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V(x,t) =
Z
/(Z
+Z
) { [V
0
0
S
S
+ t
[V
(t–t
(2L–x)) H[t–t
L
S
pd
+ t
T
[V
(t–t
L
S
S
pd
+...............}
This can be further explained by an example.
9
= sin(2 π 10
Let:
V
S
Z
= 35 ohms
S
Z
= 20 ohms
L
Z
= 50 ohms
0
L = 14 in
x = 6 inches
t
= 2 ns/ft = .17 ns/in
pd
v = [2 ns/ft) = .5 ft/ns = 6 in/ns
t
= (20 – 50)/(20 + 50) = .43
L
t
= .18
S
at t = .5 ns
V(x,t)
= V(6 in, .5 ns)
= 50/(50+35){[sin (2 π 10
+ (–0.43) {sin (2 π 10
= .59 {sin (-1.04 π ) +0} at t = .5 ns
Voltage at A with the transmission line properties accounted for. There is no reflection yet.
V(x,t)
= V(6 in, 5 ns)
= [50/50 + 35] {sin[2 π 10
+ (–.43) {sin [2 π 10
+ (–.43)( –.18) {sin [2 π 10 (5 – 17 (28 + 6))] H [5 –.17(28 + 6)]}
= .59 {sin (–1.04 π ) –.43 sin (2.52 π ) + .08 sin (–1.56 π )}
The lattice diagram is a convenient visual tool for calculating the total voltage due to reflections
as described in the previous equations. Two vertical lines are drawn to represent points A and B
on the horizontal dimension, x. The vertical dimension represents time.
A waveform travels back and forth between points A and B of the transmission line in time, pro-
ducing the lattice diagram shown in
the individual reflected voltages up to that time. Notice that at each endpoint, two waves are con-
verging, the incident wave and the reflected wave. Therefore, the voltage at the end points A and
B at the time of the waveform reflection are calculated by summing both the incident and reflect-
ed waves up to and including the point in question.
(t–t
x) H(t
–tx)]
pd
pd
(2L–x))]
pd
(2L+x))H(t–t
(2L+x))]
pd
t)
9
(0.5–0.17ns/in(6in))ns)}
9
(0.5–0.17(6))ns)H(0.5–0.17(6))}
9
(5 – (.17)(6)]
9
(5 – .17 (28 – 6))] H [5 – .17 (28 – 6)]}
Figure
10-11. The voltage at a given point is the sum of all
PHYSICAL DESIGN AND SYSTEM DEBUGGING
10-15
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