signals which have themselves as fourier transform

[smslca]

We know that the Fourier transform of a Gaussian function is Gaussian function itself. We can also say that periodic impulse train also satisfies this property. Can anyone give one or more signals which have themselves as FT?

 

[miketest123]

Let \[x(t) \] be even and let \[x(t)\Longrightarrow X(f)\], then duality says \[X(t)\Longrightarrow x(f) \] Thus \[x(t)+X(f) \Longrightarrow X(f) + x(f) \]. Thus we have infinite of them.

miketest123

 

[pangpangfish]

You probably mean to type X(t).

Let's an interesting observation. Have you constructed several such functions yet?

Let [tex:804eaf0979]x(t) [/tex:804eaf0979] be even and let [tex:804eaf0979]x(t)\Longrightarrow X(f)[/tex:804eaf0979], then duality says [tex:804eaf0979]X(t)\Longrightarrow x(f) [/tex:804eaf0979] Thus [tex:804eaf0979]x(t)+X(f) \Longrightarrow X(f) + x(f) [/tex:804eaf0979]. Thus we have infinite of them.

miketest123

 

[permute]

in general, don't you need to conjugate X?

eg, if x(t) = e^-jwt, then the forward transform using x(t) e^-jwt in the integral will give one value. The inverse transform uses e^jwt in the integral to go from X back to x.

I'd have to look at this more for the more general complex case.

 

[htg]

Hermite functions are eigenfunctions of the Fourier transform, which means that up to a constant factor (in this case it can be +i, -i, +1 or -1) they are equal to their Fourier transform. If you want the constant factor to be equal to +1, you have to choose every fourth of the Hermite functions. The zeroth Hermite function is the Gaussian function, others are products of polynomials time Gaussian.