References: <3E244AFE.4F3E@sneakemail.com> <3E26E3E2.email@example.com> <3E2922B4.C3B@sneakemail.com> <3E2A07F4.firstname.lastname@example.org> <3E2A5081.4C97@sneakemail.com> <3E2AE941.76BD@sneakemail.com>
Subject: Re: Binary Sampler
X-Newsreader: Microsoft Outlook Express 6.00.2800.1106
Date: Sun, 19 Jan 2003 21:29:54 GMT
NNTP-Posting-Date: Sun, 19 Jan 2003 16:29:54 EST
Organization: Cox Communications
"Mike Monett" wrote in message
> Mike wrote:
> > "Mike Monett" wrote in message
> > news:3E2A5081.4C97@sneakemail.com..
> > > Mike wrote:
> > >
> > > [...]
> > >
> > > > Ugh. That's not good. Okay, then, what is your sample rate for the
> > 100MHz
> > > > sine and square waves?
> > > >
> > > > -- Mike --
> > >
> > > I'll have to check the software, but I'm pretty certain the sampling
> > > at the same frequency.
> > Okay then, let's try this: you input a sine wave at frequency fo, and
> > sample at frequency fo-df, and your output frequency is df, where df <<
> > Precisely the same result would be obtained if your input frequency was
> > You get (fo-df)/df samples per cycle of the df frequency sine wave. The
> > samples are passed through a limiter and low pass filtered to average
> > the noise.
> > -- Mike --
> I'm having a bit of trouble following your idea. Can you draw a
> schematic, and what are you using for a sampler?
It was supposed to be a yes/no question, based on your Equivalent Time
Sampling page. You put in 1GHz (fo), sample at 999,999,999Hz (fo-df), and
get 1Hz (df) out, where df (1Hz) << fo (1GHz).
This is the same as what you'd get if you put a 1Hz signal into your sampler
input instead of 1GHz, and kept the sample rate the same (999,999,999Hz).
You can see this by looking at Figure 4, "Frequency Difference Sampling," on
your Equivalent Time Sampling page. When you put in a fast signal and sample
at a nearby frequency, aliasing results in the output being shifted from fo
to df. The sine wave at df is the blue trace in figure 4. A sampler - any
sampler - can't tell the difference between the black trace and the blue
trace. So, if your input was the blue trace, the sample values would look
exactly the same.
The point is that sampling the fo input at fo-df is identical to sampling a
df input at fo-df. The reason that's important is that the df signal is
oversampled, and gets fed through the low pass filter (the window I was
referring to in an earlier post) to average the +/-1 DFF output.
-- Mike --