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Re: Fourier transforms



On 10/15/2003 12:05 PM, Ludwik Kowalski wrote:
> [the idea] that for any even function
> f(x) (i.e. symmetric with respect to the y axis)
> the phases are "restricted to 0 and pi" might
> not be immediately obvious.

It's not quite obvious, because it's not quite
true ... but with one exception it's true, and
not too hard to prove.

The definition of phase(A(k)) is atan2(IP(A(k)), RP(A(k)))

For an even function, by symmetry the imaginary
part of A vanishes, so the phase is either zero,
or pi, or undefined (according to whether the RP
is positive, negative, or zero respectively).

> Suppose the sum representing the
> f(x) in the "k space" contains a pair of terms with
> phases equal to +20 and -20 degrees.

Never gonna happen for f(x)=f(-x).