FFT and Hilbert

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

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.

Peter K. wrote:
> 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