The J is probably a typo, unit of power is watt and not joule. To calculate signal power: \$P=\frac{1}{T}\int\limits_0^T{u(t)i(t)\,dt}=\frac{1}{T}\int\limits_0^T{\frac{u^2(t)}{R}dt}=\frac{4}{TR}\int\limits_0^{T/4}{(10t)^2dt}=\frac{4\cdot 100}{4\cdot 100}\int\limits_0^{1}{{t^2}\,dt}=\left[\frac{t^3}{3}\right]_0^1=\frac{1}{3}-0=\frac{1}{3}\,\mathrm{W}\$
Best Answer
Well, we have the following circuit:
simulate this circuit – Schematic created using CircuitLab
The input voltage is given in the following diagram:
Assuming an ideal diode, the negative parts are cut of for the voltage across the resistor. That is shown in the following diagram:
Now, we know that the power in a resistor is given by:
$$\text{P}_\text{R}\left(t\right)=\frac{\text{V}_\text{R}^2\left(t\right)}{\text{R}}\tag1$$
It is not hard to show that \$\text{V}_\text{R}\left(t\right)\$, is given by:
$$\text{V}_\text{R}\left(t\right)=\frac{\text{V}}{\text{T}}\sum_{\text{n}\ge0}\theta\left(t-\text{Tn}\right)\mathcal{I}_\text{n}\left(t,\text{T}\right)\tag2$$
Where:
$$\mathcal{I}_\text{n}\left(t,\text{T}\right)=\left(\text{T}\left(1+2\text{n}\right)-2t\right)\left(\theta\left(t-\text{T}\left(\frac{1}{2}+\text{n}\right)\right)+\theta\left(t-\text{T}\left(1+\text{n}\right)\right)\right)\tag3$$
Now, we can look at the average power and the RMS power using the following two formulas:
I used Mathematica to find them:
Average power:
RMS-power: