dvsarwate <dvsarwate@yahoo.com> writes:> On Tuesday, January 28, 2014 1:21:32 PM UTC-6, rudykeram wrote: >> Hi, >> >> I am trying to have a deeper understanding of what RMS means in terms of >> >> probability distribution function. >> >> >> >> There is a figure (toward the end) of the following article (doesn't have a >> >> figure number) >> >> http://www.mscsoftware.com/training_videos/patran/Reverb_help/index.html#page/Fatigue%20Quickstart%20Guide/fatq15_vibFAT.15.5.html >> >> >> >> It depicts the bell shape of a random normal distribution, where was >> >> generated by basically taking the histogram of a snapshot of a signal. And >> >> this makes perfect sense. >> >> However, the figure below it depicts the PDF of a sine wave, and I am just >> >> not sure why this is the case? >> >> To me, it looks like the distribution should be uniform. Any horizontal >> >> lines through the signal (sine wave), will cross the signal the same number >> >> of times! Isn't it? >> >> >> >> something more in line with this: >> >> http://en.wikipedia.org/wiki/Uniform_distribution_(continuous) >> >> >> >> >> >> Please correct me if I am wrong? >> >> >> >> Thanks, >> >> --Rudy >> >> >> >> _____________________________ >> >> Posted through www.DSPRelated.com > > > A snapshot of a random signal is modeled mathematically as > a segment of a _sample function_ of a _random process_ which > is a collection of (infinitely many) random variables > One method of estimating some of the (infinitely many) parameters > of the random process is to assume that all the random variables > have the same pdf and CDF and that the process is ergodic, meaning > that measurements made on sample functions are representative of > the random variable parameters. In particular, if the duration > of your snapshot is T seconds (where T is reasonably large so that > most of the variations of the process are captured), then the > _proportion of time_ that the random signal spends below a given > level x is a good measure of F(x), the value of the common CDF of > the random variables, at x. For many random signals, this CDF looks > like the CDF of a normal random variable with mean m and standard > deviation s. "Looks like" is important in that the measured CDF > is 0 at the minimum value of the signal in the segment and 1 at > the maximum value whereas the normal CDF approaches 0 at -infinity > and 1 at +infinity, but the match is pretty good at values between > m-3s and m+3s, and certainly good enough for gummint purposes. > Correspondingly, the _histogram_ from the snapshot looks like the > pdf of a normal random variable with mean m and standard deviation s. > > For periodic nonrandom signals such as sinusoids, calculations > like the above should be made with an integer number of periods. > Even one period will do, and in the case of sin(t), we can even > use the half-period from -pi/2 to +pi/2 during which the sinusoid > varies from -1 to +1. What will the _histogram_ look like? Well, > for any fixed x between -1 and +1, the sinusoid has value x at > t = arcsin(x) and value x+dx at time t = arcsin(x+dx) and so the > proportion of time that the signal spends between x and x+dx is > [arcsin(x+dx) - arcsin(x)]/pi. Jumping to the pdf, this is saying > that > > f(x)dx = [arcsin(x+dx) - arcsin(x)]/pi > > and so > > f(x) = (1/pi) (d/dx) arcsin(x) = 1/[pi sqrt{1-x^2}], -1 < x < 1 > > which, as many have already pointed out, approaches infinity > as x approaches +/-1. It is by no means a uniform pdf. If > you want a uniform pdf for the amplitude, use periodic ramps > or sawtooth signals.Nice analysis. -- Randy Yates Digital Signal Labs http://www.digitalsignallabs.com
RMS of a Sine Wave
Started by ●January 28, 2014
Reply by ●January 30, 20142014-01-30
