I need some help calculating this Fourier coefficient.
Periodic signal, six-steps. Odd periodic signal.
I've made the calculations myself, not using any software, and the results are these:
$$I_{\text{rms}}=\sqrt{2}\times I$$
and the fundamental is $b_1=1.909$.
Isn't that weird - the fundamental value being bigger than the $I_{\text{rms}}$?
If you can calculate and help me out, I'll be appreciative.
On
A stepped wave like this can be considered as a sum of several sawtooth waves: $\frac{I}{\pi}\Sigma_k \frac{1}{k}[2\sin(k\theta) + \sin(k(\theta-\frac{\pi}{3})) - \sin(k(\theta-\frac{2\pi}{3})) - 2\sin(k(\theta\pm\pi)) - \sin(k(\theta+\frac{2\pi}{3})) + \sin(k(\theta+\frac{\pi}{3}))] = \frac{I}{\pi}\Sigma_k \frac{\sin(k\theta)}{k}[1+\cos(\frac{k\pi}{3})-\cos(\frac{2k\pi}{3})-\cos(k\pi) ]$
(This may be off by a constant factor, I always get that muddled)
$b_k$ is zero whenever $k$ is a multiple of 2 or 3, otherwise it's $3I/k\pi$, if I haven't blundered.
We have $I_\text{rms}^2 = {1 \over \pi}{ \pi \over 3}(I^2+4 I^2 + I^2) = 2 I^2$.
As an aside, with a pure sine wave, we have $I_\text{rms} = {1 \over \sqrt{2}} I_\max$, and here we have $I_\max = 2I$, which would correspond to $I_\text{rms} = \sqrt{2} I$ and a fundamental of $2I$, so these results show that the fundamental above is consistent with this.
The $\sin$/$\cos$ Fourier series are convenient here, since $i$ is odd, we need only compute the $\sin$ part, which gives $\hat{I_k} = {1 \over \pi} \int_{-\pi}^\pi \sin(kt) i(t) dt = 2I {1 \over k \pi} (1-\cos(k\pi)+\cos(k{\pi \over 3})-\cos(k {2 \pi \over 3})) $. Then we have $i(t) = \sum_{k=1}^\infty \hat{I_k} \sin(kt)$.
We have $\hat{I_1} \approx 1.9099 I$.
To compute the THD of $i$, we can use Parseval's theorem which states $I^2_\text{rms} = {1 \over 2 \pi} \int_{-\pi}^\pi i(t)^2 dt = {1 \over 2} \sum_{k=1}^\infty \hat{I_k}^2$, hence $\text{THD} = {\sqrt{2 I^2_\text{rms} - \hat{I_1}^2} \over \hat{I_1}} \approx 0.31077$.
To compute the power factor, note that since the voltage is a pure sine wave, the average power is given by $V_\text{rms} {1 \over \sqrt{2}} \hat{I_1}$, hence $\text{pf.} = { V_\text{rms} {1 \over \sqrt{2}} \hat{I_1} \over V_\text{rms} \sqrt{2} I} \approx 0.95495$.