Can anyone please help in finding the Fourier transform of the distance equation?
The distance equation is:
(Z2-Z1)2+(Y2-Y1)2+(X2-X1)2=D2
X1,Y1,Z1 are the coordinates of the first point.
X2,Y2,Z2 are the coordinates of the second point.
D is the distance.
Best,
The Fourier transform is an operator that takes in input a function f(x) and produces as output another function F(omega), provided that the input function f(x) is square-integrable.
Said this, how do you exactly define "the Fourier transform of an equation" in 6 variables? what problem are you trying to solve?
Dear Ma'mon,
As Giuseppe above pointed out, you can evaluate the Fourier transform of one-dimensional signals of time or two-dimensional signals of space (namely, images) in order to obtain their frequency spectra. You can do that for any signal which satisfies the so-called Dirichlet conditions. But, what you gave in your question is the distance formula between two points in the three-dimensional Euclidean space. Are you missing something?
I agree with Giuseppe P, it depent on variables. I know the Fourier transform F(k1,k2,k3) and it is complex, you shoud change variables to dX = X2-X1, dY = Y2-Y1 and dZ = Z2-Z1. If variables are part of a dynamic phenomenon, you can change by velocities and it will be dependant of time.
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