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*To*: AUDITORY@xxxxxxxxxxxxxxx*Subject*: Re: Hilbert envelope bandwidth*From*: Ramin Pichevar <Ramin.Pichevar@xxxxxxxxxxxxxx>*Date*: Mon, 27 Sep 2004 11:44:02 -0400*Delivery-date*: Mon Sep 27 12:12:52 2004*Importance*: Normal*In-reply-to*: <1AF46B69-107D-11D9-A6A9-000A95D9E156@agere.com>*Reply-to*: Ramin Pichevar <Ramin.Pichevar@xxxxxxxxxxxxxx>*Sender*: AUDITORY Research in Auditory Perception <AUDITORY@xxxxxxxxxxxxxxx>

Christof, You may have a look at pp 796-798 of the book A.V. Oppenheim, and R. W. Schafer, "Discrete-time Signal Processig", Second Edition, which deals with the representation of bandpass signals in the Hilbert domain. The picture depicted there explains everything. Hope this can help ! Cheers, Ramin -----Message d'origine----- De : AUDITORY Research in Auditory Perception [mailto:AUDITORY@LISTS.MCGILL.CA]De la part de Christof Faller Envoye : 27 septembre 2004 08:02 A : AUDITORY@LISTS.MCGILL.CA Objet : Hilbert envelope bandwidth Dear list, I am struggling with the following question: Given a signal x(n) with X(f) = 0 for |f| < f1 or |f| > f2 (bandpass filtered signal with bandwidth B = f2-f1) e(n) is the Hilbert envelope of x(n) which can then be written as: x(n) = e(n)y(n), where y(n) is the "temporally flattened" version of x(n). The spectrum of e(n) satisfies: E(f) = 0 for |f| > f3 (Due to its DC offset, the evelope e(n) contains frequencies down to zero). ==> Can f3 be expressed as a function of B (the bandwidth of signal x)? Any comments/suggestions are appreciated. Thanks, Christof Faller

**References**:**Hilbert envelope bandwidth***From:*Christof Faller

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