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Re: Hilbert Transform

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  • Subject: [mg107243] Re: Hilbert Transform
  • From: "Nasser M. Abbasi" <nma at>
  • Date: Sat, 6 Feb 2010 03:26:04 -0500 (EST)
  • References: <hkgl91$70d$>

"dr DanW" <dmaxwarren at> wrote in message 
news:hkgl91$70d$1 at
> Has an efficient Hilbert Transform or Discrete Hilbert Transform been
> implemented in Mathematica?  I can code my own, but my implementations
> tend to be ham-handed and not too efficient.
> Daniel

I do not know about the "efficient" part, but one way I learned how to do it 
from a class in communication is as follows. (which I am sure you know this 

Given x(t), we want to find xx(t), where xx(t) is the Hilbert transform of 

1. find X(w), the Fourier transform of x(t)
2. Find the Fourier transform of xx(t) which -j*sgn(w)*X(w)
3. Hence xx(t) will be the inverse Fourier transform of the result of step 

sgn(f) is the signum function.

Again, do not know if this is an "efficient" way from a computational point 
of view, but my guess it is faster than the direct method using convolution 
as in

xx(t)= x(t) convolve with 1/(pi*t)


I am not sure about the discrete case now. But may be the above could be 
extended to discrete data just by using DFT and IDFT with some care?


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