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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

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

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

$\displaystyle L^2(-\infty,\infty)$ $\displaystyle \stackrel{\mathcal{F}^{-1}}{\longrightarrow} L^2(-\infty,\infty)$    
$\displaystyle \hat f(k)$ $\displaystyle \sim\leadsto \mathcal{F}^{-1}[\hat f](x)= \int_{-\infty}^\infty\frac{e^{ikx}}{\sqrt{2\pi}}\hat f(k)dk\equiv f(x)~.$ (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
$\displaystyle \int^\infty_{-\infty}\vert f(x)\vert^2 dx$ $\displaystyle =$ $\displaystyle \int^\infty_{-\infty} \overline{f}
  $\displaystyle =$ $\displaystyle \int^\infty_{-\infty}\overline f(x)\left[\int^\infty_{-\infty} \hat f(k')
\frac{e^{ik'x}}{\sqrt{2\pi}}dk'\right] dx$  
  $\displaystyle =$ $\displaystyle \int^\infty_{-\infty}\hat f(k')
\left[\int^\infty_{-\infty} \frac{e^{-ik'x}}{\sqrt{2\pi}}f(x)dx\right]
  $\displaystyle =$ $\displaystyle \int^\infty_{-\infty}
\hat f(k')\overline{\hat f}(k')dk'$  
  $\displaystyle =$ $\displaystyle \int^\infty_{-\infty} \vert \hat{f}(k)\vert^2 dk~~.$  

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

$\displaystyle \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,

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

$\displaystyle \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.

The Fourier transform, call it $ \cal F$ , is a linear one-to-one operator from the space of square-integrable functions onto itself. Indeed,

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

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

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

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

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

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

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


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

$\displaystyle \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

$\displaystyle \phi,~ {\cal{F}}\phi,~ {\cal{F}}^2\phi \equiv {\cal{F}}({\cal{F}}...
{\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

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

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)$ )
Exhibit a basis for $ W$ .
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).
Find the secular determinant, the eigenvalues and the corresponding eigenvectors of $ [{\cal{F}}]_B$ .
For $ W$ , exhibit an alternative basis which consists entirely of eigenvectors of $ {\cal{F}}$ , each one labelled by its respective eigenvalue.
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?


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

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

the equivalent width as

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

and the equivalent Fourier width as

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

Show that

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

is independent of the function $ f$ , and determine the value of this $ const.$
Determine the equivalent width and the equivalent Fourier width for the unnormalized Gaussian

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

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


Consider the auto-correlation $ h$ 22

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

of the function $ f$ whose Fourier transform is

$\displaystyle \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

$\displaystyle \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

$\displaystyle 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

$\displaystyle \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$ .
Compute the total energy

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

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

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


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

Consider the family of forcing functions

$\displaystyle \{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)

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

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

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

Lecture 14


...\space 22
Not to be confused with the convolution integral Eq.(2.56) on page  [*]

next up previous contents index
Next: Fourier Transform via Parseval's Up: The Fourier Integral Previous: The Fourier Integral Theorem   Contents   Index
Ulrich Gerlach 2010-12-09