329sp10hw9sol

329sp10hw9sol - ECE 329 Spring 2010 Homework 9 - Solution...

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Unformatted text preview: ECE 329 Spring 2010 Homework 9 - Solution Due: Mar. 30, 2010 1. a) For plane waves propagating in free space, E and H are related by E = H , where is the unit vector in the wave propagation direction. Therefore, H 1 = 1 (- y ) E 1 = 1 [- x 2cos( t + y )] ( A/m ) . Wave #1 is a travelling wave propagating towards the- y direction. The power per square meter is simply the Poynting vector: | S | = | E H | = E 1 E = 1 | E | 2 ( W/m 2 ) . Therefore, | S 1 | = 1 | E 1 | 2 = 1 4cos 2 ( t + y ) = 4 cos 2 ( t + y ) ( W/m 2 ) . The phasors are E 1 = z 2 e jy ( V/m ) . H 1 = 1 - x 2 e jy ( A/m ) . The time-average power cross a 1 m 2 area can be calculated as < P > = 1 T T | S | dt. By using | S 1 | , we can obtain < P 1 > = 2 ( W ) . To determine polarization, we notice that E 1 only has z component, so Wave #1 is linearly polarized and the unit vector in the polarization direction is z . b) H 2 = 1 y E 2 = 1 [- z 10cos( t- y )- x 10sin( t- y )] ( A/m ) . Wave #2 is a travelling wave propagating towards the y direction. | S 2 | = 1 | E 2 | 2 = 1 100cos 2 ( t- y ) + 100sin 2 ( t- y ) = 100 ( W/m 2 ) . E 2 = x 10 e- jy- z 10 e- jy- j 2 ( V/m ) . 1 ECE 329 Spring 2010 H 2 = 1 - z 10 e- jy- x 10 e- jy- j 2 ( A/m ) . < P 2 > = 100 ( W ) . E z E x =- 10 e- j 2 10 = j, Refer to Page 4 of Lecture 24, since the propagation direction is + y, we can see that the rotating direction of electric eld and the direction of propagation satisfy right-hand rule. Thus, Wave #2 is right-hand circularly polarized. c) For plane waves propagating in free space, E and H are also related by H = 1 E . Then we have E 3 = H 3 (- z ) = h y cos t + z + 3- x sin t + z- 6 i ( V/m ) . Wave #3 is a travelling wave propagating towards the- z direction. Since | S | = | E H | = H H = | H | 2 ( W/m 2 ) , we have | S 3 | = | H 3 | 2 = h cos 2 t + z + 3 + sin 2 t + z- 6 i = 2 cos 2...
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329sp10hw9sol - ECE 329 Spring 2010 Homework 9 - Solution...

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