Does it ever make sense to have more precision than accuracy? E.g. if you had a variable resistor which you could control with a resolution of 0.01 ohms but your multimeter only had 1% accuracy, then that extra resolution would be pretty useless wouldn't it?
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1\$\begingroup\$ "make sense" and "useless" are not very accurate or precise terms for defining a question. ;) \$\endgroup\$– Nick AlexeevDec 16, 2013 at 2:01
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\$\begingroup\$ Haha, fair enough, will take note for any future questions! \$\endgroup\$– jrandjDec 16, 2013 at 2:25
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3 Answers
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Absolutely. There are many applications in which you're much more interested in the relative changes in a signal than it's absolute magnitude.
Take digital audio, for example. It is the precision of the converters that gives you the high signal-to-noise ratio you're interested in. The absolute accuracy (i.e., whether full-scale is 1.0 or 1.1 V) is not generally all that important — it just means the signal overall sounds a little softer or louder, which you can compensate for by adjusting the volume control.
answered Dec 16, 2013 at 2:05
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\$\begingroup\$ measuring parameter variance of high-volume prodution hardware also comes to mind \$\endgroup\$– scldDec 16, 2013 at 13:47
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\$\begingroup\$ There is also a distinction between measurement precision and value precision. A 12-bit ADC provides 12 bits of value precision by definition but might provide less measurement precision. Deviation of a value from reality may be more or less systematic (with constant offset being a simple case) with the system being more or less easily modeled (randomness and limited measurement precision related to quantum mechanics being an extreme). \$\endgroup\$– user15426Dec 17, 2013 at 23:07
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1\$\begingroup\$ @PaulA.Clayton: I think what you're calling "value precision", I would simply call "resolution". But we're really getting into hair-splitting territory here. In terms of the actual performance of the application, it's the "measurement precision" that's relevant. \$\endgroup\$ Dec 17, 2013 at 23:20
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Yes it does, because when calibrating it allows you to measure the error with accuracy equal to the precision.
answered Dec 16, 2013 at 2:04
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\$\begingroup\$ Exactly. Accuracy is typically just how a component is specified... with characterization and calibration you can do sometimes much better. \$\endgroup\$– darronDec 17, 2013 at 14:42
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Errors add up. If I measure a circuit in production at 1%, and it tells me that I need 1R, I'd like to be able to set 1R, and be within 1% of correct value. If I set 1R +/- 1%, I'm within 2% of correct value.
