If all we did was launch signals down semi-infinite transmission
lines, we would not get very much useful work done. We really
need to have a finite length line, and put something at the
end...like a termination. So let's take a look at
a terminated transmission line Figure 1.
The line has characteristic impedance
Z
0
Z
0
and we assume that it is terminated with a load
resistor
R
L
R
L
. If we have connected a source to the other end of the
line, then we will have launched a voltage wave
V
1
+
V
1
+
and a current wave
I
1
+
I
1
+
down the line. If the line is
LL long, then it will take a time
τ=L
v
p
τ
L
v
p
, where
v
p
=1LC
v
p
1
L
C
, for the signal to get to the end of the line.
What happens when the signal gets to the load?
We can assume some voltage
V
L
V
L
will appear across the load resistor, and hence a current
I
L
I
L
will flow through it. The most logical thing to assume
would be that
V
L
=
V
1
+
V
L
V
1
+
. But, we quickly run into a
contradiction. If
V
L
=
V
1
+
V
L
V
1
+
, then
I
L
=
V
L
R
L
=
V
1
+
R
L
I
L
V
L
R
L
V
1
+
R
L
(1)
But Kirchoff says the sum of the currents into the output
terminal must equal zero, thus
I
1
+
I
1
+
must equal
I
L
I
L
. This then says
I
1
+
=
V
1
+
Z
0
=
I
L
=
V
1
+
R
L
I
1
+
V
1
+
Z
0
I
L
V
1
+
R
L
(2)
which can only be true if
Z
L
=
Z
0
Z
L
Z
0
, which while possible, will not be the case in general.
What are we to do? We have an obvious
contradiction. The telegrapher's equations permit
two solutions to the transmission line
problem: a signal going in the
+x
x
direction,
V
1
+
V
1
+
, and a signal going in the
-x
x
direction,
V
1
-
V
1
-
. The only way out of our problem
is to assume that when the
V
1
+
V
1
+
signal gets to the load, a new
signal, going in the
-x
x
direction, is created, which then heads back towards
the load. So, let's put in a
V
1
-
V
1
-
and
I
1
-
I
1
-
Figure 2.
Now there is a little problem with signs here that we have to
deal with. We can either draw
I
1
-
I
1
-
so that it points in the
-x
x
direction, and say
I
1
-
=
V
1
-
Z
0
I
1
-
V
1
-
Z
0
(3)
or we can draw
I
1
-
I
1
-
going in the
+x
x
direction, and say
I
1
-
=-
V
1
-
Z
0
I
1
-
V
1
-
Z
0
(4)
Since most of the time, it is better if we define
all currents going in the
same direction, and since
this equation says that
V
-
V
-
and
I
-
I
-
are related by
-
Z
0
Z
0
the latter choice seems to be the better one.
Going by that convention then, we equate voltage
on either side of the termination
V
1
+
+
V
1
-
=
V
L
V
1
+
V
1
-
V
L
(5)
and we sum currents into the output terminal
I
1
+
+
I
1
-
=
I
L
I
1
+
I
1
-
I
L
(6)
or
V
1
+
Z
0
-
V
1
-
Z
0
=
V
L
R
L
V
1
+
Z
0
V
1
-
Z
0
V
L
R
L
This makes
V
L
=
R
L
Z
0
V
1
+
-
V
1
-
V
L
R
L
Z
0
V
1
+
V
1
-
(7)
which we can substitute into
Equation 5 to get:
V
1
+
+
V
1
-
=
R
L
Z
0
V
1
+
-
V
1
-
V
1
+
V
1
-
R
L
Z
0
V
1
+
V
1
-
(8)
or
V
1
+
1-
R
L
Z
0
=
V
1
-
-
R
L
Z
0
-1
V
1
+
1
R
L
Z
0
V
1
-
R
L
Z
0
1
and this can be solved for
V
1
-
V
1
-
as
V
1
-
=
R
L
-
Z
0
R
L
+
Z
0
V
1
+
≡
Γ
vL
V
1
+
V
1
-
R
L
Z
0
R
L
Z
0
V
1
+
Γ
vL
V
1
+
(9)
where
Γ
vL
Γ
vL
is called the load
voltage reflection
coefficient.
We could also have solved for
I
1
+
I
1
+
in terms of
I
1
-
I
1
-
and we would have found:
I
1
-
=
Z
0
-
R
L
Z
0
+
R
L
V
1
+
≡
Γ
IL
V
1
+
I
1
-
Z
0
R
L
Z
0
R
L
V
1
+
Γ
IL
V
1
+
(10)
where
Γ
IL
Γ
IL
is the load
current reflection coefficient. Note
that
Γ
IL
=-
Γ
vL
Γ
IL
Γ
vL
Let's take a break from equation manipulating,
and think about what we have here. First of all, although the
result we have obtained is very important, the method we used to
get there was even more so. What did we do? We postulated a
voltage and current on the line, and then took a look to see if
that solution resulted in a reasonable result. In this case it
did not. We were in gross violation of Kirchoff Laws! We rescued
ourselves by taking the only possible escape route: we added an
additional voltage and current to the solution. Since
V
-
V
-
and
I
-
I
-
are related to each other in a different manner than
V
+
V
+
and
I
+
I
+
(by
-
Z
0
Z
0
rather than
+
Z
0
Z
0
), this gave us an additional degree of freedom so that
we could simultaneously satisfy both the
transmission line I-V relationships as well as the load I-V
relationship.
Let's take a look at
Γ
vL
Γ
vL
for a minute. Over what range can it vary? A glance at
Equation 9 shows that it depends on the range of
R
L
R
L
. If we exclude the possibility of negative resistance
(a reasonable exclusion) then
0≤
R
L
≤∞
0
R
L
. As
R
L
R
L
varies over this range, the voltage reflection coefficient
goes from
-1≤
Γ
vL
≤1
-1
Γ
vL
1
. When
R
L
R
L
is lessthan
Z
0
Z
0
, the reflection coefficient is negative. When
R
L
R
L
is greaterthan
Z
0
Z
0
, the reflection coefficient is positive. When
R
L
=
Z
0
R
L
Z
0
the reflection coefficient is zero, and the line is
said to be matched. In a matched transmission line,
a signal traveling down it is completely absorbed by the load,
and nothing is reflected from it. In this case we can have the
incident voltage and current signals just equal load voltage and
currents without the need to add reflected waves. For an
unmatched transmission line, a signal incident on the load is
(partially) reflected, and a new signal starts moving back down
the line towards the source. An example of a transmission line
is a buss in a computer. What would be the implications of
unmatched terminations on various connections between the buss
and the computer circuitry?
Two special cases of terminated transmission
lines that are of interest are
R
L
=0
R
L
0
(a shorted line) and
R
L
=∞
R
L
(an open line). For
R
L
=0
R
L
0
,
Γ
vL
=-1
Γ
vL
-1
and so a signal with the same amplitude but opposite
sign as the incident wave, is reflected back down the line. For
R
L
=∞
R
L
, a signal with the same amplitude
and sign gets reflected back down the line. It is easy to
remember which is which, and also to make sure you have the
right order in the equation for the reflection coefficient, when
you keep in mind that the voltage across the load is the sum of
V
1
+
V
1
+
and
V
1
-
V
1
-
. If the line is terminated with a short circuit,
V
L
V
L
must equal 0. Since
V
1
-
=
Γ
vL
V
1
+
V
1
-
Γ
vL
V
1
+
(11)
and
Γ
vL
=
R
L
-
Z
0
R
L
+
Z
0
Γ
vL
R
L
Z
0
R
L
Z
0
From these equations we can see that with
R
L
=0
R
L
0
(a short),
Γ
vL
Γ
vL
does indeed
=-1
-1
, and so
V
1
-
=-
V
1
+
V
1
-
V
1
+
and hence
V
L
=
V
1
+
+
V
1
-
=0
V
L
V
1
+
V
1
-
0
as it should.
Now, let's go back to our terminated
transmission line. What is going to happen to
V
1
-
V
1
-
after it leaves the load? It obviously travels back
down the line towards the generator. What happens when it gets
back there? Time to look at Figure 3.
At the generator end of the line we can not leave out
V
1
+
V
1
+
, as it is still here, so long as the source remains
connected. Once it is launched, it stays on the line forever. We
just have to add in the new
V
1
-
V
1
-
. What happens now, if we test Kirchoff's voltage law
here at the generator end?
V
S
-
I
S
R
S
-
V
1
+
-
V
1
-
=0
V
S
I
S
R
S
V
1
+
V
1
-
0
(12)
Substituting
I
S
=
I
1
+
+
I
1
-
I
S
I
1
+
I
1
-
and then using the impedance relationship between the
current and the voltages we get:
V
S
-
R
S
Z
0
V
1
+
-
V
1
-
-
V
1
+
-
V
1
-
=0
V
S
R
S
Z
0
V
1
+
V
1
-
V
1
+
V
1
-
0
(13)
Note that again, we used
V
1
+
Z
0
V
1
+
Z
0
for
I
1
+
I
1
+
and
-
V
1
-
Z
0
V
1
-
Z
0
for
I
1
-
I
1
-
.
We know
V
1
+
=
Z
0
Z
0
+
R
S
V
S
V
1
+
Z
0
Z
0
R
S
V
S
(14)
which if we substitute into
Equation 13
and re-arrange a little bit we get
V
S
1-
R
S
Z
0
+
R
S
-
Z
0
Z
0
+
R
S
+
R
S
Z
0
V
1
-
-
V
1
-
=0
V
S
1
R
S
Z
0
R
S
Z
0
Z
0
R
S
R
S
Z
0
V
1
-
V
1
-
0
(15)
The stuff inside the big parentheses sums to zero (as in fact it
should, this is just the solution to the initial
V
1
+
V
1
+
generation problem) and we are left with the
uncomfortable conclusion that
V
1
-
V
1
-
must be zero! What are we going to do? We will just
have to add
another wave
V
2
+
V
2
+
to the solution! (
Figure 4)
Now we have two ways to proceed from here. The first would be
the dumb way, and try to solve this whole problem. But then we
could also be smart and note that all of the equations we have
relating voltages and currents in this problem are
linear
equations and hence we can use the principle of
superposition Figure 5.
Thus, the top sketch in
Figure 5 can be broken into
the
two circuits (a) and (b) at the bottom
of the figure. Then, everything is trivial. We have already
solved (a) - that was just our initial
V
1
+
V
1
+
launching problem. Part (b) is just a reflection problem again,
with a wave of
V
1
-
V
1
-
incident on a load of value
R
S
R
S
through a transmission line with characteristic impedance
Z
0
Z
0
. The reflected wave in this instance is
V
2
+
V
2
+
, which is why we had to use numbered subscripts in the first
place.
If you do not believe using superposition is
valid, you can try doing the problem over again, but it should
be pretty obvious that we can write
V
2
+
=
R
S
-
Z
0
R
S
+
Z
0
V
1
-
≡
Γ
vS
V
1
-
V
2
+
R
S
Z
0
R
S
Z
0
V
1
-
Γ
vS
V
1
-
(16)
with
Γ
vS
=
R
S
-
Z
0
R
S
+
Z
0
Γ
vS
R
S
Z
0
R
S
Z
0
(17)
