Why? question about optical encoder

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The outermost ring of an optical encoder represents the least significant bit due to its larger circumference, allowing for more lines to be scribed, which enhances sensor resolution. While there is a relationship to frequency, the primary focus is on the resolution capabilities of the encoder. The discussion also touches on the use of Gray code in optical encoders, noting that the two least significant bits will have the same frequency but be 90 degrees out of phase. This design is reminiscent of earlier mouse encoders. Understanding these principles is crucial for optimizing encoder performance.
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Why is the outermost ring of an optical encoder the least significant bit? Is this somehow related to frequency?
 
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There is more length in the outside ring.
For a sensor of any given resolution the wider the disk the more lines that can be scribed on it.
While in a way this relates to frequency, it is more about sensor resolution.
 
just curious, does your optical encoder use Gray code? it should. if it does, the two LSBs (of Gray code) will have the same "frequency" but will be 90o outa phase. just like the encoders of the earlier mice.
 

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