Re: Hilbert transform

[Ben Abbott]

On Jul 6, 2012, at 12:44 PM, Sergei Steshenko wrote:

> Hello,
> 
> i am talking about 'hilbert' function from from 'signal-1.1.3/hilbert.m' 
> file, so Octave help list purists are welcome to send me with my 
> uncomfortable questions to octave-dev list.
> 
> But I'll ask my questions here - from my reading (and recollections of what I 
> learned a long long time ago) the issue is mathematical/computational.
> 
> First a couple of references:
> 
> 1) http://w3.msi.vxu.se/exarb/mj_ex.pdf - I think put together really nicely;
> 2) http://en.wikipedia.org/wiki/Hilbert_transform
> .
> 
> Wherever we look, we find that the definition of Hilbert transform is through 
> integral of a _real_ function, i.e.
> 
> hilbert(u(t)) == integral_from_minus_to_plus_inf("u(tau) / (t - tau)", "dtau")
> 
> and as such it should be a _real_ function of 't' provided u(t) is a real 
> function of 't'.
> 
> 
> Also, it is proven that
> 
> hilbert(hilbert(u(t))) == -u(t)
> .
> 
> Now, here is Octave and its package reality:
> 
> 
> "
> octave:1> hilbert([1 2 3 4])
> ans =
> 
>    1 + 1i   2 - 1i   3 - 1i   4 + 1i
> 
> octave:2> hilbert(hilbert([1 2 3 4]))
> warning: HILBERT: ignoring imaginary part of signal
> ans =
> 
>    1 + 1i   2 - 1i   3 - 1i   4 + 1i
> 
> octave:3>
> ".
> 
> Three violations already:
> 
> 1) output is complex rather than real;
> 2) the transform is not invertible;
> 3) since Hilbert transform is linear, complex input should be accepted 
> according to
> 
> hilbert(foo + i * bar) == hilbert(foo) + i * hilbert(bar)
> .
> 
> To put things politically correctly, Hilbert transform is a canine female to 
> calculate - because of the above "/ (t -tau)", and it's problematic to 
> calculate in discrete domain.
> 
> So, my first practical question is: "What does Matlab do ?".
> 
> Thanks,
>   Sergei.

Matlab's online doc for hilbert() is at the link below.

http://www.mathworks.com/help/toolbox/signal/ref/hilbert.html

The example below is included on that page.

        hilbert ([1 2 3 4])
        ans =   1 + 1i   2 - 1i   3 - 1i   4 + 1i

It appears that the hilbert() function is *not* a direct implementation of the 
Hilbert Transform, but the version in the signal package does appear to be 
consistent with the one which is part of the Matlab Signals toolbox.

After a quick look, my impression is that the hilbert() function's output is 
hilbert(x) = 1i * H(x) + x.  Where, H(x) is the Hilbert Transform.  I don' t 
know why the author (mathworks?) decided to restrict the input to real values.

Ben