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# app-c - Appendix C Integral transforms C.1 Fourier...

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Appendix C Integral transforms C.1 Fourier transform Given a real-valued function f ( x ) on the real line, define the Fourier transform of f ( x ) to be ˜ f ( k ) = integraldisplay −∞ f ( x ) exp( ikx ) dx (C.1) Then it can be shown that f ( x ) = integraldisplay −∞ dk 2 π ˜ f ( k ) exp( ikx ) (C.2) known as the inverse Fourier transform . Precise factors of 2 π vary from source to source; what is important is that in a Fourier transform followed by an inverse Fourier transform, there should be an overall factor of 1 / (2 π ). A useful identity is the following expression for the Dirac delta function: δ ( x ) = 1 2 π integraldisplay −∞ exp( ikx ) dk (C.3) One way to derive this expression is as the inverse Fourier transform of the Fourier transform of the Dirac delta function. A useful identity is known as Parseval’s theorem : integraldisplay −∞ | f ( x ) | 2 dx = 1 2 π integraldisplay −∞ | ˜ f ( k ) | 2 dk This can be derived using inverse Fourier transforms: integraldisplay −∞ | f ( x ) | 2 dx = integraldisplay −∞ dx bracketleftbigg 1 2 π integraldisplay −∞ ˜ f ( k ) exp(+ ikx ) dk bracketrightbiggbracketleftbigg 1 2 π integraldisplay −∞ ˜ f ( k ) exp( ik x ) dk bracketrightbigg = 1 (2 π ) 2 integraldisplay −∞ dk integraldisplay −∞ dk ˜ f ( k ) ˜ f ( k ) integraldisplay −∞ dx exp( i ( k k ) x ) 573

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APPENDIX C. INTEGRAL TRANSFORMS 574 = 1 (2 π ) 2 integraldisplay −∞ dk integraldisplay −∞ dk ˜ f ( k ) ˜ f ( k )(2 π ) δ ( k k ) = 1 2 π integraldisplay −∞ | ˜ f ( k ) | 2 dk There is also a notion of convolution . Given functions f ( x ), g ( x ), define ( f g )( x ) = integraldisplay −∞ g ( y ) f ( x y ) dy It is straightforward to check that the Fourier transform of the convolution of f ( x ), g ( x ), is the ordinary product of the Fourier transforms: tildewidest ( f g )( k ) = integraldisplay −∞ ( f g )( x ) exp( ikx ) dx = integraldisplay −∞ dx exp( ikx ) integraldisplay −∞ g ( y ) f ( x y ) dy = integraldisplay −∞ g ( y ) exp( iky ) dy integraldisplay −∞ f ( x y ) exp( ik ( x y )) dx = bracketleftbiggintegraldisplay −∞ g ( y ) exp( iky ) dy bracketrightbiggbracketleftbiggintegraldisplay −∞ f ( x ) exp( ikx ) dx bracketrightbigg = ˜ f ( k g ( k ) Similarly, if we define the convolution ( ˜ f ˜ g )( ω ) = 1 2 π integraldisplay −∞ ˜ g ( k ) ˜ f ( k k ) dk then its inverse Fourier transform is the ordinary product of f ( x ) and g ( x ): integraldisplay −∞ ( ˜ f ˜ g )( k ) exp( ikx ) dk 2 π = integraldisplay −∞ dk 2 π integraldisplay −∞ dk 2 π ˜ g ( k ) ˜ f ( k k ) exp( ikx ) = integraldisplay −∞ dk 2 π ˜ g ( k ) exp( ik x ) integraldisplay −∞ dk 2 π ˜ f ( k k ) exp( i ( k k ) x ) = bracketleftbiggintegraldisplay −∞ dk 2 π ˜ g ( k ) exp( ik x ) bracketrightbiggbracketleftbiggintegraldisplay −∞ dk ′′ 2 π ˜ f ( k ′′ ) exp( ik ′′ x ) bracketrightbigg = f ( x ) g ( x ) ***** Should I say something about causality and the Titchmarsh theorem? (see p 193 of
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app-c - Appendix C Integral transforms C.1 Fourier...

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