Derivation: Sign and Reciprocal

## Sign and Reciprocal

Derivation of the forward transform. The sign function can be implemented by breaking the transform integral into a term dealing with negative time (- to 0) and a term dealing with positive time (0 to +).

display='block'> φ(f)= sign(t) e i2πft dt = 0 e i2πft dt + 0 e i2πft dt MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharuqqYLwySbqee0evGueE0jxyaibaieYti9Grpfeu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@733A@

Because the exponents are imaginary, the two integrals on the right are not well-conditioned. To obtain an easily solved integral, replace the sign function with the following expression,

display='block'> lim a0 [ e a|t| sign(t) ] MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharuqqYLwySbqee0evGueE0jxyaibaieYti9Grpfeu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaaiaacaqabeaadmabbuaaaOqaamaaxababaGaciiBaiaacMgacaGGTbaaleaacaWGHbGaeyOKH4QaaGimaaqabaGcdaWadaqaaiaadwgadaahaaWcbeqaaiabgkHiTiaadggacaGG8bGaamiDaiaacYhaaaGccaaMc8Uaae4CaiaabMgacaqGNbGaaeOBaiaacIcacaWG0bGaaiykaaGaay5waiaaw2faaaaa@4DCF@

which is an odd exponential. As a 0, the exponential term approaches unity, and the overall function approaches the sign function. Using this approximation the exponential within the integral changes from imaginary to complex.

display='block'> φ(f)= lim a0 [ e a|t| sign(t) e i2πft dt ] MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharuqqYLwySbqee0evGueE0jxyaibaieYti9Grpfeu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaaiaacaqabeaadmabbuaaaOqaaiabeA8aMjaacIcacaWGMbGaaiykaiabg2da9maaxababaGaciiBaiaacMgacaGGTbaaleaacaWGHbGaeyOKH4QaaGimaaqabaGcdaWadaqaamaapehabaGaamyzamaaCaaaleqabaGaeyOeI0IaamyyaiaacYhacaWG0bGaaiiFaaaakiaaykW7caqGZbGaaeyAaiaabEgacaqGUbGaaiikaiaadshacaGGPaGaaGPaVlaadwgadaahaaWcbeqaaiabgkHiTiaadMgacaaIYaGaeqiWdaNaamOzaiaadshaaaGccaWGKbGaamiDaaWcbaGaeyOeI0IaeyOhIukabaGaeyOhIukaniabgUIiYdaakiaawUfacaGLDbaaaaa@63DA@

display='block'> φ(f)= lim a0 [ 0 e at e i2πft dt + 0 e at e i2πft dt ] MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharuqqYLwySbqee0evGueE0jxyaibaieYti9Grpfeu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@6AA6@

display='block'> φ(f)= lim a0 [ 0 e (ai2πf)t dt + 0 e (a+i2πf)t dt ] MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharuqqYLwySbqee0evGueE0jxyaibaieYti9Grpfeu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaaiaacaqabeaadmabbuaaaOqaaiabeA8aMjaacIcacaWGMbGaaiykaiabg2da9maaxababaGaciiBaiaacMgacaGGTbaaleaacaWGHbGaeyOKH4QaaGimaaqabaGcdaWadaqaaiabgkHiTmaapehabaGaamyzamaaCaaaleqabaGaaiikaiaadggacqGHsislcaWGPbGaaGOmaiabec8aWjaadAgacaGGPaGaamiDaaaakiaadsgacaWG0baaleaacqGHsislcqGHEisPaeaacaaIWaaaniabgUIiYdGccqGHRaWkdaWdXbqaaiaadwgadaahaaWcbeqaaiabgkHiTiaacIcacaWGHbGaey4kaSIaamyAaiaaikdacqaHapaCcaWGMbGaaiykaiaadshaaaGccaWGKbGaamiDaaWcbaGaaGimaaqaaiabg6HiLcqdcqGHRiI8aaGccaGLBbGaayzxaaaaaa@6919@

The first integral needs to be transformed into an expression with a negative exponential. To do this define a new variable of integration, T = -t, making dT = -dt.

display='block'> 0 e (ai2πf)t dt = 0 e (ai2πf)T dT = 0 e (ai2πf)T dT MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharuqqYLwySbqee0evGueE0jxyaibaieYti9Grpfeu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaaiaacaqabeaadmabbuaaaOqaaiabgkHiTmaapehabaGaamyzamaaCaaaleqabaGaaiikaiaadggacqGHsislcaWGPbGaaGOmaiabec8aWjaadAgacaGGPaGaamiDaaaakiaadsgacaWG0baaleaacqGHsislcqGHEisPaeaacaaIWaaaniabgUIiYdGccqGH9aqpdaWdXbqaaiaadwgadaahaaWcbeqaaiabgkHiTiaacIcacaWGHbGaeyOeI0IaamyAaiaaikdacqaHapaCcaWGMbGaaiykaiaadsfaaaGccaWGKbGaamivaaWcbaGaeyOhIukabaGaaGimaaqdcqGHRiI8aOGaeyypa0JaeyOeI0Yaa8qCaeaacaWGLbWaaWbaaSqabeaacqGHsislcaGGOaGaamyyaiabgkHiTiaadMgacaaIYaGaeqiWdaNaamOzaiaacMcacaWGubaaaOGaamizaiaadsfaaSqaaiaaicdaaeaacqGHEisPa0Gaey4kIipaaaa@6E21@

Now the expression for φ(f) can be rewritten using this modification.

display='block'> φ(f)= lim a0 [ 0 e (ai2πf)T dT + 0 e (a+i2πf)t dt ] MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharuqqYLwySbqee0evGueE0jxyaibaieYti9Grpfeu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaaiaacaqabeaadmabbuaaaOqaaiabeA8aMjaacIcacaWGMbGaaiykaiabg2da9maaxababaGaciiBaiaacMgacaGGTbaaleaacaWGHbGaeyOKH4QaaGimaaqabaGcdaWadaqaaiabgkHiTmaapehabaGaamyzamaaCaaaleqabaGaeyOeI0IaaiikaiaadggacqGHsislcaWGPbGaaGOmaiabec8aWjaadAgacaGGPaGaamivaaaakiaadsgacaWGubaaleaacaaIWaaabaGaeyOhIukaniabgUIiYdGccqGHRaWkdaWdXbqaaiaadwgadaahaaWcbeqaaiabgkHiTiaacIcacaWGHbGaey4kaSIaamyAaiaaikdacqaHapaCcaWGMbGaaiykaiaadshaaaGccaWGKbGaamiDaaWcbaGaaGimaaqaaiabg6HiLcqdcqGHRiI8aaGccaGLBbGaayzxaaaaaa@68D9@

Both terms can be solved using a tabulated integral (Gradshteyn and Ryzhik, EQ 3.310).

display='block'> 0 e px dx = 1 p Re(p)>0 MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharuqqYLwySbqee0evGueE0jxyaibaieYti9Grpfeu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaaiaacaqabeaadmabbuaaaOqaauaabeqabiaaaeaadaWdXbqaaiaadwgadaahaaWcbeqaaiabgkHiTiaadchacaWG4baaaOGaamizaiaadIhaaSqaaiaaicdaaeaacqGHEisPa0Gaey4kIipakiabg2da9maalaaabaGaaGymaaqaaiaadchaaaaabaGaciOuaiaacwgacaGGOaGaamiCaiaacMcacqGH+aGpcaaIWaaaaaaa@4ADE@

Using this solution the expression for φ(f) reduces to an imaginary, reciprocal function.

display='block'> φ(f)= lim a0 [ 1 ai2πf + 1 a+i2πf ]= 1 iπf = i πf MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharuqqYLwySbqee0evGueE0jxyaibaieYti9Grpfeu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabiGaaiaacaqabeaadmabbuaaaOqaaiabeA8aMjaacIcacaWGMbGaaiykaiabg2da9maaxababaGaciiBaiaacMgacaGGTbaaleaacaWGHbGaeyOKH4QaaGimaaqabaGcdaWadaqaaiabgkHiTmaalaaabaGaaGymaaqaaiaadggacqGHsislcaWGPbGaaGOmaiabec8aWjaadAgaaaGaey4kaSYaaSaaaeaacaaIXaaabaGaamyyaiabgUcaRiaadMgacaaIYaGaeqiWdaNaamOzaaaaaiaawUfacaGLDbaacqGH9aqpdaWcaaqaaiaaigdaaeaacaWGPbGaeqiWdaNaamOzaaaacqGH9aqpcqGHsisldaWcaaqaaiaadMgaaeaacqaHapaCcaWGMbaaaaaa@5FEF@

The above derivation was modeled after that given by Bracewell, page 130.

Derivation of the inverse transform. Start with the definition of the inverse transform.

F( t )= iA πf e i2πft df

The integral cannot be solved directly because the righthand term goes to infinity when f = 0. To avoid this problem, break the integral into two halves,

F( t ) C = R ε 1 f e iat df + ε R 1 f e iat df

where C = -iA and a = 2πt. As R  andε0,   the two integrals on the right have the following solution [T.W. Körner, Fourier Analysis, Cambridge University Press, Cambridge, 1989, p. 301].

Since the variable t determines the sign of a (when t is negative, a < 0),

F( t )=Ciπsign( t )= iA π iπsign( t )=Asign( t )

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