If you try to take the Fourier transform integral of a exp(j*omega_0*t),
it will not converge in the sense, how an improper integral's
convergence is usually understood. You will need to employ something
like Cauchy principal value or Cesaro convergence to make it converge to
zero at omega!=omega_0. At omega=omega_0 the integral diverges no matter
in which sense you take it. So, strictly speaking, Fourier transform of
a sine doesn't exist.
An equivalent look to this from the inverse transform's side is that the
spectrum of the sine is a Dirac delta function, which is not a function
in the normal sense.
So, none of the statements of the Fourier transform theory (including
the sampling theorem, which assumes the existence of the Fourier
transform), taken rigorously, seem to apply to the sinusoidal signals.
Regards,
Vadim
On 08-Jun-15 10:35, Victor Lazzarini wrote:
Not sure I understand this sentence. As far as I know the FT is defined as an
integral between -inf and +inf, so I am not quite
sure how it cannot capture infinite-lenght sinusoidal signals. Maybe you meant
something else? (I am not being difficult, just
trying to understand what you are trying to say).
========================
Dr Victor Lazzarini
Dean of Arts, Celtic Studies and Philosophy,
Maynooth University,
Maynooth, Co Kildare, Ireland
Tel: 00 353 7086936
Fax: 00 353 1 7086952
On 8 Jun 2015, at 08:19, vadim.zavalishin
<vadim.zavalis...@native-instruments.de> wrote:
It might seem that such signals are unimportant, however even the infinite
sinusoidal signals, including DC, cannot be treated by the sampling theorem,
since the Dirac delta (which is considered as their Fourier transform) is not a
function in a normal sense and strictly speaking Fourier transform doesn't
exist for these signals.
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