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A PLL with digital PFD (e.g. 4046 phase comparator 2) can potentially work over a large frequency range if no fast frequency variations are intended. But switching the loop filter per octave or at least decade would be preferred.I doubt a PLL is stable for the whole frequency range. At least I have no experience with that huge frequency range and PLL.
IBut all complete designs for a LIA I have seen online besides low-speed analog stuff have been of the variety that is completely or next to completely digital.
The AD9954 has a sync_out signal. Maybe it is possible to use this signal to generate (or synchronize) the comparator signals (use clocked comparators?).
Klaus
I don´t say the LIA is an easy task.. one must take care of signal integrity, phase angles and amplitudes. But it is not more difficult than with a high speed ADC solution.
Maybe I´m too focussed on the signal precision. Maybe generating the quadrature signal with the help of comparators is good enough for the OP. I don´t know.
Klaus
Example: a sine with 3.5V amplitude and +30° phase shift.In the mean time I thought I would ask if someone here could help me understand how to simulate the data streams coming from the complete and functioning circuit in Matlab?
It's not the problem to generate two 90° shifted digital signals.Using D-type flip flops you can produce square waves at 180 and 90 phase differences with little effort.They simply act as frequency dividers.
Example: a sine with 3.5V amplitude and +30° phase shift.
Imagine a coordinate system.
Draw a circle with radius 3.5cm around zero (0/0).
(0 degree phase shift is a line from (0/0) to the right side.)
Now draw a line from (0/0) but 30° shifted ccw. (First quadrant, x is positive, y is positive)
The line crosses the previously drawn circle.
Now calculate (measure) x and y of the crossing point.
X = 3.5 × cos(35°)
Y = 3.5 × sin(35°)
These are the two DC values output from the filtered LIA.
Klaus
It's not the problem to generate two 90° shifted digital signals.
But they need to have the EXACT same frequency as the sine, and they need to have a VERY CONSTANT phase relation to the sine.
Klaus
Example: a sine with 3.5V amplitude and +30° phase shift.
... ...
X = 3.5 × cos(35°)
Y = 3.5 × sin(35°)
These are the two DC values output from the filtered LIA.
Why 35; I believe you mean 30, right?