Contents
Problem
Calculate the energy $ E_\infty $ and the average power $ P_\infty $ for the CT signal $ x(t)=2t^2 $
Solution 1
$ E_{\infty} $
$ E_{\infty}=\int_{-\infty}^\infty |x(t)|^2\,dt $
$ E_{\infty}=\int_{-\infty}^\infty |2t^2|^2\,dt $
$ E_{\infty}=2\int_{-\infty}^\infty t^4\,dt $
$ E_{\infty}=\frac{2}{5}t^5|_{-\infty}^\infty $
$ E_{\infty}=\infty $
$ P_{\infty} $
$ P_{\infty}=\lim_{T \to \infty} \ \frac{1}{2T}\int_{-T}^{T}|x(t)|^2\,dt $
$ P_{\infty}=\lim_{T \to \infty} \ \frac{1}{2T}\int_{-T}^{T}|2t^2|^2\,dt $
$ P_{\infty}=\lim_{T \to \infty} \ \frac{1}{T}\int_{-T}^{T}t^4\,dt $
$ P_{\infty}=\lim_{T \to \infty} \ \frac{1}{5T}t^5|_{-T}^{T} $
$ P_{\infty}=\lim_{T \to \infty} \ \frac{1}{5T}[T^5 - (-T^5)] $
$ P_{\infty}=\lim_{T \to \infty} \ \frac{2T^5}{5T} $
$ P_{\infty}=\lim_{T \to \infty} \ \frac{2T^4}{5} $
$ P_{\infty}=\infty $
Both of your final answer are correct, but there are small mistakes (constant multipliers) in your computation.
Solution 2
$ E_{\infty}=\int_{-\infty}^\infty |x(t)|^2\,dt =\int_{-\infty}^\infty |2t^2|^2\,dt= 4\int_{-\infty}^\infty t^4\,dt =\infty $
$ P_{\infty}=\lim_{T \to \infty} \ \frac{1}{2T}\int_{-T}^{T}|x(t)|^2\,dt= \lim_{T \to \infty} \ \frac{1}{2T}\int_{-T}^{T}|2t^2|^2\,dt =\lim_{T \to \infty} \ \frac{2}{T}\int_{-T}^{T}t^4\,dt =\lim_{T \to \infty} \ \frac{2}{5T}t^5|_{-T}^{T}=\lim_{T \to \infty} \ \frac{2}{5T}[T^5 - (-T^5)] =\lim_{T \to \infty} \ \frac{4T^4}{5}=\infty $
Looks pretty good!
Solution 3
$ E_{\infty}=\int_{-\infty}^\infty |x(t)|^2\,dt =\int_{-\infty}^\infty |2t^2|^2\,dt= 4\int_{-\infty}^\infty t^4\,dt = 4 \frac{t^5}{5}=\infty. $
$ P_{\infty}=\lim_{T \to \infty} \ \frac{1}{2T}\int_{-T}^{T}|x(t)|^2\,dt= \lim_{T \to \infty} \ \frac{1}{2T}\int_{-T}^{T}|2t^2|^2\,dt =\lim_{T \to \infty} \ \frac{2}{T}\int_{-T}^{T}t^4\,dt =\frac{\infty}{\infty}=1. $
The energy computation looks good. But in the power computation you distributed the limit too early.
