Exam 1 - Fall 2010 - EE3340

Exam 1 - Fall 2010 - EE3340 - 13113334019me 1 Texas State...

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Unformatted text preview: 13113334019me 1_ Texas State University—San Marcos School of Engineering EE 3340 Spring 2010 Exam 1 - Feb. 11, 2010 Closed book and notes. You may use only pen or pencil, calculator, and the formula sheet provided. Express all numerical answers as a number between 1 and 999999. followed by the appropriate engineering unit (mA, A, mV, etc.) as needed. _ For 1this exam you may also use the attached Smith chart and linear scale by folding the linear scale as described in the formula—sheet section at the back. NM:_§M£_~IJ‘_§_~ .Multigle Choice: Put the letter corresponding to the fist answer in the blank provided. 1. (4 pts.) In the time- and spacedependent'wave expression - V(x, t) - V0 (205611 - fiz + it), the phase constant B has the dimensions of (A) radians (B) meters (C) Crucible)“1 (D) meters per second Answer: C 2. (4 pts) If the reference characteristic impedance 20 = 50 Q, the reflection coefficient I‘ of a short circuit (Z = O) is (A) +1 @ ‘1 (C) 0 (D) Cannotbedetermined from information given I” —» Z -2 -—-~ 1... Answer: '5 ' a "‘ BE 3340 Exam 1 2 3. (4 pts.) Two coaxial cables are fabricated with identical dimensions, except that in Cable #1, the dielectric between the inner and outer conductors is foam plastic (3, =13), _ L i and in Cable #2, the dielectric is solid PTFE plastic (3, =22). Which of the following Z; statements is true? C I (A) #1 has a h' (meanest Answer: B , r 2 3259+ 5 o 3 4. (4 pts.) A quarter—wave mat%hing section is designed with a 3114 section of 5042 transmission line so that a 25—9 resistive load is transformed to Zn 2 100 +j0 Q at f z 1 GHz. Suppose everything remains the same except that the frequency of measurement is raised to j‘ = 1.05 GHz. What is true about the new input impedance 2m"? (A) The real part remains about the same but the imaginary part is now inductive. kl t (B) The imaginary part remains about the same but the real part rises above 100 Q. ‘l-WN‘M . ... 333 v a? at Mew—3‘ 1st??? asses: 333333 .3...” s -.....3 Eg/P‘lefil l‘ (D) The imaginary isaiimmmains about the same but the real part Answer: ______C_r____ Problems: When working the following problems, show as much work g m cm! I will try to give you partial credit for partially correct answers, but if all you write down is an answer and it's wrong, that is difficult For problems involving the Smith chart, please use the attached Smith charts at the end of this exam, folding the radially-scaled parameter scales as instructed. Note in the problem that your work is shown on the Smith chart and leave it attached to the exam. 5. (20 points) Design of 75—9 coaxial cable. Suppose that a certain type of plastic—foam dielectric has a relative dielectric cons _ __=l.3. Its relative permeability ,u, = l. The outer conductor gshield) has an‘ « fan‘s . .- ‘ . memenpmsg Standard AWG wire gauges in mm are given in the following table for 12—gauge to 20— gauge wire: gr: (:3 EE3340Exam 1%,; [ ZO'QOE: Q‘s}; ‘ 3 Ci M 3 Choose a wire gauge for the center conductor so that the characteristic impedance of e resulting coaxial cable is as close as possible to 20 = 75 9. Once you have selected the wire gauge. calculate the actual impedance of the cable and the wave velocity u in m s". .. ‘ ‘M at; to {2: lug) :79» 75“50%.MC%:) a- slag is u- !) S‘Yi-nk 3,31v-H- “E” t 2°": Ea} .1 MB 7' we? H.251!“ e y“ .- w :9, 45st - WM _ c. __ My ire gauge 110.. _, Me 45 :m {.5 cuatedimpedance: “ALA Wave velocity: 1 - ‘3 '19 g M/S 6. (24 points) VSWIR to impedance. A long, low—loss transmission line whose impedance is 30 = 50 S2 is connected between a 50—9 generator on the left and an unknown load on the right. Before the load is counected, the line is connected to a perfect short circuit (R = 0) and the standing—wave voltage V}(x) is measured. The distance x in meters is measured from the generator end and goes from 0 to the length L = 7 20 m of the line, Wen the unknown load is connected, a different standing— wave pattern V2(x) results. These measurements are shown in the following figure: .xam 1 Vollaga magnitude (U) _ \ (1.4 = ," - ’i i 5 ‘ V2 (unknown) 02? . '. .' 1 7 “gm—“Wig: in; mm-“ K?“ 4:0 “57;, . """""" 330' “WW-"1733‘: Distancefr rum generator tmeIers} 5% o (a) (4 points) Calculate the VSWR ofthe unknown load by estimating the aximum and minimum values of its standing wave. ~ I \/ .ngficéimim/O USMR: Wul" -—;( o 90A ) l— : Z5329 3 a] 2/ 2&2» *“ B VSWR of unknown load = W (b) (5 points) Estimate by how man. neters the-+ioeation- of thevoltage minimum shifted in going from the short to the un own load: _. ‘7 b D V"\ 170 A,” ' Minimum shifted N meters (0) (10 points) Using the information you found 1n (a) and (b), calculate the complex- valued reflecnon coefficient Fof the Load. You may use the Smith charts attached to the end of this exam, but indicate your work on them if you do. Express the angle of the reflection _oefficient1n degrees. r - g4 79/ A, _ . ; l H. mm, C5 { N .500 L? Th1, Complete Smith Chart Black Mag1c_: Design ".3121 11 . ' {Wt-317$ 0.14 037 0.36 0.15 80 0.35 N :9 ’0 ~ 0, v '5 0 1) ~ 50 q] 33 0 '5“ "01 0.1 :9 ”a? 15b a? 15: 05 9.5 $0 0 a '\ 0 . o. o 1:. ‘1‘ ’ 1“, 1 ‘10 D 0 . ‘6': 9.0 3i -, ‘3 c9. 2 111 1n ‘51 o ($1 5'0 '11 a $3- 9 P . m E, 20 Z q a H «a . =19 a: D. N. . ‘9. =1 0. c1. 5 E G O D C: v-v F! H n N r?) I" «4 Z n C O _ 8 8 ’ 1 mmmctcowoNENT1R/m,oxco1~10u ANCECCMPONENT (c.1110 g E .... a 1 0.1 E Q ' .. a :5; 3.. .. g .1 -— E 5 53 ' a 09‘ c1 5 t1 . 1. 1:18 ’ ‘r‘ 81’ a. fi ' c: u E “3’ g P; a. Y r ‘2 0., 9'0 '0 .9. #. A. a .9 - C“. P. 7‘. ‘ _. 05‘ at: 0‘ P1.) z-a 001- 00- 91'“ ff 5 ’ n11 06' “'0 c.“ fl ,2 ‘3 :10 310 9 . 59- ‘0 - 0 1190 100 ‘E - Mummy SCALED mum-115115 TOWARDLOAD—> c—TOWARDGENIERA'IUR 1.2 1.1 1 1s 10 7 5 4 3 2 1 +1—4er- .' 111.- l‘H—H‘ 1 WW1 1 ‘1-‘1-‘1—‘1 '. I. nae—1‘ 1'. 1 1-11 1..‘. .+—11 2 1.1 11 13 1.4 1.5 111 2 a 1 5 111 20 .. 20 0.1 02 0.4 0.6 0.5 1 15 3 4 5 e. 1015- 1"r—F1'1'1—’.‘-'1‘"111HH"‘1""}‘*1’"" WWW 0.9 011 0.7 0.6 05 0.4 . 001 1.1 12 13 1.11 15 11s 1: 111 1.9 2 15 3 1 5 10» 0.9 0.3 0.7 01, 05 0.4 0.3 02 0.1 0.99 0.95 0.9 03 0.7 0.6 05 0.4 0,3 02 0.1 0 I1]l_*_*_l“L_j_llll11L_‘1_|_.II|‘LL..I_‘III|1|_J||\LI_L“|_LJI |_1ILJ_A_A.‘|“ lu—"L_I_'IL_‘|VL_I1|II_.J CENTER 0.0 0.1 0 2 113 01 0.5 0.5 0.7 0.11 0.9 1 1.1 1.2 13 1.4 15 111 1.7 1.0 1.9 I1L441:M111M1M1IIJW‘MIIMI'M'IJ £41111_1_1.1111_L.L_.1_11111_i_1_1s,_111_.1_.1_1_11111_1_1..11 ORIGIN exam 1 5 0.3 ,2) m Angle of F(degrees)= . (31% / Magnitude IN = (d) (5 points) Using your result in (c), calculate the complex—valued impedance of the unknown load in ohms. Express it in rectangular coordinates as a sum of a real part R . and an imaginary part jX. _ M . 31 49 We " Z L {.0 {5 ”:OSIJJCL X504“ ‘355 @2950 :17er ”O U0 Unknown load 2: 2 Eli " $3 2 TH : 7. (22 points) Shunt-stub impedance matching. At a certain frequency, a load impedance has the value ZL z 34.51862 82. Design a shunt—stub matching network with 50-82 transmission lines that will match this load impedance so that 23”,, = 50 $2 in the shunt-stub matching circuit below. Use the shortest possible lengths of line, make the shunt stub shorted at the far end, and express line lengths Lu and LPA in terms of wavelengths it (e. g. a quarter—wave line has L z 0.25 A). You may use the Smith charts attached to the back of this exam, but label your work on those charts and indicate in this problem that part of your work is there. Why-756.2 I Z '50 _ " l .I‘ “r «4 1-..v "-32-. .- LFVG‘W- [3.2%] EB 3340 Exam 1 6 (a) (3 points). Normalize the actual load impedance ZL to 50 9 so that normalized 7 (dimensionless) impedance 2‘; = 2,150 S2. - so “-“J “ {foo—(.121, (b) (4 points) Using either a calculator or the Smith chart, convert the normalized load impedance to normalized load admittance yL — 1/24,. {pm a J. : 1 : 0.3 35¢+51i°~fl0§+ o. ; I __.... FL 4"“ Fifi-Trait" )7 j S )4“ or $6: Sm‘eh Laure firm 31L: D_l+1 6.5 (c) (5 points) Perform the operations on the Smith chart to move the normalized load admittance to the g z 1 unit circle with L”. mn too». A‘: MEMO a: one: mute-l grew" h: 3.0-1: we GA" YLT— l +j2lfi A;'l$“0-°%=lD-i\3 L!:L96 LSE-‘M (d) (5 points) Then find the normalized admittance ym that must be presented by the shunt stub to move the impedance to 3’0 = 1.7 e ml 7% 2 ~51!” 10(1de +j1~0i EB 3340 Exam 1 Y» :- am y“: “F“ (9)13?) (c) (5 points) Finally, find the length of the shunt stub Ln needed to produce the ad h t d “311 requir sun ami‘ TBYM ’ 20M) [F l :0 (“\wa (“kaukwk . ?\M:2 EsTJEVk‘W\ LL: lo ”fonkéfl) (3* {VIM ‘- 3+I~AQQJ A: Y“ _ __-____ :Ej1‘0‘f \pkx‘c‘u {5 15?1:[/0?LL1 3“ L”: Mus Sw—\-€"h (~th . 5“ #1 A: balm—0.1»: 0.6a“ —§L __./ 8. (18 points) Transients on transmission lines. The circuit above consists of a +1 VDC ideal voltage source, an ideal switch 31, a 50-9 resistor, and at 43cm section of 50—9. transmission line shorted at the right-hand end. Before I = O the switch is open. At 2‘ = 0, the switch is closed and remains closed. The propagation velocity it along the cable is u z: 2 x 103 m s'l. Draw a "bounce diagram" in the figure above, showing the progress of the voltage wave along the transmission line in the graph below it, indicating the wave‘s position with a dashed line. Then, use the information from the bounce diagram to determine the voltage as a function of time at two fixed points: ' (a), (9 points) Atx = 20 cm (in the center of the line). (b) (9 points) At x = 0 cm (at the left-hand end of the line). l a _ _L_-__,* ._ “tout Txmc ti. £°" fitrwbt “f0 MVBU'SK ""13 0 L€ 61“ LA. .— m5:2h So l"\"€\ul Vaguu [\e\"'\* +{U ”2/:— U S“ get: 750:» 5° 0% “‘4’ (”I U;+O-$v ’ h 0“ (’Lof’f’ {'5 .— Zp‘néo v t—ip .- v{ 80 “‘ Awo" as ‘“ EE 3340 Exam 1 V(20 cm) 1 .0 V 0.5 V O t o a) to m a: II E C E C .,. 1. N 0') tr Draw your waveforms to scaie on the two graphs provided above. the total current I (1) (sum of forwaxd and reverse Balms Question (5 points): What is t at the point x = 20 cm (in the center of the line)? current waves) going from left to sigh Draw the total current wavefonn in the graph beiow. 1(20 cm) 20 mA " —- .. 10 mA 0 t Q U) C: V 1ns 2ns ans ...
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