Fourier Transform via Parseval's Relation

The Fourier transform is so robust that it can also be applied to functions which are not square-integrable. In fact, it can even be applied to generalized functions (``distributions''), i.e. to entities which are defined by the fact that they are linear functionals on the linear space of the familiar locally integrable functions. If $f(x)$ is such a function, then a generalized function, say $g(x)$ , is defined by the fact that

$$\int^\infty_{-\infty} \overline{f}(x)g(x)dx \equiv\langle f,g \rangle <\infty$$

is finite. The Dirac delta function is an example of a generalized function because

$$\int^\infty_{-\infty} \overline{f}(x)\delta(x-x')dx\equiv \overline{f}(x') <\infty~~.$$

Recall that whenever one has square-integrable functions $f$ and $g$ (whose Fourier transforms are $\hat f$ and $\hat g$ ) then the reasoning which lead to Parseval's identity leads to

$$\langle f,g \rangle= \langle \hat f,\hat g \rangle$$

One now turns this relation around and uses it to define the Fourier transform $\hat g$ of a given generalized function $g$ . In other words, start with the set of locally integrable functions $\hat{f}(k)$ and their inverse Fourier transforms

$$f(x)=\int^\infty_{-\infty} \frac{e^{ikx}}{\sqrt{2\pi}}\hat{f}(k)dk~~.$$

Next define the Fourier transform of $g$ as follows: Let it be that generalized function $\hat g$ which is determined by the compatibility (between functions and their transforms) requirement that

$$\langle \hat f,\hat g \rangle= \langle f,g \rangle$$ (240)

hold for all locally integrable functions $\hat f$ . This equality is now our fundamental requirement. It determines $\hat g$ uniquely. Indeed, for every $\hat f$ one readily determines $f$ and hence $\langle f,g \rangle$ . This implies that Eq.(2.40) is the equation which defines the linear functional $\hat g$ , the sought-after Fourier transform of $g$ . This linear functional is unique and is denoted by

$$\hat g(k)=\int^\infty_{-\infty} \frac{e^{-ikx}}{\sqrt{2\pi}}g(x)dx ~~,$$

even though $g(x)$ may not be integrable in the standard sense.

Example 1(Fourier transform of a ``ticking clock'' signal)

Consider the generalized function

$$g(x)=\sum_{n=-\infty}^\infty \delta (x-n)~~,$$

the train of evenly spaced delta function impulses. What is its Fourier transform?

We know that for any continuous function $\hat f\in L^2(-\infty,\infty)$ one can determine its inverse Fourier transform $f$ and hence

$$\langle f,g\rangle \equiv \int _{-\infty}^\infty \overline{f}(x) ... ...fty}^\infty \delta (x-n)~dx =\sum_{n=-\infty}^\infty \overline{f}(n) <\infty~~.$$ (241)

The Fourier transform of $g$ is determined by the requirement that for all appropriate $\hat f$

$$\int _{-\infty}^\infty \overline{\hat f}(k)\hat g(k)dk \stackrel{1}{=} \langle f,g\rangle$$

$$\stackrel{2}{=} \sum_{n=-\infty}^\infty \overline{f}(n)$$

$$\stackrel{3}{=} \sqrt{2\pi} \sum_{m=-\infty}^\infty \overline{\hat f}(2\pi m)$$

$$\stackrel{4}{=} \int _{-\infty}^\infty \overline{\hat f}(k)\sum_{m=-\infty}^\infty \sqrt{2\pi}~ \delta(k-2\pi m)~dk~~.$$

Equality 1 is the fundamental consistency relation, Eq.(2.40); 2 holds because of Eq.(2.41); 3 holds because of Poisson's sum formula, Eq.(2.16) on page ,

$$\sum_{n=-\infty}^\infty {f}(n)= \sum_{m=-\infty}^\infty \sqrt{2\pi}{\hat f}(2\pi m)$$

with $\hat f=\sqrt{2\pi}F$ :

$$\hat{f}(k)=\int^\infty_{-\infty} \frac{e^{-ikx}}{\sqrt{2\pi}}f(x)dx~~;$$ (242)

4 takes advantage of the sampling property of the Dirac delta function $\delta(k-2\pi m)$ . Thus one finds that

$$\int _{-\infty}^\infty \overline{\hat f}(k)\hat g(k)dk = \int _{-\infty}^\infty \overline{\hat f}(k)\sum_{m=-\infty}^\infty \sqrt{2\pi}~ \delta(k-2\pi m)~dk$$

holds for all integrable functions $\hat f(k)$ . This fact is reexpressed by the statement

$$\hat g(k)=\sum_{m=-\infty}^\infty \sqrt{2\pi}~\delta(k-2\pi m)~~.$$

This is the desired result. It says that the Fourier transform of a periodic train of infinitely sharp pulses (with period $\Delta x=1$ ) is another periodic train of pulses (with period $\Delta k=2\pi$ ) in the Fourier domain.

Example 2 (Fourier transform of a periodic function) Consider a periodic function

$$g(x)=g(x+a)~$$

whose Fourier series representation is

$$g(x)=\sum^\infty_{-\infty} c_n \frac{e^{i2\pi n x/a}}{\sqrt{a}}~,~ \textrm{with}~~c_n=\int_0^a \frac{e^{-i2\pi n x/a}}{\sqrt{a}}g(x)~dx~.$$

What is its Fourier transform?

Note that for any integrable function $f$ and its Fourier transform, Eq.(2.42), one has

$$\langle f,g\rangle = \sum^\infty_{-\infty} c_n \int^\infty_{-\infty} \overline f (x) \frac{e^{i2\pi n x/a}}{\sqrt{a}} ~dx$$

$$= \sum^\infty_{-\infty} c_n \sqrt{\frac{2\pi}{a}} ~\overline{\hat f} \left( \frac{2\pi n}{a}\right)$$

$$= \int^\infty_{-\infty}\overline{\hat f}(k)\sum^\infty_{-\infty}c_n \sqrt{\frac{2\pi}{a}} \delta(k-\frac{2\pi n}{a})\,dk$$

Using the stipulated Parseval requirement,

$$\langle f,g\rangle=\langle \hat f,\hat g\rangle \equiv \int^\infty_{-\infty} \overline{\hat f}(k) \hat g(k)~dk ~,$$

which holds for all functions $\overline {\hat f}$ , one sees that

$$\hat g(k)\equiv {\mathcal{F}}\left[ g\right](k)= \sum^\infty_{n=-\infty} c_n \sqrt{\frac{2\pi}{a}} \delta(k- n\frac{2\pi}{a})$$

Thus we have the following result: The Fourier transform of a function periodic on the given domain is a periodic train of Dirac delta functions on the Fourier domain, each one weighted by the respective Fourier coefficient.

Conversely, the Fourier transform of a periodic train of weighted Dirac delta functions is a periodic function.

What happens if all the weight are equal? In that case the periodic function $g(x)$ turn out to be a generalized function, namely

$$g(x) = \sum^\infty_{n=-\infty}\frac{e^{i2\pi n x/a}}{\sqrt{a}}$$ (243)

$$= \sqrt{a}\sum^\infty_{m=-\infty}\delta(x-ma)~.$$ (244)

The above Parseval-relation-based method for calculating the Fourier transforms applies to periodic and generalized functions as well. Consequently, one has

$$\hat g (k)\equiv {\mathcal{F}}\left[ g\right](k)= \sum^\infty_{n=-\infty} \sqrt{\frac{2\pi}{a}} \delta(k- n\frac{2\pi}{a})~.$$ (245)

What happens if one takes the Fourier transform again? Without much ado one finds that the Fourier transform of the function $\hat g$ is

$${\mathcal{F}}\left[ \sum^\infty_{n=-\infty} \sqrt{\frac{2\pi}{a}} \delta(x- n\frac{2\pi}{a}) \right](k) = \sum^\infty_{m=-\infty} \frac{e^{-i2\pi m k/a}}{\sqrt{a}}$$ (246)

$$= \sum^\infty_{n=-\infty} \sqrt{a}~\delta(k- na)~.$$ (247)

In other words, one recovers the original function $g$ .

Exercise 23.6 (EIGENFUNCTIONS OF ${\mathcal{F}}^2$ )
It is evident from Eqs.(2.47) and (2.44) that the function $g$ is an eigenfunction of the Fourier transform taken twice, i.e. of the operator ${\mathcal{F}}^2$ , with eigenvalue $\lambda=+1$ . Are there any other such functions, and if so, characterize them by a simple criterion.

Exercise 23.7 (FOURIER TRANSFORM: BASIC PROPERTIES)
Let $\hat g (k)={\mathcal{F}}\left[g(x)\right](k)$ and $H(k)={\mathcal{F}}\left[h(x)\right](k)$ be the Fourier transforms of $g(x)$ and $h(x)$ . Find

(i)

${\mathcal{F}}\left[\alpha g(x)+\beta h(x)\right](k)$ , where $\alpha$ and $\beta$ are constants.

(ii)

${\mathcal{F}}\left[g(x-\xi)\right](k)$

(iii)

${\mathcal{F}}\left[e^{ik_0x}g(x)\right](k)$

(iv)

${\mathcal{F}}\left[g(ax)\right](k)$

(v)

${\mathcal{F}}\left[ \displaystyle\frac{dg(x)}{dt}\right](k)$

(vi)

${\mathcal{F}}\left[ xg(x)\right](k)$

in terms of $\hat g (k)$ and $\hat{f}(k)$ .