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Fourier Transforms

Consider a function $ F(x)$ that is periodic in $ x$ with period $ L$ . In other words,

$\displaystyle F(x+L) = F(x)$ (698)

for all $ x$ . Recall, from Section 6.4, that we can represent such a function as a Fourier series: that is,

$\displaystyle F(x)= \sum_{n=1,\infty} \left[C_n \cos(n \delta k x) + S_n \sin(n \delta k x)\right],$ (699)

where

$\displaystyle \delta k = \frac{2\pi}{L}.$ (700)

[We have neglected the $ n=0$ term in Equation (699), for the sake of convenience.] Equation (699) automatically satisfies the periodicity constraint (698), because $ \cos(\theta+n\,2\pi)=\cos\theta$ and $ \sin(\theta+n 2\pi)=\sin\theta$ for all $ \theta$ and $ n$ (with the proviso that $ n$ is an integer). The so-called Fourier coefficients, $ C_n$ and $ S_n$ , appearing in Equation (699), can be determined from the function $ F(x)$ by means of the following readily demonstrated (see Exercise 1) results:

$\displaystyle \frac{2}{L}\int_{-L/2}^{L/2}\cos(n \delta k x) \cos(n' \delta k  x) dx$ $\displaystyle = \delta_{n,n'},$ (701)
$\displaystyle \frac{2}{L}\int_{-L/2}^{L/2}\sin(n \delta k x) \sin(n' \delta k  x) dx$ $\displaystyle = \delta_{n,n'},$ (702)
$\displaystyle \frac{2}{L}\int_{-L/2}^{L/2}\cos(n \delta k x) \sin(n' \delta k  x) dx$ $\displaystyle =0,$ (703)

where $ n$ , $ n'$ are positive integers. Here, $ \delta_{n,n'}$ is a Kronecker delta function. In fact,

$\displaystyle C_n$ $\displaystyle =\frac{2}{L}\int_{-L/2}^{L/2} F(x) \cos(n \delta k x) dx,$ (704)
$\displaystyle S_n$ $\displaystyle =\frac{2}{L}\int_{-L/2}^{L/2} F(x) \sin(n \delta k x) dx.$ (705)

(See Exercise 1.) Incidentally, any periodic function of $ x$ can be represented as a Fourier series.

Suppose, however, that we are dealing with a function $ F(x)$ that is not periodic in $ x$ . We can think of such a function as one that is periodic in $ x$ with a period $ L$ that tends to infinity. Does this mean that we can still represent $ F(x)$ as a Fourier series? Consider what happens to the series (699) in the limit $ L\rightarrow\infty$ , or, equivalently, $ \delta k\rightarrow 0$ . The series is basically a weighted sum of sinusoidal functions whose wavenumbers take the quantized values $ k_n=n\,\delta k$ . Moreover, as $ \delta k\rightarrow 0$ , these values become more and more closely spaced. In fact, we can write

$\displaystyle F(x) = \sum_{n=1,\infty} \frac{C_n}{\delta k}\,\cos(n\,\delta k\,...
...lta k+ \sum_{n=1,\infty} \frac{S_n}{\delta k} \,\sin(n\,\delta k\,x)\,\delta k.$ (706)

In the continuum limit, $ \delta k\rightarrow 0$ , the summations in the previous expression become integrals, and we obtain

$\displaystyle F(x)=\int_{-\infty}^\infty C(k) \cos(k x) dk + \int_{-\infty}^\infty S(k) \sin(k x) dk,$ (707)

where $ k=n\,\delta k$ , $ C(k)=C(-k)= C_n/(2\,\delta k)$ , and $ S(k)=-S(-k)=S_n/(2\,\delta k)$ . Thus, for the case of an aperiodic function, the Fourier series (699) morphs into the so-called Fourier transform (707). This transform can be inverted using the continuum limits (i.e., the limit $ \delta k\rightarrow 0$ ) of Equations (704) and (705), which are readily shown to be

$\displaystyle C(k)$ $\displaystyle =\frac{1}{2\pi}\int_{-\infty}^\infty F(x) \cos(k x) dx,$ (708)
$\displaystyle S(k)$ $\displaystyle =\frac{1}{2\pi}\int_{-\infty}^\infty F(x) \sin(k x) dx,$ (709)

respectively. (See Exercise 5.) The previous equations confirm that $ C(-k)=C(k)$ and $ S(-k)=-S(k)$ . The Fourier-space (i.e., $ k$ -space) functions $ C(k)$ and $ S(k)$ are known as the cosine Fourier transform and the sine Fourier transform of the real-space (i.e., $ x$ -space) function $ F(x)$ , respectively. Furthermore, because we already know that any periodic function can be represented as a Fourier series, it seems plausible that any aperiodic function can be represented as a Fourier transform. This is indeed the case. Equations (707)-(709) effectively enable us to represent a general function as a linear superposition of sine and cosine functions. Let us consider some examples.

Figure 50: Fourier transform of a top-hat function.
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Consider the ``top-hat'' function,

$\displaystyle F(x)=\left\{ \begin{array}{ccc} 1 &\mbox{\hspace{1cm}}& \vert x\vert\leq l/2 0 && \vert x\vert>l/2 \end{array}\right..$ (710)

(See Figure 50.) Given that $ \cos(-k\,x)=\cos(k\,x)$ and $ \sin(-k x)=-\sin(k x)$ , it follows from Equations (708) and (709) that if $ F(x)$ is even in $ x$ , so that $ F(-x)=F(x)$ , then $ S(k)=0$ , and if $ F(x)$ is odd in $ x$ , so that $ F(-x)=-F(x)$ , then $ C(k)=0$ . Hence, because the top-hat function (710) is even in $ x$ , its sine Fourier transform is automatically zero. On the other hand, its cosine Fourier transform takes the form

$\displaystyle C(k) = \frac{1}{2\pi}\int_{-l/2}^{l/2} \cos(k  x) dx = \frac{l}{2\pi} \frac{\sin(k l/2)}{k l/2}.$ (711)

Figure 50 shows the function $ F(x)$ , together with its associated cosine transform, $ C(k)$ .

Figure 51: A Gaussian function.
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As a second example, consider the so-called Gaussian function,

$\displaystyle F(x) =\exp\left(-\frac{x^2}{2\,\sigma_x^{\,2}}\right).$ (712)

As illustrated in Figure 51, this is a smoothly varying even function of $ x$ that attains its peak value $ 1$ at $ x=0$ , and becomes completely negligible when $ \vert x\vert\gtrsim 3 \sigma_x$ . Thus, $ \sigma_x$ is a measure of the ``width'' of the function in real space. By symmetry, the sine Fourier transform of the preceding function is zero. On the other hand, the cosine Fourier transform is readily shown to be

$\displaystyle C(k) = \frac{1}{(2\pi\,\sigma_k^{\,2})^{1/2}}\,\exp\left(-\frac{k^2}{2\,\sigma_k^{\,2}}\right),$ (713)

where

$\displaystyle \sigma_k=\frac{1}{\sigma_x}.$ (714)

(See Exercise 2.) This function is a Gaussian in Fourier space of characteristic width $ \sigma_k=1/\sigma_x$ . The original function $ F(x)$ can be reconstructed from its Fourier transform using

$\displaystyle F(x)=\int_{-\infty}^\infty C(k) \cos(k x) dk.$ (715)

This reconstruction is simply a linear superposition of cosine waves of differing wavenumbers. Moreover, $ C(k)$ can be interpreted as the amplitude of waves of wavenumber $ k$ within this superposition. The fact that $ C(k)$ is a Gaussian of characteristic width $ \sigma_k=1/\sigma_x$ [which means that $ C(k)$ is negligible for $ \vert k\vert\gtrsim 3 \sigma_k$ ] implies that in order to reconstruct a real space function whose width in real space is approximately $ \sigma_x$ it is necessary to combine sinusoidal functions with a range of different wavenumbers that is approximately $ \sigma_k=1/\sigma_x$ in extent. To be slightly more exact, the real-space Gaussian function $ F(x)$ falls to half of its peak value when $ \vert x\vert\simeq \sqrt{\pi/2} \sigma_x$ . Hence, the full width at half maximum of the function is $ \Delta x \simeq 2 \sqrt{\pi/2} \sigma_x=\sqrt{2\pi} \sigma_x$ . Likewise, the full width at half maximum of the Fourier-space Gaussian function $ C(k)$ is $ \Delta k \simeq \sqrt{2\pi} \sigma_k$ . Thus,

$\displaystyle \Delta x \Delta k \simeq 2\pi,$ (716)

since $ \sigma_k\,\sigma_x=1$ . We conclude that a function that is highly localized in real space has a transform that is highly delocalized in Fourier space, and vice versa. Finally,

$\displaystyle \int_{-\infty}^\infty \frac{1}{(2\pi\,\sigma_k^{\,2})^{1/2}}\,\exp\left(-\frac{k^2}{2\,\sigma_k^{\,2}}\right) dk = 1.$ (717)

(See Exercise 3.) In other words, a Gaussian function in real space, of unit height and characteristic width $ \sigma_x$ , has a cosine Fourier transform that is a Gaussian in Fourier space, of characteristic width $ \sigma_k=1/\sigma_x$ , and whose integral over all $ k$ -space is unity.

Consider what happens to the previously mentioned real-space Gaussian, and its Fourier transform, in the limit $ \sigma_x\rightarrow \infty$ , or, equivalently, $ \sigma_k\rightarrow 0$ . There is no difficulty in seeing, from Equation (712), that

$\displaystyle F(x)\rightarrow 1.$ (718)

In other words, the real space Gaussian morphs into a function that takes the constant value unity everywhere. The Fourier transform is more problematic. In the limit $ \sigma_k\rightarrow 0$ , Equation (713) yields a $ k$ -space function that is zero everywhere apart from $ k=0$ (because the function is negligible for $ \vert k\vert\gtrsim \sigma_k$ ), where it is infinite [because the function takes the value $ (2\pi \sigma_k)^{-1/2}$ at $ k=0$ ]. Moreover, according to Equation (717), the integral of the function over all $ k$ remains unity. Thus, the Fourier transform of the uniform function $ F(x)=1$ is a sort of integrable ``spike'' located at $ k=0$ . This unusual function is known as the Dirac delta function, and is denoted $ \delta (k)$ . Thus, one definition of a delta function is

$\displaystyle \delta (k)= \lim_{\sigma_k\rightarrow 0}\frac{1}{(2\pi\,\sigma_k^{\,2})^{1/2}}\,\exp\left(-\frac{k^2}{2\,\sigma_k^{\,2}}\right).$ (719)

As has already been mentioned, $ \delta(k)=0$ for $ k\neq 0$ , and $ \delta(0)=\infty$ . Moreover,

$\displaystyle \int_{-\infty}^\infty\delta (k) dk = 1.$ (720)

Consider the integral

$\displaystyle \int_{-\infty}^\infty F(k) \delta(k) dk,$ (721)

where $ F(k)$ is an arbitrary function. Because of the peculiar properties of the delta function, the only contribution to the previous integral comes from the region in $ k$ -space in the immediate vicinity of $ k=0$ . Furthermore, provided $ F(k)$ is well-behaved in this region, we can write

$\displaystyle \int_{-\infty}^{\infty} F(k) \delta(k) dk = \int_{-\infty}^\infty F(0) \delta (k) dk=F(0) \int_{-\infty}^{\infty} \delta(k) dk = F(0),$ (722)

where use has been made of Equation (720).

A change of variables allows us to define $ \delta(k-k')$ , which is a ``spike'' function centered on $ k=k'$ . The previous result can be generalized to give

$\displaystyle \int_{-\infty}^{\infty} F(k) \delta(k-k') dk = F(k'),$ (723)

for all $ F(k)$ . Indeed, this expression can be thought of as an alternative definition of a delta function.

We have seen that the delta function $ \delta (k)$ is the cosine Fourier transform of the uniform function $ F(x)=1$ . It, thus, follows from Equation (708) that

$\displaystyle \delta (k) = \frac{1}{2\pi}\int_{-\infty}^{\infty} \cos(k x) dx.$ (724)

This result represents yet another definition of the delta function. By symmetry, we also have

$\displaystyle 0 = \frac{1}{2\pi}\int_{-\infty}^{\infty} \sin(k x) dx.$ (725)

It follows that

$\displaystyle \frac{1}{2\pi}\int_{-\infty}^\infty \cos(k x) \cos(k' x) dx =...
...}^\infty\left\{\cos\left[(k-k') x\right]+\cos\left[(k+k') x\right]\right\}dx,$ (726)

which yields

$\displaystyle \frac{1}{2\pi}\int_{-\infty}^\infty \cos(k x) \cos(k' x) dx= \frac{1}{2}\left[\delta(k-k') + \delta(k+k')\right],$ (727)

where use has been made of Equation (724), and a standard trigonometric identity. (See Appendix B.) Likewise,

$\displaystyle \frac{1}{2\pi}\int_{-\infty}^\infty \sin(k x) \sin(k' x) dx$ $\displaystyle = \frac{1}{2}\left[\delta(k-k') - \delta(k+k')\right],$ (728)
$\displaystyle \frac{1}{2\pi}\int_{-\infty}^\infty \cos(k x) \sin(k' x) dx$ $\displaystyle =0.$ (729)

(See Exercise 4.) Incidentally, Equations (727)-(729) can be used to derive Equations (708) and (709) directly from Equation (707). (See Exercise 5.)


next up previous
Next: General Solution of 1D Up: Wave Pulses Previous: Wave Pulses
Richard Fitzpatrick 2013-04-08