- #1
jernobyl
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Homework Statement
Determine the RMS value over a period 2π/w (I can only assume this means between 0 and 2π/w) of the equation:
i = M.sin(wt + A) + N.sin(2wt + B)
Homework Equations
RMS is, I believe, the square root of 1 / b -a times the integral between b and a of the integrand squared.
Also I should mention because we're going to be dealing with sin² and cos² that in my course notes it's mentioned that to avoid dealing with squared trigonometry that sin²x should be first of all changed to 1/2(1-cos(2x)) and cos²x changed to 1/2(1+cos(2x))
The Attempt at a Solution
Well. I've been working for about three hours trying to do this darn thing.
Well, first of all I've got to remember that because of the RMS formula, I have to remember to multiply my end result by w/2π at the end, and then take the square root of my end result.
So first I tried expanding M.sin(wt + A) + N.sin(2wt + B) to (sinwtcosAcoswtsinA) and (sinwtcosBcoswtsinB)
Then of course I've got to square it before I can get to integrating, so I was ending up with something stupidly long like:
M²sin²wt.cos²A + Ncos²wt.sin²A + 2Nsinwt.cosA.coswt.sinA + N²sin²2wt.cos²B + N²cos²2wt.sin²B + 2Nsin2wtcosB
But then I'm thinking...right, I'm fairly sure I'm doing the wrong that and it should be quite as complex as that (it's about 6 times more difficult than the rest of the questions), so I must be making it more complex than it actually is.
So I try a different approach. Try making each bit into the form 1/2(1-cos(2x)) like my notes say.
So:
M.sin(wt + A) becomes M²/2.(1-cos(2wt + 2A))
and
N/2.sin(2wt + B) becomes N²/2.(1-cos4wt + 2B)
and we're left with the leftover terms
2MNsin(wt + A)sin(2wt + B)
But when I worked through that it just got ridiculously complex and again I was doubtful that I was going in the right direction at all.
I'm sure there's some way that this is all fabulously easy and I'm just not seeing it. Could someone please point me in the right direction?! Many thanks!
PS - for interest's sake, the answer is meant to come out as:
the square root of (M²+N²)/2
Apparently!