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Midtermsolution

Midtermsolution - ECE 201A Midterm Solution 1 E = ax jaz...

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ECE 201A Midterm Solution 1. ( ) ( ) 0 0 - ˆ ˆ ˆ ˆ + 2 j y j y x z x z E a ja e a ja e β β = + r (a) ( ) ( ) 0 0 - 0 1 ˆ ˆ ˆ ˆ + 2 j y j y z x z x H a ja e a ja e β β η = + + r (b) and fields propagating in +y direction are E r H r ( ) 0 - ˆ ˆ j y x z E a ja e β = + r and ( ) 0 - 0 1 ˆ ˆ j y z x H a ja e β η = + r . Hence the time average power density is ( ) ( ) ( ) ( ) 0 0 2 0 0 0 ˆ 1 1 1 ˆ ˆ ˆ ˆ ˆ Re 1 1 W/m 2 y j y j y y av z z x y a a ja e a ja e a β β η η η + + + × = + = r * 1 1 Re Re 2 2 x S E H = × = r r (c) E r and fields propagating in -y direction are H r ( ) 0 ˆ ˆ 2 j y x z E a ja e β = r and ( ) 0 0 1 ˆ ˆ 2 j y z x H a ja e β η = + r . Hence the time average power density is ( ) ( ) ( ) ( ) 0 0 2 0 0 0 1 1 1 5 ˆ ˆ ˆ ˆ ˆ ˆ 2 2 Re 4 1 W/m 2 2 j y j y y av x z z x y y S a ja e a ja e a a β β η η η × = + = − r * 1 1 Re Re 2 2 E H = × = r r (d) Net time average power density is 2 0 0 1 5 3 ˆ ˆ 1 =- W/m 2 2 y y av av av y y S S S a a η η + = + = r r r . (e) The part propagating in +y direction has two mutually orthogonal components with equal amplitude and 90 out of phase. So by definition this part is circularly polarized. The part propagating in - y direction has two mutually orthogonal components with unequal amplitude and 90 out of phase. So this part is elliptically polarized.
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