integration of a function

Question

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?

Answer 1

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

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

Answer 2

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