Solved

Posted on 2004-11-06

Hey experts!

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)

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)

19 Comments

(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)

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.

no the convolution of x with itself won't be x.........

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!

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

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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