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 +).

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,

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.

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.

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

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

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

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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