| Re: Does CONVOL convolute [message #34145] |
Tue, 25 February 2003 09:21 |
thompson
Messages: 584
Registered: August 1991
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Senior Member |
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condor@biosys.net (Big Bird) writes:
> David Fanning <david@dfanning.com> wrote in message news:<MPG.18c41ee199cd0d0b989afb@news.frii.com>...
>> Setting CENTER=1 or /CENTER is the same as leaving the CENTER
>> keyword off. (Don't ask!)
> Oy, I'll ask anyways: Who's grand idea was that?
>> If you want to perform convolution in
>> the "strictly mathematical" sense, you must explicitly set CENTER=0.
>> Is this what you were after:
>>
>> IDL> print,convol(tt,k, center=0)
>> 0.000000 0.000000 0.000000 0.000000 0.000000
>> 0.000000 0.000000 0.000000 0.000000 0.000000
>> 0.000000 0.000000 1.00000 0.000000 0.000000
>> 0.000000 0.000000 0.000000 0.000000 0.000000
>> 0.000000 0.000000 0.000000 0.000000 0.000000
> Huh? No, that isn't what I was after either.
> Maybe I'm thinking something completely wrong here somewhere but if my
> array is
> 1 0 0
> 0 0 0 ...
> 0 0 0
> .
> .
> and my convolution kernel is
> a b c
> d e f
> g h i
> then I would expect the convolution to be
> e f 0
> h i 0 ...
> 0 0 0
> .
> .
> At least for a symmetric kernel (I'd have to think about an
> unsymmetric one). ...
Actually, for an asymmetric kernel, the answer should be
e d 0
b a 0
0 0 0
What I usually do in this sort of situation is to embed my array within a
bigger array, to get around the edge effects within these IDL routines. Thus,
I would make my array
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 1 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
of which the center 3x3 portion is the real array. The convolution with the
kernal would then be
0 0 0 0 0 0 0
0 i h g 0 0 0
0 f e d 0 0 0
0 c b a 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
And taking the center 3x3 portion gives the expected answer.
William Thompson
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| Re: Does CONVOL convolute [message #34157 is a reply to message #34145] |
Mon, 24 February 2003 17:26  |
Thomas Gutzler
Messages: 44
Registered: November 2002
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Member |
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Big Bird wrote:
>
> Maybe I'm thinking something completely wrong here somewhere but if my
> array is
>
> 1 0 0
> 0 0 0 ...
> 0 0 0
> .
> .
>
> and my convolution kernel is
>
> a b c
> d e f
> g h i
>
> then I would expect the convolution to be
>
> e f 0
> h i 0 ...
> 0 0 0
> .
> .
>
> At least for a symmetric kernel (I'd have to think about an
> unsymmetric one). I expect that because the fourier transform of a
> delta function is a constant (one) which is thus the multiplicative
> neutral element. And the fourier transform of a convolution of
> functions is the multiplication of the transforms of the functions
> themselves.
>
> The longer I stare at the help, the more confused I get trying to
> figure out whether the [0,0] element of the result should be zero, as
> it seems to have a condition attached that reads (at least on my
> screen with the font the help uses) "if t >= 1-1 and u >= 1-1" . And I
> really don't think either of these "ones" could be 'lower-case ell'
> even though they use an 'ell' in the sum but without apparent
> motivation (or at least they don't seem to say what 'ell' *is*).
>
> /edge_truncate goes in the vague direction
>
> /edge_truncate,center=0 gives me something I don't understand at all
>
> /edge_wrap does what it sounds like (which is not what I would
> consider useful
> for most mathematical applications)
>
> /edge_wrap,center=0 comes closest to what I would have expected
>
> This is all rather mysterious to me - is the term "convolution" used
> differently in engineering than in math? Clearly I have to think about
> this some more ...
Maybe a bound mirror expansion combined with shrinking helps ?
IDL> a = [[1,0,0,0],[0,0,0,0],[0,0,0,0],[0,0,0,0]]
IDL> b = [[10,11,12],[13,14,15],[16,17,18]]
IDL> print, boundmirrorshrink(convol(boundmirrorexpand(a),b))
14 13 0 0
11 10 0 0
0 0 0 0
0 0 0 0
not exactly what you want, but closer, isn't it ? :)
FUNCTION BoundMirrorExpand, A
; Expand the matrix using mirror boundary condition
;
; A = [ 1 2 3 11
; 4 5 6 12
; 7 8 9 13 ]
;
; B = BoundMirrorExpand(A) =
; [ 5 4 5 6 12 6
; 2 1 2 3 11 3
; 5 4 5 6 12 6
; 8 7 8 9 13 9
; 5 4 5 6 12 6 ]
m = (size(A))[1]
n = (size(A))[2]
B = [A, A[0,*], A[0,*]]
B = [[B], [B[*,0]], [B[*,0]]] ; B and A have same type
B(1:m,1:n) = A
B(0,0) = B(2,2) ; mirror corners
B(0,n+1) = B(2,n-1)
B(m+1,0) = B(m-1,2)
B(m+1,n+1) = B(m-1,n-1)
B(1:m,0) = B(1:m,2) ; mirror left and right boundary
B(1:m,n+1) = B(1:m,n-1)
B(0,1:n) = B(2,1:n) ; mirror top and bottom boundary
B(m+1,1:n) = B(m-1,1:n)
RETURN, B
END
FUNCTION BoundMirrorShrink, A
; Shrink the matrix to remove the padded mirror boundaries
; for example
;
; A = [ 5 4 5 6 12 6
; 2 1 2 3 11 3
; 5 4 5 6 12 6
; 8 7 8 9 13 9
; 5 4 5 6 12 6 ]
;
; B = BoundMirrorShrink(A) =
; [ 1 2 3 11
; 4 5 6 12
; 7 8 9 13 ]
m = (size(A))[1]
n = (size(A))[2]
B = A(1:m-2,1:n-2)
RETURN, B
END
just my 2 ce^H^Hfunctions,
Tom
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| Re: Does CONVOL convolute [message #34158 is a reply to message #34157] |
Mon, 24 February 2003 16:50 |
condor
Messages: 35
Registered: January 2002
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Member |
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David Fanning <david@dfanning.com> wrote in message news:<MPG.18c41ee199cd0d0b989afb@news.frii.com>...
> Setting CENTER=1 or /CENTER is the same as leaving the CENTER
> keyword off. (Don't ask!)
Oy, I'll ask anyways: Who's grand idea was that?
> If you want to perform convolution in
> the "strictly mathematical" sense, you must explicitly set CENTER=0.
> Is this what you were after:
>
> IDL> print,convol(tt,k, center=0)
> 0.000000 0.000000 0.000000 0.000000 0.000000
> 0.000000 0.000000 0.000000 0.000000 0.000000
> 0.000000 0.000000 1.00000 0.000000 0.000000
> 0.000000 0.000000 0.000000 0.000000 0.000000
> 0.000000 0.000000 0.000000 0.000000 0.000000
Huh? No, that isn't what I was after either.
Maybe I'm thinking something completely wrong here somewhere but if my
array is
1 0 0
0 0 0 ...
0 0 0
.
.
and my convolution kernel is
a b c
d e f
g h i
then I would expect the convolution to be
e f 0
h i 0 ...
0 0 0
.
.
At least for a symmetric kernel (I'd have to think about an
unsymmetric one). I expect that because the fourier transform of a
delta function is a constant (one) which is thus the multiplicative
neutral element. And the fourier transform of a convolution of
functions is the multiplication of the transforms of the functions
themselves.
The longer I stare at the help, the more confused I get trying to
figure out whether the [0,0] element of the result should be zero, as
it seems to have a condition attached that reads (at least on my
screen with the font the help uses) "if t >= 1-1 and u >= 1-1" . And I
really don't think either of these "ones" could be 'lower-case ell'
even though they use an 'ell' in the sum but without apparent
motivation (or at least they don't seem to say what 'ell' *is*).
/edge_truncate goes in the vague direction
/edge_truncate,center=0 gives me something I don't understand at all
/edge_wrap does what it sounds like (which is not what I would
consider useful
for most mathematical applications)
/edge_wrap,center=0 comes closest to what I would have expected
This is all rather mysterious to me - is the term "convolution" used
differently in engineering than in math? Clearly I have to think about
this some more ...
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| Re: Does CONVOL convolute [message #34160 is a reply to message #34158] |
Mon, 24 February 2003 11:22 |
David Fanning
Messages: 11724
Registered: August 2001
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Senior Member |
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Big Bird (condor@biosys.net) writes:
> But leaving the kernel alone, and just making the original array
> unsymmetric, I get something entirely unexpected:
>
> IDL> tt = fltarr(5,5)
> IDL> tt[0,0] = 1
> IDL> print,tt
> 1.00000 0.00000 0.00000 0.00000 0.00000
> 0.00000 0.00000 0.00000 0.00000 0.00000
> 0.00000 0.00000 0.00000 0.00000 0.00000
> 0.00000 0.00000 0.00000 0.00000 0.00000
> 0.00000 0.00000 0.00000 0.00000 0.00000
> IDL> print,convol(tt,k)
> 0.00000 0.00000 0.00000 0.00000 0.00000
> 0.00000 1.00000 0.00000 0.00000 0.00000
> 0.00000 0.00000 0.00000 0.00000 0.00000
> 0.00000 0.00000 0.00000 0.00000 0.00000
> 0.00000 0.00000 0.00000 0.00000 0.00000
>
> Huh?
>
> I get the same results with or without the /center keyword to convol.
Setting CENTER=1 or /CENTER is the same as leaving the CENTER
keyword off. (Don't ask!) If you want to perform convolution in
the "strictly mathematical" sense, you must explicitly set CENTER=0.
Is this what you were after:
IDL> print,convol(tt,k, center=0)
0.000000 0.000000 0.000000 0.000000 0.000000
0.000000 0.000000 0.000000 0.000000 0.000000
0.000000 0.000000 1.00000 0.000000 0.000000
0.000000 0.000000 0.000000 0.000000 0.000000
0.000000 0.000000 0.000000 0.000000 0.000000
Cheers,
David
--
David W. Fanning, Ph.D.
Fanning Software Consulting, Inc.
Phone: 970-221-0438, E-mail: david@dfanning.com
Coyote's Guide to IDL Programming: http://www.dfanning.com/
Toll-Free IDL Book Orders: 1-888-461-0155
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