The Fourier Transform

The Fourier Transform ,we move on in our program of decomposing arbitrary functions into sinusoids. We have seen how a periodic function can be expressed as a Fourier series,
so now we seek a similar representation for nonperiodic functions. To begin with, let’s assume we are given a nonperiodic function $F\left(t\right),\ -\infty <t<\infty$ , which is, say, continuously differentiable. Then if we pick an interval of the form $\ \left(-\frac{L}{2},\frac{L}{2}\right)$ we can represent $F\left(t\right)$ by a Fourier series for in this interval:

$F\left(t\right)=\sum _{n=-\infty }^{\infty }c_ne^{\frac{in2\pi t}{L}},\ \ \ \ \ -\frac{L}{2}<t<\frac{L}{2}$ (1)

with coefficients given by

$c_n=\frac{1}{L}\int _{-\frac{L}{2}}^{\frac{L}{2}}F\left(t\right)e^{\frac{in2\pi t}{L}}$ (n=0,±1,±2,…) (2)

Actually the series in Eq. (1) defines a periodic function $F_L\left(t\right),\ \ -\infty \ <t\ <\ \infty$ , which coincides with $F\left(t\right)$ on $\left(-\frac{L}{2},\frac{L}{2}\right)$ ;

[Notice that F(t) may be discontinuous even though F(t) is smooth.]
Thus we have a sinusoidal representation of $F\left(t\right)$ over an interval of length L. If we now let $L\longrightarrow \ \infty$ it seems reasonable to conjecture that this might evolve into a sinusoidal representation of $F\left(t\right)$ valid for all t. Let’s explore this possibility. sinuot → ∞, We are going to rewrite these equations in what will seen at first like a rather bizarre form, but it will aid in interpreting them as $L\longrightarrow \ \infty$ . We define $g_n$ to be $\frac{c_nL}{2\pi }$ , and introduce the factor [(n+1)-n]= 1 into the series in Eq. (1). Then we have

$F_L\left(t\right)=\sum _{n=-\infty }^{\infty }g_ne^{\frac{in2\pi t}{L}}\ \frac{\left[\left(n+1\right)-n\right]2\pi }{L}$ (3)

Figure 1.1: Periodic replica of F(t)

and

$g_n=\frac{1}{2\pi }\int _{-\frac{L}{2}}^{\frac{L}{2}}F\left(t\right)e^{-\frac{in2\pi t}{L}}dt.\ \$ (4)

Now write $\omega _n=\frac{n2\pi }{L}$ , producing

$F_L\left(t\right)=\sum _{n=-\infty }^{\infty }G_L\left(\omega _n\right)e^{i\omega _nt}\left(\omega _{n+1}-\omega _n\right)$ (5)

where the function $G_L\left(\omega \right)$ is defined for any real $\omega$ by

$G_L\left(\omega \right)=\frac{1}{2\pi }\int _{-\frac{L}{2}}^{\frac{L}{2}}F\left(t\right)e^{-i\omega t}dt.$ (6)

As L goes to infinity, $G_L\left(\omega \right)$ evolves rather naturally into a function $G\left(\omega \right)$ which is known as the Fourier transform of F:

$G\left(\omega \right)=\frac{1}{2\pi }\int _{-\frac{L}{2}}^{\frac{L}{2}}F\left(t\right)e^{-i\omega t}dt.\ \ \$ (7)

Moreover, since $\Delta \omega _n=\omega _{n+1}-\omega _n$ goes to zero as $L\longrightarrow \infty \$ and since $\omega _n$ , ranges from $-\infty \ to\ \infty$ , Eq. (5) begins to look very much like a Riemann sum for the integral

$\int _{-\infty }^{\infty }G\left(\omega \right)e^{i\omega t}d\omega .\ \$

Thus we are led to propose the equality

$F\left(t\right)=\int _{-\infty }^{\infty }G\left(\omega \right)e^{i\omega t}d\omega$ (8)

for nonperiodic F, when G is defined by Eq. (7). Equation (8) is called the Fourier
inversion formula.
Equations (7) and (8) are the essence of Fourier transform theory. As is suggested by this discussion, it is often profitable to indulge one’s whimsy and think of the integral in Eq. (8) as a generalized “sum” of sinusoids, summed over a continuum of frequencies $\omega$ . Equation (7) then dictates the “coefficients,” $G\left(\omega \right)d\omega$ , in the sum,

Example 1

Find the Fourier transform and verify the inversion formula for the function
$F\left(t\right)=\frac{1}{t^2+4}.$

Solution.

Observe that

$F\left(t\right)=\frac{1}{t^2+4}=\frac{1}{\left(t-2i\right)\left(t+2i\right)}.$

is analytic except for simple poles at t= ±2. We shall use residue theory to evaluate the Fourier transform, interpreting the integral as a principal value:

$G\left(\omega \right)=\frac{1}{2\pi }p.v\int _{-\infty }^{\infty }\frac{e^{-i\omega t}}{t^2+4}dt.$

If $\omega > 0$ , we close the contour with expanding semicircles in the lower half-plane; we find

$\ G\left(\omega \right)=\frac{1}{2\pi }\left(-2\pi i\right)Res\left(\frac{e^{-i\omega t}}{t^2+4};\ -2i\right)$

$G\left(\omega \right)=-i.\ \lim _{t\to -2i}\ \frac{e^{-i\omega t}}{t-2i}=\frac{e^{-2\omega }}{4\ }\ \ \ \ \ \ \left(\omega \ge 0\right).$

Similarly, for $\omega <0$ we close in the upper half-plane and find
$G\left(\omega \right)=\frac{1}{2\pi }\left(2\pi i\right)Res\left(\frac{e^{-i\omega t}}{t^2+4};\ -2i\right)=\frac{e^{2\omega }}{4\ }\ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(\omega <0\right)$
In short

$G\left(\omega \right)=\frac{e^{-2\left|\omega \right|}}{4}$ .

To verify the Fourier inversion formula we compute

$\int _{-\infty }^{\infty }G\left(\omega \right)e^{i\omega t}d\omega =\int _{-\infty }^{\infty }\frac{e^{-2\left|\omega \right|}}{4}e^{i\omega t}d\omega .$

By symmetry, the imaginary part vanishes, and this integral equals

$Re\int _{-\infty }^{\infty }\frac{e^{-2\left|\omega \right|}}{4}e^{i\omega t}d\omega =2.Re\ \int _0^{\infty }\frac{e^{-2\omega }}{4}.e^{i\omega t}d\omega .$

$=\frac{1}{2}Re\ \left|\frac{e^{\left(-2+it\right)\omega }}{-2+it}\right|_{\omega =0}^{\infty }=\frac{1}{t^2+4}.$

Hence

$\frac{1}{t^2+4}=\int _{-\infty }^{\infty }\frac{e^{-2\left|\omega \right|}}{4}e^{i\omega t}d\omega$

As is the case of Fourier series, a wealth of theorems has been discovered stating conditions under which the Fourier integral representations (7) and (8) are valid. A useful one for applications deals with piecewise smooth functions F(t) ; that is, on every bounded interval F(t) is continuously differentiable for all but the finite number of values $t=\tau _1,\tau _2,\tau _3,...,\tau _n$ and at each $\tau _j$ the “one-sided limits” of F(t) and F'(t) exist. Note the principal value interpretation of the integral is called for in the theorem; this ensures that the inverse transform converges at the points of discontinuity.

Theorem 4.

Suppose that F(t) is piecewise smooth on every bounded interval and that $\int _{-\infty }^{\infty }\left|F\left(t\right)dt\right|$ exists. Then the Fourier transform, $G\left(\omega \right)$ , of F exists and

$p.v\ \int _{-\infty }^{\infty }G\left(\omega \right)e^{i\omega t}d\omega =\ \begin{cases}F\left(t\right)&\ where\ F\ is\ continuous\\ \frac{F\left(t+\right)+F\left(t-\right)}{2}&\ \ otherwise.\end{cases}$

Example 2

Find the Fourier transform of the function

$F\left(t\right)=\ \begin{cases}1,&\ -\pi \le t\le \pi \\ 0,&\ \ otherwise.\end{cases}$

and confirm the inversion formula

Figure:8.5 ‘Boxear’ function

Solution

we have

$G\left(\omega \right)=\frac{1}{2\pi }-\infty \int _{-\pi }^{\pi }\left(1\right)e^{i\omega t}d\omega =\frac{\sin \omega \pi }{\omega \pi }$ .

Hence theorem 4 tells us that

$p.v\ \int _{-\infty }^{\infty }\frac{\sin \omega \pi }{\omega \pi }e^{i\omega t}d\omega =f(x)=\begin{cases}1,&\ \ \ \ \left|t\right|<\pi \\ 0,&\ \ \ \ \left|t\right|>\pi \\ \frac{1}{2,}&\ \ \ \ \ t=\pm \pi \end{cases}$ (10)

To confirm this, rewrite the left-hand side of (10) as

$p.v\ \frac{.1}{2\pi i\ }\int _{-\infty }^{\infty }\frac{e^{i\omega \left(\pi +1\right)}-e^{i\omega \left(-\pi +1\right)}}{\omega }d\omega$ (11)

Now recall from Example 1, that

$p.v\int _{-\infty }^{\infty }\frac{e^{ix}}{x}dx=i\pi$
which, with the changes of variables $x=C\omega$ ,

generalizes to

$p.v\ \int _{-\infty }^{\infty }\frac{e^{iC\omega }}{\omega }d\omega =\begin{cases}i\pi ,&\ \ \ \ if\ C>0\\ -i\pi ,&\ \ \ \ \ if\ C<0\end{cases}$

Of course,

$p.v\ \int _{-\infty }^{\infty }\frac{1}{x}dx=0.$

Therefore we derive

if $t<-\pi ,$ then (11) becomes $\frac{1}{2\pi i}\left[-i\pi -\left(-i\pi \right)\right]=0;$

if $t=-\pi$ then (11) becomes $\frac{1}{2\pi i}\left[0-\left(-i\pi \right)\right]=\frac{1}{2};$

if $-\pi <t<\pi$ then (11) becomes $\frac{1}{2\pi i}\left[i\pi -\left(-i\pi \right)\right]=1;$

if $t=\pi$ then (11) becomes $\frac{1}{2\pi i}\left[i\pi -0\right]=\frac{1}{2};$

if $\pi <t$ then (11) becomes $\frac{1}{2\pi i}\left[i\pi -i\pi \right]=0$ .

Example 3

Find the Fourier transform of the function

$F\left(t\right)=\begin{cases}\sin t,&\ \ \ \ \left|t\right|\le 6\pi ,\\ 0,&\ \ \ \ \ otherwise.\end{cases}$

(Fig. 8.6), and confirm the inversion formula. (Physicists call this function a finite
wave train.)

Figure 8.6: Finite wave train

Solution

We have

$G\left(\omega \right)=\frac{1}{2\pi }\int _{-6\pi }^{6\pi }\left(\sin t\right)e^{-i\omega t}dt=\frac{i\sin 6\pi \omega }{\pi \left(1-\omega ^2\right)}$

Since F(t) is continuous everywhere, the inversion formula implies

$\int _{-\infty }^{\infty }\frac{i\sin 6\pi \omega }{\pi \left(1-\omega ^2\right)}e^{i\omega t}d\omega =F\left(t\right)$

To confirm this rewrite the left-hand side as

$\frac{i}{2i\pi }\int _{-\infty }^{\infty }\frac{e^{i\omega \left(6\pi +t\right)-}e^{i\omega \left(-6\pi +t\right)}}{\left(1-\omega ^2\right)}d\omega =p.v.-\frac{1}{2\pi }\int _{-\infty }^{\infty }\frac{e^{i\omega \left(6\pi +t\right)-}e^{i\omega \left(-6\pi +t\right)}}{\left(\omega -1\right)\left(\omega +1\right)}d\omega$

(because of the removable singularities at $\omega$ = ±1).
Now the integral

$p.v.\frac{-1}{2\pi }\int _{-\infty }^{\infty }\frac{e^{i\omega \left(6\pi +t\right)}}{\left(\omega -1\right)\left(\omega +1\right)}d\omega$

can be evaluated using the indented-contour techniques. For $t\ge -6\pi$ wе employ the contour shown in Fig. 8.7(a) and invoke Lemmas 3 and 4 to obtain

$p.v.\frac{-1}{2\pi }\int _{-\infty }^{\infty }\frac{e^{i\omega \left(6\pi +t\right)}}{\left(\omega -1\right)\left(\omega +1\right)}d\omega$ = $\frac{-1}{2\pi }\left(\pi i\right)\left\{Res\left(-1\right)+Res\left(1\right)\right\}\ \ \$

$p.v.\frac{-1}{2\pi }\int _{-\infty }^{\infty }\frac{e^{i\omega \left(6\pi +t\right)}}{\left(\omega -1\right)\left(\omega +1\right)}d\omega$ = $\frac{-i}{2}\left[\frac{e^{-i\left(6\pi +t\right)}}{2}+\frac{e^{i\left(6\pi +t\right)}}{2}\right]$

$p.v.\frac{-1}{2\pi }\int _{-\infty }^{\infty }\frac{e^{i\omega \left(6\pi +t\right)}}{\left(\omega -1\right)\left(\omega +1\right)}d\omega$ = $\frac{\sin \left(6\pi +t\right)}{2}=\frac{\sin t}{2}$

Similarly, for $t\le -6\pi \$ we use the contour of Fig. 8.7(b) and find

$p.v\frac{-1}{2\pi }\int _{-\infty }^{\infty }\frac{e^{-i\omega \left(6\pi +t\right)}}{\left(\omega -1\right)\left(\omega +1\right)}d\omega =\frac{-1}{2\pi }\left(-\pi i\right)\left\{Res\left(-1\right)+Res\left(1\right)\right\}$

$p.v\frac{-1}{2\pi }\int _{-\infty }^{\infty }\frac{e^{-i\omega \left(6\pi +t\right)}}{\left(\omega -1\right)\left(\omega +1\right)}d\omega =$ [latex]\frac{-\sin t}{2}[/latex]

Figure 8.7: Contours for Example 3.

By the reasoning we obtains

$p.v.\frac{-1}{2\pi }\int _{-\infty }^{\infty }\frac{e^{i\omega \left(6\pi +t\right)}}{\left(\omega -1\right)\left(\omega +1\right)}d\omega =\begin{cases}\frac{\sin t}{2}&\ if\ t\ge 6\pi .\\ -\frac{\sin t}{2}&\ \ if\ t\le 6\pi .\end{cases}$