Mathematics OpenStudy (jack1):

Maybe an odd/slightly difficult question, but would appreciate any explanation given... Why does RSM of a sine wave = peak x sqrt 2...? OpenStudy (anonymous):

RSM? ganeshie8 (ganeshie8):

RMS value i think OpenStudy (jack1):

ugh, my bad, root mean squared OpenStudy (anonymous):

RMS using the continuous definition? OpenStudy (jack1):

yeah, i guess? all i understand is how to find the RMS value of a sine wave... not the why it is so... OpenStudy (jack1):

this whole q is screwy, sorry, peak = rms x sqrt 2 my bad ... again OpenStudy (anonymous):

Start from definition of RMS $f_{RMS}=\lim_{x\to \infty}\sqrt{\int_0^x[f(t)]^2\;dt}$ OpenStudy (anonymous):

So we'll start with something as simple as $\int_0^x\sin^2(t)\;dt$ OpenStudy (anonymous):

Use $$\sin^2(t)=\frac 12[1-\cos(2t)]$$ OpenStudy (anonymous):

This doesn't look good because the limit won't converge at infinity. OpenStudy (anonymous):

So instead just do something like the period of the sine wave, I suppose. It is perfectly representative. OpenStudy (anonymous):

The period of $$\sin^2(t)$$ is $$\pi$$. OpenStudy (anonymous):

So I guess it is good enough to just do: $\sqrt{\frac 1\pi \int_0^\pi \sin^2(t)\;dt}$ OpenStudy (anonymous):

@Jack1 Does that make sense so far? OpenStudy (jack1):

yeah, im with you so far OpenStudy (anonymous):

$\int \sin^2(t)\;dt=\frac 12\int 1-\cos (2t)\;dt=\frac 12 t-\frac 14\sin(2t)+C$ OpenStudy (anonymous):

So $\int _0^\pi \sin^2 (t)\;dt = \frac 12 t-\frac 14 \sin(2t)\Bigg|_0^\pi=\frac \pi2$ OpenStudy (anonymous):

Putting it back in: $\sqrt{\frac 1\pi \times \frac \pi2} = \sqrt\frac 12$ OpenStudy (anonymous):

Peak x, is really just the amplitude. OpenStudy (anonymous):

In this case amplitude was 1. OpenStudy (jack1):

brilliant, but my integration is a bit off, so why did we take the 1/pi out in the first place again...? OpenStudy (anonymous):

Because it is in the definition. OpenStudy (anonymous):

If you're doing RMS over an interval, you have $\sqrt{\frac 1{b-a}\int _a^b[f(t)]^2\;dt}$ OpenStudy (jack1):

ahh, with u again, cheers OpenStudy (anonymous):

For a periodic function of period $$p$$, you can just let $$a=0, b=p$$ OpenStudy (anonymous):

For non periodic functions, it would seem you'd do something like $$a=0\, b\to \infty$$ OpenStudy (jack1):

thanks for that man, makes sense now cheers @wio

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