Re: Hilbert Transform

• To: mathgroup at smc.vnet.net
• Subject: [mg107243] Re: Hilbert Transform
• From: "Nasser M. Abbasi" <nma at 12000.org>
• Date: Sat, 6 Feb 2010 03:26:04 -0500 (EST)
• References: <hkgl91\$70d\$1@smc.vnet.net>

```"dr DanW" <dmaxwarren at gmail.com> wrote in message
news:hkgl91\$70d\$1 at smc.vnet.net...
> 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
method)

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

Algorithm:
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
2.

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?

--Nasser

```

• Prev by Date: Re: Glynn and Gray ebook
• Next by Date: Re: What does & mean?
• Previous by thread: Hilbert Transform
• Next by thread: Follow up to mg106646 - Selecting a range of dates?