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I found this explanation in a journal paper but I could not understand it. Can someone give me an explanation or possibly a proof that:

If $$\frac{\mathrm{d}V(t)}{\mathrm{d}t}=\sqrt{2}\sum_{h=1}^{H}h\omega V_{h}\cos\left(h\omega t+\frac{\pi }{2}\right),$$ then why integration over whole period is: $$\frac{1}{T}\int_{0}^{T} \left( \frac{\mathrm{d} V(t)}{\mathrm{d} t} \right)^{2}dt=\omega \sum_{h=1}^{H}h^{2}V_{h}^{2}.$$

I have problem with the power of $\omega$; my solution returns $\omega^2$, while the power of $\omega$ in answer is one. Here is my solution: $$\frac{1}{T}\int_{0}^{T}\ \left( \frac{dV}{dt} \right)^{2}dt=\frac{2\omega ^{2}}{T}\int_{0}^{T}\sum_{h=1}^{H}h^{2}V_{h}^{2}\sin^{2}(h\omega t)dt$$ and over whole period: $$\frac{1}{T}\int_{0}^{T}\sin^{2}(h\omega t)dt=\frac{1}{2}$$ then we will have $$\omega ^{2}\sum h^{2}V_{h}^{2} $$ not
$$\omega \sum h^{2}V_{h}^{2}$$

Why?

  • 1
    +1 for thinking about what you read and showing your work2011-02-10
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    Is the fact that $\frac{1}{T}\int_0^T\sin^2(h\omega t)\,dt = \frac{1}{2}$ a given, or something you computed? If the latter, how?2011-02-10
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    I computed it but I'm not sure. by converting $sin^{2}(ax)$ to $\frac{1}{2}(1-cos(2ax))$ and integration over whole period.2011-02-10
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    @user6856, why does $\sin(4h\omega T)=0$?2011-02-10
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    @user6856: Then I'm not sure that's correct.$$\int_0^T(\frac{1}{2}-\frac{1}{2}\cos(2hwt))dt = \frac{T}{2} - \frac{1}{4hw}\sin(2hwT).$$How did you get rid of $\frac{1}{4hw}\sin(2hwT)$? That may be the source of that $w$ in the denominator.2011-02-10
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    @Arturo Magidin, $\omega T=2\pi$2011-02-10
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    @Americo: Thanks; did I miss that in the question, or is this something standard I should have known?2011-02-10
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    @user6856, I have checked the cases $H=1$ and $H=2$ and I got the factor $\omega^2$, as you claim. What I've not yet understood is your first formula after "Here is my solution".2011-02-10
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    @Arturo Magidin, e.g. in electrical engineering $\omega=\frac{2\pi}{T}$, where $T$ is the period and $\omega$ the angular frequency (in radians/s) of a sinus wave.2011-02-10
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    @Arturo: see http://en.wikipedia.org/wiki/Angular_frequency2011-02-10
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    @Americo: Again, thank you. Context is everything; my first thought when I see $\omega$ is "the first infinite ordinal", my second is "it's just a variable, then". Don't usually think of angular frequency. (-:2011-02-10
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    @user6856, Your formula is correct because if $m\neq n$, then $\int_{0}^{2\pi }\sin nx\sin mxdx=0$2011-02-10
  • 0
    @Arturo: Not at all!2011-02-10

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