> Paul Lee wrote:
>
> > I am having problems trying to understand why would you apply a
> > Hilbert Transform after a FFT i.e. X(f)XH(f) where H denotes the
> > Hilbert transform. What are the benefits for doing this?
>
> Perhaps they're trying to find the Hilbert transform of X?
>
> X(f) = FT(x(t))
>
> Y(f) = X(f).H(f) [XH(f) was confusing)
>
> y(t) = FT^{-1}(Y(f))
>
> so that y(t) is the Hilbert transform of x(t).
Agreed, under one condition: The frequency range in the
last step is [0, Fs/2], not [-Fs/2, Fs/2].
Rune
Reply by Peter K.●April 29, 20052005-04-29
Paul Lee wrote:
> I am having problems trying to understand why would you apply a
> Hilbert Transform after a FFT i.e. X(f)XH(f) where H denotes the
> Hilbert transform. What are the benefits for doing this?
Perhaps they're trying to find the Hilbert transform of X?
X(f) = FT(x(t))
Y(f) = X(f).H(f) [XH(f) was confusing)
y(t) = FT^{-1}(Y(f))
so that y(t) is the Hilbert transform of x(t).
Ciao,
Peter K.
Reply by Paul Lee●April 29, 20052005-04-29
Hi Folks,
I am having problems trying to understand why would you apply a
Hilbert Transform after a FFT i.e. X(f)XH(f) where H denotes the
Hilbert transform. What are the benefits for doing this?
Cheers
Paul