EE5602_Assignment 1 - EE5602 Active RF and Microwave...

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Unformatted text preview: EE5602 Active RF and Microwave Techniques, Assignment 1. D Z 0 = 50 Ω Z 0Ω =5 0 Z L =50 Ω Z O U T = 1 5+ j 25 Ω S/C L Using a Smith Chart show clearly how you would design a single stub matching network to match the transistor output impedance ZOUT to 50Ω. 0 .1 1 0 .1 9 0 .0 0 .1 2 0 .3 8 0 .1 3 0 .3 7 0 .1 4 0 .3 6 80 40 0 .4 110 0 .3 9 100 90 45 0 .1 5 0 .3 5 70 60 2. 1 5 65 0. 0. o 05 14 O ), C R 70 A AP T CI 0. IV 0 30 06 0. E SC S U 1 .8 43 0 .6 0. 07 1 .6 0 12 1 .4 8 0 .0 0 .9 1 .2 55 1 0 .4 1 .0 0 .3 4 0 .1 6 50 42 0. EP Yo) jB / E (+ NC TA 0 .7 35 60 30 0 .3 3 0. 17 0. 32 18 0 .8 0 .2 50 25 0. 0. 19 0. 44 31 0. 0 .4 0 40 0 .4 jX (+ T 75 N EN 0 .0 – > W A V E L E N GTH S T O W ARD 0 .0 0 .4 9 G E N ERA 0 .4 8 ± 180 TO 170 R – 0 .4 > 7 0 .0 160 90 4 0 .4 85 6 15 0 OM PO E C IN D 80 UC T IV E R EA CT AN C 1. 0 0 .1 0 .2 0 .3 0 .4 0 .5 0 .6 0 .7 0 .8 0 .9 1 .0 1 .2 1 .4 1 .6 1 .8 2 .0 3 .0 4 .0 5 .0 10 20 50 A D < – A R D L O TOW 7 T H S 0 .4 ­1 7 0 EN G V EL W A 0 <– ­9 0 ­1 6 ­8 5 0 .2 0 .4 9 0 .1 0 .4 0 .6 0. 8 0 .4 8 jB E ( ­ NC TA ) /Yo EP SC 0 ­1 5 ­8 0 S VE U 6 0 .4 4 0 .0 I CT DU 0 .3 ­7 5 I N o) /Z 05 , O R 0. ­7 0 0. 06 ­6 0 0 .7 1 .4 0 .8 0 .9 1 .2 5 ­5 1 .0 0 ­5 5 ­4 32 0. 0 .6 1 .6 0. 1 .8 18 0 ­5 ­2 5 44 ­6 5 0 .5 0. 0 .3 7 0 .1 2. 0 3 ­3 0 ­6 0 4 0 .3 6 0 .1 ­3 ­7 0 5 0 .3 5 0 .1 5 0 .3 6 0 .1 4 ­8 0 ­4 0 0 .3 7 0 .1 3 0 .4 0 .2 ­9 0 0 .1 2 0 .3 8 0 .1 1 ­1 0 0 0 .3 9 CAP AC IT IV E R 0 .1 0 .4 ­1 1 0 EA CT A 0 .0 9 0 .4 1 NC E C ­1 2 0 OM 0 .0 8 PO 0 .4 2 N EN T (­ ­1 0. 5 0 .4 40 4 jX 5 .0 1. 0 .2 0 50 R E S IS T A N C E C O M P O N E N T ( R / Z o ), O R C O N D U C T A N C E C O M P O N E N T (G / Y o ) 20 ­1 0 10 ­1 5 4 .0 ­2 0 3. 0 1. 0 0 .8 0. 8 0 .6 0. 6 0 .4 0 .2 ­1 30 0. 07 0. 0. /Z 5 43 4 2 0. 0 .6 0 .8 3. 20 0 0. 3 1 0 .2 9 0 .2 30 0. 3 4 .0 15 1 .0 5 .0 2 0 .2 8 0 .2 0 .2 20 10 10 0 .2 4 0 .2 3 0 .2 6 0 .2 7 R E F L E C T IO N L E O F ANG IS S IO T R A N S M L E O F ANG 0 .1 20 50 0 .2 5 0 .2 6 0 .2 7 0 .2 5 0 .2 4 0 .2 3 C O E F F I C I E N T I N D E G REES N C O E F F IC IE N T I N D E GR EE S ­2 0 0 .2 2 0 .2 8 0 .2 9 0 .2 1 ­3 0 0 .2 0. ­4 0 0. 1 9 3 0. 31 ...
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This note was uploaded on 01/11/2011 for the course EE 5602 taught by Professor Wingshingchan during the Spring '10 term at City University of Hong Kong.

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