# Is the discrete Gaussian kernel an eigenfunction of the DFT?

So the Gaussian function is an eigenfunction of the Fourier transform because it transforms into itself, right?

But this isn't true for the sampled Gaussian in the DFT because the tails of the function are truncated, right?

Wikipedia describes a discrete Gaussian kernel here and here, which is different from the discretely-sampled Gaussian:

the discrete counterpart of the continuous Gaussian in that it is the solution to the discrete diffusion equation (discrete space, continuous time), just as the continuous Gaussian is the solution to the continuous diffusion equation

Does that mean it also DFT transforms exactly into itself? If not, is there a similar Gaussian-like function that does?

Since the DFT is representable by multiplication with the Fourier matrix, your question is equivalent to asking what the eigenvectors of the Fourier matrix are.

Actually, Wikipedia provides the answer (http://en.wikipedia.org/wiki/Discrete_Fourier_transform#Eigenvalues_and_eigenvectors).

However, since the eigenvalues ($1, -1, i, -i$) are not simple, the eigenvectors are not unique (i.e. linear combinations are also eigenvectors). Also no simple closed formula exists.

One formula for an eigenvector close to what you ask is provided by Wikipedia

$$F_m = \sum_{k = -\infty}^\infty \exp\left(-\frac{\pi\cdot(m+N\cdot k)^2}{N}\right) \quad\quad\quad m = 0,\ldots,N-1$$

So concluding, the Gaussian function itself is not an eigenvector, but an infinite sum of Gaussians. The infinite sum can probably be interpreted as equivalent to the discretization of the frequency and time domain when we going from the FT to the DFT. So it is not as simple, as just truncating the discrete Gaussian.

• Isnt an infinite sum of gaussians still a gaussian though? – TheGrapeBeyond Aug 22 '13 at 15:43
• No, convolution of Gaussians is still a Gaussian. The sum is Gaussian only if they are the same position and width. This function here is actally one period of a discrete Gaussian pulse train. So it doesnt even look like a Gaussian. – Andreas H. Aug 22 '13 at 15:55
• Ah, I see. In other words this sum is essentially a Gaussian train made up of Gaussians of same variance but different means? – TheGrapeBeyond Aug 22 '13 at 15:59
• exactly. The means are spaced by exactly N, the length of the DFT. – Andreas H. Aug 22 '13 at 16:01
• Ah, fascinating. One last thing, this is an infinite length vector, which means that the DFT matrix is also infinite in length, no? – TheGrapeBeyond Aug 22 '13 at 16:03