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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 δfunction but also interprets the δfunction as an extension of the Kronecker δ that appears when we examine the orthogonality of SturmLiouville eigenfunctions on finite domains (like sines and cosines). (a) For what α does the function α exp( β ( x x ) 2 ) have unit area for∞ < x < ∞ ? (b) Show that in the limit as β → ∞ ,the resulting function in part (a) satisfies the properties of the Dirac delta function δ ( x x ). (c) Obtain the Fourier transform of δ ( x x ) in two ways: i. Take the transform of part (a) and take the limit as β → ∞ . ii. Use the integration properties of the Dirac delta function. (d) Show that the transform of δ ( 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 δ ( ω ω ) = 1 2 π Z ∞∞ exp( ix ( ω ω )) dx. as a type of orthogonality relationship for the “eigenfunctions” exp( iωx ). (f) Use this orthogonality relationship to compute F ( ω ) if f ( x ) = Z ∞∞ F ( ω )exp( iωx ) dω. 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 ∞∞ α exp £ β ( x x ) 2 / dx = Z ∞∞ α exp ( βy 2 ) dy = Z ∞∞ α √ β exp ( z 2 ) dz = α √ β √ π = 1 . (1) So, α = r β π . (2) (b) Let f ( x ) ≡ r β π exp £ β ( x x ) 2 / . (3) For x 6 = x , lim β →∞ f ( x ) = 0 , (4) since exp £ β ( x x ) 2 / decays in β faster than √ β grows. Also, f ( x ) = q β π becomes unbounded as β → ∞ . Finally, by part a), Z ∞∞ f ( x ) dx = 1 (5) for all β . Thus, as β → ∞ , f ( x ) satisfies the properties of δ ( x x ). 2 (c) i. 1 2 π Z ∞∞ r β π exp £ β ( x x ) 2 / exp( iωx ) dx = r β π exp ( βx 2 ) 1 2 π Z ∞∞ exp • β x 2 2 xx i ω β x ¶‚ dx = r β π exp ( βx 2 ) 1 2 π Z ∞∞ exp " β x x i ω 2 β ¶ 2 + β x + i ω 2 β ¶ 2 # dx = r β π exp βx 2 + βx 2 + iωx ω 2 4 β ¶ 1 2 π Z ∞∞ exp " β x x i ω 2 β ¶ 2 # dx = 1 2 π exp iωx ω 2 4 β ¶ r β π Z ∞∞ exp ( βy 2 ) dy = 1 2 π exp iωx ω 2 4 β ¶ → 1 2 π exp( iωx ) as β → ∞ . (6) ii. By the properties of the Dirac δ function, 1 2 π Z ∞∞ δ ( x x )exp( iωx ) dx = 1 2 π exp( iωx ) . (7) Note (not necessary for students to state): It requires complex analysis to justify Z ∞∞ exp " β x x + i ω 2 β ¶ 2 # dx = Z ∞∞ exp ( βy 2 ) dy. (8) This justification is not required here, though. See Haberman’s “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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