[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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