Solnf98

Solnf98 - 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)
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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.
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3 SOLN.WXP Homework 2: 3. the impedance formula for the parallel combination of , , and in terms of , , and Develop VP G V = 9 U .
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4 SOLN.WXP Solutions 2: 3. Consider the parallel connection of , , and which has an impedance of VP G ^4 œ œ œ œ "V V  4 G "4 VG "4 VG ab ˆ‰ ˆ = = == "" V4 P V P 9 V "4U = = = = 9 9 9 9 = = where = 99 9 œU œ V G œ PG P È and
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5 SOLN.WXP Homework 3: 4. An -network is to be used to match a load of to and input of as shown. P "!!! "!! HH L C R 2 =1000 R t =100 R s =100 I s a. Determine and for MHz. PG 0 œ " ! b. Estimate the of the circuit. U
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6 SOLN.WXP Solutions 3: 4. a. Step 1: V œ Ê; "œ"! ;œ$ V "; > # # # or Step 2: ; œ $ œ V G Ê G œ œ %(Þ(& $ # ‚ "! ‚ "!!! = 1 9# ( pF Step 3: ; œ $ œ œ Ê P œ œ %Þ((& " P $ ‚ "!! VG V # ‚"! =1 = . 9>= > 9 ( at resonance H b. To estimate the bandwidth, use the series model as U ¸ œ œ "Þ& F ¸ œ 'Þ'( P $!! "! V  V "!!  "!! "Þ& = 9 => ( Œ MHz!!
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7 SOLN.WXP Homework 4: 5. At r/s and a , determine , , and for the circuit below: = 9" # ) œ"! Uœ"! P G G C 1 R 2 =50 500Ω C 2 R t =500 L
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8 SOLN.WXP Solution 4: 5. For this circuit, V œ ; " V Ê; "œ œ"!Ê;œ$ &!! &! ># ## ˆ‰ Thus \ œ  œ  ; ‚ V œ  "&! Ê G œ œ ''Þ( "" G "! ‚ "&! G# # 9# ) # = H pF Transforming the series combination to parallel we have \ œ  œ  "''Þ( œ  \ ² \ V ; GG P > #: " H ab For the we have U œ œ œ"! X9>+6T+<+66/6V/=3=>+8-/ T +<+66/6 P </+->+8-/ P P V ² V &!! ² &!! S86C => 99 == or P œ œ !Þ#&! \ œ #& #&! "! = .H 9 P H To obtain , we get \ G " " " "''Þ( \ \ \ "''Þ( \ œ Ê œ œ  !Þ!$%! Ê G œ $%! GPG P " pF
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9 SOLN.WXP Homework 5: 6. R 2 =100 C 2 C 3 2000 I 1000 L At 10 r/s, determine and to obtain the given and a * #$ > G G V U œ "!Þ
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10 SOLN.WXP Solutions 5: 6. In class we found V œ "!!! œ V "! œ ;" ># >> ## "# H or From the we obtained U U œ "! œ œ Ê P œ ''Þ( V²V P" ! P ''( => 9 * = nH From the reactance of of , we find P ''Þ( H ; œ œ œ "&Þ!!
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Solnf98 - EE 4605 Homework/Solutions Homework 1: 1. (1.2-1)...

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