Talk:Common mode

Latest comment: 7 years ago by Spinningspark in topic Untitled
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Untitled edit

These meanings are not independent. "Common mode" means "the mode shared between two apparently independent inputs." When the two inputs are a and y, the common mode is cm = 0.5*(a+b) and the difference is dm = 0.5*(b-a), so that b=cm+dm and a=cm-d1. Authorities differ on the factor of 0.5 in d1. It is a mistake to distinguish according to y the meanings of "x y", where y is a noun and x qualifier, in this case y = "common mode." "Common mode" has its own meaning, independent to what term it qualfies. What it means is that if the imputs x, y, and even z are x=cm+d0, y=cm+d1, z=cm+d2, and so forth, we are talking about cm independent of the d's.

o The CMMR is the ability to detect signals like d1 independent of cm.

o Common mode interference is interference that appears on cm but does not appear in the d's.

o Ths CM signal is cm itself.

o A "common mode failure" is by extension a failure such that when failure n is driven by x=cm+dn, it is cm and not the differences dn that cause the failure. Yes, algebraic notation does not represent failure modes very well, but where x=cm+d0, y=cm+d1, z=cm+d2, +ct. represent the probabilities of all the possible events, a "common mode failure" is a failure driven by cm.

This page needs to be about what the "common mode" is, rather than what the raw term means when used to qualify signal sensitivity or failures.

Pantagnosis (talk) 02:21, 1 August 2016 (UTC)Reply

@Pantagnosis: The essential thing you need to understand here is that this page is a disambiguation page. These pages do not contain any detail at all, but just point the reader to the relevant pages. We do not create articles with mixed subjects; the common mode of transmission lines has nothing to do with common failure modes despite your attempt to desribe them both with the same algebra. Describing all semantic uses of common mode together is the job of a dictionary, not an encyclopaedia, and we have Wiktionary: for that. SpinningSpark 15:07, 11 August 2016 (UTC)Reply