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au10_312ps5_sol

au10_312ps5_sol - ECE312 Autumn 2010 Problem Set 5...

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Unformatted text preview: ECE312 Autumn 2010: Problem Set 5 Solutions Problem 1 : ˜ E 1 = (ˆ x − j 2ˆ y ) e jkz , and ˜ E 2 = (2 + j )ˆ ye jkz . (a) The time average Poynting vector S av = 1 2 Re { ˜ E × ˜ H ∗ } . First find the magnetic field ˜ H = 1 η ˆ k × ˜ E = 1 η ( − ˆ z ) × (ˆ x − j 2ˆ y ) e jkz = ( − j 2ˆ x − ˆ y ) η e jkz S av = 1 2 Re { ˜ E × ˜ H ∗ } = 1 2 Re braceleftbigg (ˆ x − j 2ˆ y ) × ( j 2ˆ x − ˆ y ) η e jkz e − jkz bracerightbigg = − 5ˆ z 2 η = − ˆ z 6 . 63 × 10 − 3 W/m 2 Note this is ˆ k | ˜ E 1 | 2 / (2 η ) which applies for all single plane waves in lossless media. (b) For ˜ E 2 = (2 + j )ˆ ye jkz , S av = ˆ k | ˜ E 2 | 2 2 η = − 5ˆ z 2 η = − ˆ z 6 . 63 × 10 − 3 W/m 2 (c) ˜ E 1 + ˜ E 2 = (ˆ x + (2 − j )ˆ y ) e jkz , so S av = ˆ k | ˜ E tot | 2 2 η = − ˆ z (1 + 5) 2 η = − ˆ z 7 . 96 × 10 − 3 W/m 2 Note S tot ,av negationslash = S 1 ,av + S 2 ,av because Maxwell’s equations are linear in terms of fields not power. Fields add but powers do not....
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au10_312ps5_sol - ECE312 Autumn 2010 Problem Set 5...

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