Question

What is RMS of the input voltage and peak amplitude of the third harmonics of the circuit current response?(see attachment)

Answer 1

For the first part, by definition of RMS,\[V_{RMS} = \sqrt{\frac{1}{T}\int_0^T v_i(t)^2 dt}\]where T is the period of the function. The period of this function is 2, so:,\[V_{RMS} = \sqrt{\frac{1}{2}\int_0^2 \frac{16}{\pi^2}\left(\sin(\pi t) + \frac{1}{3}\sin(3\pi t) + \frac{1}{5}\sin(5 \pi t)\right)^2 dt}.\]
When we expand the square inside the integral, we will find terms of the form\[\sin^2(\omega t)\]and terms of the form\[\sin(\omega t) \sin(\omega' t)\]where \[\omega\]and\[\omega'\]are different frequencies. The integral from 0 to 2 of the first class of terms is always 1 and the integral from 0 to 2 of the second class of terms is always 0, so we end up with\[V_{RMS} = \sqrt{\frac{1}{2} \frac{16}{\pi^2}\left(1 + \frac{1}{9} + \frac{1}{25}\right) dt} = \sqrt{\frac{8}{\pi^2}\frac{229}{225}} = \frac{2\sqrt{518}}{15\pi} \approx 0.97.\]
For the second part, we must find the current response caused by the third harmonic. The third harmonic is simply the term with a 3 in front of the frequency in the Fourier series, so it's\[v_3(t) = \frac{4}{\pi}\frac{1}{3}\sin(3\pi t).\]We can write this term in complex notation as\[v_3(t) = \Re\left(\frac{4}{\pi}\frac{1}{3}\exp({i(3\pi t - \pi/2))}\right)\]and we can work with\[\frac{4}{\pi}\frac{1}{3}\exp({i(3\pi t - \pi/2))}\]and take the real part at the end.
The impedance of the circuit is\[Z = R + i\omega L\]so the current response is the real part of \begin{eqnarray*}\frac{\frac{4}{\pi}\frac{1}{3}\exp({i(3\pi t - \pi/2))}}{R+i\omega L} &=& -\frac{4}{3\pi}i\frac{\cos(3\pi t) + i\sin(3\pi t)}{R+i\omega L}\\&=&-\frac{4}{3\pi}i \frac{(\cos(3\pi t) + i\sin(3\pi t))(R-i\omega L)}{R^2 + \omega^2L^2}\\&=&\frac{4}{3\pi}\frac{1}{R^2+\omega^2L^2}\left(R\sin(\omega t) - \omega L\cos(\omega t) - i(R\cos(\omega t)+\omega L\sin(\omega t))\right)\end{eqnarray*}so, finally\[i_3(t) = \frac{4}{3\pi\left(R^2 + \omega^2L^2\right)}(R\sin(\omega t)-\omega L\cos(\omega t))\]and the amplitude is\[\frac{4}{3\pi\left(R^2+\omega^2 L^2\right)}\sqrt{R^2+\omega^2 L^2}=\frac{4}{3\pi\sqrt{R^2 + \omega^2 L ^2}} = \frac{4}{15\pi}.\]