Chapter 10. Fourier Transforms and the Dirac Delta Function

1 Vector Spaces in Physics 8/6/2015 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/ 2Tt 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.

2 As a starting point we rewrite the Fourier series, equation 9-39, as follows: /2/2()1nnitnnTitnTf tC enCf t edtT (10-1) The only change we have made is to add, in the upper expression, a factor of n for later use; 11nnn 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 8/6/2015 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 nnnnCC nnndn (10-13) We thus have /2/2()1in tnTin tTf tC n ednC nf t edtT (10-4) Next we change variables in the first integral from n to 2nnT : /2/2()21itTitTTf tCe dCf t edtT (10-5) Now define 2 TgC (10-6) This gives /2/21()212itTitTf tge dgf t edt (10-7) Finally, we take the limit T , giving the standard for m for the Fourier transform : 1()2itf tge d Inverse Fourier transform (10-8) 12itgf t edt Fourier transform (10-9) There are a lot of notable things about these relations.

3 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 are Vector Spaces in Physics 8/6/2015 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.

4 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( ): /4/4441212112sin4224ittTittTTTiigf t edtedteeiTTT (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.

5 B. The Dirac Delta Function (x). The Dirac Delta Function was introduced by the theoretical physicist 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 8/6/2015 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?

6 As an intellectual aid, let's compare this Function with the Kronecker Delta symbol, which (not coincidentally) has the same symbol: 1,1,031 iijijjiji (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 8/6/2015 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.

7 Its area is equal to ( )1Af x dxh 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 xe (10-16) in a form often used in statistics which is normalized so that 1)( dxxg , and so that the standard deviation of the distribution about x=0 is equal to . A graph of the Gaussian shape was given earlier in this Chapter ; the width of the curve at half maximum is about equal to 2.

8 (See figure 10-4.) It is clear that in the limit as goes to zero this Function is zero everywhere except at x = 0 (where it diverges, due to the factor 1 ), maintaining the normalization condition all the while. Properties of the Delta Function By making a change of variable one can define the Delta Function in a more general way, so that the special point where it diverges is x = a (rather than x=0): x) g(x) Figure 10-4. The Gaussian Function , becoming a Delta Function in the limit 0 . x a 1/a f(x) Figure 10-3. Rectangular Function , becoming a Delta Function in the limit a 0. Vector Spaces in Physics 8/6/2015 10 - 6 1)(,0)(dxaxaxax (10-17) Two useful properties of the Delta Function are given below: ( ) ()( )f xx a dxf a , (10-18) ( ) '()'( )f xx a dxf a , (10-19) Here the prime indicates the first derivative.

9 The property given in equation (10-18) is fairly easy to understand; while carrying out the integral, the argument vanishes except very near to x=a; so, it makes sense to replace f(a) by the constant value f(a) and take it out of the integral. The second property, Eqn. (10-19), can be demonstrated using integration by parts. The proof will be left to the problems. C. Application of the Dirac Delta Function to Fourier Transforms Another form of the Dirac Delta Function , given either in k-space or in -space, is the following: 00()0()01()21()2i k k xixedxk kedx . (10-20) We will not prove this fact, but just make an argument for its plausibility.

10 Look at the integral (10-20), for the case when k = k0. The exponential factor is just equal to 1 in that case, and it is clear that the integral diverges. On the other hand, if k is not equal to k0, it is plausible that the oscillating nature of the argument makes the integral vanish. If we accept these properties, we can interpret the Fourier transform as an expansion of a Function in terms of an orthonormal basis, just as the Fourier series is an expansion in terms of a series of orthogonal functions. Here is the picture. Basis states The functions tiee 21)( . (10-21) constitute a complete orthonormal basis for the space of ''smooth'' functions on the interval t.