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[MUG] Re: convolution operator
| [MUG] Re: convolution operator |
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Author: Maple User Group
Posted: Fri, 29 Nov 2002 09:11:32 -0500
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>> From: Maple User Group "maple_gr"
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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.
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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)
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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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