Fourier transform of a supergausian

In summary, the Fourier transform of a super-Gaussian distribution is difficult to calculate and may not converge.
  • #1
modaniel
5
0
Hi,
I was wondering if anyone might know what the analytic Fourier transform of a Super-Gaussian is?
cheers
 
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  • #2
That depends on what you mean by "super-Gaussian" distribution. Do you mean fat-tailed distributions? (distributions with tails that fall off rather slowly compared to a Gaussian).

If so, then there are several kinds of fat tailed distributions, each with its own Fourier transform. The Fourier transform of the probability density function is just the characteristic function for the distribution, which are usually listed on the wikipedia page for the distribution of interest.
 
  • #3
thanks mute for the reply. The supergaussian i refer to is the Aexp(-([itex]\frac{x}{a}[/itex])[itex]^{2n}[/itex]) where n is positive an integer and is the order of the supergaussian.
 
  • #4
Hm, that looks tricky. Looking in the integral book Gradshteyn and Ryzhik, I found two useful integrals. The first is

$$\int_0^\infty dx~\exp(-x^\mu) = \frac{1}{\mu}\Gamma\left(\frac{1}{\mu}\right),$$
which holds for ##\mbox{Re}(\mu) > 0## and can be used to find the normalization constant of your distribution. This is integral 3.326-1 in the sixth edition.

There does not appear to be an integral for ##\exp(-x^\mu+ix)##, so it's possible there may not be a closed form for the Fourier transform. However, you could expand the imaginary exponential in a power series and perform the integral term-by-term to get a power series representation of the Fourier transform. In this case, the following integral (3.326-2) is useful:

$$\int_0^\infty dx~x^m \exp(-\beta x^n) = \frac{\Gamma(\gamma)}{n\beta^\gamma},$$
where ##\gamma = (m+1)/n##.
 
  • #5
I should note that a) the series won't actually be a power series (because the moments aren't powers of anything) and b) the series looks like it will be at best an asymptotic series, as ##\Gamma((m+1)/n)## will grow quite large as ##m## gets large, and so the sum won't actually converge.
 
  • #6
Hi Mute, thanks for the replies. looks like i might have to stick to numerical transforms!
 

Related to Fourier transform of a supergausian

1. What is a supergaussian function?

A supergaussian function is a mathematical function that decays faster than a Gaussian function. It is also known as a hypergaussian function and is commonly used in signal processing and image processing.

2. What is the Fourier transform of a supergaussian?

The Fourier transform of a supergaussian is a complex-valued function that represents the frequency components of the original supergaussian function. It is a useful tool for analyzing signals and understanding their frequency content.

3. How is the Fourier transform of a supergaussian calculated?

The Fourier transform of a supergaussian can be calculated using the Fourier transform formula, which involves integrating the product of the supergaussian function and the complex exponential function. Alternatively, it can also be calculated using fast Fourier transform algorithms for faster computation.

4. What are some applications of the Fourier transform of a supergaussian?

The Fourier transform of a supergaussian has numerous applications in signal and image processing, such as image denoising, feature extraction, and image enhancement. It is also used in optics, quantum mechanics, and other areas of physics and engineering.

5. Are there any limitations or drawbacks of using the Fourier transform of a supergaussian?

One limitation of using the Fourier transform of a supergaussian is that it is not defined for all supergaussian functions. Additionally, the Fourier transform does not preserve all information about the original function, so some information may be lost in the transformation process. Finally, the Fourier transform may be computationally expensive for large datasets, requiring advanced algorithms for efficient computation.

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