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?
Electronic – About RMS in a non-sinusoidal electrical circuit
fouriervoltage
Related Solutions
If you calculate some more of them, you will come to the conclusion that almost every odd coefficient \$a_n\$ is a real positive number, except for \$a_3, a_9, a_{15}...\$, \$a_{3+6k}\$, where \$k\$ is a non-negative integer. These are all negative. (And every even coefficient is zero.) For those, which are positive numbers, the phase is 0 degrees/radians; for those, which are negative, the phase is -180°/\$-\pi\$ radians, because negating a periodic signal is equal to shifting its phase by -180°. Be careful using arctan, as it has a value range of \$\pm 90°\$ but both 0° and -180° has the same tangent value, zero.
Here I plotted the values and the phase spectrum from \$n=0\$ to \$n=21\$.
Edit: The first image is obviously not the amplitude (as it can't be negative), just the values of \$a_n\$ from 0 to 21. My bad.
If it helps, you have a sine wave multiplied by a square wave added to a phase shifted sine wave multiplied by another phase shifted square wave.
Addition translates to addition in fourier space, multiplication maps to convolution.
So what you're going to get is the FT of a square wave shifted along the w axis and with an imaginary component, I think.
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Best Answer
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
$$I_0^2 + \sum_{n=1}^{\infty} I_n^2$$