Re: Hilbert transform

[Juan Pablo Carbajal]

On Sat, Jul 7, 2012 at 12:34 AM, Sergei Steshenko <address@hidden> wrote:
>
>
>
>
> ----- Original Message -----
>> From: Ben Abbott <address@hidden>
>> To: Sergei Steshenko <address@hidden>
>> Cc: "address@hidden" <address@hidden>
>> Sent: Friday, July 6, 2012 8:53 PM
>> Subject: Re: Hilbert transform
>>
>>
>> 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
>
> Thanks to all for clarifications and explanations.
>
> I still see a problem though. First, as others have pointed out, what the 
> documentation of signal-1.1.3/hilbert.m says among other things:
>
> "
> `real(H)' contains the original signal F.  `imag(H)' contains the
>      Hilbert transform of F.
> ".
>
> So, if I want Hilbert transform proper, I need 'imag' - so far so good.
>
> Here is a quick example:
>
> "
> octave:8> imag(hilbert(imag(hilbert([-3 -1 1 3]))))
> ans =
>
>    2   2  -2  -2
> ".
>
> The input ('[-3 -1 1 3]') is a 0 DC vector:
>
> "
> center([-3 -1 1 3])
> ans =
>
>   -3  -1   1   3
> "
>
> - which makes life easier.
>
> So, above I expected '3 1 -1 -3' answer according to
>
> Hilbert(Hilbert(u(t))) == -u(t)
>
> property ('Hilbert' is meant to be true Hilbert transform).
>
>
> Obviously the answer I got: "2   2  -2  -2" is not the expected one. And the 
> difference is not just scaling coefficient.
>
>
> Any ideas ?
>
> Thanks,
>   Sergei.
>
> P.S. It looks like signal-1.1.3/hilbert.m follows what Matlab documentation 
> describes, but the question is whether what is described in the documentation 
> (the part returned with 'imag') is true Hilbert transform.
>
>
>
>
>
>
>>
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You are working with a discrete transformation very sensitive to the
length of the signal. If you check with a longer signal everything is
fine.

t  = linspace (0,1,100);
y  = sin (2*pi*8*t) - sin (2*pi*10*t);
yg= imag ( hilbert ( imag ( hilbert (y))));
plot (t,y,t,-yg,'o')

Cheers


-- 
M. Sc. Juan Pablo Carbajal
-----
PhD Student
University of Zürich
http://ailab.ifi.uzh.ch/carbajal/