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*To*: AUDITORY@xxxxxxxxxxxxxxx*Subject*: HILBERT TRANSFORM*From*: Fred Herzfeld <herzfeld@xxxxxxxxxxxx>*Date*: Mon, 27 Nov 2006 15:00:23 -0500*Delivery-date*: Mon Nov 27 15:24:29 2006*List-archive*: <http://lists.mcgill.ca/scripts/wa.exe?LIST=AUDITORY>*List-help*: <http://lists.mcgill.ca/scripts/wa.exe?LIST=AUDITORY>, <mailto:LISTSERV@LISTS.MCGILL.CA?body=INFO AUDITORY>*List-owner*: <mailto:AUDITORY-request@LISTS.MCGILL.CA>*List-subscribe*: <mailto:AUDITORY-subscribe-request@LISTS.MCGILL.CA>*List-unsubscribe*: <mailto:AUDITORY-unsubscribe-request@LISTS.MCGILL.CA>*Reply-to*: Fred Herzfeld <herzfeld@xxxxxxxxxxxx>*Sender*: AUDITORY - Research in Auditory Perception <AUDITORY@xxxxxxxxxxxxxxx>*User-agent*: Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US; rv:1.0.2) Gecko/20030208 Netscape/7.02 (VAUSSU03)

Hello list,

s(t) = cos(2*pi*f-phi) where f and phi are constant with time and s(t)

is sampled at time intervals so that s(t) falls exactly into the second FFT bin. If I perform the FFT, modify the output by 90 degrees and perform an inverse FFT, the output of the FFT will be exactly the Hilbert Transform of the original series s(t) namely HT[s(t)].

(1) Can I really compute HT[S1(t)] correctly ? (2) If I can, how should the output of the FFT of S1(t) be modified

Any thoughts on this will be very much appreciated.

Fred

-- Fred Herzfeld, MIT '54 78 Glynn Marsh Drive #59 Brunswick, Ga.31525 USA

**Follow-Ups**:**Re: HILBERT TRANSFORM***From:*Ramdas Kumaresan

**Re: HILBERT TRANSFORM***From:*Richard F. Lyon

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