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2D convolution

Posted on 2004-11-06
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x(n1, n2) = 1      for n1≥0, n2≥n1
       0             otherwise      


I want to find the colvolution of x with itself (x is a 2D signal)
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Question by:mte01
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by:mte01
ID: 12513166
"and ≥" would mean greater than or equal
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by:mte01
ID: 12513171
Here is the question written properly:

(n1, n2) = 1     for n1>=0, n2>=n1
      0             otherwise    


I want to find the colvolution of x with itself (x is a 2D signal)
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by:GwynforWeb
ID: 12513486
mte01,  
    This function is 1 in the upper right quadrant and zero else where, ie non-zero over an infnite  domain, the overlap region will always be infnite or empty so giving an infinite  convolution integral. Is there some other assumption that I am not aware of with signal convolutions?  I will look again later today, a bit busy now.
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by:GwynforWeb
ID: 12513539
I mean upper right octant (ie a triangular region )
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by:mte01
ID: 12513736
Yes the result of the convolution integral will be also infinite, but it can be defined in a summation form or something like that.....
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by:aburr
ID: 12514260
I think that if you want the convolution to be a series of terms, you must express the function itself in a series of terms
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by:aburr
ID: 12514272
I may not be visulizing this correctly, but in this case will not the self-convolution just be itself?
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by:mte01
ID: 12514286
you can say that x = sum(from n1=0 to inf) sum(from n2=n1 to inf) (1)

no the convolution of x with itself won't be x.........
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by:moorhouselondon
ID: 12514930
With convolution aren't you reversing one of the functions and then comparing the two?  If the two functions being compared are identical then you are comparing two identical signals, which is the auto-correlation of the two?  In the frequency domain you will get a spike at 0, followed by spikes at specific frequencies if there is any periodicity in the signal.  (It's years since I studied this so apologies for my naivety).
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by:mte01
ID: 12514955
Yes right, and what I want to know are these fequencies at which the "spikes" occur, that is, the values at which the convolution rseult will be 1 (in a summation form)
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by:moorhouselondon
ID: 12515079
I can't picture the shape you have there.  If it is a triangle then you will have exponentially decaying spikes, the first one at the origin which is of infinite size (because you've stepped from 0 to 1 infinitely quickly).  

You should be able to plug the figures into a Fourier Transform of your function to get the sizes of the rest of the spikes.  It will have e in it somewhere!
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by:moorhouselondon
ID: 12515093
...exponentially decaying positive and negative spikes....
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by:mte01
ID: 12515094
The Fourier Transform of this function does not exist, so we have to do a convolution in the classical addition-multiplication way, or we should try the 2D Z-transform of this function (I am right now thinking of doing that method)
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by:GwynforWeb
ID: 12515442
This is what I reckon    (using ** for convlolution operator)

                   1     for n1>=0, n2>=n1
x(n1, n2) =      
                   0     otherwise

The convoultion of x(n1, n2) with its self is
 
                x(n1, n2)**x(n1, n2) = sum(l=0,inf)sum(m=0,inf){ x(l,m)*x(l-n1,m-n2) }

The non-zero region of x(n1, n2) is the upper right hand triangle of the plane,

The non-zero region of the function x(-n1,-n2) which is the the reflection of this region about both axises (ie lower left central traingle).

For for n1>=0, n2>=n1 the sumation x(n1, n2)* x(n1, n2) is over the intersection the lower triangle shifted up to the point  (n1, n2).

This region is a parallelogram with its lower point at the origin and upper point at (n1,n2) of height n2-n1.

By my estimation the number of values in this region is n1*(n2-n1) each of value 1 so giving
           

                                        n1*(n2-n1)     for n1>=0, n2>=n1
   x(n1, n2)** x(n1, n2)  =      
                                             0               otherwise
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by:GwynforWeb
ID: 12515445
shld read the sumation x(n1, n2)**x(n1, n2)
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GwynforWeb earned 2000 total points
ID: 12515471
A couple of errors above

 Here are  errors corrected,


                   1     for n1>=0, n2>=n1
x(n1, n2) =      
                   0     otherwise

The convoultion of x(n1, n2) with its self is

   
                x(n1, n2)* x(n1, n2) = sum(l=0,inf)sum(m=0,inf){ x(l,m)x(l-n1,m-n2) }

The non-zero region of x(n1, n2) is the upper right hand triangle of the plane,

The non-zero region of the function x(-n1,-n2) which is the the reflection of this region about both axises (ie lower left central triangle).

For for n1>=0, n2>=n1 the sumation x(n1, n2)**x(n1, n2) is over the intersection the lower triangle shifted up to the point  (n1, n2).

This region is a parallelogram with its lower point at the origin and upper point at (n1,n2) of height n2-n1.

By my estimation the number of values in this region is (n1+1)(n2-n1+1) each of value 1 so giving

               


                                     (n1+1)(n2-n1+1)     for n1>=0, n2>=n1
   x(n1, n2)* x(n1, n2) =      
                                               0                 otherwise


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by:GwynforWeb
ID: 12515836
x(n1, n2)* x(n1, n2)    shld be    x(n1, n2)**x(n1, n2)  ie the convolution
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by:mte01
ID: 12516683
Amazing!!!!!......Excellant!!!!!!...this is exactly what I wanted.......
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by:GwynforWeb
ID: 12517685
Thanks for the points, it's a suprising result. For the archival purposes I have a further typo, ie shld read  x(n1, n2)**x(n1, n2) = sum(l=0,inf)sum(m=0,inf){ x(l,m)x(n1-l,n2-m),  the analysis is correct though and based on the correct formula.
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