[time-nuts] Fast frequency counting question
Richard (Rick) Karlquist
richard at karlquist.com
Tue May 6 11:34:17 EDT 2008
Now we have gotten to the crux of the matter. You have
to decide on one of two approaches:
1. The zero crossing is what is to be measured.
In this case, you have to retain all harmonics. You cannot
use most of the traditional frequency counting techniques.
You must sample the zero crossing directly in the
full bandwidth of the digital signal. The digital signal
will have some phase noise floor. This noise floor,
over the measurement bandwidth, will imply some jitter
number by itself. For example, -150 dBc/Hz is equivalent
to several ps of jitter.
You have to take a huge number of samples for two reasons:
First, only a few samples will be near a zero crossing.
Second, you have to average away the jitter.
2. The phase noise model: a carrier with phase modulation noise.
In this case you bandpass filter the signal to be measured.
This gets rid of harmonics, producing a sine wave and also gets
rid of the phase noise floor. You can now measure phase vs time
and characterize your chirp. The bandpass has to be able to pass
enough bandwidth to capture the chirp sidebands. For this approach,
you can use the many known techniques of frequency measurement.
Rick Karlquist N6RK
Bruce Griffiths wrote:
> Luis Cupido wrote:
>> Hi Bruce,
>>
>> I don't know what you mean by low resolution
>> but I can easily think of more than 12 bit at
>> higher than 150Ms/s.
>>
>>
> An ENOB of 12 is probably inadequate for useful single shot
> measurements of a 40MHz sinewave .
>> > will tend to limit the effective resolution in determining the zero
>> > crossing locations.
>>
>> by no means zero crossing is to be used. we do have
>> much more information than that... phase info is all over
>> in all sampled points not just on zero crossings.
>> This is why I've said some transform to produce a spectrogram.
>>
>>
> The application is probably more sensitive to zero crossing jitter than
> anything else.
> If one directly samples the oscillator output directly the waveform may
> not be even approximately sinusoidal.
> With a logic level output oscillator a much higher sampling rate than
> 100Msps may be required to adequately sample the transitions adequately
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