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Soln96

# Soln96 - EE 4605 Homework-Solutions 1 db(uv21 0 0 DB(S21...

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1 SOLN.WXP EE 4605 Homework-Solutions 1. 1e3 -24 -18 -12 -6 100 10e3 -30 0 Frequency (Hz) db(uv21) 100 -72 -54 -36 -18 80 160 240 320 10 1e3 -90 0 0 400 Frequency (MHz) DB(S21) ang(s21) 0.2 0.5 1 2 5 10 0.2 0.5 1 2 5 10 0 -0.2 -0.5 -1 -2 -5 -10 Sweeping C2:C (pF) @ 100 MHz s11 1 1: 15 pF 109.5, -66.81 <Rs, Xs> RC Filter 5-pole Filter with Angle 5-pole Filter with Autolines 2. ^ œ V  4\ œ ] œ œ V 4\ V  4\ V 4\ V  4\ " " " = = : : : : : : = = or ^ V \ V \ "!!!n"! *)%Þ) "(\$Þ' "!"&Þ% &(&)Þ) "!!!n%& (!(Þ" (!(Þ" "%"%Þ# "%"%Þ# "!!!n)! "(\$Þ' *)%Þ) &(&)Þ) "!"&Þ% = = : : ° ° ° Note: For the small angles, remains almost constant. Similarly for large angles, remains V \ almost constant. 3. a. For the transformation, V œ "!! œ œ Ê ; œ  " œ \$Þ!! V "!!! "!!! ;  " ;  " "!! > # # # Ê Thus \ œ  œ  \$\$\$Þ\$ Ê G œ %((Þ( \ œ ; ‚ V œ \$!! Ê P œ %(Þ(( G V ; P > # H H . pF H L C R 2 =1000 R t r L b. For a lossy inductor with , we have in series U œ \$! P < œ œ œ "! \ \$!! U \$! P P P H c. With the additional resistance, we must transform the 1000 down to 90 instead to get the H H

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2 SOLN.WXP input of 100 . Thus, H ; œ  " œ \$Þ") "!!! *! \ œ  \$"%Þ& Ê G œ &!'Þ\$ \ œ *!; œ #)'Þ# Ê P œ %&Þ&( Ê G P H H . pF H NOTE: It is not difficult to assume the Q is constant and solve for the and together. The P P < P solution requires the solving of a quadratic. For a , this gives us U œ \$! P *!" <  )!! <  *!!!! œ ! P # P for < œ *Þ&'!% ; œ \$Þ"(" \ œ  \$"&Þ\$ Ê G œ &!& \ œ #)'Þ) Ê P œ %&Þ'( P G P pF H . 4.(8/30)a. We first solve for as ; ; œ  " œ %Þ\$&* "!!! &! \ œ %Þ\$&* ‚ &! œ #"(Þ* Ê G œ \ œ œ ##*Þ% Ê P œ "!!! %Þ\$&* Ê G P H H (\$Þ!' pF \$Þ'&\$ . H b. We determine the as though the circuit is a parallel equivalent with U U œ œ œ #Þ*!' œ Ê F œ V ² V \ ##*Þ% F "! = > P #!!!‚"!!! \$!!! c d MHz \$Þ%%" MHz 5.(9/2) ^ œ V  4 P  œ V "  4 œ V "  4 " P " P 4 G V VG V VG œ V "  4 œ V "  4U " VG GP = = = = == = == = = = = = = = = = = = Œ Œ Œ Œ a b 9 9 9 9 9 9 9 9 # 9 9 where = = 9 9 œ U œ " " PG VG È and . 6.(9/2)a. Starting with the transformation we have ; œ  " œ !!! ! \ œ &!; œ \$"#Þ# Ê P œ \ œ  œ  \$#!Þ\$ Ê G œ %*(Þ# #!!! ; Ê 2 5 6.245 pF P ² >9>+6 H H %*Þ(# . H Subtracting the 10pF, we have G œ %)(Þ# pF
3 SOLN.WXP b. The bandwidth of the circuit is obtained from the as U U œ V G œ 'Þ#) ‚ "! ‚ "!!! ‚ %*(Þ# ‚ "! œ \$Þ"## Ê F œ œ 0 U = 9 >9>+6 >9>+6 ' "# 9 !Þ\$#!\$ MHz 7.(9/4) To begin the solution, simply look at the circuit as an -network transforming 100 to an input L H impedance of 1000 . The transformation for this change is given by . H ; ; œ "!!!Î"!!  " œ \$ È The second requirement is the bandwidth specification which has a to give U œ "! U œ œ Ê \ œ #!!! P œ œ M=96+>/. M8.?->[email protected]/ V/+->+8-/ \ #!!! X9>+6 </=3=>+8-/ =//8 "!!  "!! "! P P ) H or #! . H Thus we have \ œ  œ  \$\$\$Þ\$ Ê G œ œ "!!! " \$ "! ‚ "!!Î\$ \ œ \  \ œ \$ ‚ "!! œ \$!! Ê \ œ  "(!! G œ G " ) W P G G # " # # H H H \$!Þ! &Þ)) pF pF or 8.(9/6) We begin by determining the equations required for the solution, similar to the last problem. For the transformation, we have V œ "! œ ;  " V ;  " > > # # # # H ˆ a b and for the bandwidth U œ œ œ #! œ Ê œ œ #& Ê P œ œ 0 "! "!!! #& F !Þ& \ # ‚ #! # "! V Î# 9 # P P ( X H 1 \$*) nH since sees in parallel with a matching resistance with the same value (CONJUGATE \ V P # MATCH). From the value of we may obtain as \ ; P # ; œ œ œ %!

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Soln96 - EE 4605 Homework-Solutions 1 db(uv21 0 0 DB(S21...

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