Dear Manoj,  Let's consider a special Gaussian function expressed by f(x)=Exp[-Pi x], of which infinite integral NIntegrate[Exp[-Pi x^2],{x,-Infinity,Infinity}]=1.  Next, we consider the effective width of the function, which is defined by the area (infinite integral) of the function divided by its central ordinate f(0).  Known they are equally 1, the effective width should be 1.   In general, the effective width of a function is equal to the reciprocal of the effective width of its (Fourier) transfom.  Therefore, the effective width of the Fourier transfom should be 1, which should have function form of Exp[-Pi f^2], in general.  You can easily generalize this result.     Regards, Shigeo.

The result itself is not from duality, but a calculation of the integral.

It is also not a general case that 'the' Gaussian equals 'the' Gaussian after the Fourier transform. More precisely, a Gaussian function will give you another Gaussian function after the Fourier transform, yet their distribution can and will in most cases vary.

The example supplied by the first answer is essentially a unique case for a specific sigma of the Gaussian.

In general, if you have a Gaussian function x(t) with a broad distribution, i.e. high sigma, your Fourier transform will be a Gaussian function with a small sigma and vice versa.

Linear transformation does not change the distribution. But what do you mean? Fourier transformation is the amplitude or squares? The amplitudes of a Gaussian distribution. Squares is a different division. This chi-squared distribution.

Hi,

Omkar gave the answer.

Best of luck.