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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.
 Spring '11
 Kudeki

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