Orthogonality of Periodic Sinc function

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Background

The Weyl-Heisenberg set $\mathcal{S}$ of sinc functions forms an orthonormal basis for $L_2(R)$, where we have

$$\mathcal{S}=\{s_{km}(t)=\frac{sin(\pi (t-m)}{\pi (t-m)}\exp(j2\pi kt)|m,k\in\mathbb{Z}\}$$

i.e. $\mathcal{S}$ contains time- and frequency shifted versions of the sinc. For example, two elements for different $k$ are orthogonal to each other:

$$ \left<s_{k,m},s_{k+1,m}\right>=\left<\mathcal{F}\{s_{k,m}(t)\},\mathcal{F}\{s_{k+1,m}(t)\}\right>=0 $$

where $\mathcal{F}$ is the Fourier Transform. We can directly see this from the fact that $\mathcal{F}\{s_{km}(t)\}(f)$ is a Rectangular function in $f$ of width $1$. Then, when multiplying to different frequency shifts of the $s_{k,m}(t)$ function, the rects do not overlap (or only overlap at a set of Lebesgue measure 0 (i.e. at the discontinuity of the rects), and hence the scalar product is 0.

Problem

Consider the periodic sinc $\tilde{s}_{k,m}(t)=\sum_{l\in\mathbb{Z}}s_{k,m}(t-lM)$. The WH-Set of periodic sincs forms an orthogonal basis for $L_2([0,M])$. Since it's periodic, it's Fourier Transform is discrete (i.e. consists of Dirac impulses).

  • What happens, if the Dirac impulse occurs at the discontinuity of the rectangular spectrum?
    • We can define the value at the discontinuity to be 1/2. But then, it seems the orthogonality between two frequency-shifted functions is lost, because the Diracs at the discontinuity overlap. And the integral over the frequency domain would not be zero.
    • We can define the rectangular function to be 1 at one side of the rectangle, but 0 at the other side. Then, orthogonality is given, but the corresponding time-domain function would be complex (not symmetric in frequency domain).

How can this paradox be resolved? I have the feeling that it relates to the problem that the periodization actually does not converge in the common sense, but I'm not enough into these delicate mathematical problems.