Let's look at just one little section of the line, and define
some voltages and currents Figure 1.
For the section of line
Δx
Δ
x
long, the voltage at its input is just
Vxt
V
x
t
and the voltage at the output is
Vx+Δxt
V
x
Δ
x
t
. Likewise, we have a current
Ixt
I
x
t
entering the section, and another current
Ix+Δxt
I
x
Δ
x
t
leaving the section of line. Note that both the voltage and the
current are functions of
time as well as
position.
The voltage drop across the inductor is just:
V
L
=LΔx∂∂tIxt
V
L
L
Δ
x
t
I
x
t
(1)
Likewise, the current flowing down through the capacitor is
I
C
=CΔx∂∂tVx+Δxt
I
C
C
Δ
x
t
V
x
Δ
x
t
(2)
Now we do a
KVL
around the outside of the section of line and we get
Vxt-
V
L
-Vx+Δxt=0
V
x
t
V
L
V
x
Δ
x
t
0
(3)
Substituting
Equation 1 for
V
L
V
L
and taking it over to the RHS we have
Vxt-Vx+Δxt=LΔx∂∂tIxt
V
x
t
V
x
Δ
x
t
L
Δ
x
t
I
x
t
(4)
Let's multiply by -1, and bring the
Δx
Δ
x
over to the left hand side.
Vx+Δxt-VxtΔx=-L∂∂tIxt
V
x
Δ
x
t
V
x
t
Δ
x
L
t
I
x
t
(5)
We take the limit as
Δx→0
Δ
x
0
and the LHS becomes a derivative:
∂∂xVxt=-L∂∂tIxt
x
V
x
t
L
t
I
x
t
(6)
Now we can do a
KCL at the node where the inductor and
capacitor come together.
Ixt-CΔx∂∂tIx+Δxt-Vx+Δxt=0
I
x
t
C
Δ
x
t
I
x
Δ
x
t
V
x
Δ
x
t
0
(7)
And upon rearrangement:
Ix+Δxt-IxtΔx=-C∂∂tVx+Δxt
I
x
Δ
x
t
I
x
t
Δ
x
C
t
V
x
Δ
x
t
(8)
Now when we let
Δx→0
Δ
x
0
, the left hand side again becomes a derivative, and on
the right hand side,
Vx+Δxt→Vxt
V
x
Δ
x
t
V
x
t
, so we have:
∂∂xIxt=-C∂∂tVxt
x
I
x
t
C
t
V
x
t
(9)
Equation 6 and
Equation 9 are so important
we will write them out again together:
∂∂xVxt=-L∂∂tIxt
x
V
x
t
L
t
I
x
t
(10)
∂∂xIxt=-C∂∂tVxt
x
I
x
t
C
t
V
x
t
(11)
These are called the
telegrapher's equations and
they are all we really need to derive how electrical signals
behave as they move along on transmission lines. Note what they
say. The first one says that at some point
xx along the line, the incremental
voltage drop that we experience as we move down the line is just
the distributed inductance
LL times the time derivative of
the current flowing in the line at that point. The second
equation simply tells us that the loss of current as we go down
the line is proportional to the distributed capacitance
CC times
the time rate of change of the voltage on the line. As you
should be easily aware, what we have here are a pair of
coupled linear differential equations in time and
position for
Vxt
V
x
t
and
Ixt
I
x
t
