I am trying to compute the root mean square current with the main definition $$ \sqrt{\frac{1}{T}\int_0^T i^2(t)dt}$$ where the fourier expansion is given as: $$i(t)=I_0+\sum_{n=1}^{\infty}\sqrt{2}I_n\sin(n\omega t +\gamma_n)$$ so focusing only on squaring these $$I_0^2+2\sqrt{2}I_0 \sum_{n=1}^{\infty} I_n\sin(n\omega t +\gamma_n)+(\sum_{n=1}^{\infty}\sqrt{2}I_n\sin(n\omega t +\gamma_n))^2$$ Now when Integrating the first term would produce $$I_0^2$$ the second will vanish and well for the third I guess I need to extract only those that are square and the rest would be 0, but I dont really know how. I have thought of Cauchy product: $$\left(\sum_{n=0}^\infty a_n \right)\left(\sum_{n=0}^\infty b_n \right) = \sum_{n=0}^\infty \sum_{k=0}^n a_nb_{n-k}$$ But I am unsure since the sum starts from 1. Could you perhaps help me finish this?

  • \$\begingroup\$ Can’t you solve it numerically? \$\endgroup\$ – PDuarte Jun 6 '18 at 13:21
  • \$\begingroup\$ My teacher gave the answer whitout a proof to be $$\sqrt{I_0^2+\sum_{n=1}^{\infty} I_n^2}$$ and I wish to show that.. \$\endgroup\$ – Zacky Jun 6 '18 at 13:32
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    \$\begingroup\$ I find the use of \$i(t)\$ as a voltage quantity slightly misleading. Of course it doesn't matter which letter is used but there is a strong bias of \$i\$ being a current... \$\endgroup\$ – Arsenal Jun 6 '18 at 14:02
  • \$\begingroup\$ oh boy. I wrote voltage instead of current. I am sorry. \$\endgroup\$ – Zacky Jun 6 '18 at 14:09

Expanding the sum will result in

$$\left( \sum_{n=1}^{\infty} \sqrt{2}I_n\sin(n\omega t + \gamma_n)\right)^2 \\ \begin{align} &= \sum_{n_1=1}^{\infty}\sum_{n_2=1}^{\infty}\left[2I_{n_1}I_{n_2}\sin(n_1\omega t + \gamma_{n_1})\sin(n_2\omega t + \gamma_{n_2})\right] \end{align}$$

If \$n_1 = n_2 = n\$, then we simply get

$$\int_0^T 2I_n^2\sin^2(n\omega t + \gamma_n) dt = I_n^2$$

For the mixed terms where \$n_1 \neq n_2\$, we can calculate the integral using the following equality:

$$\sin(\alpha)\sin(\beta) = \frac{1}{2}\left( cos(\alpha-\beta) - cos(\alpha+\beta) \right)$$

$$\begin{align} & \int_0^T \sin(n_1\omega t + \gamma_{n_1})\sin(n_2\omega t + \gamma_{n_2})dt \\ &= \int_0^T \frac{1}{2} \left( \cos \left[ (n_1 - n_2)\omega t + (\gamma_{n_1} - \gamma_{n_2}) \right] - \cos \left[ (n_1 + n_2)\omega t + (\gamma_{n_1} + \gamma_{n_2}) \right] \right)dt \\ &= \frac{1}{2} \int_0^T \cos( (n_1-n_2)\omega t + (\gamma_{n_1} - \gamma_{n_2})) dt \\ &- \frac{1}{2} \int_0^T \cos( (n_1 + n_2)\omega t + (\gamma_{n_1} + \gamma_{n_2})) dt \end{align} $$

Both of these integrals are of the form:

$$\int_0^T cos(m\omega t + \phi)dt$$

with \$m\$ a non-zero integer (else \$n_1 = n_2\$), meaning that \$T\$ is a multiple of the period of this cosine. The latter means that the integral will result to 0.

So your formula can be simplified to

$$I_0^2 + \sum_{n=1}^{\infty} I_n^2$$

  • \$\begingroup\$ Could you please clarify more about $n_1=n_2$ case? \$\endgroup\$ – Zacky Jun 6 '18 at 14:10
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    \$\begingroup\$ I hope my edit makes things more clear. \$\endgroup\$ – Sven B Jun 6 '18 at 14:28

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