How to do 3-to-2 analog selection?
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As the title suggests, how to implement 3-to-2 selection for analog signals? What should the logic be? Who can tell me?
There’s no such thing as a two-out-of-three decision for analog signals. Generally, it means that an analog value is preferably converted into a boolean value, and then a majority vote is performed.
Yes. My original intention was that if 2 out of the 3 transmitters are working properly, we would consider it acceptable. Or take the median of 3. That’s roughly what it means.
It should be 3 choose 1; there are many strategies, such as removing those with poor signal quality, taking the average of the remaining ones, or removing the highest and lowest values and choosing the one in the middle
Add the three numbers together and divide by 3 to get the average; choose the number that is closest to this value.