I want to calculate the summation over integers of the product of two shifted sinc functions:
$$ \sum_{n=-\infty}^{\infty} \operatorname{sinc}(n-a)\operatorname{sinc}(n-b) $$
where $a,b \in \mathbb{R}$ and $\operatorname{sinc}(x)$ is the normalized sinc function:
$$ \operatorname{sinc}(x) = \begin{cases} \frac{\operatorname{sin}(\pi x)}{\pi x} & \text{if } x \ne 0 \\ 1 & \text{otherwise}\end{cases}$$
I'm quite sure that the sum evaluates to $\operatorname{sinc}(b-a)$ (I checked this numerically), but I cannot figure out how to prove it.
The following are some approaches I tried:
- Writing the sum as $\sum_{n=-\infty}^{\infty} \operatorname{sinc}(n)\operatorname{sinc}(n-(b-a))$; only the term at $n=0$ is non-zero. But how can I show that both sums are equivalent?
- Writing the sum as an integral sampled at integers, i.e. $\int_{-\infty}^{\infty} \operatorname{sinc}(x-a)\operatorname{sinc}(x-b) \Delta(x) dx$, where $\Delta(x)=\sum_{n=-\infty}^{\infty}\delta(x-n)$ is the Dirac comb. Using the Fourier transform the integral can be written as $\int_{-\infty}^{\infty} \mathcal{F}^{-1}\{e^{-ia\omega}\operatorname{rect}(\frac{x}{2\pi}) * e^{-ib\omega}\operatorname{rect}(\frac{x}{2\pi}) * \mathcal{F}\{\Delta(x)\}\} dx$, where $*$ represents convolution. How do I evaluate the integral?
- It's easy to see that the integral version of the sum is $\operatorname{sinc}(b-a)$ (as shown here), but how do I relate the discrete sum to the integral? Can I use the Euler-MacLaurin formula for this, and if so, how? (I am not familiar with the topic.)
Best Answer
Let $\chi$ be the characteristic function on the unit interval $T=[-1/2,1/2]$, and take the Fourier transform to be the form $$F(f)(\xi)=\int_{\mathbb R}f(x)e^{-2\pi i\xi x}dx$$ then $$F(\chi)(\xi)=\mathrm{sinc}(\xi).$$ With the modulation to translation property of the Fourier transform you get $$F(e^{2\pi iax}\cdot\chi)(\xi)=\mathrm{sinc}(\xi-a),\quad F(e^{2\pi ibx}\cdot\chi)(\xi)=\mathrm{sinc}(\xi-b),$$ which means $\{\mathrm{sinc}(n-a)\}_{n\in\mathbb Z}$,$\{\mathrm{sinc}(n-b)\}_{n\in\mathbb Z}$ are Fourier coefficients of $e^{2\pi iax}\cdot\chi$ and $e^{2\pi ibx}\cdot\chi$ in their Fourier expansions respectively.
Now since the exponential basis $\{e^{2\pi inx}\}_{n\in\mathbb Z}$ is an ONB on $L^2(T)$, the sum in OP is an $\ell^2$ inner product with respect to this ONB, and with the Plancherel theorem it leads to $$\sum_{n\in\mathbb Z}\mathrm{sinc}(n-a)\mathrm{sinc}(n-b)=\int_Te^{2\pi i(a-b)x}dx=F(e^{2\pi i(a-b)x}\cdot\chi)(0)=\mathrm{sinc}(b-a)=\mathrm{sinc}(a-b)$$