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Fourier transforms

Motivation and denition

Up to now, we've been expressing functions on nite intervals (usually the interval

0?x?Lor?L?x?L) as Fourier series:

f?x? ?a0??? n?1a ncos?nxL ?bnsin?nxL where a

0?12L?

L ?Lf?x?dx and a n?1L L ?Lf?x?cos?nxL dx; b n?1L L ?Lf?x?sin?nxL dx: We also occasionally thought about the complex exponential version of Fourier series:

Sinceei?cos?isinande?i?cos?isin, or equivalently

cos?ei?e?i2 and sin?ei?e?i2i; we can rewrite the above series as: f?x? ?a0e0ix??? n?1a nenix?L?e?nix?L2 ?bnenix?L?e?nix?L2i ?a0e0ix??? n?1a n?ibn2 enix?L?an?ibn2 e?nix?L n???c ne?nix?L where c n?? ?12 ?an?ibn?forn?0 a

0forn?0

12 ?a?n?ib?n?forn?0

2fourier transforms

Using the formulas foranandbngiven above, we see that, forn?0. c n?12 ?an?ibn? 12L? L ?Lf?x?cos?nxL dx?i2L? L ?Lf?x?sin?nxL dx 12L? L ?Lf?x?? cos?nxL ?isin?nxL dx 12L? L ?Lf?x?enix?Ldx:

Ifn?0 we have

c n?12 ?a?n?ib?n? 12L? L ?Lf?x?cos? ?nxL dx?i2L? L ?Lf?x?sin? ?nxL dx 12L? L ?Lf?x?? cos?nxL ?isin?nxL dx 12L? L ?Lf?x?enix?Ldx because cosine is an even function and sine is odd. So the same formula works for all the coecients (evenc0) in this case and we have f?x? ??? n???c ne?nix?Lwherecn?12L? L ?Lf?x?enix?Ldx: What we want to do here is letLtend to innity, so we can consider problems on the whole real line. To see what happens to our Fourier series formulas when we do this, we introduce two new variables:!?n?Land !??L. Then our complex

Fourier series formulas become

f?x? ??? n???c ne?i!xwherecn?!2? L ?Lf?x?ei!xdx and thenin the formula forcnis hiding in the variable!. Now, if we letec!?cn?!, we can rewrite these as f?x? ??? n???ec!e?i!x!whereec!?12? L ?Lf?x?ei!xdx: The variable!?n?Ltakes on more and more values which are closer and closer together asL? ?, soec!begins to feel like a function of the variable!dened math 4253 for all real!. Likewise, the sum on the left looks an awful lot like a Riemann sum approximating an integral. What happens in the limit asL? ?is: f?x? ?? c?!?e?i!xd!wherec?!? ?12? f?x?eix!dx: The formula on the right denes the functionc?!?as theFourier transformoff?x?, and the formula on the left denesf?x?as theinverse Fourier transformofc?!?. These formulas hold true (and the inverse Fourier transform of the Fourier transform off?x?isf?x?| the so-calledFourier inversion formula) for reasonable functions f?x?that decay to zero as?x? ? ?in such a way so that?f?x??and/or?f?x??2has a nite integral over the whole real line. There are many standard notations for Fourier transforms (and alternative de- nitions with the minus sign in the Fourier transform rather than in the inverse, and with the 2factor in dierent places, so watch out if you're looking in books other than our textbook!), including b f?!? ?F?!? ?F?f?x???!? ?12? f?x?eix!dx and ?F?x? ?f?x? ?F?1?F?!???x? ??

F?!?e?ix!d!:

Properties and examples.

The Fourier transform is an operation that maps a function ofx, sayf?x?to a function of!, namelyF?f??!? ??f?!?. It is clearly alinearoperator, so for functions f?x?andg?x?and constantsandwe have

F?f?x? ?g?x?? ?F?f?x?? ?F?g?x??:

Some other properties of the Fourier transform are

1.Translation(or shifting):F?f?x?a???!? ?ei!aF?f?x???!?. And in the other

direction,F?eiaxf?x???!? ?F?f?x???!?a?.

2.Scaling:F?1a

f?xa ???!? ?F?f?x???a!?, and likewiseF?f?ax???!? ?1a

F?f?x???!a

3.Operational property(derivatives):F?f??x???!? ? ?i!F?f?x???!?, and

F?xf?x???!? ? ?idd!

?F?f?x???!??. The operational property is of essential importance for the study of dierential equa- tions, since it shows that the Fourier transform converts derivatives to multiplication

4fourier transforms

{ so it converts calculus to algebra (or might reduce a partial dierential equation to an ordinary one). Here are the proofs of the rst of each of the three pairs of formulas to give a sense of how to work with Fourier transforms, and leave the other three as exercises. For the rst shifting rule, we make the substitutiony?x?a(sody?dxandx?y?a) to calculate

F?f?x?a???!? ?12?

f?x?a?ei!xdx 12? f?y?ei!yei!adx ?ei!aF?f?x???!? For the rst scaling rule, we make the substitutiony?x?a(sodx?ady) and get F ?1a f?xa ?!? ?12? ??1a f?xa e i!xdx 12? f?y?eia!ydy ?F?f?x???a!? For the operational property we rst point out that since the Fourier transforms of bothf??x?andf?x?exist, we must have thatf?x? ?0 andf??x? ?0 asx? ??. Therefore the endpoint terms will vanish when we integrate by parts (withu?ei!x anddv?f??x?dx, sodu?i!ei!xandv?f?x?):

F?f??x???!? ?12?

f??x?ei!xdx

12ei!xf?x?????x??

x????12? i!f?x?ei!xdx ?0?i!2? f?x?ei!xdx ? ?i!F?f?x???!? Let's calculate a few basic examples of Fourier transforms:

Example 1. LetSa?x?be the function dened by

S a?x? ??1 if?x? ?a

0 otherwise

math 4255 Then F ?Sa?x???!? ?12? a ?aei!xdx?ei!a?e?i!a2i!?sina!! Example 2. Letu?x? ?e?ax2?2, so the graph ofu?x?is a \Gaussian" or \bell-shaped curve". Thenu?x?satises the dierential equationu??axu?0. We can use this fact and the properties of the Fourier transform to calculatebuas follows: Take the Fourier transform of the dierential equation and use linearity and both parts of property (3) above to get

0?F?u??axu??!? ?F?u?? ?aF?xu?

? ?i!F?u? ?aidF?u?d!

ThereforeF?u?satises the dierential equation

F?u???1a

!F?u? ?0 the solution of which is F ?u? ?Ce?!2??2a?:

The constantCis the value ofF?u??0?, i.e.,

C?12? e?ax2?2dx?12?2 a e?y2dy?1?2a; using the substitutiony??a 2 xand the familiar (or at least accessible) fact that?? ??e?y2dy??. Therefore F e ?ax2?2? ?!? ?1?2ae?!2??2a?; so the original Gaussian is transformed into a dierent one. An interesting observation is what happens fora?1: Then we have F e ?x2?2? ?1?2e?!2?2; so the specic Gaussiane?x2?2is aneigenfunction of the Fourier transform with eigenvalue1??2.

Example 3. Letf?x? ?e?a?x?, so

f?x? ??e?axifx?0 e axifx?0:

6fourier transforms

Then F ?e?a?x???!? ?12? e?a?x?ei!xdx 12? 0 eaxei!xdx?12? 0 e?a!ei!xdx 12? 0quotesdbs_dbs17.pdfusesText_23