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# Pr zc 2 2 8rez rez 2rez rez 2 7 c pt

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Unformatted text preview: )\$=*(\$.'\$>'-,+'2'>\$)0\$*\$<*)=&'>\$-0*>?\$1&,=&\$,5\$)&'\$=*5'\$&'2' power absorbed by the load is given by PL = 1 Re{L |L } . Then, using the voltage 2 Z division rule, we write 2 ! "# ! %! "& 2 \$ " \$)&'2'302'\$ 1 '( |VT |Re{ZL } )2 \$1 # = |Zc I1 | . PL = ( 6 7'8" ! | Re{ZL } 2 |ZL + ZT 9 6 7'8" ! 9 8Re{Z2 } " ! Note that the power PL = Pr . \$A01\$,5\$)&,5\$50?\$*))*=&\$)&'\$-0*>\$*(>\$=0(5,>'2\$)&'\$+0-)*B'\$*(>\$=;22'()\$*=2055\$)&'\$-0*>@\$ c) After dividing Pr by Pt , we have |Z c I 1 |2 ! Pr # |Zc | 2 # 2} 8Re{Z%# " = %# &Re{Z }2Re{Z } Re{Z 2 } = 7' (# c Pt ! 1 |I1 |2 # 1 ! 7'82 9 4Re{Z1 }Re{Z2 } "# 2 2Re{Z2 } 2 2Re{Zc } . Therefore, we can # recast the above equation as # ! 7'8" # 9 ! %! # # ! " \$ &# ! " ! "! "! 4. |Zc |2 4Re{Z1 }Re{Z2 } ! " # Pr!= Pt'8" # 9 "7 2 2Re{Zc } \$ . a) Looking at Figure 1 which shows the possible orientations and locations fo...
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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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