Now this new
V
2
+
V
2
+
will head back towards the load and ...Hmmm... things are going
to get kind of messy and complicated. Fortunately for us,
transmission line engineers came up with a scheme for keeping
track of all of the waves bouncing back and forth on the
line. The scheme is called a bounce diagram. A
bounce diagram consists of a horizontal distance line, which
represents distance along the transmission line, and a vertical
time axis, which represents time since the battery was first
connected to the line. Just to keep things conceptually clear,
we usually first start out by showing the line, the battery, the
load and a switch, S, which is used to connect the source to the
line. It doesnt hurt to make a little sketch like Figure 1 , and write down the length of the line,
Z
0
Z
0
and
v
p
v
p
,
along with the source and load resistances. Now we draw the
bounce diagram, which is shown in Figure 2
Normally, you would not put the formula for
Γ
v
S
Γ
v
S
and
Γ
v
L
Γ
v
L
by 0 and
LL in the diagram, but
rather their values. This will become clear when we do an
example. The next thing we do is calculate
V
1
+
V
1
+
and draw a straight line on the bounce diagram (nominally at a
slope of
1
v
p
1
v
p
)
which will represent the initial signal going down the line. We
mark a
τ=L
v
p
τ
L
v
p
on the vertical axis to show how long it takes for the wave to
reach the end of the line
Figure 3.
Once the initial wave hits the load, a second, reflected wave
V
1
-
=
Γ
v
L
V
1
+
V
1
-
Γ
v
L
V
1
+
is sent back the other way. So we add it to the bounce
diagram. This is shown in
Figure 4. Since all of
the waves move with the same phase velocity, we should be
careful to draw all of the lines with the same slope. Note that
the time when the reflected wave hits the generator end is a
total round trip time of
2τ
2
τ
. (This simple concept is one which students often
forget come test time, so be forewarned!)
We saw that the next thing that happens is that another wave is
reflected from the generator, so we add that to the bounce
diagram as well. This is shown in
Figure 5.
Finally, one last wave, as we are almost bounced right off the
diagram, as shown in
Figure 6!
OK, so we've got a bounce diagram, so what? Having the diagram
is only part of the solution. We still have to see what good
they are. Let's do a numerical example, as it is maybe a little
more illustrative, and certainly will be easier to write out
than all these ratios all the time. We will just pick some
typical numbers, and then work out the answers. Let's let
V
S
=
40
V
V
S
40
V
,
R
S
=
150
Ω
R
S
150
Ω
,
Z
0
=
50
Ω
Z
0
50
Ω
and
R
L
=
16.7
Ω
R
L
16.7
Ω
.
The line will be 100m long, and
v
p
=2×108ms
v
p
2
10
8
m
s
Figure 7.
First we calculate the reflection coefficients
Γ
v
L
=
R
L
-
Z
0
R
L
+
Z
0
=16.7-5016.7+50=-0.50
Γ
v
L
R
L
Z
0
R
L
Z
0
16.7
50
16.7
50
-0.50
(1)
and
Γ
v
S
=
R
S
-
Z
0
R
S
+
Z
0
=150-50150+50=0.50
Γ
v
S
R
S
Z
0
R
S
Z
0
150
50
150
50
0.50
(2)
The initial voltage signal
V
1
+
V
1
+
is
V
1
+
=5050+15040=
10
V
V
1
+
50
50
150
40
10
V
(3)
and the propagation time is
τ=L
v
p
=
100
m
2×108ms=
0.5
μ
s
τ
L
v
p
100
m
2
10
8
m
s
0.5
μ
s
(4)
So we draw the bounce diagrams seen in
Figure 8.
Now, here's how we use a bounce diagram, once we have
it. Suppose we want to know what
Vt
V
t
, the voltage as a function of time, would look like
half-way down the line. We draw a vertical line at the place we
are interested in (the dotted line in
Figure 8) and
then just go up along the line, adding voltage to whatever we
had before whenever we cross one of the "bouncing"
signal lines. Thus for the line as shown we would have for
Vt
V
t
what we see in
Figure 9.
For the first 0.25μs we have no voltage, because
V
1
+
V
1
+
has not reached the half-way point yet. The voltage then jumps
to +10V when
V
1
+
V
1
+
comes by. It stays like that until the -5V
V
1
-
V
1
-
comes by 0.5μs later. The voltage then remains constant at 5V
until the -2.5V
V
2
+
V
2
+
comes along to drop the total voltage down to only 2.5
volts. When
V
2
-
V
2
-
comes along, it has been switched back to a positive voltage
wave by the negative load reflection coefficient, and so now the
voltage jumps back up to 3.75V. It will keep oscillating back
and forth until it finally settles down to some asymptotic
value.
What will that asymptotic value be? One approach
is to write down the following equation.
Vx∞=
V
1
+
1+
Γ
L
+
Γ
L
Γ
S
+
Γ
L
2
Γ
S
+…
V
x
V
1
+
1
Γ
L
Γ
L
Γ
S
Γ
L
2
Γ
S
…
(5)
Which we can re-write as
V
1
+
1+
Γ
L
Γ
S
+
Γ
L
Γ
S
2+…+
Γ
L
V
1
+
1+
Γ
L
Γ
S
+
Γ
L
Γ
S
2+…
V
1
+
1
Γ
L
Γ
S
Γ
L
Γ
S
2
…
Γ
L
V
1
+
1
Γ
L
Γ
S
Γ
L
Γ
S
2
…
(6)
Now, remembering the infinite sum relationship:
∑n=0∞xn=11-x
n
0
x
n
1
1
x
(7)
for
|x|<1
x
1
(which is
always the case for a reflection
coefficient). We can substitute
Equation 7 for the
terms inside the parentheses in
Equation 6 and we get
Vx∞=
V
1
+
11-
Γ
L
Γ
S
+
Γ
L
1-
Γ
L
Γ
S
=
V
1
+
1+
Γ
L
1-
Γ
L
Γ
S
V
x
V
1
+
1
1
Γ
L
Γ
S
Γ
L
1
Γ
L
Γ
S
V
1
+
1
Γ
L
1
Γ
L
Γ
S
(8)
We will leave it as an exercise to the reader to show that if we
substitute
(Reference),
(Reference) and finally
(Reference) into
Equation 8 we
will eventually get:
Vx∞=
R
L
R
L
+
R
S
V
S
V
x
R
L
R
L
R
S
V
S
(9)
Look back at
Figure 1 and see if
Equation 9 makes any sense. It should. If we wait long
enough, it is reasonable to expect that any "transmission line"
effects should go away, and we would be back to the same
situation we would have if the line was just some wire
connecting the source to the load. In this case, the load
resistor and the source resistor would form a voltage divider,
and we would expect the voltage across the load to be determined
by the voltage divider equation. That's all
Equation 9 is saying!
What do we do if we want, say, the voltage
across the load with time? To do this we move up the RHS of the
bounce diagram, and count voltage waves as we move across
them. We start out at zero, of course, and do not see anything
until we get to 0.5ms. Then we cross the 10V
V
1
+
V
1
+
wave and we cross the -5V
V
1
-
V
1
-
wave at the same time, so the voltage only goes up to
+5V. Likewise, another 1ms later, we cross both the -2.5V
V
2
+
V
2
+
and the +1.25V
V
2
-
V
2
-
wave, and so the voltage ends up at the 3.75V position Figure 10.
We can also use the bounce diagram to find the voltage as a
function of position, for some fixed time,
t
0
t
0
Figure 11.
To do this, we draw a horizontal line at the time we are
interested in, say 0.75μs. Now, for each position
xx, we go from the bottom of the
diagram, up to the horizontal line, adding up voltage as we
go. Thus for the example: we get what we see in
Figure 12 . For the first half of the line, we cross the
+10V
V
1
+
V
1
+
, but that's it. For the second half of the line we
cross
both the +10V line as well as -5V
V
1
-
V
1
-
wave, and so the voltage drops down to 5V.
Of particular interest to many of you will be the way in which a
pulse moves down a line and is reflected
etc. This is also quite easy to do with a
reflection diagram, if we simply break the pulse into two waves,
one which has a
positive swing at
t=0
t
0
and another which is a
negative going wave
at
t=
τ
p
t
τ
p
,
where
τ
p
τ
p
is the pulse width of the pulse being generated. The way we do
this is suggested in
Figure 13 . We replace the
pulse generator with two battery/switch combinations. The first
circuit is just like we have seen so far, with a battery equal
to the open circuit pulse height of the generator, and a switch
which closes at
t=0
t
0
. The second circuit has a battery with an amplitude of
minus the pulse height, and a switch which
closes at
t=
τ
p
t
τ
p
, the pulse width of the pulse itself.
By superposition, we can just add these two generators, one
after the other, and see how the pulse goes down the
line. Suppose
V
p
V
p
is 10 volts,
τ
p
=
0.25
μ
s
τ
p
0.25
μ
s
,
R
S
=
50
Ω
R
S
50
Ω
,
Z
0
=
50
Ω
Z
0
50
Ω
and
R
L
=
25
Ω
R
L
25
Ω
. With the numbers, we find that
V
1
+
=
25
V
V
1
+
25
V
.
Γ
v
L
=-13
Γ
v
L
-1
3
and
Γ
v
S
=0
Γ
v
S
0
.
Let's assume that the propagation time on the line is still
0.5μs to get from one end of the line to the other.
We draw the bounce
diagram, and launch two waves, one
which leaves at
t=0
t
0
has an amplitude of
V
1
+
=
5
V
V
1
+
5
V
. The second wave leaves at a time
τ
p
τ
p
, later, and has an amplitude of -5V.
Now when we want to see what the voltage as a function of time
looks like, we again draw a line up the middle, and add voltages
as we cross them. Here we see, again, no voltage until we cross
the first wave at 0.25μs, which pops us up to +5V. At a
time 0.25μs later however, the -5V wave comes along, and
we go back down to zero. At
t=
0.75
μ
s
t
0.75
μ
s
, the reflected -1.67V pulse comes along, and so we see
that. Since the source is matched to the line,
Γ
v
S
=0
Γ
v
S
0
and so this is the end of the story
Figure 15.
You can get somewhat more interesting waveforms if you go
someplace where the two pulses at least partially overlap. Let's
look at say,
x=
87.5
m
x
87.5
m
.
Here is the bounce
diagram.
And
here is the voltage waveform we
get.
This time the 1.67V pulse gets to us before the +5V pulse has
completely passed, and so we drop from 5V to 3.33V. Then, when
the -5V wave goes by, we drop down to -1.67V for a little while,
until the +1.67V wave comes along to bring us back to
zero.
