329fall09hw13sol

329fall09hw13sol - ECE 329 Fall 2009 Homework13 Solution...

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Unformatted text preview: ECE 329 Fall 2009 Homework13 Solution Due: Dec. 8, 2009 1. a) (i) z L = Z L /Z = 2 . Locating the point z L on the Smith Chart, we can nd the corresponding L = 0 . 333 . (ii) (iii) Rotating along the constant | | circle towards the generator (clockwisely) by a distance l = 0 . 2 , we nd ( l ) = 0 . 333 , where =- . 2 . 25 180 o =- 144 o , and z ( l ) = 0 . 54- j . 24 . Then, Z ( l ) = Z z ( l ) = 27 .- j 12 . 0 () . b) V ( l ) = Z ( l ) Z ( l ) + Z g V g = 27 .- j 12 . 27 .- j 12 . 0 + 50 10 = 3 . 66- j . 99 = 3 . 79 - 15 . 1 o ( V ) . c) V ( l ) = V + e jl + L e- jl = V + ( e j . 4 + 0 . 333 e- j . 4 ) , V + = 1 . 54- j 4 . 77 = 5 . 01 - 72 . 1 o ( V ) . d) V (0) = V + (1 + L ) = V + (1 + 0 . 333) = 2 . 05- j 6 . 36 = 6 . 68 - 72 . 1 o ( V ) . e) I (0) = V (0) Z L = 0 . 0205- j . 0636 = 0 . 0668 - 72 . 1 o ( A ) . 1 ECE 329 Fall 2009 2. In this case, Z L = Z (matched), so we actually don't need a Smith Chart. a) L = Z L- Z Z L + Z = 0 , ( l ) = L e- j 2 d = 0 , Z ( l ) = Z = 50 () . b) V ( l ) = Z ( l ) Z ( l ) + Z g V g = 50 50 + 50 10 = 5 ( V ) ....
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329fall09hw13sol - ECE 329 Fall 2009 Homework13 Solution...

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