Nov 20, 2012 #1 MamadJun Junior Member level 2 Joined Jan 11, 2011 Messages 24 Helped 0 Reputation 0 Reaction score 0 Trophy points 1,281 Location Belgium Activity points 1,612 can anyone explain and say the reason of this sentence? Any uniformly distributed random variable in the range (-a/2 to a/2) has an average power(variance) given by a^2/12. is it ok?
can anyone explain and say the reason of this sentence? Any uniformly distributed random variable in the range (-a/2 to a/2) has an average power(variance) given by a^2/12. is it ok?
Nov 22, 2012 #2 M Mityan Full Member level 5 Joined Jul 11, 2012 Messages 275 Helped 48 Reputation 96 Reaction score 45 Trophy points 1,308 Activity points 2,835 Yes. Variable - v. Mean-square value = integral (or sum) of v^2 divided by a (i.e. in interval from -a/2 to a/2). It means (v^3)/3/a. From -a/2 to a/2. After placing in this formula the integration limits instead of v, you get a^2/12.
Yes. Variable - v. Mean-square value = integral (or sum) of v^2 divided by a (i.e. in interval from -a/2 to a/2). It means (v^3)/3/a. From -a/2 to a/2. After placing in this formula the integration limits instead of v, you get a^2/12.