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. ^œV 4\ œ œ V4\ 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  $$$Þ$ Ê G œ %((Þ( \ œ ; ‚ V œ $!! Ê P œ %(Þ(( G V ; P> # H H. pF H L C R 2 =1000 R t r L b. For a lossy inductor with , we have in series $ ! P œ œ " ! \ $!! U$ ! P P P H c. With the additional resistance, we must transform the 1000 down to 90 instead to get the HH
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2 SOLN.WXP input of 100 . Thus, H "œ$ Þ") "!!! *! \ œ  $"%Þ& Ê G œ &!'Þ$ \ œ *!; œ #)'Þ# Ê P œ %&Þ&( Ê G P H H. pF H NOTE: It is not difficult to assume the Q is constant and solve for the and together. The PP <P solution requires the solving of a quadratic. For a , this gives us $ ! P *!" <  )!! <  *!!!! œ ! P # P for < œ *Þ&'!% ; œ $Þ"(" \ œ  $"&Þ$ Ê G œ &!& \ œ #)'Þ) Ê P œ %&Þ'( P GP pF H . 4.(8/30)a. We first solve for as ; ; œ  " œ %Þ$&* "!!! &! \ œ %Þ$&* ‚ &! œ #"(Þ* Ê G œ \ œ œ ##*Þ% Ê P œ "!!! %Þ$&* Ê G P H H ($Þ!' pF $Þ'&$ . H b. We determine the as though the circuit is a parallel equivalent with U U œ œ œ #Þ*!' œ Ê F œ V²V \ ##*Þ% F "! => P #!!!‚"!!! $!!! cd MHz $Þ%%" MHz 5.(9/2) ^œV4 P œV "4 "P " P 4G V VG V œV "4U " = == = = = = = = = = = ”• Œ Œ ŒŒ ab 99 9 9 # where = = 9 9 œU œ "" PG È and . 6.(9/2)a. Starting with the transformation we have "œ !!! ! \ œ &!; œ $"#Þ# Ê P œ \ œ  œ  $#!Þ$ Ê G œ %*(Þ# #!!! ; Ê 2 5 6.245 pF P ² >9>+6 H H %*Þ(# . H Subtracting the 10pF, we have G œ %)(Þ# pF
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3 SOLN.WXP b. The bandwidth of the circuit is obtained from the as U U œ V G œ 'Þ#) ‚ "! ‚ "!!! ‚ %*(Þ# ‚ "! œ $Þ"## Ê F œ œ 0 U = 9 >9>+6 >9>+6 ' " # 9 !Þ$#!$ MHz 7.(9/4) To begin the solution, simply look at the circuit as an -network transforming 100 to an input L H impedance of 1000 . The transformation for this change is given by . H ; ; œ "!!!Î"!!  " œ $ È The second requirement is the bandwidth specification which has a to give Uœ"! U œ œ Ê \ œ #!!! P œ œ M=96+>/. M8.?->3@/ V/+->+8-/ \ #!!! X9>+6 </=3=>+8-/ =//8 "!!  "!! "! P P ) H or #! . H Thus we have \ œ  œ  $$$Þ$ Ê G œ œ "!!! " $ "! ‚ "!!Î$ \ œ \  \ œ $ ‚ "!! œ $!! Ê \ œ  "(!! G œ G" ) WPG G # " ## H HH $!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 Vœ" ! œ ;" V ;" > > # # # # H ˆ‰ ab and for the bandwidth U œ œ œ #! œ Ê œ œ #& Ê P œ œ 0 "! "!!! #& F! Þ & \ # # ! # " ! VÎ# 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 \; œ œ% ! V "!!! \# & # # P and thus ; œ %!  "  " œ $Þ)(% Ê \ œ  œ  #Þ&) Ê G œ "! "! "!!! $Þ)(% >G # # Êa b # H '"'' pF The last step is to determine the center element, . The reactance is determined by converting the G " output to series and adding to . Thus we have \ G " \ œ \  \ œ ; ‚ V  ; ‚ V œ $Þ)(%  %! œ  ##Þ&( "!!!
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This note was uploaded on 01/31/2012 for the course EE 4002 taught by Professor Scalzo during the Fall '06 term at LSU.

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

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