Help with Exam Question on Optical Position Encoder

[rlspin]

Hello everyone, I have an exam in a few days and need help with a question on one of the past exam papers.

An optical position encoder used on a robot axis has a 10:1 gear ratio, an optical disc with 72 slits, and a 12 bit binary counter. Determine:

(i)
The resolution of the encoder i.e. the minimum angular movement of the measured shaft that can be detected?

(ii)
The maximum allowable shaft motion to ensure that the counter never over-ranges?

(iii)
The amount of shaft movement represented by a binary output of 0101 1111 0010?

For (i), the encoder has a 12 bit binary count, so the resolution will be:
360/2^12 = 0.088°

Im not sure what to do for (ii) and for (iii), I understand that the binary output corresponds to the numbers 5 15 2 but what do I do with them, if i even need them?

 

[berkeman]

For (i), you need to use the 10:1 gear ratio (wish they were more explicit about which way the X10 goes...) and the number of slits, not the max count of the counter.

It's in (ii) that the max count comes into play, along with your answer for (i).

On (iii), that number is not 5 15 2. The spaces are just there for readability. What is that binary number in decimal?

 

[rlspin]

Ok cool, so am I right in saying that (i) the resolution will be 360/(10*72)?

Im still not 100% sure about (ii)

and for (iii), 0101 1111 0010 in decimal would be:
2^1 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 + 2^10 = 1522

 

[berkeman]

Ok cool, so am I right in saying that (i) the resolution will be 360/(10*72)?

Im still not 100% sure about (ii)

and for (iii), 0101 1111 0010 in decimal would be:
2^1 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 + 2^10 = 1522

Good. For (ii), you know how many pulses per revolution you get from (i), and you should be able to say what the maximum count is out of the 12 bit counter...