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*To*: AUDITORY@xxxxxxxxxxxxxxx*Subject*: Re: Hilbert envelope bandwidth*From*: Tarun Pruthi <tpruthi@xxxxxxxxxxxx>*Date*: Mon, 27 Sep 2004 10:34:45 -0400*Comments*: To: Christof Faller <cfaller@AGERE.COM>*Delivery-date*: Mon Sep 27 11:05:29 2004*In-reply-to*: <1AF46B69-107D-11D9-A6A9-000A95D9E156@agere.com>*Reply-to*: Tarun Pruthi <tpruthi@xxxxxxxxxxxx>*Sender*: AUDITORY Research in Auditory Perception <AUDITORY@xxxxxxxxxxxxxxx>

I would think that it would simply be (f2-f1)/2. It should atleast be true if the spectrum is symmetrical around (f2+f1)/2. I am not sure what would happen if the spectrum is not symmetrical around (f2+f1)/2. Please correct me if I am wrong. Tarun On Mon, 27 Sep 2004, Christof Faller wrote: > 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 > ----------------------------------------- Tarun Pruthi Graduate Research Assistant, ECE Room 3180, A V Williams Building University of Maryland, College Park MD 20742 USA Email: tpruthi@glue.umd.edu Web: www.ece.umd.edu/~tpruthi Ph: 301-405-1365 ----------------------------------------

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

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