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xarqi, Generalist
Category: Homework
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# Instrumentation

### Customer Question

Shown above is an optical encoder that can be used to measure position. The dark regions indicate opaque areas and the light sections are transparent regions. The circles represent LED/photodiode reading pairs. (As the disk rotates, readings under these fixed location components will vary.) Let a dark region produce binary ones (1) and the transparent areas generate binary zeros (0). (a) Produce a table that shows the codes as a function of the 16 positions shown on the encoder. (b) Describe a serious limitation with this arrangement. (Make sure that you provide a complete and detailed answer, not just a general response.) (c) What can you do to overcome this deficit?

Submitted: 11 years ago.
Category: Homework
Expert:  xarqi replied 11 years ago.
a)
Position Code
0             1111
1             1110
2             1101
3             1100
4             1011
5             1010
6             1001
7             1000
8             0111
9             0110
10           0101
11           0100
12           0011
13           0010
14           0001
15           0000

b) Resolution is only 4 bits, that is, the rotational position can
only be determined to a precision of 1 part in 16.

c)
More concentric patterns would allow increased precision. 16
would allow 1 part in 65536.
Customer: replied 11 years ago.

can you expand a little bit on parts b and c ?

Thank you

Expert:  xarqi replied 11 years ago.
I can try.

The encoder is divided into just 16 zones, so whatever the
physical rotation, the value it gives is one of just 16 possibilities.
That means that we can only know the real angle of rotation to a
precision of 22.5°, or 1/16 x 360°.

If the number of distinct zones were increased to 256 (using 8
concentric rings), we could tell the real position with a precision
of 1/256 x 360° = 1.4°. With 16 rings, the precision would
become 0.0055°.
Customer: replied 11 years ago.

ok man got it thanks

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