329fall08hw5sol

329fall08hw5sol - ECE-329 Fall 2008 Homework 5 — Solution 1 Let us consider the following four plane waves in free space E 1 = 2cos ωt βz)ˆ x

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Unformatted text preview: ECE-329 Fall 2008 Homework 5 — Solution September 30, 2008 1. Let us consider the following four plane waves in free space, E 1 = 2cos( ωt- βz )ˆ x V m E 2 = E o (cos( ωt- βz )ˆ x- sin( ωt- βz )ˆ y ) V m H 3 = cos( ωt + βz + π 3 )ˆ x + sin( ωt + βz- π 6 )ˆ y A m H 4 = cos( ωt- βx )ˆ z + sin( ωt- βx )ˆ y A m . a) The electric and magnetic fields ( E and H ) of uniform plane waves are perpendicular to each other and to the direction of propagation, therefore, it can be verified that such fields satisfy the following relation E = η H × ˆ β, where ˆ β is the unit vector parallel to the propagation direction and η is the intrinsic impedance. Using this relation, we can find the expressions for H or E that accompany the waves given above, ˆ β 1 = ˆ z → H 1 = 2 η o cos( ωt- βz )ˆ y A m ˆ β 2 = ˆ z → H 2 = E o η o (cos( ωt- βz )ˆ y + sin( ωt- βz )ˆ x ) A m ˆ β 3 =- ˆ z → E 3 = η o ( cos( ωt + βz + π 3 )ˆ y- sin( ωt + βz- π 6 )ˆ x ) V m ˆ β 4 = ˆ x → E 4 = η o (cos( ωt- βx )ˆ y- sin( ωt- βx )ˆ z ) V m . b) The instantaneous power flow density is given by the Poynting vector P = E × H . Therefore, the instantaneous power that crosses some surface S is given by P = ´ S P · d S . In the case of uniform plane waves, this expression simplifies to...
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This note was uploaded on 04/14/2009 for the course ECE 329 taught by Professor Franke during the Spring '08 term at University of Illinois at Urbana–Champaign.

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329fall08hw5sol - ECE-329 Fall 2008 Homework 5 — Solution 1 Let us consider the following four plane waves in free space E 1 = 2cos ωt βz)ˆ x

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