# 2D convolution

Hey experts!

x(n1, n2) = 1      for n1&#8805;0, n2&#8805;n1
0             otherwise

I want to find the colvolution of x with itself (x is a 2D signal)
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Author Commented:
"and &#8805;" would mean greater than or equal
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Author Commented:
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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Commented:
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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Commented:
I mean upper right octant (ie a triangular region )
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Author Commented:
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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Commented:
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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Commented:
I may not be visulizing this correctly, but in this case will not the self-convolution just be itself?
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Author Commented:
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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Commented:
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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Author Commented:
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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Commented:
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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Commented:
...exponentially decaying positive and negative spikes....
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Author Commented:
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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Commented:
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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Commented:
shld read the sumation x(n1, n2)**x(n1, n2)
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Commented:
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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Commented:
x(n1, n2)* x(n1, n2)    shld be    x(n1, n2)**x(n1, n2)  ie the convolution
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Author Commented:
Amazing!!!!!......Excellant!!!!!!...this is exactly what I wanted.......
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Commented:
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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