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Quizzes2002 - Gnu-t 2 1(10 pts In the system shown above...

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Unformatted text preview: Gnu-t 2. 1. (10 pts) In the system shown above, the propagation velocity VI, = 2 x 108 m/s and the characteristic impedance is Z0 = 50 £2. The generator voltage vG(t) is given by , V6 an V t (as) O 5 '0 is The load voltage vL(t) is observed to be as follows: VI. ’0 V I .. 0. ’33 V t (as) O 5 lo I 5' 2 o z 5 3g 3 S (a) Find the length of the transmission line, d: . g _ ‘1 A_ :. LVP)(T)=Q-K\O )(lO‘UO 3 d= & (m) (b) Find RL. —— Z: =l .. ”Q .. l .2 .. ’1’ x q a- ° m" 7:“ RL = 83 3 (9) (c) Find the inductance per unit length, L Urs—L—)'Z>’ i“- §3=L= 5" =2.S'xlo'q' HIM a. W: 2. Ma‘ L = Q50 k(H/m) (d) Find the capacitance per unit length C \ .. Cc. 7‘ \ X l0 “9 F ‘M is“? C = loo ?(F/m) (c) Find and plot Vin over the time interval 0 < t < 40: vk‘ 9 2. 6"!- r1 I (a (“3) 3 == ‘5 O (a 20 39 ‘10 \l‘b=%)v-= 2-) V+‘0.6?) V—= 0J4? 2. (10 pts) In the lossless transmission line system shown, the lines are initially uncharged and the switch closes at t=0. loom. __§’t=.o,l+ 7?(I._ E5—;=o 0.1, (a) Find the total line voltages indicated on the reflection diagram. Vfla—l'b V2.7: +2—Q Vt“: 3.84 U2?=|,qq vB = 4 VC = 4 VD = 7 - 8 Li (b) Find VL as t —) oo VL = I 0 (0) Plot the input power versus t for 0 < t < 4T: ,?.* [,0 W 0.16 W t o T 21‘ 3" «T 3. (10 pts) The transmission line shown below is charged for a long time with the switch in position A. At time t=0, the switch is moved to position B. (a) Determine the total load voltage VL before the switch is moved from A to B (IL: 01969? A) VL= (0-97 V (b) Determine the steady-state values V55+ and Vss_ before the switch is moved from A to B v++v‘= (0.665? V85: 50 +_.U—=-.. [email protected]?‘ = 3.3 - V (0 )%O 33 V88 = l‘ a? (c) Determine the total load voltage VL at t=0+, after the switch is moved to position B 5° ' LS'JL ZVE= ‘0 9” VL= 3.33 (d) Fill in the reflection diagram below, to determine the total line voltages V1, V2, and V3, at the times and locations indicated in the diagram. o Vt {5 V ’3 T 11‘ b ' V1= 1m“? ", V2 = 3. 3 3 V3 = 3. 3?) (e) Plot the load voltage versus t for 0 < t < 4T: V._ '3 . 33 >13 0 1- zr 3r «1- 4. (4 pts) The switch in the transmission line system shown below is closed at t=0. The load consists of a nonlinear device with the I-V characteristic shown on the following page. \[+: s" ’7 Ik: \D-VL. VL= 7-0 V so IL = [Z 0 MA 5. (6 pts) A lossless air-filled transmission line has Z0 = 100 Q. (a) A 10 (cm) length of this line is terminated in an open circuit. Find the input impedance of the line at f=25 MHz. 7.5 \D 1:ch ‘ 6) (0.0:- 0.0524 3 x \o 6 ‘ . Zm= ,3 gout(/¥L), In: (13) (C) 34;»: J3 \°l 0% .0. Is there a shorter length of this transmission line that has thesame input impedance as the 10 (cm) section? If so, what is its length? % xsmv =s2M —» L.= 0+3.) cc *1; a. 25:04:" Sketch the voltage standing wave pattern on the 10 cm line, clearly showing the t ng location of minima and maxima: Wt) )1- : I0- 6 3’ to “I“ In). -]%?X:E€ 1i:”' 11;ii:t:¥;:j:“r i igifl' EEE ii? I E-EEiiEI IE ii.§:£-E 2-4 '-> 1412-11 7 » flEEIEEEEEEiEEE E.EEE EEE EEEEE.EE EEEEE- EE :1 13;: _‘ 11:1; :3. " 1 EEE- EEE: EEE: :iiiii iii- in ~- a 111I111... 1 111 :i:1:1 1:11;." " -‘.E I: Iii! EEEEIEEEEE EEE E: :Egiifl Him: . -L. ., ._ _ - ._- . ,. - ,- T1f11f1_' gt E E I: “I: '" :J“i“'1ia_ EEEEEEIE Efi fiiflliigfi fiii:§iiiiiiiii E.“ ;'1::.f;“_x;1:t;. 1: ;_:__|_ “EE: EEEE'EE EEE! r '-1-‘-':1i=~i_?-5 E11; 2;»?3 E E! EEEEEEEEEEEEEEEE EEEE EEE EEEEEEEEE 1 ‘ 1 T“ M I I- -I:I .111. 1 -5. ,- fli1fif-1$1' 1-1~ ' 11 .1 1 -1..-’ - f- .-1- 1 , . t" . 11% 11' -~ ' ' ' “1:“? 11:11; -1: .1,_“_'f§_. =- g“! '1 1' F L - If _ __ _. - - , 1.1111111?! 11 7' ‘1 1 T 1' l'" '— III 1 11111111111111!!! 111111151111151111111111: E EEEEEEEEE' EEEEE_ 'EEEE EEE =EEEE EEEE EE EEEEEEEEEE.EE EE': 1111 1111.1 -1111=111111=11-111 11 11111111111 11111111111111 111111 .111 fifi 1111111151111 11111111: 1'1—E: 15511111 11111111 1511111 11 EEEEEEEEEE EEE ..EEEEE .EEEEEEE EEEEEE'EEEE IEEE ' 1' 1115111i 111111111111 '111111'"11= 11.... 1111- '111111 111.11" * l - --.—lo. 1 ’ ' ' 1 i. ,, :1? ' 111E111“ 1E»- 111111111-"1111111.11111111111111111111 ~ .11111-1141 114 I EEEEEEIEEEEEEEEEE 'EEEEE EEEEEEEEEEEEEEEEEE EE EEEE _ 1511115111111111'111111'1111111111111111111111 11111111111111'E' 1-1 "11111111111"1 111.11 111111111111111111111111 aura: i— Sumku. 2.007.. 1. (10 pts) The curve C defined by the equation '3x2—2=y3=4z+1 and the straight line L defined by the equation . x=y=22+1 intersect at the point (1,1,0). (a) Find a unit vector tangent to C at (1,1,0). bxdx == 33:45:444; —) afiClfip) de=3a¢3=q¢u 41- : ALI): +0936 +6473- APHZM £4 daft-E. : fl:b¢+(§_&‘)g J .. +9-4 M,L+h3L¢i/defia V=ZQ+HCS+3$ (4!) OLD-wet “ind-w: iA+_L4 3 4 4W: 2'“! 2134-.7‘2. (MSRMM-‘F 32. fig’t‘Kfi, gym!) (b) Find a differential length vector tangent to L and expressed in terms of dz. dx = ‘3': cad-e 4:02: (24:39 +(2Aeflg + 4% €— (lZISZ. apt “KM. M) (c) Find a vector that is perpendicular to C m L at (1,1,0). A a A, X 3 1. VHF 01% 2 4 3 =4e(—29+4g-qg) 2 z I 2. (10 pts) A volume charge density (C/m3) is given in spherical coordinates by 2 gg-B'd‘ssl-lfifzbv‘ l—r O<r<2 5‘ 9V“): 1'4 2<r<4 0 r>4 Using the integral form of Gauss’s law“,r find the 15 —field as a function of r v (a) intheregionr<2 Gum" - is f1 Saw-2) $181. 9 chi-A9149 ¢ao 95° van: 5' = V‘ "L. I-mE ,5 7 'IDvr. 62““; = 3.2—3.1 04"“?- Atrrb 3 5' ______________________ M (b) intheregion2<r<4 & __ «VLF.- “7'1'4‘75' ‘r A!" rag 1-32. -m if = «W [ 3 ? '1,— + ?] 5 Que -2.‘2..Ol‘l ‘1‘... 4. 29v- 4“,”: v; + =r 24v ‘4 ._____—______________ H (c) intheregionr>4 Qu‘ qwfl—ifi; ‘13:? 440384} 3. (10 pts) The static electric field in a region of free space (8 = so) is given by 1.3: (3x2 + yz) §+ (4y3 + xz) 9+ (423 + xy);. (a) The associated electric scalar potential V(x,y,z) is (circle one) — (x3yz + y4xz + z4xy) —xyz(1+x3+y4+z4) — (xyz + x3 + y4+ 24) — (yz + 6x + xz +12y2 + xy +1222) none of these "VZ~-(13E+s<3+3‘+ 2‘5} : a? (3x1+3%) + fi (433+ ‘9 + 209443.) (b) Determine the volume charge density as a function of x, y, and z in that region of space. {N = v.55 =-. v. (55):: éo{ 62x +l131+12ezf 4. (10 pts) Region 1, comprising the half-space x < 0, is a dielectric region with permittivity 81:86 and permeability in = 110, whereas region 2, the half-space x > 0, has permittivity 22 = 280 and permeability M2 = 200%. The fields at x=0+ in region 2 are given by Eq=10§+15§—202, H2=2£—5§+2 Find the electric and magnetic fields at x = 0' in region 1. ...
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