Timothy Stockman Wrote: 
> 
> If you have a 20 KHz square wave that's bandwidtch limited to 1 MHz and
> pick a Nyquist frequency of 22.05 KHz, you've picked too low a Nyquist
> limit to accurately reproduce the waveform. But if you pick a Nyquist
> frequency of above 1 MHz, the waveform can be reproduced exactly.
> 
Pat Farrell;231200 Wrote: 
> 
> Right. That's real likely.
> 

?!?!?

Pat, I don't really get what you're saying here. Is it that you believe
the Shannon Nyquist theorem is incorrect, or do you believe that a
band-limited square wave can not be expressed as a finite sum of sine
waves, all below said band limit? This is pretty fundamental stuff...
the latter statement is true simply by _definition_ of "band limited".

Pat Farrell;231200 Wrote: 
> 
> Round the numbers to make it easier to see.
> 20kHz square wave, sample at 40kHz.
> result is 1, -1, 1, -1, 1, -1, ...
> 

You've chosen an especially poor example here because in this case
20KHz is at exactly the Nyquist frequency. For waves at this exact
frequency there are in fact an infinite number of different
phase/amplitude combinations which could yield a given set of samples.


But that aside, the important point here is that you've forgotten the
step where said 20KHz square wave is filtered before sampling. When you
remove content above 20KHz, what you end up with is a perfect 20KHz sine
wave. That is really ALL there is left. A square wave is comprised of a
sine at its fundamental frequency, plus a sine at every odd harmonic
above that:  sin(x) + sin(3x)/3 + sin(5x)/5 + sin(7x)/7  etc. Try
graphing it.

This is the basis of what opaqueice stated earlier. The ONLY difference
between a 20KHz square and a 20KHz sine is at 60KHz and up. Band limit
them and they're identical.

> Feed it through a DAC and you get a 20kHz sine wave.

And that is exactly what went into the ADC!


-- 
seanadams
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