0125 watts 2 120 4 2 e making use of pr found in part

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Unformatted text preview: its matched load is given by Pr ⌘ Sinc A(✓, ) where Sinc = 2 4⇡ G ( ✓ , ˜ |Einc |2 2⌘o and A(✓, ) = ), we can write Pr = 4 (2⇡ )2 3 sin2 45 = 0.0125 Watts. 2 ⇥ 120⇡ 4⇡ 2 e) Making use of Pr found in part (d), we can calculate the open circuit voltage as | Vo | = q 8Pr Re{Zth } = p 8 ⇥ 0.0125 ⇥ 1 ⇡ 0.316 V. f) When the load is replaced by ZL = 1, the mismatch loss factor will be q= which implies that 4R 2 R L = 0.8 | ZL + Z2 | 2 PL = qPr = 0.01 Watts. 3. !"! # a) Following from the below figure, "% "$ " ! &$ "! "! "# " $% & "% # '"! ()*+,-.#/0.#-./1)23#/)#/0.#2,40/#)-#/0.#5622.-/#7)625.#'7#'-#.86,9':.-/#,*;.<'-5."# the equivalent impedance Zeq is given by Zeq = ZL + (Z2 ZC ) || ZC + (Z1 ZC ). Thus, we write 2 2 Z ( + ⇤= " $% " # Z " %= " !1> Z2" ! Z2 ) "$ Zc" ! > Z1 2Re{Z2 } Zc . = = eq ⇤ Z2 + Z2 2Re{Z2 } Then, the average = " " is calculat...
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This note was uploaded on 02/03/2014 for the course ECE 350 taught by Professor Kudeki during the Spring '11 term at University of Illinois, Urbana Champaign.

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