyericsson.com> wrote in message news:<xxpu0z1mgfo.fsf@usrts005.corpusers.net>...
>> The DFT, BY DEFINITION, produces a *FINITE* set of values as output.
>
> On a finite set of input values. There is no need for these input
> samples to have any assumptions whatsoever.

Proof by assertion?

>> These are, BY DEFINITION, frequency-domain output samples. It is
>> an absolute IRREFUTABLE property that discrete-time samples in the
>> frequency REQUIRE a periodic time-domain function. Period. End of
>> argument.
>
> Then you must argue that  IFFT (FFT (x) ) != x

No. I would agree that F^(-1)[FFT(x)] != x, however, where F^(-1) is
the inverse Fourier transform.

> Do you so argue?

I don't see your point. No, I don't argue any such thing. 

> An FFT or DFT, taken by itself, on its limited-time data, with no
> other considerations, constitutes a fully functional, precisely
> orthonormal linear-algebraic transform.

I use the FFT to glean frequency information from the signal, and I
bet you do too. That's a far cry from viewing it as some cold, meaningless
transform. This is the crux of the matter. I don't think anyone's arguing
its transform character.

> That's a fact.
>
> It's one that ALSO can be explained by the properties of the integral
> transform that originally created it, the period-interpretation
> Fourier Integral.  One interpretation does not nullify the other.

I already agreed with Bob that IF you don't interpret the FFT output
as frequency domain samples, then you can drop the implication that
the input is periodic. I never said dropping such an assumption made
any practical sense, however. 
-- 
%  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