Fourier Series of a Function

Consider a periodic function, $f(x) = f(x+2\pi )$ and its $N$ th partial Fourier sum

$$S_N(x) = \frac{a_0}{2}+\sum^N_{n=1} a_n\cos nx +\sum^N_{n=1} b_n \sin nx$$

$$= \int^\pi_{-\pi}\delta_N (t-x)f(t)dt\,.$$

Here $\delta_N(t-x)$ is the familiar Dirichlet kernel and the integration limits have been shifted downward without affecting the integral. This can always be done when integrating a periodic function. In fact, the shift can be any real amount:

$$\int^{2\pi}_0{\textrm{periodic~function}\choose \textrm{with~peri... ...\pi +a}_a{\textrm{periodic~function}\choose \textrm{with~ period}~2\pi}\,dt\,.$$

(Verify that this identity holds for any real $a$ .)

Question: What is $\lim\limits_{N\to\infty} S_N(x)$ ?

Answer: $S_N(x)\to\frac{1}{2}[f(x^+)+f(x^-)]$ as $N\to\infty$ .

One arrives at this answer by means of a four step argument.

1. Shift the integration limit by the amount $a=x$ and obtain
   $$S_N(x)=\int^{x+\pi}_{x-\pi} \delta_N(t-x)f(t)\,dt \quad .$$
   
   
   

   This places $x$ at the center of the integration interval. Now break up the integral into two parts

   $$S_N(x) = \underbrace{\int^x_{x-\pi}\delta_N(t-x)f(t)dt}_{\display... ...J_N(x)} +\underbrace{\int^{x+\pi}_x \delta_N(t-x)f(t)dt}_{\displaystyle I_N(x)}$$ (214)

   
   
   and show that
   $$J_N(x)\to \frac{1}{2}f(x^-) ~\hbox{and}~I_N(x)\to \frac{1}{2}f(x^+) ~\hbox{as} ~ N\to\infty \quad .$$

2. Look at each integral in turn. Let $u=t-x$ and obtain
   $$I_N(x)=\int^\pi_0 f(x+u)\delta_N(u)du \quad .$$
   Figure 2.2 depicts the graphs of the two factors making up the integrand.

   Figure 2.2: The graph of $\delta _N (u)$ and the function $f(x+u)$ around $u=0$ , where it has the indicated jump discontinuity.

   
   Using
   $$\delta_N(u)=\frac{1}{2\pi} \frac{\sin(N+\frac{1}{2})u}{\sin \frac{u}{2}}$$
   we obtain
   $$\lim_{N\to\infty} I_N(x)=\frac{1}{2} f(x^+)\,.$$
   The details are as follows:
   $$I_N(x)=\frac{1}{2\pi}\int^\pi_0 \underbrace{ \frac{f(x+u)-f(x^+)}... ...ert \\ \displaystyle \frac{1}{2}\\ \displaystyle \textrm{(indep.'t~of}~N)}}.$$
   Note that $G(u)$ is piecewise continuous. Why? Because, assuming that $f$ has one-sided derivatives $f'_L(x)$ and $f'_R(x)$ at $x$ , we have
   $$G(0^+)\equiv \lim_{u\to 0^+} \frac{f(x+u)-f(x^+)}{\sin\frac{u}{2}} = 2f'_R(x)\,.$$
   Thus we see that the integrand is piecewise continuous throughout $[0,\pi]$ , even at $u=0$ , where the denominator $\sin \frac{u}{2}$ vanishes. The total integral in question,
   $$I_N(x)=\frac{1}{2\pi} \int^\pi_0 G(u)\sin \left[ \left( N+\frac{1}{2}\right) u \right] \, du+\frac{1}{2} f(x^+)\,,$$
   is, therefore, well-defined.

   As $N\to\infty$ the integrand is a function $G(u)$ , piecewise continuous and finite on $[0,\pi]$ , multiplied by a rapidly oscillating function.

   Such an integral averages to zero as $N\to\infty$ . The vanishing of such an integral is also known as the Riemann-Lebesgue lemma. See the ensuing exercise on page 

   Conclusion: $\lim\limits_{N\to\infty}I_N(x)=0+\frac{1}{2} f(x^+)$ .

3. Similarly, with $u=x-t$ and $\delta_N(-u)=\delta_N(u)$ one has
   $$J_N(x)=\frac{1}{2\pi}\int^\pi_0 f(x-u)\delta_N(u)\,du$$
   and
   $$\lim_{N\to\infty} J_N(x)=0+\frac{1}{2} f(x^-)\,.$$

4. Consequently, the limit of the partial sum, Eq.(2.14), is
   $$\lim_{N\to\infty} S_N(x)=\frac{1}{2}[ f(x^-)+f(x^+)]\,.$$

To summarize, one has the following

Theorem 21.2 (Fourier Series Theorem)
.

1. Let $f(x)$ be a function which is piecewise continuous on $[-\pi , \pi ]$ .

   Its Fourier series is given by

   $$\frac{1}{2\pi}\int^\pi_{-\pi}f(t)\,dt+\frac{1}{\pi}\sum^\infty_{n=1} \int^\pi_{-\pi} f(t)\cos n(t-x)\,dt = \frac{1}{2}[f(x^-)+f(x^+)]$$
   at each point $-\pi<x<\pi$ where the one sided derivatives $f'_R(x)$ and $f'_L(x)$ both exist.

2. If $f$ is continuous the result is
   $$\frac{1}{2\pi}\sum^\infty_{n=-\infty}\int^\pi_{-\pi} e^{in(x-t)} f(t)\,dt =\frac{1}{2}[f(x^-)+f(x^+)]=f(x)\quad \forall\, f\in C[-\pi ,\pi ]\,.$$

Exercise 21.3 (RIEMANN-LEBESGUE LEMMA)
The Riemann-Lebesgue lemma is a well-known fact among radio amateurs and electrical engineers. There it is the time average of an amplitude modulated signal,

$$G(u)\, \sin(N+\frac{1}{2} )u~,$$

whose carrier is $\sin(N+\frac{1}{2})u$ , a rapidly oscillating function whose modulation amplitude is $G(u)$ . The higher the carrier frequency $(N+\frac{1}{2})$ , the more closely that average approaches zero.

Given that $G(u)$ is piecewise continuous on $[0,\pi]$ and has left and right hand derivatives at each point in $[0,\pi]$ , show that

$$\lim_{N\to\infty} \int_0^\pi G(u)\, \sin(N+\frac{1}{2} )u \,du~=~0~.$$

Lecture 13

Poisson's summation formula

The periodicity of the Dirichlet kernel guarantees that Fourier's theorem holds also when $x$ lies outside the interval $[-\pi , \pi ]$ , even if the function $f$ is not periodic. Let us therefore apply the Fourier theorem to the new (continuous) function $f(t+2\pi m)$ :

$$\frac{1}{2\pi}\sum^\infty_{n=-\infty}\int^\pi_{-\pi} e^{in(x-t)} f(t+2\pi m)\,dt =f(x+2\pi m)\quad.$$

Now

(i)

shift the integration variable $t$ by $2\pi m$ ,

(ii)

make use of the periodicity of the exponential,

(iii)

sum over all the integral values of $m=0,\pm 1,\pm 2 ,\cdots$ , and

(iv)

use the fact that $\sum_{-\infty}^\infty \int_{-\pi +2\pi m}^{\pi +2\pi m}\cdots =\int_{-\infty}^\infty \cdots$ .

The result is

$$\frac{1}{2\pi}\sum^\infty_{n=-\infty}\int^\infty_{-\infty} e^{in(x-t)}f(t)\,dt=\sum^\infty_{m=-\infty}f(x+2\pi m).$$ (215)

Setting $x=0$ , one obtains

$$\boxed{ \sum^\infty_{n=-\infty} F(n)=2\pi \sum^\infty_{m=-\infty}f(2\pi m), }$$ (216)

where

$$F(n)=\int^\infty_{-\infty} e^{-int}f(t)\,dt$$ (217)

is the Fourier transform of the function $f$ . This equality is known as the Poisson sum formula.
An alternative equally useful form of this formula is

$$\boxed{ \sum^\infty_{m=-\infty}f(m)=\sum^\infty_{n=-\infty} F(2\pi n)~. }$$ (218)

Example (Closed form via Poisson summation)

Employ the Poisson summation formula to find the value of the sum

$$\sum_{m=-\infty}^\infty f(m)=\sum_{m=-\infty}^\infty \frac{1}{m^2+a^2}$$

in terms of elementary functions.
The first step in finding this value is to determine the Fourier amplitude

$$F(k)\equiv \int_{-\infty}^\infty e^{-ikx} \frac{1}{x^2 +a^2}dx$$

of the given function $f$ . Using Cauchy's integral formula one finds that

$$F(k)=\frac{\pi}{a} e^{-\vert k\vert a}~,$$

where without loss of generality $a>0$ .
The second step consists of evaluating the right hand side of Eq.(2.18).

$$\sum^\infty_{n=0} F(-2\pi n)+\sum^\infty_{n=1}F(2\pi n) =\frac{\pi}{a}\left[1+2\sum^\infty_{n=1}e^{-2\pi an}\right]$$

$$=\frac{\pi}{a}\left[1+\frac{2}{e^{-2\pi a}-1}\right]$$

$$=\frac{\pi}{a}\frac{e^{-2\pi a}+1}{e^{-2\pi a}-1}$$

$$=\frac{\pi}{a}\coth \pi a~.$$ (219)

One has therefore the following result.

$$\sum_{m=-\infty}^\infty \frac{1}{m^2+a^2}= \frac{\pi}{a}\coth \pi a~.$$

The simplicity of the Poisson summation formula is enhanced if one does not refer to the function $f$ explicitly. Reexpressing the right hand side of Eq.(2.15) in terms of equally spaced Dirac delta functions,

$$\sum^\infty_{m=-\infty}f(x+2\pi m)\\ =\sum^\infty_{m=-\infty}\int _{-\infty}^\infty \delta (t-x-2\pi m)f(t)~dt ,$$

and observing the fact that Eq.(2.15) holds for all continuous functions $f$ whose infinite sum of sampled values converges, we leave these individual functions unspecified and simply write Eq.(2.15) in the form

$$\frac{1}{2\pi}\sum^\infty_{n=-\infty} e^{inu} \equiv \lim_{N\to\infty} \delta_N(u) = \sum^\infty_{m=-\infty} \delta (u-2\pi m)$$ (220)

$$= \textrm{\lq\lq comb}_{2\pi} u\textrm{''}~;u=t-x ~~,$$

or equivalently,

$$\sum^\infty_{n=-\infty} e^{i\frac{2\pi}{a}nu}=\vert a\vert\sum^\infty_{m=-\infty} \delta (u- ma),$$

whenever $a$ is positive. (Question: what happens when $a<0$ ?) This is obviously an alternate form of the Poisson sum formula. It says that as the number of terms becomes very large, the Dirichlet kernel approaches a periodic train of delta function impulses. However, it needs to be emphasized that this relation is based on the tacit understanding that one first multiply by some appropriate function $f(t)$ and do the $t$ -integration before one compares the sums on the two sides of this relation.

What happens if one first rescales the domain of the function $f(x)$ by a non-zero real factor before shifting it by the amount $2\pi m$ ? In that case one applies the Fourier theorem to the function

$$f\left(\frac{x+2\pi m}{a}\right)$$

and the Poisson sum formula, Eq.(2.16), becomes

$$\sum^\infty_{n=-\infty} F(na) = \frac{2\pi}{a}\sum^\infty_{m=-\infty}f\left( \frac{2\pi m}{a}\right).$$ (221)

Exercise 21.4 (POISSON FORMULA AND ORTHONORMALITY)

Stephane G. Mallat in his article ``A Theory of Multiresolution Signal Decomposition: The Wavelet Theory'' (IEEE Transactions on Pattern Analysis and Machine Intelligence, Vol. 11, p. 674-692, 1989) makes the following claim in Appendix B of his article:

Let $\hat \phi (\omega)$ be the Fourier transform of $\phi(x)$

$$\hat \phi (\omega) = \frac{1}{\sqrt {2\pi}} \int^\infty_{-\infty} \phi(x) e^{-i\omega x} dx.$$

With the Poisson formula one can show that the family of functions

$$\lbrace \phi (x-k): k = 0, \pm 1, \pm 2, \cdots \rbrace$$

is orthonormal if and only if

$$\sum^\infty_{k = - \infty} \vert \hat \phi (\omega + 2 \pi k)\vert^2 =const.$$

Prove or disprove this claim. If the claim is true, what would be the value of ``$const.$ ''

Exercise 21.5 (PHASE SHIFTED POISSON FORMULAS)
Using the Fourier transform notation

$$F(n)=\int^\infty_{-\infty} e^{-int}f(t)\,dt~,$$

prove or disprove that

$$\sum_{m=-\infty}^\infty f\left([2m+1]\pi\right) =\frac{1}{2\pi} \sum_{n=-\infty}^\infty (-1)^n F(n)$$

$$\sum_{n=-\infty}^\infty F(n+\frac{1}{2}) =2\pi \sum_{m=-\infty}^\infty (-1)^m f(2m\pi)$$

$$\frac{1}{2\pi}\sum_{n=-\infty}^\infty (-1)^n e^{inu} = \sum_{m=-\infty}^\infty \delta\left(u-[2m+1]\pi\right)~,$$
and more generally that

$$\sum_{m=-\infty}^\infty f\left(\frac{[2m+1]\pi}{a}\right) = \frac{a}{2\pi} \sum_{n=-\infty}^\infty (-1)^n F(na)$$

$$\frac{1}{2\pi}\sum_{n=-\infty}^\infty (-1)^n e^{inua} = \sum_{m=-\infty}^\infty \frac{1}{\vert a\vert}\delta\left(u-\frac{[2m+1]\pi}{a}\right)~.$$