AD2S1210 Equation 5 (LOS angular latency) missing square root?

Hi David,

I guess the Equation (5) is not derived from the monitor signal, but from the basic resolver calculation "angle = arc tan(SIN/COS)", and that the "arc cos" in Equation (5) should be an "arc tan".

My thinking is that during operation it is completely normal that one of the signals SIN or COS is below the LOS-threshold (i.e. at the 90° boundaries). To detect a LOS, both voltages must be below the LOS threshold. So there are two regions: One where both voltages are below the LOS threshold, and an error can be triggered, and another region which looks like a signal at the 90° boundary. The size of the regions depends on the set LOS threshold. If set lower, the region where an error is triggered is smaller and the region which looks like normal operation gets bigger, and vice versa. Equation (5) would give you an idea how big the region of "looks like normal operation to me" would be with a specific set threshold.

But, just a guess. Maybe @skowalik is still around, and can correct me.

Regards

@rehfi @KallinD  I'm still here!!!!!

The answer to your question on the monitor signal is described in the paragraphs just below equation 4.   Assuming the no gain errors between the resolve inputs to the device the magnitude term A1 for the sin channel will equal A2 the amplitude.  If the tracking error is zero then the monitor signal simplifies to 

A\times(\sin^2(\theta)+\cos^2(\theta))=A

So there is no need for a square root as the trigonometric identity for the sum of sine squared and cosine squared is unity.

Sean

Thanks @skowalik , but isn't that the no-fault case? My understanding is that Equation (5) applies when one channel is lost.

Thanks, rehfi . I think you're onto something, although I can't see where the arctan term comes from when I try to derive it myself.

Let's assume the cosine channel fails and remains at zero, while the sine channel continues operating normally. In that case, the cosine channel is always below the threshold, whereas the sine channel only falls below the threshold at arcsin(Threshold/Amplitude).

If LOS detection is triggered only when both channels are simultaneously below the threshold, then the resulting blind-zone width becomes:

2 × arccos(LOS Threshold / Amplitude)

This derivation seems to support Equation (5) being correct, although for a different reason than I originally assumed. My initial assumption was that LOS was based on the monitor signal, whereas this result follows from individual SIN/COS threshold comparisons combined with an AND condition. However, with this approach and no faults (both cosine and sine channels nominal), LOS would repeatedly be triggered at e.g.: 45, 135, degrees if the LOS threshold would be set lower than A x sin(45) = A x 1/sqrt(2), which seems odd.

To rephrase my question: Is LOS detected by comparing the monitor signal's values against the threshold, or by comparing the SIN and COS channels individually using an AND condition? Could you confirm which method is actually implemented?



Regards,
David