The Poisson Summation Formula, the Sampling Theorem, and Dirac Combs

The Poisson summation formula, the sampling theorem, and Dirac combs

Jordan Bell [email protected] Department of Mathematics, University of Toronto April 3, 2014

1 Poisson summation formula

Let S(R) be the set of all inﬁnitely diﬀerentiable functions f on R such that for all nonnegative integers m and n,

m (n) νm,n(f) = sup |x| |f (x)| x∈R is finite. For each m and n, νm,n is a seminorm. This is a countable collection of seminorms, and S(R) is a Fréchet space. (One has to prove for vk ∈ S(R) that if for each fixed m, n the sequence νm,n(fk) is a Cauchy sequence (of numbers), then there exists some v ∈ S(R) such that for each m, n we have νm,n(f −fk) → 0.) I mention that S(R) is a Fréchet space merely because it is satisfying to give a structure to a set with which we are working: if I call this set Schwartz space I would like to know what type of space it is, in the same sense that an Lp space is a Banach space. For f ∈ S(R), define Z fˆ(ξ) = e−iξxf(x)dx, R and then 1 Z f(x) = eixξfˆ(ξ)dξ. 2π R Let T = R/2πZ. For φ ∈ C∞(T), define

1 Z 2π φˆ(n) = e−intφ(t)dt, 2π 0 and then X φ(t) = φˆ(n)eint. n∈Z

1 For f ∈ S(R) and λ 6= 0, deﬁne φλ : T → C by X φλ(t) = 2π fλ(t + 2πj), j∈Z

∞ where fλ(x) = λf(λx). We have φλ ∈ C (T). For n ∈ Z, Z 2π 1 −int X φcλ(n) = e · 2π fλ(t + 2πj)dt 2π 0 j∈Z Z 2π X −int = e fλ(t + 2πj)dt 0 j∈Z Z −inx = e fλ(x)dx R = fcλ(n).

ˆ ξ ˆ n One checks that fcλ(ξ) = f λ , so φcλ(n) = f λ . Thus for t ∈ T,

X n φ (t) = fˆ eint. λ λ n∈Z

Therefore, if f ∈ S(R) then for each t ∈ T, X X n 2π f (t + 2πj) = fˆ eint, λ λ j∈Z n∈Z where fλ(x) = λf(λx). This is the Poisson summation formula.

2 Nyquist-Shannon sampling theorem

Let f ∈ S(R). Suppose that there is some L such that f(x) = 0 if |x| > L (namely, f is spacelimited). We can in fact choose λ so that all terms with j 6= 0 on the left-hand side of the Poisson summation formula are zero, which then expresses fλ(t) in terms of discrete samples of the Fourier transform of f. If there is an L such that fˆ(ξ) = 0 for |ξ| > L, we say that f is bandlimited. (For f : T → C this correspond to f being a trigonometric polynomial.) We can prove a dual Poisson formula and apply it to bandlimited functions, but instead I am going to give a direct argument following Charles L. Epstein, Introduction to the Mathematics of Medical Imaging, second ed., which is a good reference for sampling. Suppose that f ∈ S(R) and f(ξ) = 0 for |ξ| > L.1 For −π ≤ ξ ≤ π, deﬁne

1 1 In fact, if f, fˆ ∈ L (R) and f is bandlimited then it follows that f is inﬁnitely diﬀerentiable, although it does not follow that f is Schwartz.

2 F : T → C by2 ξL F (ξ) = fˆ . π For n ∈ Z, 1 Z π Fb(n) = e−inξF (ξ)dξ 2π −π 1 Z π ξL = e−inξfˆ dξ 2π −π π Z L 1 −in πt = e L fˆ(t)dt. 2L −L

On the other hand, for x ∈ R, using the fact that f is bandlimited, 1 Z 1 Z L f(x) = eixξfˆ(ξ)dξ = eixξfˆ(ξ)dξ. 2π 2π R −L Therefore for n ∈ Z, π −nπ Fb(n) = f . L L As F ∈ C∞(T), for t ∈ T we have

X X π −nπ F (t) = Fb(n)eint = f eint. L L n∈Z n∈Z Hence for −L ≤ ξ ≤ L, πξ X π −nπ inπξ fˆ(ξ) = F = f e L . L L L n∈Z

For 0 ≤ η < L, let φ : R → R satisfy ( 1 |ξ| ≤ L − η φˆ(ξ) = 0 |ξ| > L.

For instance, for η = 0 let φˆ be the characteristic function on [−L, L]. But if η > 0 we can choose ηˆ to be smooth rather than a characteristic function. If fˆ(ξ) = 0 for |ξ| > L − η (rather than just for |ξ| > L), then fˆ = fˆφˆ, because φ

2We are deﬁning a periodic function F from a function fˆ that is not periodic, but rather which becomes 0 after L.

3 is equal to 1 on the support of fˆ. Then for x ∈ R, 1 Z f(x) = eixξfˆ(xi)φˆ(ξ)dξ 2π R Z 1 ixξ X π −nπ inπξ = e f e L φˆ(ξ)dξ 2π L L R n∈Z Z 1 X −nπ ixξ+ inπξ = f e L φˆ(ξ)dξ 2L L n∈Z R 1 X −nπ nπ = f · 2π · φ x + 2L L L n∈Z π X −nπ nπ = f φ x + . L L L n∈Z ˆ nπ φ can be chosen so that φ x + L quickly goes to 0 as n → ∞. This means that fewer terms of the above series need to be calculated to get a good approximation of f(x). For φˆ the characteristic function on [−L, L], we have 1 Z L sin(xL) L φ(x) = eixξdξ = = sinc(xL), 2π −L πx π sin x where sinc(x) = x . Then we obtain, for x ∈ R, X −nπ nπ f(x) = f sinc x + , L L n∈Z or X nπ nπ f(x) = f sinc x − . L L n∈Z Thus if f is bandlimited, we can sample f at a sequence of points and using these values reconstruct f(x) for any x ∈ R. This fact is the Nyquist-Shannon sampling theorem, and the actual formula above is the Whittaker-Shannon in- terpolation formula. A pleasant exposition of sampling is given in Sampling: What Nyquist Did- nÕt Say, and What to Do About It, by Tim Wescott, 2010.

3 Dirac combs, pulse trains

Deﬁne a tempered distribution X on R by X X(t) = δ(t − n). n∈Z This is called a Dirac comb or pulse train. If f ∈ S(R), X Xf = f(n). n∈Z

4 The convolution of f and X is the function X (X ∗ f)(t) = f(t − n). n∈Z Define 1 t X (t) = X . T T T Because δ(t/a) = |a|δ(t) (the scaling property of the Dirac delta distribution; we have an absolute value sign because if a is negative then doing a change of variables in the definition of δ(t/a) we get two negative signs), if T > 0 then X 1 t X X (t) = · δ − n = δ(t − nT ), T T T n∈Z n∈Z and we have X XT f = f(nT ) n∈Z and X (XT ∗ f)(t) = f(t − nT ). n∈Z XT (t + T ) = XT (t), so XT is a periodic distribution on R. It follows that XT is a distribution on T. Folland talks about periodic distributions around p. 298 of his Real Analysis, second ed. ∞ R 1 Let γ ∈ C ( ) such that γ(x)dx = , and let ω = γ ∗ χ[0,2π). If we have c R R 2π a 2π-periodic distribution F on R and ψ ∈ C∞(T), we define F ψ to be F (ωψ) (ψ is viewed in this instance as a 2π-periodic function on R). For ψ ∈ C∞(T) we have, viewing ψ as a 2π-periodic function on R (thus ψ(2nπ) = ψ(0)), X X2π(ωψ) = δ(t − 2nπ)(ωψ) n∈Z X = ω(2nπ)ψ(2nπ) n∈Z X Z = ψ(0) γ(x)χ[0,2π)(2nπ − x)dx n∈Z R Z X = ψ(0) γ(x) χ[0,2π)(2nπ − x)dx R n∈Z Z = ψ(0) γ(x) R ψ(0) = . 2π ∞ ψ(0) Thus as a distribution on T, for ψ ∈ C (T) we have X2πψ = 2π . For n ∈ Z, −int 1 X[2π(n) = X2πe = 2π . Thus the Fourier series of X2π is 1 X eint. 2π n∈Z