We can use bounce diagrams to handle somewhat
more complicated problems as well.
Arnold Aggie decides to add an additional
ethernet interface to the one already connected to his
computer. He decides just to add a "T" to the terminal where the
cable is connected to his "thin-net" interface, and add on some
more wire. Unfortunately, he is not careful about the coaxial
cable he uses, and so he has some
75Ω
75
Ω
TV co-ax instead of the
50Ω
50
Ω
ethernet cable. He ends up with the situation shown
here. This kind of problem is called
a cascaded line problem because we have two
different lines, one hooked up after the other. The analysis is
similar to what we have done before, just a little more
complicated is all.
We will have to do a little more thinking before
we can draw out the bounce diagram for this problem. The driver
for ethernet cable coming to Arnold's computer can be modeled as
a 10V (open circuit) source with a
50Ω
50
Ω
internal impedance. Since the source does not
(initially) know anything about how the line it is driving is
terminated, the first signal
V
1
+
V
1
+
will be the same as in our initial problem, in this
case just a +5V signal headed down the line.
Let's focus on the "T" for a minute Figure 2.
V
1
+
V
1
+
is incident on the junction. When it hits the
junction, there will be a reflected wave
V
1
-
V
1
-
and also now, a transmitted wave
V
T
1
+
V
T
1
+
. Since the incident wave can not tell the difference
between a
75Ω
75
Ω
resistor and a
75Ω
75
Ω
transmission line, it
thinks it
is seeing a termination resistor equal to a
50Ω
50
Ω
resistor
(
R
L
1
R
L
1
) in parallel with a
75Ω
75
Ω
resistor (the second line).
50Ω
50
Ω
in parallel with
75Ω
75
Ω
is
30Ω
30
Ω
. Let's call this "apparent" load resistor
R
L
'
R
L
'
), so that we can then calculate
Γ
V
1
2
Γ
V
1
2
, the first voltage reflection coefficient in going
from line 1 to line 2 as:
Γ
V
1
2
=
R
L
'
−
Z
0
1
R
L
'
+
Z
0
1
=30−5030+50=.25
Γ
V
1
2
R
L
'
Z
0
1
R
L
'
Z
0
1
30
50
30
50
.25
(1)
Note that we
could have started from
scratch and written down KVLs and KCLs for the junction
V
1
+
+
V
1
-
=
V
T
1
+
V
1
+
V
1
-
V
T
1
+
(2)
and
I
1
+
+
I
1
-
=
I
R
L
+
I
T
1
+
I
1
+
I
1
-
I
R
L
I
T
1
+
(3)
Then, by re-writing
Equation 3 in terms of voltage
and impedances we have:
V
1
+
Z
0
1
−
V
1
-
Z
0
1
=
V
T
1
+
Z
0
2
+
V
T
1
+
R
L
V
1
+
Z
0
1
V
1
-
Z
0
1
V
T
1
+
Z
0
2
V
T
1
+
R
L
(4)
We now have two equations with two unknowns
(
V
1
-
V
1
-
and
V
T
1
+
V
T
1
+
). By solving
Equation 4 for
V
T
1
+
V
T
1
+
and then plugging that into
Equation 2, we
could get the ratio of
V
1
-
V
1
-
to
V
1
+
V
1
+
, or the voltage reflection coefficient. The interested
reader can confirm that indeed, you get the very same result
this way.
In order to completely solve this problem, we
also need to know
V
T
1
+
V
T
1
+
, the transmitted wave as well. Since Equation 2 says
V
T
1
+
V
T
1
+
is just the sum of the incident and reflected waves on
the first line
V
T
1
+
=
V
1
+
+
Γ
V
1
2
V
1
+
V
T
1
+
V
1
+
Γ
V
1
2
V
1
+
(5)
We can thus write
V
T
1
+
V
L
+
=1+
Γ
V
1
2
=
R
L
'
+
Z
0
1
R
L
'
+
Z
0
1
+
R
L
'
−
Z
0
1
R
L
'
+
Z
0
1
=2
R
L
'
R
L
'
+
Z
0
1
=6030+50=0.75≡
T
V
1
2
V
T
1
+
V
L
+
1
Γ
V
1
2
R
L
'
Z
0
1
R
L
'
Z
0
1
R
L
'
Z
0
1
R
L
'
Z
0
1
2
R
L
'
R
L
'
Z
0
1
60
30
50
0.75
T
V
1
2
(6)
An important thing to note is that
T
V
=1+
Γ
V
T
V
1
Γ
V
(7)
NOT
T
V
+
Γ
V
=1
T
V
Γ
V
1
(8)
We do not "conserve" voltage at a termination, in the sense that
the reflected and transmitted voltage have to add up to be the
incident voltage. Rather, the transmitted voltage is the
sum of the incident voltage and the
reflected voltage, so that we can obey Kirchoff's voltage law.
We can now start to make our bounce diagram. We
propagate a +5V wave and a -5V wave (separated by 100ns) down
towards the junction. Since the line is 40m long, and the waves
move at
2×108ms
2
10
8
m
s
, it takes 200ns for them to get to the
junction. There, a -1.25V wave is reflected back towards the
source, and a +3.75V wave is transmitted into the second
transmission line Figure 3.
Since the load for the second line is
50Ω
50
Ω
, and the characteristic impedance,
Z
0
2
Z
0
2
for the second line is
75Ω
75
Ω
, we will have a reflection coefficient,
Γ
V
2
=
R
L
2
−
Z
0
2
R
L
2
+
Z
0
2
=50−7550+75=-0.2
Γ
V
2
R
L
2
Z
0
2
R
L
2
Z
0
2
50
75
50
75
-0.2
(9)
Thus a -0.75V signal is reflected off of the second load
Figure 4.
What is the magnitude of the voltage which
is developed across the second load?
What happens to the 0.75V pulse when it gets to the "T"? Well
there is another mismatch here, with a reflection coefficient
Γ
V
2
1
Γ
V
2
1
given by
Γ
V
2
1
=25−7525+75=-0.5
Γ
V
2
1
25
75
25
75
-0.5
(10)
(The
50Ω
50
Ω
resistor and the
50Ω
50
Ω
transmission line look like a
25Ω
25
Ω
termination to the
75Ω
75
Ω
line) and a transmission coefficient
T
V
2
1
=1+
Γ
V
2
1
=0.5
T
V
2
1
1
Γ
V
2
1
0.5
(11)
and so we add to the bounce diagram
Figure 5.
We
could keep going, but the voltage
reflected off of the second load will only be 75mV now, and so
let's call it a day.
There are a couple of other interesting
applications of bounce diagrams and the transient behavior of
transmission lines that we might look at before we move on to
other things. The first is called the Charged Line
Problem. Here it is:
We have a transmission line with characteristic impedance
Z
0
Z
0
and phase velocity
v
p
v
p
. It is
LL long, and for
some time has been connected to a battery of potential
V
g
V
g
Figure 7. At time
t=0
t
0
, the switch S, is thrown, which removes the battery
from the circuit, and connects the line to a load resistor
R
L
R
L
. The question is: what does the voltage across the
load resistor,
V
L
V
L
, look like as a function of time? This is
almost like what we have done before, but
not quite.
In the first place, we now have non-zero initial conditions. For
t<0
t
0
we will have both voltages and current on the line. In
order to match boundary conditions, we must do more than have
one voltage and one current, because the voltage on the line
must be
V
g
V
g
, while the current flowing down the line must be
0. So, we will put in both a
V
+
V
+
and a
V
-
V
-
and their corresponding currents. Note that
x
x
is going to the left this time. Let's forget about the
switch and the load resistor for a minute and just look at the
line and battery. We have two equations we must satisfy
V
0
+
+
V
0
-
=
V
g
V
0
+
V
0
-
V
g
(12)
and
I
0
+
+
I
0
-
=0
I
0
+
I
0
-
0
(13)
We can use the impedance relationship to change
Equation 13 to:
V
0
+
Z
0
−
V
0
-
Z
0
=0
V
0
+
Z
0
V
0
-
Z
0
0
(14)
I hope
most of you can then see by
inspection that we must have
V
0
+
=
V
0
-
=
V
g
2
V
0
+
V
0
-
V
g
2
(15)
OK, the switch S is thrown at
t=0
t
0
. Now the end of the line looks like
this.
We have anticipated the fact that we are going to need another
voltage and current wave if we are going to be able to match
boundary conditions when the load resistor is connected, and
have added a
V
1
+
V
1
+
and a
V
1
-
V
1
-
to the line. These are new voltage and current
waves which originate at the load resistor position in order
to satisfy the new boundary conditions there. Now we do KVL
and KCL again.
V
0
+
+
V
0
-
+
V
1
+
=
V
L
V
0
+
V
0
-
V
1
+
V
L
(16)
and
V
0
+
Z
0
−
V
0
-
Z
0
+
V
1
+
Z
0
=−
V
L
R
L
V
0
+
Z
0
V
0
-
Z
0
V
1
+
Z
0
V
L
R
L
(17)
We have already made the impedance substitution for the current
equation in
Equation 17. We know what the sum and
difference of
V
0
+
V
0
+
and
V
0
-
V
0
-
are, so let's substitute in.
V
g
+
V
1
+
=
V
L
V
g
V
1
+
V
L
(18)
and
V
1
+
Z
0
=−
V
L
R
L
V
1
+
Z
0
V
L
R
L
(19)
From this we get
V
L
=−(
R
L
Z
0
V
1
+
)
V
L
R
L
Z
0
V
1
+
(20)
which we substitute back into
Equation 18
V
g
+
V
1
+
=−(
R
L
Z
0
V
1
+
)
V
g
V
1
+
R
L
Z
0
V
1
+
(21)
which we can solve for
V
1
+
V
1
+
V
1
+
=−
V
g
1+
R
L
Z
0
=−(
Z
0
R
L
+
Z
0
V
g
)
V
1
+
V
g
1
R
L
Z
0
Z
0
R
L
Z
0
V
g
(22)
The voltage on the load is given by
Equation 18 and
is clearly just:
V
L
=
V
g
−
Z
0
R
L
+
Z
0
V
g
V
L
V
g
Z
0
R
L
Z
0
V
g
(23)
and in particular, when
R
L
R
L
is chosen to be
Z
0
Z
0
(which is usually done when this circuit is used), we have
V
L
=
V
g
2
V
L
V
g
2
(24)
Now what do we do? We build a bounce diagram! Let us stay with
the assumption that
R
L
=
Z
0
R
L
Z
0
, in which case the reflection coefficient at the
resistor end is 0. At the open circuit end of the transmission
line
ΓΓ is +1. So we have
this.
Note that for
this bounce diagram, we have
added an additional voltage,
V
g
V
g
, on the baseline, to indicate that there is an initial
voltage on the line, before the switch is thrown, and
tt starts on the bounce diagram.
If we concentrate on the voltage across the load,
we add
V
g
V
g
and
−
V
g
2
V
g
2
and find that the voltage across the load resistor
rises to
V
g
2
V
g
2
at time
t=0
t
0
Figure 10. The
−
V
g
2
V
g
2
voltage wave travels down the line, hits the open
circuit, reflects back, and when it gets to the load resistor,
brings the voltage across the load resistor back down to
zero. We have made a pulse generator!
In today's digital age, this might seem like a strange way to go
about creating a pulse. Imagine however, if you needed a pulse
with a very large potential (100s of thousands or even millions of
volts) for say, a particle accelerator. It is unlikely that a
MOSFET will ever be built which is up to the task! In fact, in a
field of study called
pulsed power electronics just
such circuits are used all the time. Sometimes they are built with
real transmission lines, sometimes they are built from discrete
inductors and capacitors, hooked together just as in the
distributed parameter
model. Such circuits are called
pulse forming
networks or PFNs for short.
Finally, just because it affords us a good
opportunity to review how we got to where we are right now,
let's consider the problem of a non-resistive load on the end of
a line. Suppose the line is terminated with a capacitor! For
simplicity, let's let
R
s
=
Z
0
R
s
Z
0
, so when S is closed a wave
V
1
+
=
V
g
2
V
1
+
V
g
2
heads down the line Figure 11. Let's
think about what happens when it hits the capacitor. We know we
need to generate a reflected signal
V
1
-
V
1
-
, so let's go ahead and put this in the figure, along with its companion current
wave.
The capacitor is initially uncharged, and we know we can not
instantaneously change the voltage across a capacitor (at least
without an infinite current!) and so the
initial voltage across the capacitor should
be zero, making
V
1
-
0=−
V
1
+
V
1
-
0
V
1
+
, if we make time
t=0
t
0
be when the initial wave just gets to the
capacitor. So, at
t=0
t
0
,
Γ
V
0=-1
Γ
V
0
-1
. Note that we are making
ΓΓ a function of time now, as
it will change depending upon the charge state of the capacitor.
The current into the
capacitor,
I
C
I
C
is just
I
1
+
+
I
1
-
t
I
1
+
I
1
-
t
.
I
C
0=
I
1
+
+
I
1
-
0=
V
g
Z
0
I
C
0
I
1
+
I
1
-
0
V
g
Z
0
(25)
since
I
1
+
=
V
1
+
Z
0
=
V
g
2
Z
0
I
1
+
V
1
+
Z
0
V
g
2
Z
0
(26)
and
I
1
-
0=−
V
1
Z
0
=
V
g
2
Z
0
I
1
-
0
V
1
Z
0
V
g
2
Z
0
(27)
How will the current into the capacitor
I
C
t
I
C
t
behave? We have to remember the capacitor equation:
I
C
t=Cd
V
C
td
t
=C∂(
V
1
+
+
V
1
-
t)∂
t
=Cd
V
1
-
td
t
I
C
t
C
t
V
C
t
C
t
V
1
+
V
1
-
t
C
t
V
1
-
t
(28)
since
V
1
+
V
1
+
is a constant and hence has a zero time
derivative. Well, we also know that
I
C
t=
I
1
+
+
I
1
-
t=
V
1
+
Z
0
−
V
1
-
t
Z
0
I
C
t
I
1
+
I
1
-
t
V
1
+
Z
0
V
1
-
t
Z
0
(29)
So we equate
Equation 28 and
Equation 29
and we get
Cd
V
1
-
td
t
=
V
1
+
Z
0
−
V
1
-
t
Z
0
C
t
V
1
-
t
V
1
+
Z
0
V
1
-
t
Z
0
(30)
or
d
V
1
-
td
t
+1
Z
0
C
V
1
-
t=1C
V
1
+
Z
0
t
V
1
-
t
1
Z
0
C
V
1
-
t
1
C
V
1
+
Z
0
(31)
which gets us back to
another Diff-E-Q!
The homogeneous solution is easy. We have
d
V
1
-
td
t
+1
Z
0
C
V
1
-
t=0
t
V
1
-
t
1
Z
0
C
V
1
-
t
0
(32)
for which the solution is obviously
V
1
homo
-
t=
V
0
e−t
Z
0
C
V
1
homo
-
t
V
0
t
Z
0
C
(33)
After a long time, the derivative of the homogeneous solution is
zero, and so the particular solution (the constant part) is the
solution to
1
Z
0
C
V
1
part
-
=1C
V
1
+
Z
0
1
Z
0
C
V
1
part
-
1
C
V
1
+
Z
0
(34)
or
V
1
part
-
=
V
1
+
V
1
part
-
V
1
+
(35)
The complete solution is the sum of the two:
V
1
-
t=
V
1
homo
-
t+
V
1
part
-
=
V
0
e−t
Z
0
C+
V
1
+
V
1
-
t
V
1
homo
-
t
V
1
part
-
V
0
t
Z
0
C
V
1
+
(36)
Now all we need to do is find
V
0
V
0
, the initial condition. We know, however, that
V
1
-
0=−
V
1
+
V
1
-
0
V
1
+
, so that makes
V
0
=-2
V
1
+
V
0
-2
V
1
+
! So we have:
V
1
-
t=2
V
1
+
e−t
Z
0
C+
V
1
+
=
V
1
+
(1−2e−t
Z
0
C)
V
1
-
t
-2
V
1
+
t
Z
0
C
V
1
+
V
1
+
1
2
t
Z
0
C
(37)
Since
V
1
+
=
V
g
2
V
1
+
V
g
2
we can plot
V
1
-
t
V
1
-
t
as a function of time from which we can make a
plot of
Γ
V
t
Γ
V
t
The capacitor starts off looking like a short circuit, and
charges up to look like an open circuit, which makes perfect
sense. Can you figure out what the shape would be of a pulse
reflected off of the capacitor, given that the time constant
Z
0
C
Z
0
C
was short compared to the width of the pulse?
