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$$ \begin{align} & \mathrm{sinc}(As + .5)\sum_{n=-\infty}^{\infty} \delta (s - n/A)\ \star \\ & \mathrm{sinc}(Bs + .5)\sum_{n=-\infty}^{\infty} \delta (s - n/B) \end{align} $$

How to compute?

Attempt so far

Convolution theorem. Take in parts

$$ \mathcal{F}\{\mathrm{sinc}(As + .5)\} = \frac{\pi}{|A|} e^{j(.5/A) \omega} \Pi\left(\frac{\omega}{2A}\right) $$ $$ \mathcal{F}\left\{\sum_{n=-\infty}^{\infty} \delta (s - n/A)\right\} = 2\pi A\sum_{k=-\infty}^{\infty} \delta \left( \omega - 2\pi A \cdot k \right) $$

so we need $\mathcal{F}^{-1}$ of (disregard constants)

$$ \begin{align} & e^{j(.5/A) \omega} \Pi\left(\frac{\omega}{2A}\right) \sum_{k=-\infty}^{\infty} \delta \left( \omega - 2\pi A \cdot k \right) \star \\ & e^{j(.5/B) \omega} \Pi\left(\frac{\omega}{2B}\right) \sum_{k=-\infty}^{\infty} \delta \left( \omega - 2\pi B \cdot k \right) \end{align} $$

which is

$$ \begin{align} & \sum_{k=-\infty}^{\infty} e^{j(.5/A)(\omega - 2\pi A \cdot k)} \Pi\left(\frac{\omega - 2\pi A \cdot k}{2A}\right) \star \\ & \sum_{k=-\infty}^{\infty} e^{j(.5/B)(\omega - 2\pi B \cdot k)} \Pi\left(\frac{\omega - 2\pi B \cdot k}{2B}\right) \end{align} $$

Does Fubini apply here? If no, then we have $(a + b) \star (c + d) = \text{distribute}$. For just boxcars, $\Pi_A \star \Pi_B$, we have a bunch of signums. However, if we include the $e^j$... for both, I don't know, but just for one

$$ e^{j(1/A)\omega} \Pi(\omega/A) \star \Pi(\omega / B), $$

WA gives

Didn't get farther. Overcomplicating?

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    $\begingroup$ Doesn’t the multiplication of the sinc by the impulse train in the $s$ domain lead to their convolution in the Fourier domain? So the “by parts” won’t work that way. $\endgroup$
    – Peter K.
    Jun 6, 2022 at 1:03
  • $\begingroup$ @PeterK. woops... thanks. $\endgroup$ Jun 6, 2022 at 1:16

1 Answer 1

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The question title asks about convolution of sinc (pulse?) trains, but the written expression is about the convolution of sampled sinc functions. I'll only address the written expression here.

But first some ...

Preliminaries

Some definitions that I'll use later on.

$$\mathscr{F}\left\{f(x)\right\} = F(s) = \int_{-\infty}^\infty f(x) e^{-2\pi i xs} dx$$

$$\mathrm{sinc}(x) = \dfrac{\sin(\pi x)}{\pi x}$$

$$\Pi\left(x\right) = \begin{cases} 1 & |x| < \frac{1}{2} \\ 0 & |x| > \frac{1}{2} \\ \end{cases}$$

$$\mathrm{III}(ax) = \dfrac{1}{|a|}\sum_{n=-\infty}^{\infty} \delta\left(x -\dfrac{n}{a}\right)$$

Convolution of two sinc() functions

As a baseline it is helpful to know that the convolution of two sinc() functions is just the wider of the two sinc() functions, with a vertical scaling and a different shift caused by the narrower of the two sinc() functions. This is easy to prove with Fourier Transforms.

Staring with

$$\mathrm{sinc}(Ax - \alpha) * \mathrm{sinc}(Bx - \beta) = ? , \quad A > B > 0$$

and taking the Fourier Transform $$\begin{align*}\mathscr{F}\left\{\mathrm{sinc}(Ax - \alpha) * \mathrm{sinc}(Bx - \beta)\right\} &= \mathscr{F}\left\{\mathrm{sinc}\left(A\left[x - \dfrac{\alpha}{A}\right]\right)\right\}\mathscr{F}\left\{\mathrm{sinc}\left(B\left[x - \dfrac{\beta}{B}\right]\right)\right\}\\ \\ &= \dfrac{e^{-2\pi i \frac{\alpha}{A}s}}{|A|}\Pi\left(\dfrac{s}{A}\right)\dfrac{e^{-2\pi i \frac{\beta}{B}s}}{|B|}\Pi\left(\dfrac{s}{B}\right)\\ \\ &= \dfrac{e^{-2\pi i \left(\frac{\alpha}{A}+\frac{\beta}{B}\right)s}}{|A|}\dfrac{1}{|B|}\Pi\left(\dfrac{s}{B}\right)\\ \end{align*}$$

Thus

$$\begin{align*}\mathrm{sinc}(Ax - \alpha) * \mathrm{sinc}(Bx - \beta) &= \mathscr{F}^{-1}\left\{ \dfrac{e^{-2\pi i \left(\frac{\alpha}{A}+\frac{\beta}{B}\right)s}}{|A|}\dfrac{1}{|B|}\Pi\left(\dfrac{s}{B}\right)\right\}\\ \\ &= \dfrac{1}{|A|}\mathrm{sinc}\left(B\left[x - \left(\dfrac{\alpha}{A}+\dfrac{\beta}{B}\right)\right]\right) \\ \\ &= \dfrac{1}{|A|}\mathrm{sinc}\left(Bx - \beta - \dfrac{B}{A}\alpha \right) \\ \end{align*}$$

Convolution of two sampled sinc() functions

The convolution of two sampled sinc() functions, both sampled at the same rate, would yield a result that was a sampled version of what was just proved above: a sampled version of the wider of the two sinc() functions with a vertical scaling and additional shift caused by the narrower of the two sinc() functions.

When sampled at a different rate, the results aren't so nice. Unless one sample period is a multiple of the other, the periodic trains of rectangle functions in the frequency domani don't regularly overlap fully or often, leaving a much sparser spectrum. So to derive this result, I won't go to the frequency domain.

Again assuming $A >B >0$, and also $a > b > 0$

$$\begin{align*}& \mathrm{sinc}(Ax - \alpha)|a|\mathrm{III}(ax) * \mathrm{sinc}(Bx - \beta)|b|\mathrm{III}(bx) \\ \\ &= \int_{-\infty}^\infty \mathrm{sinc}(A\tau - \alpha)|a|\mathrm{III}(a\tau) \mathrm{sinc}(B[x-\tau] - \beta)|b|\mathrm{III}(b[x-\tau]) d\tau\\ \\ &= \int_{-\infty}^\infty \mathrm{sinc}(A\tau - \alpha)\sum_{n=-\infty}^\infty \delta\left(\tau-\dfrac{n}{a}\right) \mathrm{sinc}(B[x-\tau] - \beta)\sum_{m=-\infty}^\infty \delta\left(x-\tau-\dfrac{m}{b}\right) d\tau\\ \\ &= \sum_{n=-\infty}^\infty \mathrm{sinc}\left(A\dfrac{n}{a} - \alpha\right)\mathrm{sinc}\left(B\left[x-\dfrac{n}{a}\right] - \beta\right)\sum_{m=-\infty}^\infty \delta\left(x-\dfrac{n}{a}-\dfrac{m}{b}\right)\\ \\ &= \sum_{n=-\infty}^\infty \mathrm{sinc}\left(A\dfrac{n}{a} - \alpha\right)\sum_{m=-\infty}^\infty \mathrm{sinc}\left(B\dfrac{m}{b}- \beta\right)\delta\left(x-\dfrac{n}{a}-\dfrac{m}{b}\right)\\ \\ &= \sum_{n=-\infty}^\infty \sum_{m=-\infty}^\infty \mathrm{sinc}\left(A\dfrac{n}{a} - \alpha\right)\mathrm{sinc}\left(B\dfrac{m}{b}- \beta\right)\delta\left(x-\dfrac{n}{a}-\dfrac{m}{b}\right)\\\end{align*}$$

And that's about as simple and symmetric as it gets: a weird summation of samples of the sinc() functions, with the resultant samples only at points $x = \dfrac{n}{a} + \dfrac{m}{b}$.

This does collapse down into something simpler for the case of $a=b$, which corresponds to the the original sampled sinc() functions both being sampled at the same rate.

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