# RE: [SI-LIST] : Nyquist Sampling Rate

From: AA ([email protected])
Date: Fri Mar 30 2001 - 17:42:30 PST

Tom,
Thanks for the feedback. I know that the sampling
rate has to be at least twice that hight frequency
component in the signal. I.e to recover a 60 HZ
sinwave it needed to be sampled by 120sample/sec min.
How do we know a sine wave produced these samples not
a triangulare wave or other periodic wave form.

Thanks

--- Thomas Jackson <[email protected]> wrote:
>
> The sampling theorem assumes that you are sampling a
> band-limited signal.
> Therefore, the highest possible frequency signal
> through any two points
> would be a sinewave at 1/2 the sampling rate.
> Anything else would have
> frequency components above the Nyquist rate and that
> violates the first
> assumption.
>
> By the way, it should be obvious that the two
> samples cannot occur at the
> zero crossings.
>
> Tom
>
> Thomas L. Jackson, P.E.
> Staff VLSI Design Engineer
> Network Access Development
> Systems Solutions Group
> FUJITSU MICROELECTRONICS, INC.
> 3545 North First Street
> San Jose, CA  95134-1804
> telephone: (408) 922-9574
> facsimile: (408) 922-9618
> http://www.fujitsumicro.com
>
>
> -----Original Message-----
> From: AA [mailto:[email protected]]
> Sent: Friday, March 30, 2001 4:43 PM
> To: [email protected]
> Subject: [SI-LIST] : Nyquest Sampling Rate
>
>
> DEAR SI list subscribers,
> Can any one explain to me how you can recover a
> periodic signal form only 2 samples. I can
> understand
> the math but I am having difficulty visualizing
> this.
> Draw me any 2 points in the time domain and I can
> make
> endless number of periodic signal go through them?
>
> I know I am missing a key point but I can quite put
> my
> finger on it.
>
> Your input is very well appreciated.
>
>
>
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