Root-mean-square of a sine wave

1. Nov 6, 2006

xlq

rms of sine wave = peak * 1/SQRT(2)

how is this derived from the rms equation?

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2. Nov 6, 2006

ObsessiveMathsFreak

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.

3. Nov 6, 2006

xlq

OK thanks. One more question: how did you integrate the sine-squared?

4. Nov 6, 2006

Tomsk

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)) :)

5. Nov 6, 2006

mathman

sin^2 + cos^2 =1. The rest is trivial.

6. Nov 6, 2006

quasar987

how do u use sin^2 + cos^2 =1 to evaluate the integral of sin² mathman?

7. Nov 7, 2006

ObsessiveMathsFreak

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.

$$\int_0^{\pi} \sin^2{\theta} d\theta = \int_0^{\pi} \frac{1}{2} \left(1-\cos{2\theta} \right) d\theta$$
$$= \frac{1}{2} \int_0^{\pi}1 d\theta -\frac{1}{2} \int_0^{\pi} \cos{2\theta} d\theta$$
$$= \frac{1}{2} \left[ \theta |_0^{\pi}\right] + \frac{1}{4}\left[ \sin{2\theta}|_0^{\pi}\right]$$
$$= \frac{1}{2} ( \pi - 0) \frac{1}{4} ( 0 - 0) = \frac{\pi}{2}$$

Last edited: Nov 7, 2006
8. Nov 7, 2006

StatusX

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:

$$\pi=\int_0^\pi dx = \int_0^\pi (\sin^2 x + \cos^2 x) dx = 2 \int_0^\pi \sin^2 x dx$$

9. Nov 7, 2006

mathman

Integral of sin^2 = integral of cos^2, so each must be 1/2 of the integral of 1.

10. Feb 5, 2010

opitts2k

Re: Integral Sine Squared Schaums - Theory and Problems of Electric Circuits 1.3

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:

$$P_{avg} = ( \frac{ W_{T} }{ \frac { 2\pi}{\omega}} ) = 56.25mW$$

Last edited: Feb 5, 2010
11. May 24, 2010

agouti

Re: Integral Sine Squared Schaums - Theory and Problems of Electric Circuits 1.3

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 $$f_{avg}=\frac{1}{2}$$

12. May 24, 2010

halawaliama

I have a problem that has been bugging me all year. Yes, all year - which i know is sad. can anyone tell me how to solve the following problem?

What is the power applied to a string that is 10 meters long, has a hertz of 440?

I can't really find a formula that i can understand. i don't know much about physics and this was a problem from last semesters final, as extra credit on a algebra test - go figure. anyway it is still bugging me.

thanks,

13. May 25, 2010

agouti

Are you sure that is the whole question? That can't possibly be enough information to come up with an answer, surely...

Without knowing anything about the string, we could assume that it was a ideal (lossless) carrier of the wave, so the "power" would be the energy taken to excite it - however we aren't given the amplitude of the wave.

I'm assuming that the system is a string attached to a fixed point, being excited by the other end?

The question seems far to vague to have a numeric answer, and if they just wanted proof of understanding why did they give exact length/frequency instead of length of L, frequency of F Hz or such...

14. May 25, 2010

halawaliama

i think i might have left off a part...something about the wavelength being 2 meters...woudl that help? if not then i will just have to try to get the question again.

thanks for looking at this for me ;-)