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From: jbuck@galileo.berkeley.edu (Joe Buck)
Newsgroups: comp.dsp
Subject: Re: Discreet Sampling
Message-ID: <11588@pasteur.Berkeley.EDU>
Date: 1 Mar 91 23:07:28 GMT
References: <1991Feb28.194203.27097@alzabo.ocunix.on.ca>
Sender: news@pasteur.Berkeley.EDU
Reply-To: jbuck@galileo.berkeley.edu (Joe Buck)
Lines: 29

In article <1991Feb28.194203.27097@alzabo.ocunix.on.ca>, rob@alzabo.ocunix.on.ca (Robert Hilchie) writes:
|> I hope this question isn't too elementary for this newsgroup.
|> 
|> It is claimed that any signal sampled at a rate of 2s can be reproduced exactly,
|> provided that the original signal did not contain frequencies above s.
|> 
|> Now suppose the sampling rate is 40 kHz and the signal being sampled is a
|> sine wave of constant amplitude at 19 999 Hz. At some point the samples will
|> occur near the peaks and troughs of the sine wave, while, half a second later,
|> the samples will occur at the midpoints between the peaks and troughs. Thus,
|> the reproduced signal will fluctuate in amplitude every second.
|> 
|> How can this "beating" be avoided? Thanks in advance,

You will get no "beating" at all if you reconstruct the signal by filtering
impulses with heights equal to the samples through an ideal lowpass filter
with cutoff at 20 kHz.  You're imagining a "connect the dots and smooth"
type of reconstruction, and that's not what the sampling theorem calls for.

Of course, in the real world there's no such thing as an ideal
lowpass filter, so you're simply not going to be able to reproduce accurately
a signal that close to the Nyquist frequency.  That's why CDs run at 44.1 KHz
and only attempt to reproduce signals up to 20 KHz.  The extra 2.1 KHz
bandwidth is to allow for filter rolloff (and even then you need to get
fancy to get that steep a rolloff).

--
Joe Buck
jbuck@galileo.berkeley.edu	 {uunet,ucbvax}!galileo.berkeley.edu!jbuck