## Problem

Calculate the energy $E_\infty$ and the average power $P_\infty$ for the CT signal $f(t)=5 j \sin (t)$

## Solution 1

$E_\infty = \int_{-\infty}^\infty |5sin(t)|^2\,dt$

$E_\infty = \int_{-\infty}^\infty 25sin(t)^2 \,dt$

$E_\infty = \int_{-\infty}^\infty 25(.5 + .5cos(2t)),dt$

$E_\infty =\frac{25t}{2} + \frac{25sin(2t)}{4}\bigg]_{-\infty}^\infty)$

$E_\infty =\infty-0 = \infty$ (*)

$P_\infty calculation$

$P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}\int_{-T}^T|5sin(t)|^2dt$

$P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}\int_{-T}^T \frac{25}{2} + \frac{25cos(2t)}{2}dt$

$= lim_{T \to \infty} \frac{1}{2T}*(\frac{25t}{2} + \frac{25sin(t)}{4}|_{-T}^T)$ (*) (*)

$= lim_{T \to \infty} \frac{1}{2T}*(25T + \frac{25sin(T)}{4}-\frac{25sin(-T)}{4}$ (*)

$= lim_{T \to \infty} \frac{1}{2T}*(25T + \frac{25sin(T)}{2})$ (*)

$P\infty= \frac{25}{2} + 0$

• * I would not use the star symbol to denote multiplication here. It is usually reserved for convolution in electrical engineering.
• * You are missing a 2 inside the sine.
• *You should not put a zero here, as the limit when t goes to infinity of of sin(t) is not defined.

## Solution 2

$E_\infty = \int_{-\infty}^\infty |5sin(t)|^2\,dt = \int_{-\infty}^\infty 25sin(t)^2 dt = \infty,$ since the function integrated is always non-negative and does not decrease as t approaches $\pm \infty$.

$P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}\int_{-T}^T|5sin(t)|^2dt= lim_{T \to \infty} \ \frac{1}{(2T)}\int_{-T}^T25 \sin^2 (t) dt = lim_{T \to \infty} \ \frac{1}{(2T)}\int_{-T}^T \frac{25}{2} + \frac{25cos(2t)}{2}dt$

$= lim_{T \to \infty} \frac{1}{2T}\left.\left( \frac{25t}{2} + \frac{25sin(2t)}{4} \right)\right|_{-T}^T \,$

$= lim_{T \to \infty} \frac{1}{2T}\left( 25T + \frac{25sin(2T)}{4}-\frac{25sin(-2T)}{4}\right)$

$= lim_{T \to \infty} \frac{25}{2}+ \frac{25sin(2T)}{4T}$

$= \frac{25}{2} + 0$, since $\sin (2T)$ is bounded by (-1) and 1

$= \frac{25}{2}$

• Looks good!

## Alumni Liaison

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett