11
$\begingroup$

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
  • 0
    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
  • 0
    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
  • 0
    @user6856, why does $\sin(4h\omega T)=0$?2011-02-10
  • 0
    @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
  • 0
    @Arturo Magidin, $\omega T=2\pi$2011-02-10
  • 0
    @Americo: Thanks; did I miss that in the question, or is this something standard I should have known?2011-02-10
  • 0
    @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
  • 2
    @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
  • 0
    @Arturo: see http://en.wikipedia.org/wiki/Angular_frequency2011-02-10
  • 1
    @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
  • 1
    @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

2 Answers 2

7

Your solution is right. It should be a typo in the paper. Here is my evaluation confirming yours. Since

$$\begin{eqnarray*} \frac{dV(t)}{dt} &=&\sqrt{2}\sum_{h=1}^{H}h\omega V_{h}\cos \left( h\omega t+% \frac{\pi }{2}\right) \\ &=&-\sqrt{2}\sum_{h=1}^{H}h\omega V_{h}\sin h\omega t, \end{eqnarray*}$$

and assuming $\omega$ is the angular frequency given by

$$\omega =\frac{2\pi }{T},$$

we have

$$\left( \frac{dV(t)}{dt}\right) ^{2}=2\omega ^{2}\left( \sum_{h=1}^{H}hV_{h}\sin \left( h\omega t\right) \right) ^{2}$$

and

$$\begin{eqnarray*} \frac{1}{T}\int_{0}^{T}\left( \frac{dV(t)}{dt}\right) ^{2}dt &=&\frac{\omega }{2\pi }\int_{0}^{2\pi /\omega }\left( \frac{dV(t)}{dt}\right) ^{2}dt \\ &=&\frac{\omega }{2\pi }\int_{0}^{2\pi /\omega }2\omega ^{2}\left( \sum_{h=1}^{H}hV_{h}\sin \left( h\omega t\right) \right) ^{2}dt \\ &=&\frac{\omega ^{3}}{\pi }\int_{0}^{2\pi /\omega }\left( \sum_{h=1}^{H}hV_{h}\sin \left( h\omega t\right) \right) ^{2}dt. \end{eqnarray*}$$

The integrand $\left( \sum_{h=1}^{H}hV_{h}\sin \left( h\omega t\right) \right) ^{2}$ is a sum of terms of two different types:

i) $h^{2}V_{h}^{2}\sin ^{2}\left( h\omega t\right) $ and

ii) $k\left( pV_{p}\sin \left( p\omega t\right) \cdot qV_{q}\sin \left( q\omega t\right) \right) \,$, with $p\neq q$ and $p,q,k\in\mathbb{N}$.

The second type terms do not contribute to the last integral, because the $\sin nx$ ($n\in\mathbb{N}$) functions form an orthogonal system over $[0,2\pi ]$:

$$\int_{0}^{2\pi /\omega }k\left( pV_{p}\sin \left( p\omega t\right) \cdot qV_{q}\sin \left( q\omega t\right) \right) dt=0\quad p\neq q$$

The sum of the first type ones is $\sum_{h=1}^{H}h^{2}V_{h}^{2}\sin ^{2}\left( h\omega t\right) $. Thus

$$\begin{eqnarray*} \frac{1}{T}\int_{0}^{T}\left( \frac{dV(t)}{dt}\right) ^{2}dt &=&\frac{\omega ^{3}}{\pi }\int_{0}^{2\pi /\omega }\sum_{h=1}^{H}h^{2}V_{h}^{2}\sin ^{2}\left( h\omega t\right) dt \\ &=&\frac{\omega ^{3}}{\pi }\sum_{h=1}^{H}h^{2}V_{h}^{2}\int_{0}^{2\pi /\omega }\sin ^{2}\left( h\omega t\right) dt \\ &=&\frac{\omega ^{3}}{\pi }\sum_{h=1}^{H}h^{2}V_{h}^{2}\cdot \frac{\pi }{% \omega } \\ &=&\omega ^{2}\sum_{h=1}^{H}h^{2}V_{h}^{2}, \end{eqnarray*}$$

because

$$\begin{eqnarray*} \int \sin ^{2}\left( h\omega t\right) dt &=&\frac{1}{h\omega }\left( -\frac{1% }{2}\cos h\omega t\sin h\omega t+\frac{1}{2}h\omega t\right) \\ \int_{0}^{2\pi /\omega }\sin ^{2}\left( h\omega t\right) dt &=&\frac{\pi }{% \omega }. \end{eqnarray*}$$

  • 0
    Thank you for confirmation, but I don't think that it is a typo, because he substituted this result into another equation which is correct. is it possible that I send the paper for you?2011-02-11
  • 0
    @ user6856: Yes. Please see the email-address shown on my profile page.2011-02-11
  • 0
    @ user6856: Is $h$ dimensionless? If it is, the formula makes sense only in case it has $\omega ^2$ on the RHS (assuming $\omega$ is an angular frequency, $V$ a voltage and $t$ the time).2011-02-11
  • 0
    @Américo: Dear Américo, only moderators can see the email address entered in your profile.2011-02-11
  • 0
    @Akhil: Dear Akhil, thanks for the information.2011-02-11
  • 0
    @Américo: h is exactly dimensionless and ω is an angular frequency, V a voltage and t the time. I could not find your Email address. Mine is phantomlord70200@yahoo.com, Please send an email for me and I'll send the paper. or if you have access to IEEE explorer, the paper name is "a time domain load modeling technique and harmonic analysis", Eq. 12. the same thing append for Eq.20.2011-02-11
3

I agree, it should be $\omega^2$. The whole thing is in fact just the Pythagorean theorem: the functions $\sqrt{2} \cos(\dots)$ are orthonormal in the space $L^2([0,T])$, and the integral is the square of the $L^2$ norm of $dV/dt$, hence the sum of the squares of the coefficients: $\sum (h\omega V_h)^2$.

  • 0
    Somewhat related stuff: http://math.stackexchange.com/q/7391/1242 and http://math.stackexchange.com/q/19876/1242.2011-02-10