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hwset10solution - [8-13(a L in ZL ZL S11 0.6 Z in S11(b Av...

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[8-13] (a) 11 9 50 500 50 500 0 0 ± ² ± ² * Z Z Z Z L L L L L in S S S S * ² * ± * 22 21 12 11 1 (8.41) 11 9 30 1 . 0 1 11 9 105 175 . 0 100 6 . 0 q ² ² q ± q ² q ² 33 . 107 4615 . 0 in in in Z Z * ² * ± 1 1 0 q ² ² q ² ± 33 . 107 4615 . 0 1 33 . 107 4615 . 0 1 50 608 . 29 447 . 26 j ² ? 500 500 ' 11 ± ² in in Z Z S 500 608 . 29 447 . 26 500 608 . 29 447 . 26 ± ² ² ² j j q ² 20 . 173 900 . 0 (b) ) 1 )( 1 ( ) 1 ( 22 21 1 2 in L L v S S V V A * ± * ² * ± ) 33 . 107 4615 . 0 1 )( 11 9 30 1 . 0 1 ( ) 11 9 1 ( 60 50 . 3 q ² ± q ² ² ± q q 54 . 84 065 . 7 500 2 2 2 1 2 1 1 2 2 ' 21 ± in in S S Z Z V V V V V V V V S 608 . 29 447 . 26 500 608 . 29 447 . 26 54 . 84 065 . 7 2 j j ² ± ² q q 56 . 39 069 . 1
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[8-18] S 11 = 0.35 S 12 = 0.1 S 21 = 3.0 S 22 = 0.5 (a) 21 19 50 1000 50 1000 = + = Γ L Note that the “load” the first 2-port sees is the input impedance of the second 2-port. Thus 8457 . 0 1 22 21 12 11 ) 2 ( = Γ Γ + = Γ L L in S S S S and 265 . 425 7896 . 0 1 ) 1 ( ) 2 ( 22 ) 2 ( 21 12 11 ) 1 ( = = Γ Γ + = Γ in in in in Z S S S S From this, we can get P in . in two different ways. i) Note that ) 1 ( ) 1 ( ) 1 ( in avs T P P G G = or in dB ’s P in (1) ( dBm ) = P avs ( dBm ) – G (1) ( dB ) + G T (1) ( dB ) where dB N D S S S G in in in 1 . 13 ) Re( 2 ) | | | (| | | ) | | 1 ( ) | | 1 ( | | port - 2 first the to delivered Power port - 2 second the to delivered Power ) 2 ( 2 2 22 2 ) 2 ( 2 11 2 ) 2 ( 2 21 ) 1 ( = Γ Γ + Γ = = and
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dB S S S S S G S in in S in S T 86 . 8 ) 1 )( 1 ( ) | | 1 )( | | 1 ( | | source from available Power port - 2 second the to delivered Power 2 ) 2 ( 21 12 ) 2 ( 22 11 2 ) 2 ( 2 2 21 ) 1 ( = Γ Γ Γ Γ Γ Γ = = Therefore, P in (1) = –10 ( dBm ) – 13.1 + 8.86 = –14.24 ( dBm ) ii) Note that input mismatch factor is ) ( 24 . 4
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