comb filters and fourier transforms for splitting sound into frequencies

Stan Pawlukiewicz wrote:

> Multirate filters generalize filters so that the number of input points 
> don't have to equal the number of output points.  

But, nonetheless, there is a rate involved.  In the same 
sense, what is the rate of a DFT?

> I don't think Bob asked any question.  

We agree on that much.  :-)


Bob
-- 

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

                                              A. Einstein

Bob Cain wrote:
> 
> 
> Stan Pawlukiewicz wrote:
> 
>> Multirate filters generalize filters so that the number of input 
>> points don't have to equal the number of output points.  
> 
> 
> But, nonetheless, there is a rate involved.  In the same sense, what is 
> the rate of a DFT?

Depends on how often you use it.  What's the rate on filtering a single 
block of data in memory?  Fundamentally this is semantics. Do you 
actually have a mathematical argument?  Can you find a book that 
disagrees with Strang's?
> 
>> I don't think Bob asked any question.  
> 
> 
> We agree on that much.  :-)
> 
> 
> Bob

Stan Pawlukiewicz wrote:

>> But, nonetheless, there is a rate involved.  In the same sense, what 
>> is the rate of a DFT?
> 
> 
> Depends on how often you use it.  

How about once?

> What's the rate on filtering a single 
> block of data in memory?  Fundamentally this is semantics. 

Not at all!  The most common use is the one-off because it 
contains the information relevant to many analyses.  Of what 
use is one cycle of what is usually called a filter?

While you can construct a filter bank of sorts from 
sequential DFTs if you are so inclined, the DFT operation 
itself is not a filter, it's a domain transformation.  Do 
filters transform domains?  None that I've seen do that; 
what comes in is time domain and what comes out is time domain.

> Do you 
> actually have a mathematical argument?  

For what?

> Can you find a book that 
> disagrees with Strang's?

Why?  I can think for myself.


Bob
-- 

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

                                              A. Einstein

Bob Cain wrote:
> 
> 
> Stan Pawlukiewicz wrote:
> 
>>> But, nonetheless, there is a rate involved.  In the same sense, what 
>>> is the rate of a DFT?
>>
>>
>>
>> Depends on how often you use it.  
> 
> 
> How about once?
> 
>> What's the rate on filtering a single block of data in memory?  
>> Fundamentally this is semantics. 
> 
> 
> Not at all!  The most common use is the one-off because it contains the 
> information relevant to many analyses.  Of what use is one cycle of what 
> is usually called a filter?
> 
> While you can construct a filter bank of sorts from sequential DFTs if 
> you are so inclined, the DFT operation itself is not a filter, it's a 
> domain transformation.  Do filters transform domains? 

A DFT is a matrix vector multiplication.  I can express a filter in 
precisely the same terms.


What's the distinction between an invertible mapping of a metric space 
and a transform?  Isn't a matched filter a projection of an input onto a 
particular waveform?
  None that I've
> seen do that; what comes in is time domain and what comes out is time 
> domain.

Perhaps you should equate yourself with Bacon's idols.

> 
>> Do you actually have a mathematical argument?  
> 
> 
> For what?

The distinction between a mapping and a transform,  What else?
> 
>> Can you find a book that disagrees with Strang's?
> 
> 
> Why?  I can think for myself.

So you disagree with Strang.
> 
> 
> Bob

Stan Pawlukiewicz wrote:

   ...

> A DFT is a matrix vector multiplication.  I can express a filter in 
> precisely the same terms.

And the motion of a planet in its orbit, and the polymerization of 
isoprene, and ...   Lot's of food for thought there!

Jerry
-- 
Engineering is the art of making what you want from things you can get.
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Stan Pawlukiewicz <spam@spam.mitre.org> writes:
> [...]
> A DFT is a matrix vector multiplication.  I can express a filter in
> precisely the same terms.

Stan, humor me for a moment. 

A vector in the sense you refer to regarding the DFT is finite. A
vector for a "signal" may be in-finite. How *do* you go from one to
the other without trying to consider things like operating on a subset
of the signal, end effects due to truncation of the signal, etc.,
which are all very pertinent to this discussion and not so necessarily
straight-forward to resolve?
-- 
Randy Yates
Sony Ericsson Mobile Communications
Research Triangle Park, NC, USA
randy.yates@sonyericsson.com, 919-472-1124

Randy Yates wrote:
> Stan Pawlukiewicz <spam@spam.mitre.org> writes:
> 
>>[...]
>>A DFT is a matrix vector multiplication.  I can express a filter in
>>precisely the same terms.
> 
> 
> Stan, humor me for a moment. 
> 
> A vector in the sense you refer to regarding the DFT is finite. A
> vector for a "signal" may be in-finite. How *do* you go from one to
> the other without trying to consider things like operating on a subset
> of the signal, end effects due to truncation of the signal, etc.,
> which are all very pertinent to this discussion and not so necessarily
> straight-forward to resolve?

An image is a finite object. Is there some reason that we can't filter 
one of these?

  Wrt to time, we can in theory have signals that extend to infinity in 
both directions but they are finite in practice.  If you do the 
decimation correctly you can neglect start up and end transients in a 
decimated filter bank. If I have finite sequence of time samples stored 
away some place in memory, there is nothing that says you have to 
process it sequentially, particularly for non-recursive filter. There is 
nothing that requires a filter the same number of outputs as inputs. It 
can be a single point, as an example, the classic textbook replica 
correlator.

There have been numerous posts in the past where people have taken the 
liberty of extending the DFT in the limit to infinite duration 
sequences.  I don't recall anyone saying, "Can't do that, its not a DFT 
then, has to be finite".  A matrix is an abstraction, there isn't 
anything that says you can't consider one, as its dimension approaches 
infinity.