# Root-mean-square of a sine wave

by xlq
Tags: rootmeansquare, sine, wave
 P: 2 rms of sine wave = peak * 1/SQRT(2) how is this derived from the rms equation? (Search engines have returned no useful results.)
 P: 406 The original sine wave is take to represent a current. This current is squared to get the power. P=I^2. Integrating sine squared between from time 0 to pi gives; $$\int_0^{\pi} \sin^2{t} dt = \frac{\pi}{2}$$ This is the same a multiplying the constant power 1/2 by the time pi seconds. So 1/2 is the "average" power level, and since I = sqrt(P), 1/sqrt(2) is the "average" current level needed to obtain a meaningful powerlevel result.
 P: 2 OK thanks. One more question: how did you integrate the sine-squared?
P: 227

## Root-mean-square of a sine wave

use the trig identity sin^2(t)=1/2(1-cos(2t) (it might be +, i can't remember. Derive it from the double angle formula for cos(2t)) :)
P: 5,768
 OK thanks. One more question: how did you integrate the sine-squared?
sin^2 + cos^2 =1. The rest is trivial.
 PF Patron HW Helper Sci Advisor P: 4,755 how do u use sin^2 + cos^2 =1 to evaluate the integral of sin² mathman?
P: 406
 Quote by mathman sin^2 + cos^2 =1. The rest is trivial.
No. That was the trivial part. The rest is hard.

Using the double angle cossine formula;

$$\cos{2\theta} = cos^2{\theta} - sin^2{\theta} = 1 - 2\sin^2{\theta}$$
Therefore;
$$2\sin^2{\theta}= 1-\cos{2\theta}$$
$$\Rightarrow \sin^2{\theta}= \frac{1}{2} \left(1-\cos{2\theta} \right)$$

So that means the integral can be evaluated as follows.

$$\int_0^{\pi} \sin^2{\theta} d\theta = \int_0^{\pi} \frac{1}{2} \left(1-\cos{2\theta} \right) d\theta$$
$$= \frac{1}{2} \int_0^{\pi}1 d\theta -\frac{1}{2} \int_0^{\pi} \cos{2\theta} d\theta$$
$$= \frac{1}{2} \left[ \theta |_0^{\pi}\right] + \frac{1}{4}\left[ \sin{2\theta}|_0^{\pi}\right]$$
$$= \frac{1}{2} ( \pi - 0) \frac{1}{4} ( 0 - 0) = \frac{\pi}{2}$$
 HW Helper P: 2,566 sin2(x) is just cos2(x) shifted by pi/2, and each have a period of pi, so their integrals over 0 to pi are equal. Thus: $$\pi=\int_0^\pi dx = \int_0^\pi (\sin^2 x + \cos^2 x) dx = 2 \int_0^\pi \sin^2 x dx$$
 P: 1 Schaums - Theory and Problems of Electric Circuits (4nd ed) 1.3- Solution $$\copyright$$ Sin squared = 1 - cosine 2x 1.3 A certain circuit element has a current i = 2.5 sin $$\omega$$t (mA), where $$\omega$$ is the angular frequency in rad/s, and a voltage difference v = 45 sin $$\omega$$t (V) between terminals. Find the average power $$P_{avg}$$ and the energy $$W_{T}$$ transferred in one period of the sine function. RECALL THAT : VOLTAGE(v) X CURRENT(i) = WORK($$W_{T}$$) 2.5 X 45 = 112.5 so.... $$W_{T} = \int_{0}^{\frac{2\pi}{\omega}} vi dt = W_{T} = 112.5 \int_{0}^{\frac{2\pi}{\omega}} \sin x^2 dt$$ RECALL TRIGONMETRY IDENTITY $$sin^2x = 1/2(1- cos2x)$$ $$W_{T} = \int_{0}^{\frac{2\pi}{\omega}} \sin x^2 dt = W_{T} = \frac{1}{2} \int_{0}^{\frac{2\pi}{\omega}} (1 - \cos 2x) dt$$ WE NOW HAVE .... $$W_{T} = \frac{1}{2} \int_{0}^{\frac{2\pi}{\omega}} (1 - \cos 2x) dt$$ CALCULUS IDENTITY FOR $$\int\cos(x) dx = \sin x + C$$ PROOF with U Substitution : Let u = 2x THEN $$\frac{du}{dx} = 2$$ AND dx = $$\frac{1}{2}du$$ $$\int\cos\2x$$ = $$\int\cos u\frac{1}{2}du$$ $$\frac{1}{2}\int\cos u du$$ $$\frac{1}{2}\sin u + C$$ $$\frac{1}{2}\sin2x + C$$ SO NOW .... $$\left[ \frac{1}{2} ( x - \frac{1}{2} \sin 2x ) \right]_{0}^{\frac{2\pi}{\omega}}$$ NOW UPPER AND LOWER INTEGRATION GIVES $$\left[ \frac{1}{2} ( x - \frac{1}{2} \sin 2x ) \right]_{}^{\frac{2\pi}{\omega}} - \left[ \frac{1}{2} ( x - \frac{1}{2} \sin 2x ) \right]_{0}^{\frac{}}$$ SUBSTITUTION HERE .... AFTER SUBSTITUTION RECALL TRIGONOMETRY SINE $$\frac{\pi}{\theta}$$ = 0 $$\left[ \frac{1}{2} ( ( \frac{2\pi}{\omega} ) - \frac{1}{2} \sin 2 ( \frac{2\pi}{\omega} ) ) \right]_{}^{\frac{2\pi}{\omega}} - \left[ \frac{1}{2} ( ( 0 ) - \frac{1}{2} \sin 2 ( 0 ) ) \right]_{0}^{\frac{}}$$ OK ... THIS IS WHAT'S LEFT $$W_{T} = 112.5 \left[ ( \frac{1}{2} ) ( \frac{2\pi}{\omega} ) - 0 \right]$$ THEREFORE: $$W_{T} = 112.5 \left[ ( \frac{\pi}{\omega} ) \right] = ( \frac{ 112.5\pi}{\omega} ) \right]$$ FINALLY: $$P_{avg} = ( \frac{ W_{T} }{ \frac { 2\pi}{\omega}} ) = 56.25mW$$
 P: 2 The answers given above by ObsessiveMathsFreak, namely that the average is $$\frac{\pi}{2}$$ is wrong. The integral over a period will *not* give the average, merely the area under the curve, which should be obvious to everyone who provided the answers - as proof consider that the average of $$\sin^2(\theta)$$ over any whole period should be the same, be it from 0 to $$\pi$$ or from 0 to $$500\pi$$. Yet, with the above answers this is clearly not the case. A (or The) correct way to find the average is via average value theorem. $$f_{avg}(x)=\frac{1}{b-a}\int_{a}^{b}f(x)$$ Which in the case of $$\sin^2(\theta)$$ yields $$f_{avg}=\frac{1}{2}$$