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

RSM?

RMS value i think

ugh, my bad, root mean squared

RMS using the continuous definition?

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

this whole q is screwy, sorry, peak = rms x sqrt 2 my bad ... again

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

So we'll start with something as simple as \[ \int_0^x\sin^2(t)\;dt \]

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

This doesn't look good because the limit won't converge at infinity.

So instead just do something like the period of the sine wave, I suppose. It is perfectly representative.

The period of \(\sin^2(t)\) is \(\pi\).

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

@Jack1 Does that make sense so far?

yeah, im with you so far

\[ \int \sin^2(t)\;dt=\frac 12\int 1-\cos (2t)\;dt=\frac 12 t-\frac 14\sin(2t)+C \]

So \[ \int _0^\pi \sin^2 (t)\;dt = \frac 12 t-\frac 14 \sin(2t)\Bigg|_0^\pi=\frac \pi2 \]

Putting it back in: \[ \sqrt{\frac 1\pi \times \frac \pi2} = \sqrt\frac 12 \]

Peak x, is really just the amplitude.

In this case amplitude was 1.

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

Because it is in the definition.

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

ahh, with u again, cheers

For a periodic function of period \(p\), you can just let \(a=0, b=p\)

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

thanks for that man, makes sense now cheers @wio

Join our real-time social learning platform and learn together with your friends!