keywords: energy, phasor, exponential

# HW1, ECE301, Fall 2008, Prof. Boutin

## Problem

Given complex signal $f(t) = \cos(t) + j \sin(t)$, find $E_\infty$ and $P_\infty$.

## Background Knowledge

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

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

## Solution 1

• $|x(t)| = |\cos(t) + j \sin(t)| = \sqrt{(\cos(t))^2+(\sin(t))^2} = \sqrt{1} = 1$
• $E_\infty(x(t)) = \int_{-\infty}^\infty |x(t)|^2\,dt = \int_{-\infty}^\infty 1\,dt = t|_{-\infty}^{\infty} = \infty$
• $P_\infty(x(t)) = \lim_{T \to \infty} (\frac{1}{2T} \int_{-T}^T |x(t)|^2\,dt) = \lim_{T \to \infty} (\frac{1}{2T} \int_{-T}^T 1\,dt) = \lim_{T \to \infty} (\frac{1}{2T} t|_{-T}^T)$

$= \lim_{T \to \infty} (\frac{1}{2T} (2T)) = \lim_{T \to \infty} (1) = 1$

## Solution 2

$|x(t)|=|\cos(t)+\jmath\sin(t)|=\sqrt{\cos^2(t)+\sin^2(t)}=1$

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

   $E\infty=\int_{-\infty}^\infty |1|^2\,dt=t|_{-\infty}^\infty$
$E\infty=\infty$


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

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


*I would be careful using the start symbol to denote multiplication in this context. The start symbol is typically used for denoting convolution in electrical engineering.

## Solution 3

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

$E\infty=\int_{-\infty}^\infty |\cos(t)+j\sin(t)|^2dt$

$E\infty=\int_{-\infty}^\infty |e^{j*t}|^2dt$

$E\infty=\int_{-\infty}^\infty |e^2*e^{j*t}|dt$

$E\infty=e^2/j*|e^{j*t}|_{-\infty}^{\infty}$

$E\infty=e^2/j*(\infty-0)$

$E\infty=\infty$

Compute $P\infty$

$x(t)=\cos(t)+j*\sin(t)$

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

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

$P\infty=\lim_{T \to \infty}\int_{-T}^T|e^{j*t}|^2dt$

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

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

$P\infty=\lim_{T \to \infty}\frac{\int_{-\infty}^\infty |e^2*e^{j*t}|dt}{2*T}$

$P\infty=\frac{d(\int_{-\infty}^\infty |e^2*e^{j*t}|dt)}{d(2*T)}$

$P\infty=\frac{e^{2*j*t}}{2}$

$P\infty=\infty$

This answer is problematic. You are computing a quantity which is always real (not complex) and non-negative. This is because you are integrating the norm of something, and a norm is always real and non-negative. Why do you these "j" coming out of your integrals?

## Solution 4

Magnitude

$|x(t)|=|\cos(t)+\jmath\sin(t)|=\sqrt{\cos^2(t)+\sin^2(t)}=\sqrt{2}$ Oh, oh...

$E\infty$

   $E\infty=\int_{-\infty}^\infty |\sqrt{2}|^2\,dt=2t|_{-\infty}^\infty$

   $E\infty=\infty$


$P\infty$

   $P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}\int|2|^2dt$

   $P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}*2|_{-T}^T$

   $P\infty=lim_{T \to \infty} \ \frac{1}{(2T)}*2(T-(-T))$

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

   $P\infty=2$


You need to brush up on the concept of magnitude of a complex number. The norm of a phasor is always one.

## Alumni Liaison

Have a piece of advice for Purdue students? Share it through Rhea!

Alumni Liaison