Vector Spaces in Physics 8/16/2014 10 - 1 Chapter 10. Fourier Transforms and the Dirac Delta Function A. The Fourier transform. The Fourier-series expansions which we have discussed are valid for functions either defined over a finite range (/2 /2T t T , for instance) or extended to all values of time as a periodic function. This does not cover the important case of a single, isolated pulse. But we can approximate an isolated pulse by letting the boundaries of the region of the Fourier series recede farther and farther away towards , as shown in figure 10-1. We will now outline the corresponding mathematical limiting process. It will transform the Fourier series, a superposition of sinusoidal waves with discrete frequencies n, into a superposition of a continuous spectrum of frequencies . As a starting point we rewrite the Fourier series, equation 9-39, as follows: /2/2()1nnitnnTitnTf t C e nC f t e dtT (10-1) The only change we have made is to add, in the upper expression, a factor of n for later use; 11n n n is the range of the variable n for each step in the summation. We now imagine letting T get larger and larger. This means that the frequencies Figure10-1. Evolution of a periodic train of pulses into a single isolated pulse, as the domain of the Fourier series goes from [-T/2, T/2] to [-, ].Vector Spaces in Physics 8/16/2014 10 - 2 2nnT (10-2) in the sum get closer and closer together. In the large-n approximation we can replace the integer variable n by a continuous variable n, so that nnnnC C nnn dn (10-13) We thus have /2/2()1i n tnTi n tTf t C n e dnC n f t e dtT (10-4) Next we change variables in the first integral from n to 2 nnT: /2/2()21itTitTTf t C e dC f t e dtT (10-5) Now define 2TgC (10-6) This gives /2/21()212itTitTf t g e dg f t e dt (10-7) Finally, we take the limit T , giving the standard for m for the Fourier transform: 1()2itf t g e dInverse Fourier Transform (10-8) 12itg f t e dtFourier Transform (10-9) There are a lot of notable things about these relations. First, there is a great symmetry in the roles of time and frequency; a function is completely specified either by f(t) or by g(). Describing a function with f(t) is sometimes referred to as working in the "time domain," while using g() is referred to as working in the "frequency domain." Second, both of these expressions have the form of an expansion of a function in terms of a set of basis functions. For f(t), the basis functions areVector Spaces in Physics 8/16/2014 10 - 3 12ite; for g(), the complex conjugate of this function, 12ite, is used. Finally, the function g() emerges as a measure of the "amount" of frequency which the function f(t) contains. In many applications, plotting g() gives more information about the function than plotting f(t) itself. Example - the Fourier transform of the square pulse. Let us consider the case of an isolated square pulse of length T, centered at t = 0: 1,()440 otherwiseTTtft (10-10) This is the same pulse as that shown in figure 9-3, without the periodic extension. It is straightforward to calculate the Fourier transform g(): /4/4441212112sin4224ittTittTTTiig f t e dte dteeiTTT (10-11) Here we have used the relation sin2iieei. We have also written the dependence on in the form sinsincxxx. This well known function peaks at zero and falls off on both sides, oscillating as it goes, as shown in figure 10-2. B. The Dirac delta function (x). The Dirac delta function was introduced by the theoretical physicist P.A.M. Dirac, to describe a strange mathematical object which is not even a proper mathematical function, but which has many uses in physics. The Dirac delta function is more properly referred to as a distribution, and Dirac played a hand in developing the theory of distributions. Here is the definition of (x): 1)(0,0)(dxxxx (10-12)Vector Spaces in Physics 8/16/2014 10 - 4 Isn't this a great mathematical joke? This function is zero everywhere! Well, almost everywhere, except for being undefined at x=0. How can this be of any use? In particular, how can its integral be anything but zero? As an intellectual aid, let's compare this function with the Kronecker delta symbol, which (not coincidentally) has the same symbol: 1,1,031iijijjiji (10-13) There are some similarities. But the delta function is certainly not equal to 1 at x = 0; for the integral over all x to be equal to 1, (x) must certainly diverge at x = 0 In fact, all the definitions that I know of a Dirac delta function involve a limiting procedure, in which (x) goes to infinity. Here are a couple of them. The rectangular delta function Consider the function Figure10-2. The Fourier transform of a single square pulse. This function is sometimes called the sync function.Vector Spaces in Physics 8/16/2014 10 - 5 01/ x2( ) lim0x2aaxa (10-14) This function, shown in figure 10-3, is a rectangular pulse of width a and height h = 1/a. Its area is equal to ( ) 1A f x dx h a , so it satisfies the integral requirement for the delta function. And in the limit that a 0, it vanishes at all points except x = 0. This is one perfectly valid representation of the Dirac delta function. The Gaussian delta function Another example, which has the advantage of being an analytic function, is . 222011( ) lim2xxe (10-15) The function inside the limit is the Gaussian function, 22211()2xg x e (10-16) in a form often used in statistics which is normalized
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