This preview shows pages 1–6. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: ﬂax/We“ ado/6%? 0‘2 .— Jo/oﬁ/a/c—«s 26. 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 25. (a) Bounce diagram and V503) vetSus t. (b) V36) versus
t, obtained via superposition. _ RL—Zo_100Zo—Zo N
‘ RL+Z0 ' 10020420 ‘
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 26a. 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 26b. Bounce diagram and ‘VL versus t for tw = 15.1.
no}
A (v)
1 2 3 Fig. 2.12. Figure for Problem 26c. Bounce diagram and ‘VL versus t for t“, = id / 2. ti: i
l 28. 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. 213. Digital IC chips. (a) The oneway 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 213a. 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 2131). 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 2143. “VI, “V2, and “V3 versus t. 32.. 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 32. Sketch of V(z) and I(z)[ versus 2. 33. 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 shortcircuited 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 opencircuited 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 ...
View
Full
Document
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.
 Summer '08
 STAFF

Click to edit the document details