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This is a question from this application note:
http://www.atmel.com/images/doc8003.pdf

In section 3.2 on noise it says that adding some amount noise into the signal helps in increasing the resolution of the oversampling converter. It says if LSB toggles due to noise and we are averaging 4 of the samples then the quantization step is halved. I do not understand how is the quantization step becomes half. Could anyone explain this?

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    \$\begingroup\$ Dithring is explained in this answer. \$\endgroup\$
    – CL.
    Commented Sep 2, 2015 at 13:21

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Suppose you have a signal which is 1/4 they way between LSBs. With four noiseless samples you would just get four zeros.

With a slightly noisier signal, one fourth of those samples (1 on average) would actually get high enough to become a 1. The average of three zeros and a one becomes 1/4.

Similarly if it were closer to 1, you would get some smaller fraction of 0s.

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  • \$\begingroup\$ Of course, the trick is to add just the right amount of noise, so it doesn't show up too much at higher levels. \$\endgroup\$ Commented Sep 2, 2015 at 14:06

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