# solution03 - 2 1 1 1 1 ML ML MS Max T S S G Γ − Γ −...

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Suggested Solutions to Assignment 3 Problem 1 (a) (b) If , Max LU Max SU Max TU G G G G , 0 , , = 2 22 2 21 2 11 1 1 1 1 S S S = dB 21 . 16 79 . 41 = = o 89 38 . 0 = Γ L o 132 52 . 0 = Γ S 2 22 2 2 21 2 11 2 1 1 1 1 L L S S TU S S S G Γ Γ Γ Γ = dB 9 . 15 12 . 39 = =

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Problem 2 (a) As S 12 = 0, (b) (c) o 167 5 . 0 11 = = Γ S IN o 99 47 . 0 22 = = Γ S OUT o 7 . 16 6 . 17 1 1 0 = Γ Γ + = IN IN IN Z Z o 50 3 . 44 1 1 0 = Γ Γ + = OUT OUT OUT Z Z 0 50 50 50 50 0 0 = + = + = Γ Z Z Z Z S S S 33 . 0 50 100 50 100 0 0 = + = + = Γ Z Z Z Z L L L 2 22 2 2 21 2 11 2 1 1 1 1 L L S S TU S S S G Γ Γ Γ Γ = dB 31 . 10 74 . 10 = = Max LU Max SU Max TU G G G G , 0 , , = 2 22 2 21 2 11 1 1 1 1 S S S = dB 46 . 13 18 . 22 = =
Problem 3 (a) (b) Max LU Max SU Max TU G G G G , 0 , , = 2 22 2 21 2 11 1 1 1 1 S S S = dB 46 . 13 19 . 22 = = 172 . 0 186 . 0 j + = 965 . 0 1 = B 907 . 0 2 = B 161 . 0 46 . 0 1 j C + = 523 . 0 096 . 0 2 j C + = 52 . 0 2 4 1 2 1 2 1 1 = ± = Γ C C B B MS 49 . 0 2 4 2 2 2 2 2 2 = ± = Γ C C B B ML 2 22 2 2 21

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Unformatted text preview: 2 , 1 1 1 1 ML ML MS Max T S S G Γ − Γ − ⋅ ⋅ Γ − = dB 53 . 13 53 . 22 = = (c) ( ) ( ) 0434 . 1 1 2 22 2 11 22 11 21 12 = − − = S S S S S S U 2 , 2 ) 1 ( 1 ) 1 ( 1 U G G U Max TU T − < < + dB G G dB Max TU T 3858 . 3692 . , < − < − ⇒ dB G dB T 85 . 13 09 . 13 < < ⇒...
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## This note was uploaded on 05/01/2011 for the course ELECTRICAL EE5602 taught by Professor Xuequan during the Spring '11 term at City University of Hong Kong.

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solution03 - 2 1 1 1 1 ML ML MS Max T S S G Γ − Γ −...

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