In the previous post, we looked at an elegant equation involving integrals of the sinc function and computed the corresponding integrals for the jinc function.

It turns out the analogous equation holds for sums as well:

As before, we’d like to compute these two sums and see whether we can compute the corresponding sums for the jinc function.

The Poisson summation formula says that a function and its Fourier transform produce the same sums over the integers:

Recall from the previous post that the Fourier transform of sinc is the function π box(π *x*) where the box function is 1 on [-1/2, 1/2] and zero elsewhere. The only integer *n* with π*n* inside [-1/2, 1/2] is 0, so the sum of sinc(*n*) over the integers equals π. A very similar argument shows that the sum of jinc(*n*) over the integers equals its Fourier transform at 0, which equals 2.

Let tri(*x*) be the triangle function, defined to be 1 – |*x*| for -1 < *x* < 1 and 0 otherwise. Then the Fourier transform of tri(*x*) is sinc^{2}(π ω) and so π tri(π *x*) and sinc^{2} are Fourier transform pairs. The Poisson summation formula says the sum of sinc^{2} over the integers is the sum of π tri(π *x*) over the integers, which is π.

I don’t know the Fourier transform of jinc^{2} and doubt it’s easy to compute. Maybe the sum could be computed more easily without Fourier transforms, e.g. using contour integration.

Haven’t worked this one out, but the jinc^2 sum, at least numerically, looks to be the same as the integral result, namely 8/(3*pi).

BTW, similarly to the integrals, the sum of sinc(n-a)*sinc(n-b) is over all integers n is pi * sinc(p-q).

The Fourier transform of jinc^2 is the self-convolution of the Fourier transform of jinc. The FT of jinc, per John’s previous post, is 2 sqrt(1-2pi|w|) except that the thing inside the sqrt is clipped at zero. Its support is therefore [-1/2pi,+1/2pi] and the support of its self-convolution is [-1/pi,+1/pi]. In particular, the only integer at which it’s non-zero is 0.

So, sum jinc^2(n) = sum F[jinc^2](n) = F[jinc^2](0) which we already established is equal to the corresponding integral, and we’re done.

In passing we’ve noticed a sufficient condition for the sum over integers to equal the integral over all x: the Fourier transform needs to be zero outside the open interval (-1,+1). In other words, you can integrate band-limited functions by summing them at regular intervals.

The Fourier transform on jinc^2 can be written using elliptic integrals.

(2/3)*(abs(phi)+2)*((abs(phi)^2+4)*E((-2+abs(phi))/(abs(phi)+2))-4*K((-2+abs(phi))/(abs(phi)+2))*abs(phi))

for -2 <= phi <= 2.

Found using Maple.

how will we compute the sum of two different sinc functions, lets say sinc(2.pi.B.t)+sinc(2.pi.B.t-pi) ??