If, on the other hand, we have a finite line,
terminated with some load impedance, we have a somewhat more
complicated problem to deal with Figure 1.
There are several things we should note
before we head off into equation-land
again. First of all, unlike the transient problems we looked at
in a
previous chapter, there can
be no more than
two voltage and current
signals on the line, just
V
+
V
+
and
V
-
V
-
, (and
I
+
I
+
and
I
-
I
-
). We no longer have the luxury of having
V
1
+
V
1
+
,
V
2
+
V
2
+
, etc., because we are talking now about a
steady state system. All of the transient
solutions which built up when the generator was first connected
to the line have been summed together into just two waves.
Thus, on the line we have a single total
voltage function, which is just the sum of the positive and
negative going voltage waves
Vx=
V
+
ⅇ-ⅈβx+
V
-
ⅇ+ⅈβx
V
x
V
+
β
x
V
-
β
x
(1)
and a total current function
Ix=
I
+
ⅇ-ⅈβx+
I
-
ⅇ+ⅈβx
I
x
I
+
β
x
I
-
β
x
(2)
Note also that until we have solved for
V
+
V
+
and
V
-
V
-
, we do not know
Vx
V
x
or
Ix
I
x
anywhere on the line. In particular, we do not know
V0
V
0
and
I0
I
0
which would tell us what the apparent impedance is
looking into the line.
Z
in
=Z0=
V
+
+
V
-
I
+
+
I
-
Z
in
Z
0
V
+
V
-
I
+
I
-
(3)
Until we know what kind of impedance the generator is seeing, we
can not figure out how much of the generator's voltage will be
coupled to the line! The input impedance looking into the line
is now a function of the load impedance, the length of the line,
and the phase velocity on the line. We have to solve this
before we can figure out how the line and
generator will interact.
The approach we shall have to take is the
following. We will start at the load end of
the line, and in a manner similar to the one we used
previously, find a relationship between
V
+
V
+
and
V
-
V
-
, leaving their actual magnitude and phase as something
to be determined later. We can then propagate the two voltages
(and currents) back down to the input, determine what the input
impedance is by finding the ratio of (
V
+
+
V
-
V
+
V
-
) to (
I
+
+
I
-
I
+
I
-
), and from this, and knowledge of properties of the
generator and its impedance, determine what the actual voltages
and currents are.
Let's take a look at the load. We again use KVL
and KCL (Figure 2) to match voltages and currents
in the line and voltages and currents in the load:
V
+
ⅇ-ⅈβL+
V
-
ⅇ+ⅈβL=
V
L
V
+
β
L
V
-
β
L
V
L
(4)
and
I
+
ⅇ-ⅈβL+
I
-
ⅇ+ⅈβL=
I
L
I
+
β
L
I
-
β
L
I
L
(5)
Now, we
could substitute
±V
Z
0
±
V
Z
0
for the two currents on the line and
V
L
Z
L
V
L
Z
L
for
I
L
I
L
, and then try to solve for
V
-
V
-
in terms of
V
+
V
+
using
Equation 4 and
Equation 5
but we can be a little clever at the outset, and make our
(complex) algebra a good bit cleaner
Figure 3. Let's make a change of variable and let
s≡L−x
s
L
x
(6)
This then gives us for the voltage on the line (using
x=L−s
x
L
s
)
Vs=
V
+
ⅇ-ⅈβLⅇⅈβs+
V
-
ⅇⅈβLⅇ-ⅈβs
V
s
V
+
β
L
β
s
V
-
β
L
β
s
(7)
Usually, we just fold the (constant) phase terms
ⅇ±ⅈβL
±
β
L
terms in with the
V
+
V
+
and
V
-
V
-
and so we have:
Vs=
V
+
ⅇⅈβs+
V
-
ⅇ-ⅈβs
V
s
V
+
β
s
V
-
β
s
(8)
Note that when we do this, we now have a
positive exponential in the first term
associated with
V
+
V
+
and a
negative exponential
associated with the
V
-
V
-
term. Of course, we also get for
Is
I
s
:
Is=
I
+
ⅇⅈβs+
I
-
ⅇ-ⅈβs
I
s
I
+
β
s
I
-
β
s
(9)
This change now moves our origin to the
load end of the line, and reverses the
direction of positive motion.
But, now when
we plug into
ⅇⅈβs
β
s
the value for
ss at the load
(
s=0
s
0
), the equations simplify to:
V
+
+
V
-
=
V
L
V
+
V
-
V
L
(10)
and
I
+
+
I
-
=
I
L
I
+
I
-
I
L
(11)
which we then re-write as
V
+
Z
0
−
V
-
Z
0
=
V
L
Z
L
V
+
Z
0
V
-
Z
0
V
L
Z
L
(12)
This is beginning to look almost exactly like a
previous chapter. As a reminder, we
solve
Equation 12 for
V
L
V
L
V
L
=
Z
L
Z
0
V
+
−
V
-
V
L
Z
L
Z
0
V
+
V
-
(13)
and substitute for
V
L
V
L
in
Equation 10
V
+
+
V
-
=
Z
L
Z
0
V
+
−
V
-
V
+
V
-
Z
L
Z
0
V
+
V
-
(14)
From which we then solve for the reflection coefficient
Γ
ν
Γ
ν
, the ratio of
V
-
V
-
to
V
+
V
+
.
V
-
V
+
≡
Γ
ν
=
Z
L
−
Z
0
Z
L
+
Z
0
V
-
V
+
Γ
ν
Z
L
Z
0
Z
L
Z
0
(15)
Note that since, in general,
Z
L
Z
L
will be complex, we can expect that
Γ
ν
Γ
ν
will also be a complex number with both a magnitude
|
Γ
ν
|
Γ
ν
and a phase angle
θ
Γ
θ
Γ
. Also, as with the case when we were looking at
transients,
|
Γ
ν
|<1
Γ
ν
1
.
Since we now know
V
-
V
-
in terms of
V
+
V
+
, we can now write an expression for
Vs
V
s
the voltage anywhere on the line.
Vs=
V
+
ⅇⅈβs+
V
-
ⅇ-ⅈβs
V
s
V
+
β
s
V
-
β
s
(16)
Note again the change in signs in the two exponentials. Since
our spatial variable
ss is going in
the opposite direction from
xx, the
V
+
V
+
phasor now goes as
+ⅈβs
β
s
and the
V
-
V
-
phasor now goes as
-ⅈβs
β
s
.
We now substitute in
Γ
ν
V
+
Γ
ν
V
+
for
V
-
V
-
in Equation 16, and for reasons that will
become apparent soon, factor out an
ⅇⅈβs
β
s
.
Vs=
V
+
ⅇⅈβs+
Γ
ν
V
+
ⅇ-ⅈβs=
V
+
ⅇⅈβs+
Γ
ν
ⅇ-ⅈβs=
V
+
ⅇⅈβs1+
Γ
ν
ⅇ-2ⅈβs
V
s
V
+
β
s
Γ
ν
V
+
β
s
V
+
β
s
Γ
ν
β
s
V
+
β
s
1
Γ
ν
2
β
s
(17)
We could have also written down an equation for
Is
I
s
, the current along the line. It will be a good test
of your understanding of the basic equations we are developing
here to show yourself that indeed
Is=
V
+
ⅇⅈβs
Z
0
1−
Γ
ν
ⅇ-2ⅈβs
I
s
V
+
β
s
Z
0
1
Γ
ν
2
β
s
(18)
"This course offers an introduction to solid state device including field effect and bipolar transistors. Properties of transmission lines and propagating E&M waves are also presented. It is […]"