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Unformatted text preview: ACM 95/100c Problem Set 4 Solutions Lei Zhang April 22, 2006 Each problem is worth 10 points  each part of a multipart problem is weighted equally Problem 1 (Collaboration allowed) The following problem examines the Fourier transform for the function but also interprets the function as an extension of the Kronecker that appears when we examine the orthogonality of SturmLiouville eigenfunctions on finite domains (like sines and cosines). (a) For what does the function exp( ( x x ) 2 ) have unit area for < x < ? (b) Show that in the limit as ,the resulting function in part (a) satisfies the properties of the Dirac delta function ( x x ). (c) Obtain the Fourier transform of ( x x ) in two ways: i. Take the transform of part (a) and take the limit as . ii. Use the integration properties of the Dirac delta function. (d) Show that the transform of ( x x ) is consistent with the following idea: Transforms of sharply peaked functions are spread out (that is they contain lots of frequencies). (e) Interpret the following equation (  ) = 1 2 Z  exp( ix (  )) dx. as a type of orthogonality relationship for the eigenfunctions exp( ix ). (f) Use this orthogonality relationship to compute F ( ) if f ( x ) = Z  F ( )exp( ix ) d. This is a way of computing the Fourier coefficients of f ( x ) otherwise known as the Fourier transform of f ( x ) using the appropriate orthogonality condition and the appropriate eigenfunctions. Solution 1 (a) For the function to have unit area for < x < means that 1 Z  exp  ( x x ) 2 / dx = Z  exp ( y 2 ) dy = Z  exp ( z 2 ) dz = = 1 . (1) So, = r . (2) (b) Let f ( x ) r exp  ( x x ) 2 / . (3) For x 6 = x , lim f ( x ) = 0 , (4) since exp  ( x x ) 2 / decays in faster than grows. Also, f ( x ) = q becomes unbounded as . Finally, by part a), Z  f ( x ) dx = 1 (5) for all . Thus, as , f ( x ) satisfies the properties of ( x x ). 2 (c) i. 1 2 Z  r exp  ( x x ) 2 / exp( ix ) dx = r exp ( x 2 ) 1 2 Z  exp  x 2 2 xx i x dx = r exp ( x 2 ) 1 2 Z  exp " x x i 2 2 + x + i 2 2 # dx = r exp x 2 + x 2 + ix 2 4 1 2 Z  exp " x x i 2 2 # dx = 1 2 exp ix 2 4 r Z  exp ( y 2 ) dy = 1 2 exp ix 2 4 1 2 exp( ix ) as . (6) ii. By the properties of the Dirac function, 1 2 Z  ( x x )exp( ix ) dx = 1 2 exp( ix ) . (7) Note (not necessary for students to state): It requires complex analysis to justify Z  exp " x x + i 2 2 # dx = Z  exp ( y 2 ) dy. (8) This justification is not required here, though. See Habermans Appendix to 10.3 for details.This justification is not required here, though....
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 Spring '09
 NilesA.Pierce

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