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Subject: Re: Binary Sampler
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Date: Tue, 21 Jan 2003 15:29:47 GMT
NNTP-Posting-Date: Tue, 21 Jan 2003 10:29:47 EST
Organization: Cox Communications
"Mike Monett" wrote in message
> Mike wrote:
> > "John Larkin" wrote in
> > message news:email@example.com..
> > > On Sun, 19 Jan 2003 22:04:26 GMT, "Mike" wrote:
> > >
> > > >The only way I'd know of to generate the sample
> > > >clock for Mike's scheme is to either generate it from the same source
> > > >that used to generate the input, or to derive the clock from the
> > then
> > > >use it to sample the input.
> > >
> > > It think this has been done. It's called 'triggered sweep.'
> > I was thinking of locking a PLL to the input and using the edges of the
> > recovered clock as the triggers, rather than triggering on each edge of
> > input signal. You get some noise rejection with the PLL, but with the
> > of noise Mike's adding, I suspect the PLL wouldn't lock... Of course,
> > neither would driving a trigger directly from the noisy input signal.
> > you thinking of something different?
> > -- Mike --
> Mike, I'm sorry I haven't seen this before, and you post has dropped off
> the server so I have to guess at the content.
> If you use the heterodyne technique described on my web site, you don't
> even need a trigger. All you need to know is the frequency of the signal
> you are trying to recover.
> After the samples are in memory, it is easy to reconstruct the waveform
> and display it with whatever starting point that is desired.
I suppose it depends on how well you're trying to reconstruct it, and what
information it contains.
Since your clocks aren't synchronous, your reconstructed waveform will have
variations due to both frequency offsets and jitter accumulation. For small
frequency offsets, you will lose information conveyed by the frequency of
the aliased sine wave. For large frequency offsets, the bandwidth of your
filter has to be high, and you lose noise rejection.
It looks like the only reason your circuit works is that your signal
frequency and your sample frequency are perfect, and contain no jitter. What
would happen if the frequencies were offset by 40ppm and the cycle to cycle
jitter was 0.001UI? I think the output would be much different...
-- Mike --