RL with Circular Wave Source

Great start to the year!

First, the triangular wave signal's equation is computed, which is:

$V_S(t)= \begin{cases} \sqrt{\frac{\pi^2}{4} - \left(t - \left(2 \pi n + \frac{\pi}{2}\right)\right)^2} & 2\pi n \leq t < (2\pi n +\pi)\\ -\sqrt{\frac{\pi^2}{4} - \left(t - \left(2 \pi n + \frac{3\pi}{2}\right)\right)^2} & (2\pi n + \pi ) \leq t < (2\pi n +2\pi)\\ \end{cases}$

$n =\biggr \lfloor \frac{t}{2 \pi}\biggr \rfloor$

Having obtained $V_S(t)$ the circuit equation is:

$L\dot{I} + IR = V_S(t)$ $I(0) = 0$

I have avoided doing this analytically. So numerical integration takes care of the rest. The following plot is hopefully illustrative:

Here, I have compared the current response due to the circular wave source with that of the triangular and sinusoidal sources. This plot strengthens my assertion about the fact that the magnitude of the current response is higher as the source supplies a higher magnitude of energy per unit current.