[time-nuts] FE-5680A FAQ update: question about frequency synthesizer architecture
beale
beale at bealecorner.com
Fri Jan 27 20:01:14 UTC 2012
I appreciate those notes; the data sheet for the DDS part also explains it. It has a 12-bit LUT driving an internal 10-bit DAC (the 5.3 MHz sine is still a bit coarse, with < 4 samples per cycle, but in the 5680 it is cleaned up by an external filter).
"Because phase information maps directly into amplitude, a ROM LUT converts the phase information into amplitude. To do this, the digital phase information is used to address a sine ROM LUT. Although the NCO contains a 32-bit phase accumulator, the output of the NCO is truncated to 12 bits. Using the full resolution of the phase accumulator is impractical and unnecessary because this would require a look-up table of 2^32 entries."
...from http://www.analog.com/static/imported-files/data_sheets/AD9832.pdf
However, my question was actually about how the remainder of the circuitry in the FE-5680A combines the 5.3 MHz from the DDS (at 4 mHz tuning step size), and the 60 MHz VCXO, to reference against the 6.835 GHz Rb frequency and ultimately achieve 0.18 uHz (micro-Hz) tuning step size at the final 10 MHz output. I don't think a simple multiplier-mixer-divider chain (for example) could give you such a small tuning step size at the output, the frequency ratios don't work out. I've heard of fractional-N PLL synthesizers but I'm not sure if that's the principle here.
> -------Original Message-------
> From: Graham / KE9H <timenut at austin.rr.com>
>
> The AD9832 is an Analog Devices DDS which has a 32 bit tuning word.
> The way a DDS generates the output, is that it (effectively) has a cosine
> wave look-up table, with 2^32 entries that comprise a single cosine wave
> cycle.
>
> The tuning word tells it how many entries the DDS should advance every
> reference input clock cycle, then it pushes that amplitude value in the
> look-up table
> to the output D->A converter.
>
> So, if the input reference is 20 MHz, then the DDS can generate frequencies
> with a resolution step of
>
> Vref/2^32 = 20,000,000 / 4,294,967,296 = 0.0046566 Hz.
>
> The DDS output frequency is (tuning word /2^32) times Vref.
>
> In the actual implementation, rather than a 4 billion entry look-up table,
> I am sure they have some algorithm that calculates the amplitude of
> a cosine wave, or a much smaller table with a sophisticated interpolation
> routine.
>
> --- Graham / KE9H
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