[Scilab-users] Hilbert transform query

[Lester Anderson]

So what would the syntax be for doing a column run Hilbert, and row run Hilbert?

It does sound like the suggesting would do what I think it is meant to.

X = hilbert(data) // does compute pretty fast!

At the moment I have used the netCDF code to read in GMT (net CDF)
data, which is a 601 x 601 matrix (x,y). So in order to do the column
and row method is it necessary to transpose the matrix, e.g. x = x' ?

Thanks for any pointers

On 5 February 2016 at 19:02, Tim Wescott <tim at wescottdesign.com> wrote:
> Any time you go from 1D to 2D you suddenly end up with more than one way
> to do things, so I'm pretty sure that "how would one..." should really
> be worded "how would YOU...", or perhaps "how would someone in this
> field...".
>
> It sounds like you want to keep things rectilinear, so it may be best to
> just apply the transform column-by-column and row-by-row.  That SHOULD
> work, and if you do it as matrix operations it should be pretty fast in
> Scilab.
>
> On Fri, 2016-02-05 at 12:52 +0000, Lester Anderson wrote:
>> Hi Serge,
>>
>> I am working with grid data, so looking for the 2D Hilbert, the
>> results I think should appear similar to doing a directional
>> derivative, where X highlights features with N-S trends and E-W for Y.
>>
>> Lester
>>
>>
>>
>> On 5 February 2016 at 12:24, Serge Steer <Serge.Steer at inria.fr> wrote:
>> > Le 05/02/2016 10:56, Lester Anderson a écrit :
>> >>
>> >> Hello
>> >>
>> >> A quick query. How would one define the Hilbert transform of a grid
>> >> for X and Y directions; looking for two solutions Hx and Hy (for the
>> >> real values).
>> >
>> > Can you explain more precisely what you expect?
>> >  Do you want to apply Hilbert transform to each column and to each rows or
>> > to perform a 2D Hilbert transform?
>> > Serge
>> >>
>> >>
>> >> Thanks
>> >> Lester
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> --
>
> Tim Wescott
> www.wescottdesign.com
> Control & Communications systems, circuit & software design.
> Phone: 503.631.7815
> Cell:  503.349.8432
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