311_exam2sol_sp10

311_exam2sol_sp10 - ECE 311 - Spring Quarter 2010 Exam #2...

Info iconThis preview shows pages 1–8. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

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

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

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

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

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

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

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

View Full DocumentRight Arrow Icon
Background image of page 8
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ECE 311 - Spring Quarter 2010 Exam #2 May 17th, 2010 Write your name below and sign the honor pledge “No aid given, received, or observed” if it applies. There are 3 problems on this exam. Exam is open book and notes. Please box or underline your final answers, and remember to include units. Be sure to show all work clearly if you wish to obtain any partial credit. Try to keep your work within the provided space. Use the back of a previous sheet if necessary. A list of useful formulas is given below for your reference. Name: Sb 1) “The Pledge”: No aid given, received, or observed - 1 Losslessllnes: B = 21V?» = col/Es u = 3°. = __ P 13 fit}, Terminated lossless lines: r = (ZL—Zo)/(ZL+ZO) = V5/ V3 Z(——I) = ZO{[ZL +jZotan(Bl)]/[ZO +jZLtan(Bl)]} Time Domain: v(z, t) = Vg(t——z/up) + VE)(t+z/up) i(z,t) = [V;(t—z/uP)—Vo(t+z/uP)]/Zo rs = (ZS—Zo)/(ZS+ZO) Coulomb’s Law: Fez: = R12 qiqzz 47t80Rl2 Electric field of a point charge: EU?) = q(1_€ — R')/(4n80[1_€ — FIB) Problem 1 (4 parts, 36 points! A 15 V DC voltage source with a 45 ohm source resistance is connected at time zero to a 35 Ohm load through an 3 m long 75 Ohm transmission line (as shown). The lossless transmission line contains a material having a = 880 and u = “0. Be sure to label all axes and values of the function in any plots. 45 9 F0 W VG=15 V T; 35 Q z=—3 m 2:0 (a) Plot the current as a function of position on the line at time t = 1 nsec. (10 points) “rm ru‘u'xm A» him-k Mn \'\N is 3’2. . 93 & 3xmx/‘rg ‘ ' 3 Nu... Q‘i" “(5%: \l\ H-RJL €u\k\\&o Vmpgdu‘x 3m»? Sq ‘ 0,3 '7 (H! <-‘,_ 1.31( < .. “d: ,\1. J— 7! k —3h ’2. n 2 (b) Plot the voltage as a function of position on the line at time t = a9 nsec. (10 points) '1 M 1" Mu) him 9W4 3M“ “M “\i‘d‘du (kw-n Jci-Jm Mk 0.01“ A Lgqu .9 r. '3 Wham aA«\-$\Ms‘u QL‘K 7‘ 1"“ $14.45” 3* V5: $363651“): ~3~mw {H = “ (mm, s» -- 2n ‘- 6 .01 9‘ D, S‘? (0) Plot the current as a fiinction of position on the line at time t = 3,1 nsec. (10 points) N $8 m)w¢’w6pm_ (“gum : awn wart otémmbw mum, mm A, ‘3» ’1‘. WV). ._ (d) Find the current flowing through the 35 Ohm load a very long time after the switch is closed. (6 points) We Problem 2 (3 parts, 40 points) A 6 V amplitude voltage source at 1.26 GHz has a 125 9 source impedance and is to be used to power a load of (125 + j 62.5) 9 at this frequency. (a) Using only transmission lines, design a lossless impedance matching network that will allow maximum power dissipation in the load. Use any network type you like, and document your design by drawing a picture of the network and specifying the characteristic impedances and lengths (in meters, not wavelengths) of the transmission lines you need. Assume all lines are filled with a dielectric medium with s = 580 , and choose 1259 for the characteristic impedance of lines if the choice is arbitrary. Use the Smith chart on the next page for any graphical calculations. (30 points) Ckwk. coin.— a CWW UN“- +Ms<brnv~ v kJé-{gé‘rv‘a +W \zr 1' \“3°-{ ‘— ?\v\-°~dnw+ swam I ' 3' A 3 ° "‘1 “W D‘kau s: Dram—mum: 0.1m Li :2 C \H‘f Q } fiflbe - “w Fur MN» m. 2%“ng N09 ’k/ I h r ‘ ‘1 wk“ 1:0“— Qo§.ang) 2‘s most Nebiflq “ 6W \A My» ems“. L a fists? Wr 6v LtJSm ‘ M <\ v SW4» , - W . NJ“— WHL- ‘3~S¥°51'{10I8‘)Do\‘ Nth M SK \wegaMCoS—OMK)+ 01%: 0.3% moat“. C NDWW mfiwkrm'u \‘VEDK (Aua‘ 5% dc 0.942%. 1: (A Maven: ., lav «a “l:- I I «VMV‘». h 9. 99 ‘éfi :2": w Em ¢ O? : gwmfl ‘ ‘o’étOfigo 0: 00 no . gamma Mm -4 * nlga' w... “1:6 “ ‘Qv‘. WW 9 ‘ 1m 9‘0 fi‘fiwhfl‘“ yfiw ‘~ . 32’5""? (9/ RADIALLY SCALED PARAMETERS e 6‘ ' O TOWARD LOAD _> <— TOWARD GENERATOR 466-? . 40 20 [0 5 4 3 2.5 2 [.8 [.6 [.4 [.2 [.l [ [5 [0 7 5 4 3 Z [ Y' \a z [.‘+.P+*+H[r'r‘.[[H—‘r—l-fiHFLJFLH'EEHHIII.'1J.'.*-r‘.‘|""4‘H".'."+w-*'1-"."r‘.'4‘["¢-‘1—i'L-—‘-r*— 1' 396% 40 30 20 [5 [0 8 G 5 4 3 2 [ [ [ [.l [.2 [.3 [.4 [.6 [.8 2 3 4 5 [0 20 m L’VV i. 5 G 7 8 9 [0 [Z [4 20 30 0 0.[ I 0.2 0.4 0.6 0.8 | [.5 2 3 4 5 G [0 [5- 6 Q (5 . H . . .' . . ' ' ' L144" :y'r-i—1L4+.r1-[.r-H15n’1~.4... {H—‘v"! k-"v-L, ."r-‘r-r‘r '~‘ "3m. 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.[ 0.05 0.0[ 00 [.l [.2 [.3 [.4 [.5 LG [.7 [.8 [.9 2 2.5 3 4 5 [0- I; 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.[ 0 [ 0.90 0.95 0.0 0.8 0.1 0.6 0.5 0.4 0.3 0.2 0.1 0 “ (S MAW LI I ' L4..LA_J [ ALL—J [.4 [ | I V L—l I 4 4:_J F V ' WLA L .».__.' 1—- x .__. * l I CENIER a}; 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 [ [.l [.2 [.3 [.4 [.5 [.6 [.7 [.8 [.9 2 “ AéJ—LLg'l—‘A—JJ—A—L—IAAw—ww‘u X I l _L_J ' l '..LJ._J [ [ ‘ ‘ l ‘ l ‘ | ‘ J l l |_._.[ l ' L.L l ' |_| ' —L_.J_[ ' ’ ' ' L—l 1 __LJ (b) How much power is dissipated in the load using your network? (6 points) S'au w» aux“ PMMV E (A +\\ NM“ NW \A 4x I»; C“ 5‘- “gvnk “A? NM. {3‘1 219‘be = ‘AWAQ = 3é~mu (0) Circle True or False. (4 points) If the frequency is changed to 2 GHZ using your network, the power dissipated in the load will be the same as in part (b). True -. _ False h _ \ . Problem 3 (6 parts, 24 points): circle the best answer (1) It is known that the electric field at a point is (2 x 106)(J‘c + 5/) Wm at a point. A point charge of 10—6 C placed at this point will experience a force (in Newtons) of: “M, —..__ (a) 2(2) (b) jl(—2) (c)(—2 x 10—6)(J‘c +52) ((1) (2 x 10—9)(5c +37) (:6) none ofthe alibi/:9 (2) A point charge of -1 nC is located at the point (x=2, y=1, z=1) m. A second point charge of 5 nC is located at the point (x=2, y=1, z=2) m. The force that results on the second point charge in Newtons is: r,” (a) 2(45 x 10'8) (b) rc(—4.5 x 10—8) (c)2(4.5 x 10‘8 \_(d) 2(—4.5 x 10:?(e) none ofthe above "=1~—. ..->v- " (3) Which of the following is true about “slotted line” measurements? a t ey can be used to etermine the impedance that terminates a transmission line"’\.. ey 1nvo ve measurements of the phase of either the voltage or current versus position on the line (c) they require that a parallel plate transmission line be used (d) all of the above (6) none of the above (4) Which of the following represents an integral over the volume of a cylinder of radius 3 m and length 1t m? (a) f” f: f=0(dx)(dy)dz z=0 y=0 x= =21: = 3 _ 2 (b) f R(s1n6) (dR)(dG)d¢ — =0 R=0 (c) J: f [=3r(dr)(dz)d¢ z=0 r=0 f7: (dz)d¢ ¢=0 z=-§ (6) none of the above (5) Which of the following is NOT a vector from the origin to the point (x = 1 , y = 1 ,z = l )? ff'7h" “ (a) J? + j; + 23\(b) J5? + 23¢) J31} (d) all of the above (e) none of the above (6) Which of the following is true about the gradient? (a) the amplitude of the gradient is small if a fimction is changing rapidly _, - _{-b) "the g'radient’s direction__i_s_ the direction of maximum increase in the function) (c) the gradient‘s' direction isTangfitial to" surfaée's“3fz§b"fi§tiifitvamémtion (d) the gradient can be computed only for a vector function of 3-D space (6) none of the above ...
View Full Document

Page1 / 8

311_exam2sol_sp10 - ECE 311 - Spring Quarter 2010 Exam #2...

This preview shows document pages 1 - 8. Sign up to view the full document.

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