Solnf98

Solnf98 - EE 4605 Homework/Solutions Homework 1 1(1.2-1...

Info iconThis preview shows pages 1–11. Sign up to view the full content.

View Full Document Right Arrow Icon
1 SOLN.WXP EE 4605 Homework/Solutions Homework 1: 1. (1.2-1) 2. (1.3-1)
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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.
Background image of page 2
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 .
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 4
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
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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!!
Background image of page 6
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
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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 $%! GPG P " pF
Background image of page 8
9 SOLN.WXP Homework 5: 6. R 2 =100 C 2 C 3 2000 I 1000 L At 10 r/s, determine and to obtain the given and a * #$ > G G V U œ "!Þ
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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!!
Background image of page 10
Image of page 11
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 01/31/2012 for the course EE 4002 taught by Professor Scalzo during the Fall '06 term at LSU.

Page1 / 37

Solnf98 - EE 4605 Homework/Solutions Homework 1 1(1.2-1...

This preview shows document pages 1 - 11. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online