The Fourier Transform as a Unitary Transformation

Fourier's integral theorem expresses a linear transformation, say $\mathcal{F}$ , when applied to the space of square integrable functions. From this perspective one has

$$L^2(-\infty,\infty) \stackrel{\mathcal{F}}{\longrightarrow} L^2(-\infty,\infty)$$

$$f(t) \sim\leadsto \mathcal{F}[f](k)= \int_{-\infty}^\infty\frac{e^{-ikt}}{\sqrt{2\pi}}f(t)dt\equiv \hat f(k)~.$$ (244)

Furthermore, this transformation is one-to-one because Fourier's theorem says that its inverse is given by

$$L^2(-\infty,\infty) \stackrel{\mathcal{F}^{-1}}{\longrightarrow} L^2(-\infty,\infty)$$

$$\hat f(k) \sim\leadsto \mathcal{F}^{-1}[\hat f](x)= \int_{-\infty}^\infty\frac{e^{ikx}}{\sqrt{2\pi}}\hat f(k)dk\equiv f(x)~.$$ (245)

That $\mathcal{F}$ maps square integrable functions into square integrable functions is verified by the following computation, which gives rise to Parseval's identity: For $f\in L^2(-\infty ,\infty )$ we have

$$\int^\infty_{-\infty}\vert f(x)\vert^2 dx = \int^\infty_{-\infty} \overline{f} (x)f(x)dx$$

$$= \int^\infty_{-\infty}\overline f(x)\left[\int^\infty_{-\infty} \hat f(k') \frac{e^{ik'x}}{\sqrt{2\pi}}dk'\right] dx$$

$$= \int^\infty_{-\infty}\hat f(k') \overline{ \left[\int^\infty_{-\infty} \frac{e^{-ik'x}}{\sqrt{2\pi}}f(x)dx\right] } dk'$$

$$= \int^\infty_{-\infty} \hat f(k')\overline{\hat f}(k')dk'$$

$$= \int^\infty_{-\infty} \vert \hat{f}(k)\vert^2 dk~~.$$

Thus we obtain Parseval's identity (= ``completeness relation'', see Eq.(1.17) on page ). The only proviso is (a) that the function $f$ be square-inegrable and (b) that its Fourier transform $\hat f$ be given by the Fourier transform integral.

Remark 1: The fact that the Fourier transform is a one-to-one linear transformation from the linear space $L^2(-\infty ,\infty )$ to the linear space $L^2(-\infty ,\infty )$ is summarized by saying that the Fourier transform is an ``isomorphism''.

Remark 2: The line of reasoning leading to Parseval's identity also leads to

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

whenever $f,g \in L^2(-\infty ,\infty )$ .

Remark 3: The above two remarks imply that the Fourier transform is a unitary transformation in $L^2(-\infty ,\infty )$ . Unitary transformations are ``isometries'' because they preserve lengths and inner products. One says, therefore, that the space of functions defined on the spatial domain is ``isometric'' to the space of functions defined on the Fourier domain. Thus the Fourier transform operator is a linear isometric mapping. This fact is depicted by Figure 2.6

Figure 2.6: The Fourier transform is an isometry between $L^2(-\infty ,\infty )$ , the Hilbert space of square integrable functions on the spatial domain, and $L^2(-\infty ,\infty )$ , the Hilbert space of square integrable functions on the Fourier domain.

Note, however, that even though the Fourier transform and its inverse,

$$f(x) = \int^\infty_{-\infty} f(x') \delta (x'-x)dx'=\int^\infty_{-\infty} \hat{f}(k)\frac{e^{ikx}}{\sqrt{2\pi}} dk ~~,$$ (246)

take square integrable functions into square integrable functions, the ``basis elements'' $e^{ikx}$ are not square integrable. Instead, they are ``Dirac delta function'' normalized, i.e.,

$$\int^\infty_{-\infty} \frac{e^{ikx}}{\sqrt{2\pi}} \frac{e^{-ik\overline{x}}}{\sqrt{2\pi}} dk = \delta (x-\overline{x})~~.$$

Thus they do not belong to $L^2(-\infty ,\infty )$ . Nevertheless linear combinations such as Eq.(2.46) are square integrable, and that is what counts.

Exercise 23.1 (THE FOURIER TRANSFORM: ITS EIGENVALUES)
The Fourier transform, call it $\cal F$ , is a linear one-to-one operator from the space of square-integrable functions onto itself. Indeed,

$${\cal F}:~~L^2(-\infty,\infty) \rightarrow L^2(-\infty,\infty)$$

$$f(x) \sim \rightarrow {\cal F}[f](k) \equiv \frac{1}{ \sqrt{2\pi}} \int^\infty_{-\infty} e^{-ikx}f(x)~dx \equiv \hat{f}(k)$$

Note that here $x$ and $k$ are viewed as points on the common domain $(-\infty ,\infty)$ of $f$ and $\hat f$ .

(a)

Consider the linear operator ${\cal F}^2$ and its eigenvalue equation

$${\cal F}^2f=\lambda f.$$

What are the eigenvalues and the eigenfunctions of ${\cal F}^2$ ?

(b)

Identify the operator ${\cal F}^4$ ? What are its eigenvalues?

(c)

What are the eigenvalues of ${\cal F}?$

Exercise 23.2 (THE FOURIER TRANSFORM: ITS EIGENVECTORS)

Recall that the Fourier transform ${\cal{F}}$ is a linear one-to-one transformation from $L^2(-\infty ,\infty )$ onto itself.
Let $\phi$ be an element of $L^2(-\infty ,\infty )$ .
Let $\hat \phi = {\cal{F}}\phi$ , the Fourier transform of $\phi$ , be defined by

$$\hat \phi (p) = \int^\infty_{-\infty} \frac{e^{-i p x}}{ \sqrt {2\pi}}\phi (x) dx \quad p \in (-\infty, \infty).$$

It is clear that

$$\phi,~ {\cal{F}}\phi,~ {\cal{F}}^2\phi \equiv {\cal{F}}({\cal{F}}... ...{\cal{F}}^2\phi),~ {\cal{F}}^4\phi \equiv {\cal{F}} ({\cal{F}}^3\phi), ~\cdots$$

are square-integable functions, i.e. elements of $L^2(-\infty ,\infty )$ .
Consider the SUBSPACE $W \subset L^2(-\infty, \infty)$ spanned by these vectors, namely

$$W= \hbox {span}\{ \phi, {\cal{F}}\phi, {\cal{F}}^2\phi, {\cal{F}}^3\phi, {\cal{F}}^4\phi, \cdots \} \subset L^2(-\infty, \infty).$$

(a)

Show that $W$ is finite dimensional.
What is $\dim W$ ?
(Hint: Compute ${\cal{F}}^2 \phi (x), {\cal{F}}^3 \phi (x)$ , etc. in terms of $\phi (x), \hat \phi (x)$ )

(b)

Exhibit a basis for $W$ .

(c)

It is evident that ${\cal{F}}$ is a (unitary) transformation on $W$ .
Find the representation matrix of ${\cal{F}}, [{\cal{F}}]_B$ , relative to the basis $B$ found in part b).

(d)

Find the secular determinant, the eigenvalues and the corresponding eigenvectors of $[{\cal{F}}]_B$ .

(e)

For $W$ , exhibit an alternative basis which consists entirely of eigenvectors of ${\cal{F}}$ , each one labelled by its respective eigenvalue.

(f)

What can you say about the eigenvalues of ${\cal{F}}$ viewed as a transformation on $L^2(-\infty ,\infty )$ as compared to $[{\cal{F}}]_B$ which acts on a finite-dimensional vector space?

Exercise 23.3 (EQUIVALENT WIDTHS)

Suppose we define for a square-integrable function $f(t)$ and its Fourier transform

$$\hat{f}(\omega)=\int_{-\infty}^\infty \frac{ e^{-i\omega t} }{\sqrt{2\pi} }f(t)~d t$$

the equivalent width as

$$\Delta_t= \left\vert \frac{ \int_{-\infty}^\infty f(t)~dt}{ {f}(0) }\right\vert~,$$

and the equivalent Fourier width as

$$\Delta_\omega = \left\vert\frac{ \int_{-\infty}^\infty \hat{f}(\omega)~d\omega }{ \hat{f}(0) }\right\vert~.$$

(a)

Show that

$$\Delta_t \Delta_\omega = const.$$

is independent of the function $f$ , and determine the value of this $const.$

(b)

Determine the equivalent width and the equivalent Fourier width for the unnormalized Gaussian

$$f(t)=e^{-x^2/2b^2}$$

and compare them with its full width as defined by its inflection points.

Exercise 23.4 (AUTO CORRELATION SPECTRUM)

Consider the auto-correlation $h$ ²²

$$h(y) = \int^\infty_{-\infty} {f} (x) f (x-y)dx$$ (247)

of the function $f$ whose Fourier transform is

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

Compute the Fourier transform of the auto correlation function and thereby show that it equals the ``spectral intensity'' (a.k.a. power spectrum) of $f$ whenever $f$ is a real-valued function. This equality is known as the Wiener-Khintchine formula.

Exercise 23.5 (MATCHED FILTER)

Consider a linear time-invariant system. Assume its response to a specific driving force, say $f_0(t)$ , can be written as

$$\int^\infty_{-\infty} g(T-t)f_0(t) dt \equiv h (T).$$

Here $g(T-t)$ , the ``unit impulse response' (a.k.a. ``Green's function'', as developed in CHAPTER 4 and used in Section 4.2.1), is a function which characterizes the system completely. The system is said to be matched to the particular forcing function $f_0$ if

$$g(T)= \overline{f_0} (-T).$$

(Here the bar means complex conjugate.) In that case the system response to a generic forcing function $f(t)$ is

$$\int^\infty_{-\infty} \overline{f_0} (t-T)f(t) dt \equiv h(T).$$

A system characterized by such a unit impulse response is called a matched filter because its design is matched to the particular signal $f_0(t)$ . The response $h(T)$ is called the cross correlation between $f$ and $f_0$ .

(a)

Compute the total energy

$$\int^\infty_{-\infty} \vert h(T)\vert^2 dT$$

of the cross correlation $h(T)$ in terms of the Fourier amplitudes

$$\hat {f_0}(\omega ) = \int^\infty_{-\infty}\frac{e^{-i\omega t} }{ \sqrt {2\pi}}f_0 (t)dt$$

and

$$\hat f(\omega ) = \int^\infty_{-\infty} \frac{e^{-i\omega t} }{\sqrt {2\pi}}f(t)dt.$$

(b)

Consider the family of forcing functions

$$\{f_0(t), f_1(t), \cdots, f_N (t)\}$$

and the corresponding family of normalized cross correlations (i.e. the corresponding responses of the system)

$$h_k(T) = \frac{\int^\infty_{-\infty} \overline{f_0}(t-T)f_k(t) dt... ...^\infty_{-\infty}\vert f_k (t)\vert^2dt\right]^{1/2}}\quad k = 0, 1, \cdots, N$$

Show that

(i)

$h_0(T)$ is the peak intensity, i.e., that

$$\vert h_k(T)\vert^2 \le \vert h_0(T)\vert^2\quad k = 0, 1, \cdots$$

(Nota bene: The function $h_0(T)$ corresponding to $f_0(t)$ is called the auto correlation function of $f_0(t)$ ). Also show that

(ii)

equality holds if the forcing function $f_k(t)$ has the form

$$f_k(t) = \kappa f_0 (T)\quad \kappa = \hbox {constant}$$

Lecture 14

Footnotes

...\space ²²

Not to be confused with the convolution integral Eq.(2.56) on page