I see, most people want *more* parallelisation in their algorithms, not less ;)

Perhaps you could make a 'guess' at the first filter and then solve the problem 
of finding a second filter which gives you the desired result. Since:

f1*f2=result

...we know f1 & f2 and so can find 'result', and...

f1*fa*fb=result

so by guessing at fa (for example, just taking the first n coefficients) the 
problem is simplified to finding an fb for which the above equation is true.

Since we know that convolution is multiplication in the frequency domain, and 
assuming we were using fast fourier transforms, the coefficients would be 
calculated by:

IFFT( FFT('result') / FFT( f1*fa) )

I.e. what would we multiply f1*fa by in the frequency domain (aka convolve with 
it) in order to achieve 'result'.

Feel free to point out any big holes in my logic :I

Tom

-----Original Message-----
From: music-dsp-boun...@music.columbia.edu 
[mailto:music-dsp-boun...@music.columbia.edu] On Behalf Of Uli Brueggemann
Sent: 19 January 2011 16:30
To: A discussion list for music-related DSP
Subject: Re: [music-dsp] "Factorization" of filter kernels

Thomas,

I suppose that a decomposition of a n-taps kernel into n zero-padded
kernels would directly lead to the basics of the convolution algorithm
:-)
But your proposal also introduces a parallel computation, where the
results have to be offset and added (incl. overlap treatment).
My question is aiming a serial computation, like f1*f2 = f1*fa*fb with
f2=fa*fb. Again: f1 is given and fa, fb are searched.

Greetings, Uli

On Wed, Jan 19, 2011 at 5:07 PM, Thomas Young
<thomas.yo...@rebellion.co.uk> wrote:
> Hi Uli
>
> I don't know if this will be useful for your situation, but a simple method 
> for decomposing your kernel is to simply chop it in two. So for a kernel:
>
> 1 2 3 4 5 6 7 8
>
> You can decompose it into two zero padded kernels:
>
> 1 2 3 4 0 0 0 0
>
> 0 0 0 0 5 6 7 8
>
> And sum the results of convolving both of these kernels with your signal to 
> achieve the same effect as convolving with the original kernel. You can do 
> this because convolution is distributive over addition, i.e.
>
> f1*(f2+f3) = f1*f2 + f1*f3
>
> For signals f1,f2 & f3 (* meaning convolve rather than multiply).
>
> Obviously all those zero's do not need to be evaluated, meaning the problem 
> is changed to one of offsetting your convolution algorithm (which may or may 
> not be practical in your situation), but does allow you to use half the 
> number of coefficients.
>
> Thomas Young
>
> Core Technology Programmer
> Rebellion Developments LTD
>
> -----Original Message-----
> From: music-dsp-boun...@music.columbia.edu 
> [mailto:music-dsp-boun...@music.columbia.edu] On Behalf Of Uli Brueggemann
> Sent: 19 January 2011 14:56
> To: A discussion list for music-related DSP
> Subject: Re: [music-dsp] "Factorization" of filter kernels
>
> Hi,
>
> thanks for the answer so far.
> A polyphase filter is a nice idea but it does not answer the problem.
> The signal has to be demultiplexed (decimated), the different streams
> have to be filtered, the results must be added to get the final output
> signal.
>
> My question has a different target.
> Imagine you have two system (e.g. some convolution  boards with DSP).
> Each system can just run a 512 tap filter. Now I like to connect the
> two systems in series to mimic a desired 1024 tap filter. The 1024
> kernel is known and shall be generated by the two 512 tap filters.
> So what's a best way to decompose the known kernel into two parts ? Is
> there any method described somewhere?
>
> Uli
>
>
> 2011/1/19 João Felipe Santos <joao....@gmail.com>:
>> Hello,
>>
>> a technique that allows something similar to what you are suggesting
>> is to use polyphase filters. The difference is that you will not
>> process contiguous vectors, but (for a 2-phase decomposition example)
>> process the even samples with one stage of the filter and the odd
>> samples with another stage. It is generally used for multirate filter
>> design, but it makes sense to use this kind of decomposition if you
>> can process the stages in parallel... or at least it is what I think
>> makes sense.
>>
>> You can search for references to this technique here [1] and here [2].
>> A full section on how to perform the decomposition is presented on
>> "Digital Signal Processing: a Computer-based approach" by Sanjit K.
>> Mitra.
>>
>> [1] 
>> http://www.ws.binghamton.edu/fowler/fowler%20personal%20page/EE521_files/IV-05%20Polyphase%20FIlters%20Revised.pdf
>> [2] https://ccrma.stanford.edu/~jos/sasp/Multirate_Filter_Banks.html
>>
>> --
>> João Felipe Santos
>>
>>
>>
>> On Tue, Jan 18, 2011 at 5:46 AM, Uli Brueggemann
>> <uli.brueggem...@gmail.com> wrote:
>>> Hi,
>>>
>>> a convolution of two vectors with length size n and m gives a result
>>> of length n+m-1.
>>> So e.g. two vectors of length 512 with result in a vector of length 1023.
>>>
>>> Now let's assume we have a vector (or signal or filter kernel) of size
>>> 1024, the last taps is 0.
>>> How to decompose it to two vectors of half length? The smaller vectors
>>> can be of any arbitrary contents but their convolution must result
>>> must be equal to the original vector.
>>>
>>> It would be even interesting to "factorize"  given kernel into n
>>> smaller kernels. Again the smaller kernels may have any arbitrary but
>>> senseful contents, they can be identical but this is not a must.
>>>
>>> Is there a good method to carry out the kernel decomposition? (e.g.
>>> like calculating n identical factors x of a number y by x =
>>> Exp(Log(y)/n) with x^n = x*x*...*x = y)
>>>
>>> Uli
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