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Solnf97

# Solnf97 - EE 4605 Homework/Solutions Homework 1 1(1.2-1...

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1 SOLN.WXP EE 4605 Homework/Solutions Homework 1: 1. (1.2-1) 2. (1.3-1) 3. (1.3-3)

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2 SOLN.WXP Solutions 1: 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. a. "!! œ Ê Z œ #!! Z œ "!! #‚&! WV o r V Š‹ Z # # Wß:/+5 69+.ß :/+5 Wß:/+5 b. Z œ "!! ‚ œ !Þ**!" "5 "5  "!!5 7 V ] œ "!  4'Þ#)\$ ‚ "! Ê ^ œ  4 "5 "5 \$ % 717.0 450.5 H H ] œ "!  4'Þ#)\$ ‚ "! Ê ^ œ #&Þ\$#  4"&*" "!!5 "!!5 & % then Z œ "!! ‚ œ "!! ‚ ^] ^ ^ ]  ] œ "!! ‚ œ "!! ‚ œ \$)Þ*)Z "Þ!" ‚ "Þ'"# ‚ "! 'Þ#)% ‚ "! 7 "5 "!!5 "5 "!!5 "!!5 "5 % \$ ºº º º "!  4'Þ#)\$ ‚ "! "!  4# ‚ 'Þ#)\$ ‚ "! & % \$ % d. We would like to maintain 0.9901V vs f. Thus ] œ "!!] œ "!  4'Þ#)\$ ‚ "! "5ß-9<</->/. "!!5 \$ # 4F œ ]  ] œ 4'Þ#)\$ "!  "! œ 4'Þ##! ‚ "! - "5ß-9<</->/. "5 # % \$ ˆ‰ or Gœ ** pF across the 1k H pF total ab "!! [Note: the RC time constant for the 100k and the 1k are identical now (10 s ] ( Ñ e. The total impedance across the 50 H load is given by ^ œ ^ "  Þ!" œ #&Þ&(  4"'!( œ "Þ!" ‚ "! ²  4"'!( œ "'!( n  )*Þ" "!!5 & HH H ° The magnitude of this impedance is , giving a small error in the load kk ^ œ "'!( œ \$#Þ"\$ ‚ &! impedance.
3 SOLN.WXP 3. a. + œ .> œ !Þ&!!Z " X 9 XÎ% XÎ% ( > . # 8 œ 8 ## X # X8 X % 8 # 8 ( Œ Š‹ cos sin sin == =1 1 Thus: or + œ !Þ'\$''ß + œ !Þ!!!ß + œ  !Þ!#"##ß + œ !Þ!!!ß + œ !Þ!"#(\$ß á "# \$ % & @ œ !Þ&!!  !Þ'\$'' "! >  !Þ!#"## \$ ‚ "! >  !Þ!"#(\$ & ‚ "! >  á 9=- )) ) cos cos cos ab b. Then @ œ .300 0</; 90 !Þ'\$'' "! > ‚ "Þ!" ‚ "! > œ "! > œ !Þ\$")\$ "! > !Þ'\$'' # 9 ' ' ‘ ˆ‰ ˆ cos cos cos cos

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4 SOLN.WXP Homework 2: 4. (2.1-1) 5. C L R 2 =100 R t =1000 a. At 1MHz, determine and to obtain the given PG V Þ > b. What is the unloaded of (a)? U ? c. due to loss in the inductor. Determine the additional series resistance , , U œ &! < 38.?->9< P representing this loss. d. Assuming does not change, what are the new values of and to give < P G V œ "!!! Þ P > H C 1 ' L R 2 =100 2000Ω 2pF C 2 BJT output a. If the 100 is transformed to 2000 by this circuit, determine to give a of 10 at 10.7MHz. HH PU b. Determine and to complete the transformation. GG "# 7. R 2 =50 C 2 C 3 R t =10 10 a. For the power BJT output shown, determine , , and for a and 30MHz. G Uœ" ! #\$ b. Repaet (a) for the minimum Q possible. [Hint: this may eliminate one component] c. What is the "approximate" bandwidth for (b).
5 SOLN.WXP Solutions 2: 4. ^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. Another form for this solution is Z 1000 10° 1000 45° 1000 80° a. 984.8+j173.6 707.1+j707.1 173.6+j984.8 b. 1015 j5759 1414 j1414 5759 j1015 nnn ²²² 5. a. H 1 pF 10 ; œ  " œ \$ Ê P œ œ %(Þ(& Ð\$ ‚ !! Ñ \$ ‚ "!! #‚ " ! G œ œ %((Þ& Ð Ñ \$ ‚ "! !! # ‚ "! \$ É "!!! "!! ' \$ ' 1 .H 1 H b. 3 Uœ;œ ? c. r P> 9 > + 6 \$!! &! œ œ ' V œ "!' HH ab d. H pF ; ¸  " œ #Þ*!% Ê P œ œ %)Þ** Ð\$!(Þ) Ñ #Þ*!% ‚ "!' " ! G œ œ %'#Þ# Ð\$%%Þ% Ñ #Þ*!% ‚ "! " ! É "!!! "!' ' \$ ' 1 1 H Exact solution with H, , , < œ \ Î&! À U œ &!ß P œ %*Þ!\$ U œ #Þ*!" < œ 'Þ"' PP 3 8 . ? - > 9 < P G œ %'"Þ) pF 6. a. H U œ "! œ Ê P œ œ #*Þ(& = = 9 9 P #‚"!!

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Solnf97 - EE 4605 Homework/Solutions Homework 1 1(1.2-1...

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