Sorry, it was getting late here and I didn’t have the energy for words…
I was trying to point out the relationship between bandwidth and sample rate in response to deku’s post.
In this case the sampling is also happening on the MCU side, since you read the linear hall with the ADC at discrete points in time.
The sensor itself also has a bandwidth though (as stated in the datasheet). As an analog sensor, I don’t think it has an internal “sample rate” - but it still has an analog bandwidth, which is typically related to the RC time constants of the analog circuit. To change the voltage it has to drive charge into the capacitance of the output line and switches through the resistance. So if you add an external RC-filter to the hall’s output, you would be changing (lowering) the analog bandwidth, but it also has internal capacitances and resistances which mean the bandwidth can’t be infinite.
As the motor speed approaches the sensor’s bandwidth, I think the output signal will be both phase shifted and attenuated. But with a 20kHz bandwidth you should be ok for quite some speed. To give you an idea:
Lets say your motor has 7 pole pairs, then the frequency of the sensor’s sine or cosine waves will be
F_s = \frac{v}{2\pi} \times pp
where v is rotor velocity in rad/s.
So 10000RPM = 1166Hz commutation frequency on a 7PP motor, which should be comfortably in the 20kHz bandwidth afforded by the sensor. And 10000 RPM is a lot!
Anyways, I would focus more on the sampling by the MCU, as I think this will more likely be the limiting factor:
As you can see, as the MCU sample rate per electrical revolution goes down, your MCU will see the curve less and less as a sine wave. This will affect the accuracy of your position estimation, and introduce errors.
I think this way of thinking about it may not be numerically accurate, but is intuitively correct. The relationship isn’t exactly linear, since the effective torque will be related to the cosine of the angle error, but the idea is the right one.