Given the mathematical expression for the discrete Gabor time-frequency atom $$g_{s,u,\omega,\theta}(n) = \frac{K_{s,u,\omega,\theta}}{\sqrt{s}} e^{-\pi(n-u)^2/s^2} \cos[2\pi\omega(n-u)+\theta]$$ where the variables are described in the original paper here (page 6 of the PDF), how do I solve for the normalization constant $K_{s,u,\omega,\theta}$? My attempt was to solve for the normalization constant using the equation $$\int_{-\infty}^{\infty}|g_{s,u,\omega,\theta}(n)|^2\mathrm dn = \int_{0}^{N}|g_{s,u,\omega,\theta}(n)|^2\mathrm dn = 1 $$ where $n$ is the time index and $N$ is the total number of samples in the atom, but I have not arrived at a solution. Any help would be great. Thanks!


1 Answer 1


The discrete squared L2 norm is a sum, not an integral:

$$ ||g||^2 = \sum_{n=0}^{N-1} |g(n)|^2 = 1 $$

To norm, we set $K = 1/||g||$ (rather than $1/||g||^2$).

I don't know if there's a closed form solution, as WolframAlpha can't even do $\sum e^{-n^2}$. The context usually implies translation invariance, so the result should be independent of $u$, but I can't tell for this paper. In practice we simply do

g /= sqrt(sum(abs(g)**2))
  • $\begingroup$ Thank you for your answer. I was given the task of checking whether it can be solved theoretically or not, but to no avail. I guess this solution would suffice. $\endgroup$
    – J. Herrera
    Mar 20, 2022 at 8:08

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