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 jxj jf (x)j x2R 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 2 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 2 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 2 S(R), define Z f^(ξ) = e−iξxf(x)dx; R and then 1 Z f(x) = eixξf^(ξ)dξ: 2π R Let T = R=2πZ. For φ 2 C1(T), define 1 Z 2π φ^(n) = e−intφ(t)dt; 2π 0 and then X φ(t) = φ^(n)eint: n2Z 1 For f 2 S(R) and λ 6= 0, define φλ : T ! C by X φλ(t) = 2π fλ(t + 2πj); j2Z 1 where fλ(x) = λf(λx). We have φλ 2 C (T). For n 2 Z, Z 2π 1 −int X φcλ(n) = e · 2π fλ(t + 2πj)dt 2π 0 j2Z Z 2π X −int = e fλ(t + 2πj)dt 0 j2Z Z −inx = e fλ(x)dx R = fcλ(n): ^ ξ ^ n One checks that fcλ(ξ) = f λ , so φcλ(n) = f λ . Thus for t 2 T, X n φ (t) = f^ eint: λ λ n2Z Therefore, if f 2 S(R) then for each t 2 T, X X n 2π f (t + 2πj) = f^ eint; λ λ j2Z n2Z where fλ(x) = λf(λx). This is the Poisson summation formula. 2 Nyquist-Shannon sampling theorem Let f 2 S(R). Suppose that there is some L such that f(x) = 0 if jxj > 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 f^(ξ) = 0 for jξj > 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 2 S(R) and f(ξ) = 0 for jξj > L.1 For −π ≤ ξ ≤ π, define 1 1 In fact, if f; f^ 2 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) = f^ : π For n 2 Z, 1 Z π Fb(n) = e−inξF (ξ)dξ 2π −π 1 Z π ξL = e−inξf^ dξ 2π −π π Z L 1 −in πt = e L f^(t)dt: 2L −L On the other hand, for x 2 R, using the fact that f is bandlimited, 1 Z 1 Z L f(x) = eixξf^(ξ)dξ = eixξf^(ξ)dξ: 2π 2π R −L Therefore for n 2 Z, π −nπ Fb(n) = f : L L As F 2 C1(T), for t 2 T we have X X π −nπ F (t) = Fb(n)eint = f eint: L L n2Z n2Z Hence for −L ≤ ξ ≤ L, πξ X π −nπ inπξ f^(ξ) = F = f e L : L L L n2Z For 0 ≤ η < L, let φ : R ! R satisfy ( 1 jξj ≤ L − η φ^(ξ) = 0 jξj > L: For instance, for η = 0 let φ^ be the characteristic function on [−L; L]. But if η > 0 we can choose η^ to be smooth rather than a characteristic function. If f^(ξ) = 0 for jξj > L − η (rather than just for jξj > L), then f^ = f^φ^, because φ 2We are defining a periodic function F from a function f^ that is not periodic, but rather which becomes 0 after L. 3 is equal to 1 on the support of f^. Then for x 2 R, 1 Z f(x) = eixξf^(xi)φ^(ξ)dξ 2π R Z 1 ixξ X π −nπ inπξ = e f e L φ^(ξ)dξ 2π L L R n2Z Z 1 X −nπ ixξ+ inπξ = f e L φ^(ξ)dξ 2L L n2Z R 1 X −nπ nπ = f · 2π · φ x + 2L L L n2Z π X −nπ nπ = f φ x + : L L L n2Z ^ nπ φ can be chosen so that φ x + L quickly goes to 0 as n ! 1. This means that fewer terms of the above series need to be calculated to get a good approximation of f(x). For φ^ 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 2 R, X −nπ nπ f(x) = f sinc x + ; L L n2Z or X nπ nπ f(x) = f sinc x − : L L n2Z Thus if f is bandlimited, we can sample f at a sequence of points and using these values reconstruct f(x) for any x 2 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): n2Z This is called a Dirac comb or pulse train. If f 2 S(R), X Xf = f(n): n2Z 4 The convolution of f and X is the function X (X ∗ f)(t) = f(t − n): n2Z Define 1 t X (t) = X : T T T Because δ(t=a) = jajδ(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 n2Z n2Z and we have X XT f = f(nT ) n2Z and X (XT ∗ f)(t) = f(t − nT ): n2Z 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. 1 R 1 Let γ 2 C ( ) such that γ(x)dx = , and let ! = γ ∗ χ[0;2π). If we have c R R 2π a 2π-periodic distribution F on R and 2 C1(T), we define F to be F (! ) ( is viewed in this instance as a 2π-periodic function on R). For 2 C1(T) we have, viewing as a 2π-periodic function on R (thus (2nπ) = (0)), X X2π(! ) = δ(t − 2nπ)(! ) n2Z X = !(2nπ) (2nπ) n2Z X Z = (0) γ(x)χ[0;2π)(2nπ − x)dx n2Z R Z X = (0) γ(x) χ[0;2π)(2nπ − x)dx R n2Z Z = (0) γ(x) R (0) = : 2π 1 (0) Thus as a distribution on T, for 2 C (T) we have X2π = 2π . For n 2 Z, −int 1 X[2π(n) = X2πe = 2π . Thus the Fourier series of X2π is 1 X eint: 2π n2Z 5.