[time-nuts] DIY TimePod
John Swenson
johnswenson1 at comcast.net
Tue Jun 14 20:53:50 EDT 2016
Got it, I missed the 27 MHz low pass filter with 60 db attenuation. So
the ADC really is mostly seeing a sine wave.
I guess it's back to the drawing board and doing this with the filter
and the ADCs.
Thanks for setting me straight on this.
John S.
On 6/14/2016 3:01 PM, Chris Caudle wrote:
> On Tue, June 14, 2016 2:35 am, John Swenson wrote:
>> The idea here is around a 80MHz sample clock with a
>> maximum input/ref signal of around 25MHz.
>
> Without some pretty steep low pass filtering that will violate the Nyquist
> criterion (for 80MHz sample clock the input must be strictly limited to
> less than 40MHz). You can't even get the first odd harmonic in of a 25MHz
> square wave input.
>
>> This is based on the TimePod with ADCs, which is
>> supposed to work with square waves.
>
> The ADC's would have a low pass filter in front. Think of it in terms of
> the Shannon information capacity, the amount of information conveyed is
> determined by the bandwidth and the signal to noise ratio. The bandwidth
> is determined by the sample rate, the signal to noise ratio by the number
> of (effective) bits of the ADC.
> I forget which ADC someone mentioned recently as being in the TimePod.
> Isn't it a 16 bit converter? So that is getting around 96dB integrated
> signal to noise ratio per converter, and you are starting with 6dB.
>
>> When you feed a square wave into this you have several samples at say
>> 50, then it jumps to 50,000 stays there for several samples, then jumps
>> down to 50 again.
>
> The key thing you are missing which happens with a multi-bit ADC is that
> the signal has a finite rise time, so it doesn't "jump" to 50,000, it has
> a transition region where you get several samples of different values.
> Those samples fit an infinite number of possible signals, but only one
> signal which is limited to the Nyquist criterion bandwidth. Using those
> samples and the knowledge of the system bandwidth you can interpolate
> where the zero crossing must have been.
>
> With a single bit quantizer (and no feedback to shape the noise), you get
> very little information about the signal values in the transition region.
>
>> This still seems like a binary sample. The difference
>> is that every now and then the sample hits during a ramptime of the
>> square wave and will give some intermediate value,
>
> No, every time you will sample during the transition, because the "square"
> wave still has a finite rise time, and if you have properly bandwidth
> limited the signal as required by the Nyquist sampling criterion (input
> signal must be less than half the frequency of the sampling clock) then
> you know what the upper limit on the rise time is.
>
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