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Re: Please explain FFT normalization

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"Cliff Curry" <CliffCurry@home.com> wrote in message <ptRW6.187186$p33.3812338@news1.sttls1.wa.home.com>...
> Using the FFT to find the spectrum of what are supposed to be "continuos"
> waveforms is somewhat tricky.
> To find the Fourier Transform of a transient signal, take the FFT and
> multiply by the sampling interval. Do not divide by the number of points in
> the FFT.
> To find the Fourier Series coefficient of a periodic waveform, take the FFT
> of one period, and divide by the number of
> points in the FFT. The FFT also returns both positive and negative
> frequencies, so if you want to think of the spectrum as only positive
> frequencies, you have to multiply by two after discarding the negative
> frequencies.

A bit late, but... isn't FFT supposed to be a sampling of the actual DTFT? Which DTFT happens to match with 1/Ts the magnitude of the continuous time FT X(?). Witch FT corresponds to the coefficients of the fourier integral (which is the fourier series of a finite-length waveform T->inf).
So why don't we also multiply FFT with Ts, when we normalize it to match the FT (or else to find the exact magnitudes of the coefficients)?

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