We need to solve the telegrapher's
equations,
∂Vxt∂
x
=−(L∂Vxt∂
t
)
x
V
x
t
L
t
V
x
t
(1)
∂Ixt∂
x
=−(C∂Vxt∂
t
)
x
I
x
t
C
t
V
x
t
(2)
The way we will proceed to a solution, and the way you always
proceed when confronted with a pair of equations such as these,
is to take a spatial derivative of one equation, and then
substitute the second equation in for the spatial derivative in
the first and you end up with...well, let's try it and see.
Taking a derivative with respect to
xx of Equation 1
∂
2
Vxt∂
x
2
2
=−(L∂2Ixt∂
t
∂
x
)
x
2
2
V
x
t
L
t
x
I
x
t
(3)
Now we substitute in for
∂Ixt∂
x
x
I
x
t
from
Equation 2
∂
2
Vxt∂
x
2
2
=LC∂
2
Vxt∂
t
2
2
x
2
2
V
x
t
L
C
t
2
2
V
x
t
(4)
It should be
very easy for you to derive
∂
2
Ixt∂
x
2
2
=LC∂
2
Ixt∂
t
2
2
x
2
2
I
x
t
L
C
t
2
2
I
x
t
(5)
Oh, I know you all
love differential
equations! Well, let's take a look at these and just
think for a minute. For
either
Vxt
V
x
t
or
Ixt
I
x
t
, we need to find a function that has some rather
stringent requirements. First of all, the function must be of
the form such that no matter whether we take its second
derivative in space (
xx) or in time
(
tt), it must end up differing in
the way it behaves in
xx or
tt by no more than just a constant
(
LC
L
C
).
In fact, we can be more specific than that. First
Vxt
V
x
t
must have the same functional form for
both its xx
and tt variation. At most, the two
derivatives must differ only by a constant. Let's try a "lucky"
guess and let:
Vxt=
V
0
fx−vt
V
x
t
V
0
f
x
v
t
(6)
where
V
0
V
0
is the amplitude of the voltage, and
ff is some function, of a form yet
undetermined. Well
∂fx−vt∂
t
=−(vdfx−vtd)
t
f
x
v
t
v
f
x
v
t
(7)
and
∂(−(vfx−vt))∂
t
=v2d
2
fx−vtd
t
2
t
v
f
x
v
t
v
2
t
2
f
x
v
t
(8)
Note also, that
∂
2
fx−vt∂
x
2
2
=d
2
dt
2
fx−vt
x
2
2
f
x
v
t
t
2
f
x
v
t
(9)
Thus, we have
∂
2
Vxt∂
x
2
2
=
V
0
d
2
fx−vtd
t
2
=1LC∂
2
Vxt∂
t
2
2
=v2LC
V
0
d
2
fx−vtd
t
2
x
2
2
V
x
t
V
0
t
2
f
x
v
t
1
L
C
t
2
2
V
x
t
v
2
L
C
V
0
t
2
f
x
v
t
(10)
which works as a solution as long as
So, what is this
fx−vt
f
x
v
t
? We don't know yet what its actual functional form
will be, but suppose at some point in time,
t
1
t
1
, the function looks like
Figure 1.
At point
x
1
x
1
, the function takes on the value
V
1
V
1
. Now, let's advance to time
t
2
t
2
. We look at the function and we see
Figure 2.
If
tt increases from
t
1
t
1
to
t
2
t
2
then
xx will have to increase from
x
1
x
1
to
x
2
x
2
in order for the argument of
ff to have the same value,
V
1
V
1
. Thus we find
x
1
−v
t
1
=
x
2
−v
t
2
x
1
v
t
1
x
2
v
t
2
(12)
which can be re-written as
x
2
−
x
1
t
2
−
t
1
=ΔxΔt≡
v
p
=1LC
x
2
x
1
t
2
t
1
Δ
x
Δ
t
v
p
1
L
C
(13)
where
v
p
v
p
is the velocity with which the function is moving along
the x-axis! (We use the subscript "p" to indicated that what we
have here is what is called the
phase velocity. We
will encounter another velocity called the
group
velocity a little later in the course.)
If we had "guessed" an
fx+vt
f
x
v
t
for our function, it should be pretty easy to see that
this would have given us a signal moving in the
minus xx
direction, instead of the plus xx
direction. Thus we shall denote
V
plus
=
V
+
fx−1LCt
V
plus
V
+
f
x
1
L
C
t
(14)
the
positive going voltage function and
V
minus
=
V
-
fx+1LCt
V
minus
V
-
f
x
1
L
C
t
(15)
which is the negative going voltage function. Notice that since
we are taking the
second derivative of
ff with respect to
tt, we are free to choose either a
1LC
1
L
C
or a
−1LC
1
L
C
in front of the time argument inside
ff. Also note that these are our
only choices for a solution. As we know
from Differential Equations, a second order equation has, at
most, two independent solutions.
Since
Ixt
I
x
t
has the same differential
equation describing its behavior, the solutions for
II must also be of the exact same
form. Thus we can let
I
plus
=
I
+
fx−1LCt
I
plus
I
+
f
x
1
L
C
t
(16)
represent the current function which goes in the positive
xx direction, and
I
minus
=
I
-
fx+1LCt
I
minus
I
-
f
x
1
L
C
t
(17)
represent the negative going current function.
Now, let's take Equation 16 and Equation 14 and substitute them into Equation 1:
V
+
LCfx−1LCt=L
I
+
fx−1LCt
V
+
L
C
f
x
1
L
C
t
L
I
+
f
x
1
L
C
t
(18)
This can be solved for
V
+
V
+
in terms of
I
+
I
+
.
V
+
=LC
I
+
≡
Z
0
I
+
V
+
L
C
I
+
Z
0
I
+
(19)
where
Z
0
=LC
Z
0
L
C
is called the
characteristic impedance
of the transmission line. We will leave it as an exercise to the
reader to ensure that indeed
LC
L
C
has units of Ohms. For practice, and understanding
about just how these equations work, the reader should ensure
him/her self that
V
-
=−(LC
I
-
)≡−(
Z
0
I
-
)
V
-
L
C
I
-
Z
0
I
-
(20)
Note the "subtle" difference here, with a "-" sign in front of
the RHS of the equation!
We've been through lots of equations recently, so
it is probably worth our while to summarize what we know so far.
-
The telegrapher's equations allow two solutions for the
voltage and current on a transmission line. One moves in the
x
x
direction and the other moves in the
−x
x
direction.
- Both signals move at a constant velocity
v
p
v
p
given by Equation 21.
-
The voltage and current signals are related to one another by
the characteristic impedance
Z
0
Z
0
, with Equation 22
