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IDL hilbert() function [message #63183]

Sat, 01 November 2008 02:14

lecacheux.alain
Messages: 325
Registered: January 2008

Senior Member

Is this function actually computing the Hilbert transform ?
The Hilbert transform is known to be idempotent, i.e. H(H(x)) = -x.
However, by applying the IDL function, one get for instance :
IDL> print, hilbert (hilbert (indgen(8)))
( 6.00000, 0.000000)( 7.00000, 0.000000)
( 4.00000, 0.000000)( 5.00000, 0.000000)
( 2.00000, 0.000000)( 3.00000, 0.000000)
( 0.000000, 0.000000)( 1.00000, 0.000000)

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[Message index]

		IDL hilbert() function By: lecacheux.alain on Sat, 01 November 2008 02:14
		Re: IDL hilbert() function By: lecacheux.alain on Tue, 04 November 2008 05:45
		Re: IDL hilbert() function By: R.G. Stockwell on Mon, 03 November 2008 11:54

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