329sum09hw9sol

# 329sum09hw9sol - ECE-329 Summer 2009 Homework 9 —...

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Unformatted text preview: ECE-329 Summer 2009 Homework 9 — Solution July 24, 2009 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 a uniform plane wave are orthogonal to each other and to their direction of propagation, thus, 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 and considering η o ≈ 120 π Ω (free space), we can find the expressions for H or E that accompany the given wave fields, 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 P = P · ˆ nA, where ˆ n is the vector normal to the flat area A . Below, we are considering A = 1 m 2 and ˆ n = ˆ z. • For wave 1, we have P 1 = E 1 × H 1 = 4 η o cos 2 ( ωt- βz )ˆ z W m 2 , thus, P 1 = 4 η o cos 2 ( ωt- βz ) W . • For wave 2, we have P 2 = E 2 × H 2 = E o (cos( ωt- βz )ˆ x- sin( ωt- βz )ˆ y ) × E o η o (cos( ωt- βz )ˆ y + sin( ωt- βz )ˆ x ) = E 2 o η o ( cos 2 ( ωt- βz )ˆ z + sin 2 ( ωt- βz )ˆ z ) = E 2 o η o W m 2 , 1 ECE-329 Summer 2009 thus, P 2 = E 2 o η o W . • For wave 3, we have P 3 = E 3 × H 3 = η o cos( ωt + βz + π 3 )ˆ y- sin( ωt + βz- π 6 )ˆ x × cos( ωt + βz + π 3 )ˆ x + sin( ωt + βz- π 6 )ˆ y =- η o cos 2 ( ωt + βz + π 3 )ˆ z + sin 2 ( ωt + βz- π 6 )ˆ z =- η o cos 2 ( ωt + βz + π 3 )ˆ z + cos 2 ( ωt + βz + π 3 )ˆ z =- 2 η o cos 2 ( ωt + βz + π 3 )ˆ z W m 2 , thus, P 3 =- 2 η o cos 2 ( ωt + βz + π 3 ) W ....
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329sum09hw9sol - ECE-329 Summer 2009 Homework 9 —...

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