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Unformatted text preview: Fourier Analysis Fourier series allow you to expand a function on a finite interval as an infinite series of trigonometric functions. What if the interval is infinite? That’s the subject of this chapter. Instead of a sum over frequencies, you will have an integral. 15.1 Fourier Transform For the finite interval you have to specify the boundary conditions in order to determine the particular basis that you’re going to use. On the infinite interval you don’t have this large set of choices. After all, if the boundary is infinitely far away, how can it affect what you’re doing over a finite distance? But see section 15.6 . In section 5.3 you have several boundary condition listed that you can use on the differential equation u 00 = λu and that will lead to orthogonal functions on your interval. For the purposes here the easiest approach is to assume periodic boundary conditions on the finite interval and then to take the limit as the length of the interval approaches infinity. On L < x < + L , the conditions on the solutions of u 00 = λu are then u ( L ) = u (+ L ) and u ( L ) = u (+ L ) . The solution to this is most conveniently expressed as a complex exponential, Eq. ( 5.19 ) u ( x ) = e ikx , where u ( L ) = e ikL = u ( L ) = e ikL This implies e 2 ikL = 1 , or 2 kL = 2 nπ , for integer n = 0 , ± 1 , ± 2 ,... . With these solutions, the other condition, u ( L ) = u (+ L ) is already satisfied. The basis functions are then u n ( x ) = e ik n x = e nπix/L , for n = 0 , ± 1 , ± 2 , etc. (15 . 1) On this interval you have the Fourier series expansion f ( x ) = ∞ X∞ a n u n ( x ) , and u m ,f = u m , ∞ X∞ a n u n = a m u m ,u m (15 . 2) In the basis of Eq. ( 15.1 ) this normalization is u m ,u m = 2 L . Insert this into the series for f . f ( x ) = ∞ X n =∞ u n ,f u n ,u n u n ( x ) = 1 2 L ∞ X n =∞ u n ,f u n ( x ) Now I have to express this in terms of the explicit basis functions in order to manipulate it. When you use the explicit form you have to be careful not to use the same symbol ( x ) for two different things in the same expression. Inside the u n ,f there is no “ x ” left over — it’s the dummy variable of integration and it is not the same x that is in the u n ( x ) at the end. Denote k n = πn/L . f ( x ) = 1 2 L ∞ X n =∞ Z L L dx u n ( x ) * f ( x ) u n ( x ) = 1 2 L ∞ X n =∞ Z L L dx e ik n x f ( x ) e ik n x Now for some manipulation: As n changes by 1, k n changes by Δ k n = π/L . f ( x ) = 1 2 π ∞ X n =∞ π L Z L L dx e ik n x f ( x ) e ik n x = 1 2 π ∞ X n =∞ e ik n x Δ k n Z L L dx e ik n x f ( x ) (15 . 3) James Nearing, University of Miami 1 15—Fourier Analysis 2 For a given value of k , define the integral g L ( k ) = Z L L dx e ikx f ( x ) If the function f vanishes sufficiently fast as x → ∞ , this integral will have a limit as L → ∞ . Call that limit g ( k ) . Look back at Eq. ( 15.3 ) and you see that for large L the last factor will be approximately...
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This note was uploaded on 01/30/2012 for the course PHYS 315 taught by Professor Nearing during the Fall '08 term at University of Miami.
 Fall '08
 Nearing

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