Hw1.5 Zachary Curosh - Energy and Power of a Tangent ECE301Fall2008mboutin - Rhea

Introduction

This page calculates the energy and power of the $2\sin(t)\cos(t)$ signal.

$P = \int_{t_1}^{t_2} \! |f(t)|^2\ dt$

$P = \int_0^{2\pi} \! |2\sin(t)\cos(t)|^2\ dt$

$P = \int_0^{2\pi} \! |sin(2t)|^2\ dt$

$P = \int_0^{2\pi} \! sin^2(2t)\ dt$

$P = \int_0^{2\pi} \! {(1-\cos(4t))\over 2} dt$

$P = {1\over 2}\int_0^{2\pi} \! {(1-\cos(4t))} dt$

$P = {1\over 2}t - {1\over 8}\sin(4t) \mid_0^{2\pi}$

$P = {1\over 2}(2\pi) - {1\over 8}\sin(4*2\pi) - [{1\over 2}(0) - {1\over 8}\sin(4*0)]$

$P = \pi - {1\over8}\sin(8\pi)$

$P = \pi$

$E = {1\over(t2-t1)}\int_{t_1}^{t_2} \! |f(t)|^2 dt$

$E = {1\over(2\pi-0)}\int_{0}^{2\pi} \! |2\sin(t)\cos(t)|^2 dt$

$E = {1\over2\pi}\int_0^{2\pi} \! |sin(2t)|^2\ dt$

$E = {1\over2\pi}\int_0^{2\pi} \! sin^2(2t)\ dt$

$E = {1\over2\pi}\int_0^{2\pi} \! {(1-\cos(4t))\over 2} dt$

$E = {1\over2\pi}*{1\over 2}\int_0^{2\pi} \! {(1-\cos(4t))} dt$

$E = {1\over 4\pi} ( t - {1\over 4}\sin(4t) )\mid_0^{2\pi}$

$E = {1\over 4\pi}t - {1\over 32\pi}\sin(4t) \mid_0^{2\pi}$

$E = {1\over 4\pi}(2\pi) - {1\over 32\pi}\sin(4*2\pi) - [{1\over 4\pi}(0) - {1\over 32\pi}\sin(4*0)]$

$E = {1\over 2} - {1\over32\pi}\sin(8\pi)$

$E = {1\over 2}$