Article View: comp.sys.amiga.tech
Article #12109Re: Audio questions
From: Randell Jesup
Date: Wed, 05 Apr 1989 04:41
Date: Wed, 05 Apr 1989 04:41
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1049 bytes
In article <2097@pur-phy> murphy@newton.physics.purdue.edu.UUCP (William J. Murphy) writes:
>In article <2093@pur-phy> murphy@newton.physics.purdue.edu (William J. Murphy) writes:
>>My guess is that it will be freq = sampling rate/(428 * 16), but what's the
>>sampling rate?
>Well, to answer my own questions, I dug through an RKM at the bookstore and
>found that each tic of period = 279.365 ns. Thus the formula is
>
> frequency = 1/(279.365e-9 * Period * Repeat_of_Data)
> = 1/(279.365e-9 * 428 * 16) = 522.7 Hz
Note: only for NTSC!!!
--
Randell Jesup, Commodore Engineering {uunet|rutgers|allegra}!cbmvax!jesup
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Path: arc1-ca.pugleaf.net!utzoo!utgpu!jarvis.csri.toronto.edu!mailrus!ulowell!cbmvax!jesup From: jesup@cbmvax.UUCP (Randell Jesup) Newsgroups: comp.sys.amiga.tech Subject: Re: Audio questions Message-ID: <6502@cbmvax.UUCP> Date: 5 Apr 89 04:41:53 GMT References: <2093@pur-phy> <2097@pur-phy> Reply-To: jesup@cbmvax.UUCP (Randell Jesup) Organization: Commodore Technology, West Chester, PA Lines: 14