## 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 and so your final answer is wrong.