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

# FFT and Hilbert

Started by ●April 29, 2005

Reply by ●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 ●May 2, 20052005-05-02

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