## Problem

Calculate the energy $E_\infty$ and the average power $P_\infty$ for the CT signal $x(t)=\sqrt{2t}u(t)$

## Solution 1

Compute $E\infty$

$E\infty=\int_{-\infty}^\infty |x(t)|^2dt$

$E\infty=\int_{-\infty}^\infty |\sqrt{2t}u(t)|^2dt$ (*)

$E\infty=\int_{-\infty}^\infty |2tu(t)|dt$ (*)

$E\infty=\int_{0}^\infty 2tdt$ (*)

$E\infty=t^2|_{0}^{\infty}$

$E\infty= \infty-0$

$E\infty=\infty$

Compute $P\infty$

$P\infty=\lim_{T \to \infty}\frac{1}{2*T}\int_{-T}^T |x(t)|^2dt$

$P\infty=\lim_{T \to \infty}\frac{1}{2*T}\int_{-T}^T |\sqrt(2t)u(t)|^2dt$

$P\infty=\lim_{T \to \infty}\frac{1}{2*T}\int_{-T}^T|2tu(t)|dt$

$P\infty=\lim_{T \to \infty}\frac{1}{2*T}\int_{0}^T2tdt$

$P\infty=\lim_{T \to \infty}\frac{|t^2|_{0}^{T}}{2*T}$ (*)

$P\infty=\lim_{T \to \infty}\frac{T^2-0}{2*T}$

$P\infty=\lim_{T \to \infty}\frac{T}{2}$

$P\infty=\infty$

• It would be better not to use the start symbol to denote multiplication in this context. The star symbol in electrical engineering is usually denoting convolution.
• *This is probably not the best way to order these steps, but it worked out at the end.
• *You should not have the absolute values at this point.

## Solution 2

$E_\infty=\int_{-\infty}^\infty |x(t)|^2d=\int_{-\infty}^\infty |\sqrt{2t}u(t)|^2dt =\int_{0}^\infty | \sqrt{2t}|^2dt =\int_{0}^\infty 2tdt= \infty$

$P\infty=\lim_{T \to \infty}\frac{1}{2T}\int_{-T}^T |x(t)|^2dt =\lim_{T \to \infty}\frac{1}{2T}\int_{-T}^T \left| \sqrt{2t} u(t) \right|^2dt =\lim_{T \to \infty}\frac{1}{2T}\int_{0}^T 2t dt= \lim_{T \to \infty}\frac{ t^2|_{0}^{T}}{2 T}=\lim_{T \to \infty}\frac{T^2-0}{2T}= \infty$

Short and sweet!

## Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett