## 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!

## Alumni Liaison

Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010