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Horrendous RMS problem.

  1. Apr 10, 2007 #1
    1. The problem statement, all variables and given/known data

    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)

    2. Relevant 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))

    3. 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.


    M.sin(wt + A) becomes M²/2.(1-cos(2wt + 2A))


    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?! :cry: Many thanks!

    PS - for interest's sake, the answer is meant to come out as:

    the square root of (M²+N²)/2

  2. jcsd
  3. Apr 10, 2007 #2

    D H

    Staff: Mentor

    What is wrong with squared trigonometry? The following identities might help:

    [tex]\int\sin^2 x \, dx = \frac 1 2(x-\cos x \sin x)[/tex]

    [tex]\int \sin(x) \sin(2x) \, dx = \frac 2 3 \sin^3 x[/tex]
    Last edited: Apr 10, 2007
  4. Apr 10, 2007 #3
    Thanks for this. I understand that absolutely, but as I'm sure you can appreciate, this was coursework from waaaay back this year, and as such we were only meant to use the methods we had been taught. And at that point we had only learned that to integrate squared trigonometry we were first supposed to rearrange it in the form 1/2(1±cos(2x)), so I'm trying to get round it that way. I know I'm being awkward, but that's the way I'm supposed to do it at this point...
  5. Apr 10, 2007 #4
    I hope you understand I don't mean to turn down help, it's just I believe we were meant to do it in a certain way...
  6. Apr 10, 2007 #5

    D H

    Staff: Mentor

    First, expand [itex](M\sin(\omega t + A) + N\sin(2\omega t + B))^2[/itex]:

    [tex](M\sin(\omega t + A) + N\sin(2\omega t + B))^2 =
    M^2\sin^2(\omega t + A) +
    2MN\sin(\omega t + A)\sin(2\omega t + B) +
    N^2\sin^2(2\omega t + B)[/tex]

    Integrate the three terms separately. A change of variables will help to get rid of those nasty [itex]+A[/itex] and [itex]+B[/itex] terms in the sin2 integrals.

    I assume "a period of 2n/w" means 2n cycles or [itex]4n\pi/\omega[/itex]. (if not, the posted answer is wrong). Assuming this, you should find that the sin2 integrals come out nice and easy. The trick is to show the cross term integrates to zero.
    Last edited: Apr 10, 2007
  7. Apr 11, 2007 #6
    Okay...what did you mean by "a change of variables"? I didn't quite follow. Did you mean that you should change the sin² to 1/2(1-cos)?

    As for "a period of 2n/w", it was 2pi/w in the problem...does that mean that it means 2pi cycles, or 4pi/w?

    And, um, sorry, but what did you mean the cross term? IE the 2MNsin(wt+A)sin(2wt+B)?

    Many thanks!
  8. Apr 11, 2007 #7
    Hello...? :S
  9. Apr 11, 2007 #8

    D H

    Staff: Mentor

    There is an easy way and a hard way to attack an integral like

    [tex]\int \sin(\omega t + \theta) dt[/tex]

    The hard way is to expand [itex]\sin(\omega t + \theta)[/itex] and then integrate. The easy way is to change variables: [itex]x=\omega t + \theta[/itex]

    [tex]\int \cos(\omega t + \theta) dt = \frac 1{\omega} \int \cos x dx = \frac {\sin x}{\omega} = \frac {sin(\omega t + \theta)}{\omega}[/tex]
  10. Apr 11, 2007 #9


    User Avatar
    Science Advisor
    Homework Helper

    I don't know if this is of any use to you, but there is an 'easy' way to do this, if you know some stuff already. f1=sin(wt+A) and f2=sin(nwt+B) for integer n>1 are orthogonal functions over the interval [0,2pi/w]. And the integral of each of their squares is 1/2. In vector space language, f1.f2=0 and f1.f1=f2.f2=1/2. So your given problem becomes sqrt((M*v1+N*v2).(M*v1+N*v2))=sqrt(M^2/2+N^2/2). Like I say, if you don't know any of this stuff it won't help you - but at least it explains why the answer is so simple.
  11. Apr 11, 2007 #10
    Sorry if I'm being dumb, but...if we're completely substituting [itex]\omega t + \theta[/itex] as x, then how come it goes to 1/w times the integral of cosx dx? Why isn't it just the integral of cosx dx? Why is there a w involved in the substituted version?
    Last edited: Apr 11, 2007
  12. Apr 11, 2007 #11

    D H

    Staff: Mentor

    The substitution is [itex]x=\omega t + \theta[/itex]. Thus [itex]dx = \omega\,dt[/itex] or [itex]dt = 1/\omega\,dx[/itex].
  13. Apr 11, 2007 #12
    So despite the fact that for ease, we're reducing [itex]\omega t + \theta[/itex] to just [itex]x[/itex], we're still remembering that there's an [itex]\omega[/itex] in there somewhere, therefore we have to add a factor of
    [tex]\frac 1{\omega}[/tex]...?
  14. Apr 12, 2007 #13
    Also, am I able to do the same thing for the sin(2wt + B) variables, just reduce them to, say, z or something?
  15. Apr 12, 2007 #14
    Okay, to be honest guys, this is starting to drive me mental, I've been working on it for an hour and I've gotten absolutely no-where.

    I've changed the variables and this is where I am. I don't even know if this is right.

    Having substituted (wt + A) for a, and (2wt + B) for b:

    [tex]\frac{\omega}{4\pi} \int_{0}^{\frac{4\i}{\omega}} M^2\sin^2 a + 2MN(\sin a)(\sin b) + M^2 \sin^2 b dt[/tex]

    [tex]\frac{\omega}{4\pi} \int_{0}^{\frac{4\i}{\omega}} \frac{M^2}{2}(1-\cos 2a) + 2MN\sina \sinb + \frac{N^2}{2}(1-\cos 2b)[/tex]

    [tex]\frac{\omega}{4\pi} \int_{0}^{\frac{4\i}{\omega}} \frac{M^2}{2} - \frac{M^2}{2}\cos2a + 2MN\sina\sinb + \frac{N^2}{2} - \frac{N^2}{2}\cos2b[/tex]

    Is that even right? If so, I know I have to integrate that but to be honest I haven't got a clue where to go from here. This really is driving me round the bend.
    Last edited: Apr 12, 2007
  16. Apr 12, 2007 #15

    D H

    Staff: Mentor

    What are you integrating against (I don't see a dwhatever)? Omitting this is sloppy, will lead to errors, may even turn you into a physicist.

    You have three terms to integrate with respect to time: [itex]M^2 \sin^2(\omega t + A)[/itex], [itex]2MN \sin (\omega t + A)\sin (2\omega t + B)[/itex], and [itex]N^2 \sin (2\omega t + B)[/itex].

    Some hints:
    • Integrate the three terms separately, then combine the results at the end. Doing so will make the problem a lot less daunting.
    • Use integration by parts to solve the cross term ([itex]2MN \sin (\omega t + A)\sin (2\omega t + B)[/itex]) integral. A trick: integrate by parts twice.
  17. Apr 12, 2007 #16
    There should be a dt at the end.

    Okay, split it up into the threee terms and integrate. Fine.

    What IS integration by parts?
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