RMS Integral Calculus

RMS Integral Calculus

Postby Guest » Wed Jan 08, 2020 7:17 am

Calculate the mean and RMS values over half a cycle from the following function

Phi = 3SIN omega t
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Re: RMS Integral Calculus

Postby HallsofIvy » Thu Feb 06, 2020 10:19 am

Do you know what "mean" and "RMS" ("root mean square") are?

For a finite collection of numbers, the "mean" ("average value") is the total of the numbers divided by now many numbers there are. The "mean" of the numbers 2, 5, 8 is (2+ 5+ 8 )/3= 15/3= 5. For the "mean" of a continuous function over a given interval, the "sum" is replaced by the integral over the interval and "how many numbers" by the length of the interval.
The sine function, sin(x), has period (so cycle length) [tex]2\pi[/tex] so [tex]sin(\omega x)[/tex] has period [tex]\omega x= 2\pi[/tex] or [tex]x= \frac{2\pi}{\omega}[/tex]. Half the cycle is [tex]\frac{\pi}{\omega}[/tex].
The "mean" of the function [tex]f(t)= 3 sin(\omega t)[/tex] over the interval [tex][0, \pi/\omega][/tex] is [tex]\frac{3\int_0^{\pi/\omega} sin(\omega t) dt}{\pi/\omega}[/tex].

The "root mean square" is the square root of the mean of the square. For [tex]f(t)= 3 sin(\omega t)[/tex] that is [tex]\sqrt{\frac{9\int_0^{\pi/\omega} sin^2(\omega t)dt}{\pi/\omega}}[/tex].

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