ps6sol

# ps6sol - ACM 95b/100b Problem Set 6 Solutions Prepared by:...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ACM 95b/100b Problem Set 6 Solutions Prepared by: Lei Zhang February 22, 2007 Total: 136 points Include grading section: 2 points Problem 1 (2 × 5 points) Shifting properties for the Fourier transform. a) Derive an expression for F{ f ( x- a ) } in terms of F ( k ) ≡ F{ f ( x ) } . b) Derive an expression for F- 1 { F ( k- ib ) } in terms of f ( x ) ≡ F- 1 { F ( k ) } . Solution 1 a) Write, F { f ( x- a ) } = Z ∞-∞ f ( x- a ) e- ikx dx = e- ika Z ∞-∞ f ( x- a ) e- ik ( x- a ) dx. (1) Now make the substitution u = x- a . Then (1) becomes, e- ika Z ∞-∞ f ( u ) e- iku du = e- ika F ( k ) . So, F { f ( x- a ) } = e- ika F ( k ) . b) This problem turned out to be harder than intended because the stated form of the shifting theorem requires some additional assumptions to be valid. Nonetheless, we proceed and state the assumptions where necessary. At the end, we mention a form of the shifting theorem that is easier to prove. Note that, F- 1 { F ( k- ib ) } = 1 2 π Z ∞-∞ F ( k- ib ) e ikx dk = e ix ( ib ) 2 π Z ∞- ib-∞- ib F ( κ ) e ixκ dκ, where κ = k- ib . Since we have moved the integration off the real axis, we need to investigate whether this integral converges. A sufficient condition for it to converge is that we assume F ( k ) is analytic on and in the box- b < Im( k ) < 0 and-∞ < Re( k ) < ∞ . Then we consider a contour that goes from-∞ to ∞ along the real axis, goes from ∞ to-∞ at- ib , and connects at the ends. Since we have assumed F ( k ) is analytic in that region, the integral around the contour is zero. Further, the Riemann-Lebesgue lemma tells us that if f ( x ) is absolutely integrable, i.e. R ∞-∞ | f ( x ) | dx < ∞ , then lim k →±∞ F ( k ) = 0. * Thus, by the ML-inequality with L = b , the contribution from the ends at infinity vanishes. Thus, Z ∞-∞ F ( k ) e ixk dk = Z ∞- ib-∞- ib F ( k ) e ixk dk = Z ∞-∞ F ( ω- ib ) e i ( ω- ib ) x dω = e bx Z ∞-∞ F ( ω- ib ) e iωx dω, where we have made the substitution, ω = k + ib . Notice that the first term is 2 πf ( x ), and the last term is 2 πe bx F- 1 { F ( ω- ib ) } . Thus, F- 1 { F ( k- ib ) } = e- bx f ( x ) . (2) * H. Weinberger. A First Course in Partial Differential Equations with Complex Variables and Transform Methods , 1965, p. 316....
View Full Document

## This note was uploaded on 11/23/2009 for the course ACM 95b taught by Professor Nilesa.pierce during the Winter '09 term at Caltech.

### Page1 / 7

ps6sol - ACM 95b/100b Problem Set 6 Solutions Prepared by:...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online