[MUG] convolution operator

[MUG] Re: convolution operator
Author: Maple User Group    Posted: Fri, 29 Nov 2002 09:11:32 -0500
>> From: Maple User Group "maple_gr" -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- From: Karshi "karshi.hasanov" To: "maple-list" Subject: convolution operator Date: Wed, 27 Nov 2002 09:09:36 -0500 My best recomendation is to write your own code in Matlab. The matlab has a "conv" function but it's very slow when you reach 30x30x30. You can look at the example in Matlab "Help" how to use the Convolution Theorem (1d only) , and than write your own code. My code has only few lines and works very fast even for 100x100x100 points. -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- From: Brad Camroux "prickles" To: "maple-list" Subject: convolution operator Date: Wed, 27 Nov 2002 12:52:04 -0700 Assuming you are working with two time series, say f(t) and g(t), then I would use the Fourier Transform to convert to the frequency domain, resulting in F(w) and G(w). Then convolution is simply multiplication in the frequency domain. Don't forget to back-transform (apply the FFT again) the convolved sequence to get back to the time domain and see what the convolved signal actually looks like. Hope this helps, Brad Camroux Geophysics Student, University of Calgary, Canada -- "Even the smallest person can change the course of the future." -- From the Lord of the Rings (The Fellowship of the Ring) -=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=-=*=- Date: Wed, 27 Nov 2002 22:14:23 +0100 (CET) From: Bertfried Fauser "fauser" To: "maple-list" Subject: convolution operator This depends rather strongly on the type of convolution you seek for. If you just mean convolution of two integrable functions you may define the convolution as a simple integral. If you are seeking for series, things start to get more complicated, since the definition of 'convolution' then depends strongly on the type of series you want to manipulate, e.g. formal polynomial rings, exponentially generated series or even Dirichlet series, which are the hardest here. A convolution of morphisms f : C -> A can be very generally defined on any pair (C,A) of a coalgebra C and an algebra A using the coproductc \Delta and product m as (f \star g)(x) = m 0 (f \otimes g) 0 \Delta(x) which is also an morphisms from C -> A. However, if you drop the argument you will see that this is an algebra itself under the convolution product. Such a structures lives in a 2-category. A convolution is hence not an operator in the normal sense, but a morphism on the monoid of C -> A morphisms under composition. It can be addressed as multiplication. * It seems to be to much for a CAS like maple to provide all these cases by 'a' convolution operator. * Unfortunately Maple is a CAS (computer _algebra_ systems) and does know very few things about coalgebras, coproducts etc. * If you are interested in convolutions in geometry you may check the web-site http:/math.tntech.edu/rafal/cliff8/ where the package BIGEBRA is located, which makes good use of convolution products, coalgebras etc. best BF. % Bertfried Fauser Fachbereich Physik Fach M 678 % Universit"at Konstanz 78457 Konstanz Germany % Phone : +49 7531 883786 FAX : +49 7531 88-4864 or 4266 % E-mail: "Bertfried.Fauser" % Web : http://clifford.physik.uni-konstanz.de/~fauser

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