Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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
$$\omega \sum h^{2}V_{h}^{2}$$


share|cite|improve this question
+1 for thinking about what you read and showing your work – Ross Millikan Feb 10 '11 at 14:12
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? – Arturo Magidin Feb 10 '11 at 14:28
@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. – Américo Tavares Feb 10 '11 at 16:39
@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. (-: – Arturo Magidin Feb 10 '11 at 16:51
@user6856, Your formula is correct because if $m\neq n$, then $\int_{0}^{2\pi }\sin nx\sin mxdx=0$ – Américo Tavares Feb 10 '11 at 18:38

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

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


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

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

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


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

share|cite|improve this answer
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? – user6856 Feb 11 '11 at 5:50
@ user6856: Yes. Please see the email-address shown on my profile page. – Américo Tavares Feb 11 '11 at 9:56
@ 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). – Américo Tavares Feb 11 '11 at 11:04
@Américo: Dear Américo, only moderators can see the email address entered in your profile. – Akhil Mathew Feb 11 '11 at 14:33
@Akhil: Dear Akhil, thanks for the information. – Américo Tavares Feb 11 '11 at 18:44

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: $\sum (h\omega V_h)^2$.

share|cite|improve this answer
Somewhat related stuff: and – Hans Lundmark Feb 10 '11 at 20:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.