From: Daniel, Erik S. ([email protected])
Date: Sat Mar 31 2001 - 06:21:39 PST
Adam,
Just to throw some numbers in, suppose you are sampling a periodic 60 Hz
"waveform" sampled by a 120 Hz clock. If that 60 Hz waveform were anything
except for a sine wave (e.g., triangular, square, sawtooth, ...), it would
be composed of a 60 Hz sine wave and a linear combination of sine waves at
the harmonic frequencies (120 Hz, 180 Hz, 240 Hz, ...). These harmonics are
above the first Nyquist band (60 Hz bandwidth).
If they are not filtered out, these higher frequencies "fold back" into the
first Nyquist band. For example, suppose you were sampling a signal, again
at a 120 Hz sample rate, and the signal consisted of a superposition of a 40
Hz sine wave and a 100 Hz sine wave. The 40 Hz sine wave is within the
Nyquist band and will therefore be accurately reflected. The 100 Hz sine
wave is above the 60 Hz Nyquist band and will actually show up as a 20 MHz
sampled sine wave (120 Hz - 100 Hz).
Also, as was pointed out by someone else, realize that there are some
underlying assumptions of periodicity (as someone else already pointed out).
- Erik
==================================================================
Erik Daniel, Ph.D. Voice: (507) 538-5461
Mayo Foundation Fax: (507) 284-9171
200 First Street SW E-mail: [email protected]
Rochester, MN 55905 Web: www.mayo.edu/sppdg/
==================================================================
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:
> Adam,
>
> 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.
>
> Adam
>
>
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