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329fall08he3sol - ECE 329 Introduction to Electromagnetic...

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Unformatted text preview: ECE 329 Introduction to Electromagnetic Fields Fall 08 University of Illinois Oelze, Kim, Waldrop, Kudeki Exam 3 Thursday, Nov 20, 2008 — 7:00-8:15 PM Name: _ - 64am; -| 8 AM 10 AM 12 Noon 1 PM Please clearly PRINT your name in CAPITAL LETTERS and circle your section in the above boxes. This is a closed book exam and calculators are not allowed. You are allowed one formula sheet of 8.5 by 11 inch dimensions —— both sides of the sheet may be used. Please show all your work and make sure to include your reasoning for each answer. All answers should include units wherever appropriate. Problem 1 (25 points) Problem 2 (25 points) Problem 3 (25 points) Problem 4 (25 points) '- TOTAL (100 points) - 1. Consider a pair of parallel infinite conducting plates which carry equal and opposite surface charge densities and are separated in the 2 direction by a distance d— — 1 m. In the region between the sheets (0 < z < d), the displacement field is constant at D— — 3602 C/mg, While the electric permittivity varies with distance as 6(2): 3—3 F/rn. a) (5 pts.) What is the electric field E in the region between the plates? ’- 54; =E(2—%>/%\ ZZ’ & 2—2: =<3*'g%1/Z\ X— % =§ 2 b) (5 pts.) The top plate (at z = d) is held at a constant electric potential (Dd = 2 V. Will the electrostatic potential on the bottom plate, say Q30, be smaller or larger than (Pd? 5mm. ‘5‘ mel'i M 52:. chmofiM/ at) 513% c) (10 pts.) What is the electrostatic potential <I>o= _ (MO) of the bottom plate? (Hint: you will need to derive <I>(z) as the appropriate integral of E ) - -47 sfia E: _€:a>2_ 4) 275% .. :1 -—- 1‘ 2 £2321~3a4< CF 39: jg. +K ) CFC) z; .. _ __ _._ = 2.__ ¢(1)=§—*3”K*% ~> 1c 3 —> fit) 2: any tama=évflg _é#,,/ R“ d) (5 pts.) What is the volumetric charge density p in the region between the plates? (Be sure to justify your answer.) 2V E ‘0 S’lVW—ea ’1‘) l5 (“Sb—Cl 0V9 $141.36ka {0‘wa /, 2. A parallel—plate transmission line is made up of perfect conductors of width, 111 = 5 cm, in y direction, and separation distance, 2 1 cm, in a: direction. The medium between the plates is a perfect dielectric with e = 460 and ,u = no. Neglecting fringing effects, for a uniform plane wave propagating down the line along +2 with H(z, t) = 10 cos(37r x 10st — 27rz)gj A/m between the plates, determine the following: 00" F“ K I‘ a) (4 pts) The propagation speed 2),, along +73, ._ X g U“ =__‘-_L—"3“)“° :lfxm M F (5 Z'fi' ; “up; Hi, (M3 2%: x (ow(wt~($=t):_zl anode—(42) AZ gm 0) (5 pts) Electric field E(z, t) between the plates that accompanies H(z, t), /\ X 17:! (Owflrtafaz); X l 1291?" lo am(w'l:—($T> /’ /\ ng'VLHB: d) (4 pts) Waveform v(z, t) corresponding to the voltage drop measured from a: 2 0 plane, Welsh/(:1 15,421+.) - Siam? Cuwalz—(b’l): 61,—“, (mgcfi) V/ 40“ e) (4 pts) Characteristic impedance of the line, 20, in s 1 :_ 6f: ”—7.“. —.——~ ; vZ o f) (4 pts) Power p(z, t) transported down the line. F(%,t): (112,9). flex) r 61; ontopl—age) 3. A system of two connected transmission lines driven by a voltage source and terminated by a resistive load of 509 is shown in the figure below. A switch is closed at t = 0 and the positive voltages are measured for 5 its giving the bounce diagram in the figure. S t=0 Z1 Z2 T‘ T2 R5509 Rg V0 The impedance and transmission time of line 1 are Z1 and T1 and those of line 2 are Z2 and T2. Using the figures, identify the following parameters in appropriate units: a) Transmission time T1: b) Transmission time T2: Ta: L f f/ c) Impedance Z2: EtsQLJS-O 9-— Glace; kc rcfl-eeficm pg». {3L % d) Reflection coefficient F12 between lines 1 and 2: Note Alovaéslckv MM-i~I«/F‘ce. image; :3 Hf: 4 / e) Impedance 21: g) Source voltage V2,: AoVsz EJ— s-D V9=4°gflffa4osgfl§: [2° \1/ 2561-, t, 74' h) Transmitted voltage V‘++ on line 2: V4: 153v (we—sir) 23' i) Steady state voltage V1 on line 1 as t —> oo: Its—L940) V"; Va EL.- ,. duv)£°i’:_ :CDV amou— 7"» j) Steady state voltage V; on line 2 as t —> oo: V‘Lav‘ a4 4x54) :5) V1772, we“ 4. Consider a transmission line segment of propagation velocity up = %c = 2 X 108 m/s, characteristic impedance Z, = 50 Q, and length l. As shown in figure (a) below, the segment is connected in parallel with a 50 S] resistor and an ideal current source t(t) = Re{IeY“’t}, where I = 240 A is the source current phasor and w = 7r X 108 rad/s; figure (b) depicts an equivalent circuit in terms of input impedance of the transmission line at d = l, namely Z (l) E 115% I(l) I (1/2) 1(0) —-> —-> —> v l 210 A 0 V(0) 210 A o {I Z“) = 7% <—l—l——l d 1 1/2 0 (a) Transmission line circuit (b) Equivalent circuit In answering the following questions assume that the circuit above is in sinusodial steady—state: a) (2 pts) What is the signal wavelength /\ (in 1n) on the transmission line? H” ——-—> /\=IL =3fl‘12. =L+w r at My, / n 2% b) (2 pts) Since an “open termination” is located on the line at d = 0, what is the pertinent. “boundary condition” involving the phasor 1(0) for all possible line lengths l? I(°)=O/ c) (3 pts) Does the transmission line support a standing wave in the above circuit for all values of non-zero l? Justify your answer. Y”, hawk, In):i WH’K bf“: Wax/41m. M‘évtd W4 [NM 1"“. fat—Me. Mrbkaotgéfl Mandi—n." WOW‘Q/ WHOL- ’5 2W5,“ fa Slaw? A moths W< - d) (3 pts) What is the smallest non-zero value of l (in m) if phasor I R = 240 A? Explain. j&:2_fi' a) iCaCl=O . 34““ Shulec Lou/MT our“; 64¢. Lark/1') 2 5“““m” “”1” <=é = 2'“/. e) (2 pts) For l determined in part (d), what is phasor V(l/2)? Explain. Flatw- I(§L)=o E; «9+ (Damiolt {h This Cm Since. 1:) . VI g) (4 ptS) What is the smallest non—zero value of l if I R = 0? i =0 :2 V 1-)=O , $futt‘Vo —€ M4! Mfg—l M“! K 7 ( 64.: A 5317/ Wallql rum-3,2» 7* c > 4w 1 ,4 s» I. h) (4 pts) Given that I (l /2) = 0, what is the smallest possible value of Z? Explain :[(_£:>=O & .T(0>=o 2) £32. 91) {gA’A'w/ (MW’HV;J&9 2.61 i) (3 pts) For 1 determined in part (h), what is V(l)? Explain. ...
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