Problem
Calculate the energy $ E_\infty $ and the average power $ P_\infty $ for the CT signal
$ x(t)=tu(t) $
Solution 1
$ E_\infty = \int_{-\infty}^\infty |tu(t)|^2\,dt = \int_{0}^\infty t^2\,dt=\infty $
$ P_\infty = lim_{T \to \infty} \ \frac{1}{2T} \int_{-T}^T |tu(t)|^2\,dt = lim_{T \to \infty} \ \frac{1}{2T} \int_{0}^T t^2\,dt =\frac{\infty}{\infty}=1 $
Your energy is correct, but you distributed the limit too early when you computed the average power, so your answer came out wrong.
Solution 2
$ E_\infty = \int_{-\infty}^\infty |tu(t)|^2\,dt) $
$ E_\infty = \int_{0}^\infty t^2\,dt) $
$ E_\infty =\frac{t^3}{3}\bigg]_0^\infty) $
$ E_\infty =\infty-0 = \infty $
Calculating $ P_\infty $
$ P_\infty = lim_{T \to \infty} \ \frac{1}{2T} \int_{-T}^T |tu(t)|^2\,dt $
$ P_\infty = lim_{T \to \infty} \ \frac{1}{2T} \int_{0}^T t^2\,dt $
$ P_\infty = lim_{T \to \infty} \ \frac{1}{2T} \frac{t^3}{3}\bigg]_0^T $
$ P_\infty = lim_{T \to \infty} \ \frac{1}{2T} \frac{T^3}{3} $
$ P_\infty = lim_{T \to \infty} \ \frac{T^2}{6} $
$ P_\infty = \infty $
Looks pretty good!
