Fourier Transforms

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f̂k a−a1 e ikx dxa 1ike ikx −a2sinkakNote that the function has a discontinuity at x a but its FT is still defined.Example 3Suppose that fx is a real-valued even function, then determine its FT and show that it alsoeven.By definitionf̂k − −fx e ikx dxfx coskx isinkx dx −fx coskx dx i −fx sinkx dx 2−fx coskx dx 2f̂ckwhere the properties of coskx and sinxk being even and odd respectively has beenemployed together with the knowledge that fx is even. It is clear that f̂k is even, sincecoskx cos−kx and hence f̂−k f̂k.3.2.2 Derivatives and other useful identitiesFT of Derivative. The notation f̂k employed in the definition of the complex FTessentially defines a function of k. It is also possible to consider the definition as anoperator on fx, sothatFfx f k −fx e ikx dx .and F indicates the FT operator. This enables us to determine the FT of the derivatives offx,Ff ′ x −f ′ x e ikx dx fx e ikx −− ik −by definitionfx e ikx dxintegrating by parts Ff ′ x − ik Ffx − ik f̂k . (3.2.3)In this derivation , the property that fx → 0as|x| → if fx is absolutely integrable hasbeen employed.6
FT of Higher Derivatives. The higher derivatives can be obtained in a similar manner andit is straightforward to show thatFf ′′ x −ik 2 Ffx − k 2 f̂kFf n x −ik n Ffx −ik n f̂k (3.2.4)Derivative of FT. AstheFT f k of fx is a function of k, it can be differentiated in theusual way to gived f kdk d dk −−fx e ikx dx ix fx e ikx dx−ddk fx eikx dx d f kdk iFxfx or Fxfx − i d f kdk. (3.2.5)Thus if the FT of fx is known, then the FT of xfxcan be determined by differentiation.This can easily be extended to giveF x n fx −i n d n f kdk n .Both the result in (3.2.5) and the more general result require that x n fx is absolutelyintegrable for whatever value of n is being employed. This comes with a warning: if fx isabsolutely integrable, it should not be assumed that the same property holds for x n fx.Shift Operator. Consider the function fx e icx ,wherec is a real constant and the FT offx is known.F fx e icx −fx e icx e ikx dx −fx e ikcx dx f k c . (3.2.6)The appearance of c "shifts" the value of k at which the transform is evaluated.3.2.3 Convolution of two functionsAll of the previous discussion is solely concerned with a single function and sometimes it isnecessary to consider two or more functions. This can also arise when we considerfunctions that themselves can be considered to be the products of two or more functions.Suppose that we have two functions fx and gx, both of which are bounded andabsolutely integrable. The Convolution of fx and gx is a function, written f ∗ g, definedby7
Page 2 and 3: −xfx dx f0 (3.1.1)or equivalentl
Page 4 and 5: fx Hx a 1 − Hx − aSignum (Sig
Page 9 and 10: fx fx −2 12−f̂k e −ikx dk
Page 11 and 12: −|e −a|x| | 2 dx 2 0e −2ax d
Page 13 and 14: −−d 2 ydx 2d 2 y− 2 y e ikx
Page 15 and 16: uk,y Ak e|k|y Bk e −|k|y .Now ap
Page 17 and 18: ecause of the convolution structure
Page 19 and 20: complex FS of f on −L ≤ ≤
Page 21 and 22: 3.4 Fourier Transforms and Generali
Page 23 and 24: function and kgk 0, then gk is a c

Fourier Transforms

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