HW 2 Sol - flax/We“ ado/6%? 0‘2 .— Jo/ofi/a/c—«s...

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Unformatted text preview: flax/We“ ado/6%? 0‘2 .— Jo/ofi/a/c—«s 2-6. Pulse excitation. The reflection coefficients at the twa ends of the line are “RsHZomZo—Zo_ l" - -——-——— .. -——————— .. 0 8 R5 + Zn Z0 + Zn “V5 U) (V) 1 First half 3019) Second half Ins ' mm; a Vs=0.375V 1.5ns ‘1’: (I) 2115 I 1 0.1875V "V320.1375V ' 25“ First half 3ns ' 0.094V -. 1(ns} 'Vs=—0.'ISV 10.5115 Second half shifted by 9ns 11115 “VF—0375‘! 11.5115 -1 —0.994V 12ns - .1875V “VF—0.1 B75V 12.5115 13né (a) Fig. 2.9. Figure for Probiem 2-5. (a) Bounce diagram and V503) vetSus t. (b) V36) versus t, obtained via superposition. _ RL—-Zo_100Zo—Zo N ‘ RL+Z0 ' 100204-20 ‘ At t = 0, an incident voltage of amplitude Z Z A 0V? : .____.°___A = __0_A : .. R5 + Z0 Z0 + Z0 2 is launched at the source end of the line. When the input pulse turns off at t = tw, this change can be represented as a new voltage disturbance ‘ FL +1 A °V+*=_ +=__.. 1 V1 2 launched on the line at t = tw. One by one, the new voltage cancels out the existing voltage disturbances °if+, “V1”, ‘1”, “V2— , etc. along the line. The bounce diagram and the load voltage 1/105) versus 1% are provided below for each case corresponding to a different input pulse lengths. [It Fig. 2.10. Figure for Problem 2-6a. Bounce diagram and °VL versus t for tw = 2m, where 124 is the one—way time delay along the line. no) A (V) :de 1 2 3 Fig. 2.11. Figure for Problem 2-6b. Bounce diagram and ‘VL versus t for tw = 15.1. no} A (v) 1 2 3 Fig. 2.12. Figure for Problem 2-6c. Bounce diagram and ‘VL versus t for t“, = id / 2. ti: i l 2-8. Cascaded transmission lines. The reflection coefficients are I"S = 0 (since RS = 50(2), and FL m +1 (since RL = 00) respectively. The bounce diagram and the sketches of voitages hut) and V1175) are as shown. 2-13. Digital IC chips. (a) The one-way time delay of each interconnect can be calculated as id = l/vp=(15 cm)/(10 cm—ns‘l) :15 ns. The reflection coefficients at the source and load ends of the interconnects are T's = 0 and 1"“ = 1'}; = 0 respectively. When the incident voltage of amplitude w 50 z. = . v ‘ 50+500v) 05 (launched at the driver end of the 509 interconnect) reaches the junction between the three interconnects at t = 1.5 ns, no reflection occurs since (100 || 100) m 509. As aresult, a voltage of amplitude 0.5 V is launched on each of the 10052 interconnects. At t = 3 us, these voltages “VI (1) 113 0.25 Fig. 2.17. Figure for Problem 2- - Versus 13. 12' (a) Bounce dlagram' (bwmtages V1. V2. and “V3 r = r=0 FL: =FL2=O VLI ' "rm 0.5 (V) l 2 3 4 Fig. 2.18. Figure for Problem 2-13a. Bounce diagram and “V” and °VL2 versus 1: w mm ,.-,.:u,tx;wmmmmm;~ Si 2% g: 3': 3% reach the load ends where they are completer absorbed. The bounce diagram and the voitages “V” and “V” are sketched as shown. (b) Since the load end of one of the 1009. interconnects (point A) is now an open circuit (i.e., 112 = +1), reflection occurs. The reflected voltage "fl. = 0.5 V arrives the junction at t = 4.5 us where another reflection occurs sending back a voltage ‘V” = 11°in = —0.25 V toward point A where _ (50 u 100) — 100 FA “” (50 || 100)+ 100 = ——0.5 and launching a voltage of “VXI +°V+ = 0.5 m 0.25 = 0.25 V on each of the other two lines. This process continues on. The bounce diagram and the variation of the voltages °VL1 and V]; are as shown. Note that at steady—state, 100 °VL1 = “V1.2 = 1/55 = S V V1.1“) = G=e,-0.5 112:“ r (V) 0.75 Stcagystege _ _ _ _ _ _ _ _ _ ((ns) 123456889 6ns V1.2 U ) 0.75 gm ~ — w u — - - - - - - Steady state 0.5 [(ns) 123456389 Fig. 2.19. Figure for Problem 2-131). Bounce diagram and °Vu and °VL2 versus t. v1 (n (V) “V2 m 315 (r) (V) 1.5 (35fl7)V 1.5 (100181)V 0.5 t(ns) 1015) 0.5 Fig. 2.20. Figure for Problem 2-143. “VI, “V2, and “V3 versus t. 3-2.. Resistive load. (a) The load reflection coefficient is given by H ZL—Zo _150—50m rm _—_._._.. L ZL+ZO 150+50 0.5 Since V” = 1 V, we have Vnm =[V+|(I+p)=1+0.s = 1.5 v Vmin =|V+|(1— p) = 1 — 0.5 = 0.5 V _ Vmu _ 1.5V _ 1m _—Z;... 509 _o.o3 A _ Vnfln _ u Imm — Zo — 509 "0.01 A (b) The sketch of |V(z)| and iI(z)| versus 2 are as shown. 1.220cm 29:50!) 25129509 |V(2)LH(2:)I 1.5V (0.03A) 0.5V (0.01A) ‘dcm’ 20 15 ED 5. Fig. 3.1. Figure for Problem 3-2. Sketch of |V(z)| and |I(z)[ versus 2. 3-3. Microwave filter. Since the coaxial line is air—filled, A = c/ f = (3 x 108 m—s“)l(2.5 GHz)==12 cm. (a) The input impedance of a short-circuited transmission line is given by Zsc : jZO tan(filsc) = 323152 21:“ "" gist: : (— /\ ) in 2 1.257 rad ~—> lsc : 2.4 cm (b) The input impedance of an open-circuited line is given by Zoe = ‘jZO com/Bloc) = j231~Q 2 —> was 2 5°C 2 2.83 rad m» to: z 5.4 cm Note that the difference between the t ' - — — expected. we lengths lS lac lsc — 5.4 —~ 2.4 — 3 cm 2 M4, as 22wWWWMMWWWWWMWWWWWWmmmmmmwwwwmwwmmmmmmmmmmmmwwmeW mm.me mmmmmmmmmwmmmmmmww .- WWMMM‘WWMMWW ...
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This note was uploaded on 01/17/2011 for the course ECE 305 taught by Professor Staff during the Summer '08 term at Michigan State University.

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HW 2 Sol - flax/We“ ado/6%? 0‘2 .— Jo/ofi/a/c—«s...

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