1 db and gt 1 power delivered to the second 2 port

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Unformatted text preview: in ( 2 ) N ) = 13.1 dB and GT (1) = = Power delivered to the second 2 - port Power available from source | S 21 |2 (1− | ΓS |2 )(1− | Γin ( 2 ) |2 ) (1 − S11 ΓS )(1 − S 22 Γin ( 2 ) ) − S12 S 21 Γin ( 2 ) ΓS 2 = 8.86 dB Therefore, Pin(1) = –10 (dBm) – 13.1 + 8.86 = –14.24 (dBm) ii) Note that input mismatch factor is 4 RS Rin (1) ( RS + Rin (1) ) 2 = 0.3765 = −4.24 ( dB ) Therefore Pin(1) = –10 + (–4.24) = –14.24 (dBm) (b) From part i) of (a), the power delivered to the second 2-port is Pin(2) (dBm) = Pavs (dBm) + GT(1) (dB) = –10 dBm + 8.86 dB Pin(2) = –1.14 (dBm) (c) The power delivered to the load, PL Note that Power delivered to the load G( 2 ) = Power delivered to the second 2 - port = | S 21 |2 (1− | ΓL |2 ) (1− | S11 |2 ) + | ΓL |2 (| S 22 |2 − | D |2 ) − 2 Re( ΓL N ) = 12.81 dB Therefore, PL (dBm) = Pin(2) (dBm) + G(2) (dB) = –1.14 dBm + 12.81 dB PL = 11.67 (dBm) (d) The transducer gain for the cascaded 2-port, GT GT = or Power delivered...
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