If we are going to try to use phasors on a
transmission line, then we have to allow for spatial variation
as well. This is simple to do, if we just let the phasor be a
function of xx, so we have
V
∼
x
V
∼
x
. How the phasor varies in
xx is one of the things we now have
to find out.
Let's start with the Telegrapher's
Equations again.
∂∂xVxt=-L∂∂xIxt
x
V
x
t
L
x
I
x
t
(1)
∂∂xIxt=-C∂∂xVxt
x
I
x
t
C
x
V
x
t
(2)
For
Vxt
V
x
t
we can now substitute
V
∼
xⅇⅈωt
V
∼
x
ω
t
and for
Ixt
I
x
t
we plug in
I
∼
xⅇⅈωt
I
∼
x
ω
t
. So we get:
∂∂x
V
∼
xⅇⅈωt=-∂∂tⅇⅈωt
x
V
∼
x
ω
t
t
ω
t
(3)
and
∂∂x
I
∼
xⅇⅈωt=-∂∂tⅇⅈωt
x
I
∼
x
ω
t
t
ω
t
(4)
We take the derivative with respect to time, which brings down a
ⅈω
ω
and then we cancel the
ⅇⅈωt
ω
t
from both sides of each equation:
∂∂x
V
∼
x=-ⅈωL
I
∼
x
x
V
∼
x
ω
L
I
∼
x
(5)
and
∂∂x
I
∼
x=-ⅈωC
V
∼
x
x
I
∼
x
ω
C
V
∼
x
(6)
Viola! In one simple motion, we have
completely eliminated the time variable,
tt, from our equations! It is not
really gone, of course, for once we figure out what
V
∼
x
V
∼
x
is, we have to multiply it by
ⅇⅈωt
ω
t
and then take the real part before we can extract once
again, the actual
Vxt
V
x
t
that we want. Nonetheless, insofar as the
telegrapher's equations are concerned,
tt has disappeared from the radar
screen.
To solve these we do just as we did with the
transient problem. We take a derivative with respect to
xx of Equation 5, which
gives us a
∂∂x
I
∼
x
x
I
∼
x
on the right hand side, for which we can substitute
Equation 6, which leaves us with
∂2∂x2
V
∼
x=-
V
∼
x
x
2
2
V
∼
x
V
∼
x
(7)
(- times - is +, but
ⅈⅈ=-1
-1
and so we have a - in front of the
ω2
ω
2
). We then re-write
Equation 7 as
∂2∂x2
V
∼
x+ω2LC
V
∼
x=0
x
2
2
V
∼
x
ω
2
L
C
V
∼
x
0
(8)
The simplest solution to this equation is
V
∼
x=
V
0
ⅇ±ⅈωLCx
V
∼
x
V
0
±
ω
L
C
x
(9)
from which we can then get the actual voltage signal
Vxt=
V
∼
xⅇⅈωt=
V
0
ⅇⅈωt±ωLCx
V
x
t
V
∼
x
ω
t
V
0
±
ω
t
ω
L
C
x
(10)
Note that we could factor out an
ⅇⅈωLC
ω
L
C
, from the exponent, which, since it is just a
constant, we could include in
V
0
V
0
(and call it
V
0
'
V
0
'
, switch the order of
xx
and
tt, and write
Equation 10 as
Vxt=
V
0
'
ⅇⅈx±1LCt
V
x
t
V
0
'
±
x
1
L
C
t
(11)
which looks a lot like the "general"
fx±vt
f
±
x
v
t
solution we were talking about
earlier!
The number
ωLC
ω
L
C
is special. It is usually represented with a Greek
letter ββ and is called the
propagation coefficient. Thus we have
Vxt=
V
0
ⅇⅈωt±βx
V
x
t
V
0
±
ω
t
β
x
(12)
As previously, a point on the wave of constant phase requires that the
argument inside the parenthesis remains constant. Thus if
V
x
1
t
1
V
x
1
t
1
is going to equal
V
x
2
t
2
V
x
2
t
2
(
i.e. what was at point
x
1
x
1
at
t
1
t
1
is now at
x
2
x
2
at time
t
2
t
2
it must be that
ω
t
1
±β
x
1
=ω
t
2
±β
x
2
±
ω
t
1
β
x
1
±
ω
t
2
β
x
2
(13)
or
x
2
-
x
1
t
2
-
t
1
=ΔxΔt=±ωβ=±ωωLC=±1LC≡
v
p
x
2
x
1
t
2
t
1
Δ
x
Δ
t
±
ω
β
±
ω
ω
L
C
±
1
L
C
v
p
(14)
Which one again, defines the
phase velocity of the
wave. Other relationships to keep in mind are
β=2πλ
β
2
λ
(15)
λ=
v
p
f=ωβω2π=2πβ
λ
v
p
f
ω
β
ω
2
2
β
(16)
The first comes from the fact that the wave varies in
xx as
ⅇⅈβx
β
x
. Thus when
x=γ
x
γ
, the wavelength,
βγ
β
γ
just increases by
2π
2
, to get the phasor to go through one full
rotation. Note also, as before, the choice of the minus sign in
the ± in
Equation 12 represents a wave
going in the
+x
x
direction, while the choice of the + sign will
give a wave going in the
-x
x
direction. Clearly, by starting out taking the
x-derivative of the equation for
Ixt
I
x
t
we would end up with
Ixt=
I
0
ⅇⅈωt±βx
I
x
t
I
0
±
ω
t
β
x
(17)
Let's consider the two phasors then, and define the voltage
phasor associated with the positive going voltage wave as
V
∼
plus
x=
V
+
ⅇ-ⅈβx
V
∼
plus
x
V
+
β
x
(18)
and the negative voltage phasor as
V
∼
minus
x=
V
-
ⅇⅈβx
V
∼
minus
x
V
-
β
x
(19)
We should keep in mind that both
V
+
V
+
and
V
-
V
-
can be, and probably are, complex numbers. (From now
on we will drop the little ~ over the variables because its very
tedious to get it to show up with this word processor. You will
just have to keep in mind that any variable we do not explicitly
put inside absolute value markers (
i.e.
|
V
+
|
V
+
) is going to be, in general, a complex number). We
will, of course, have similar expressions for the positive and
negative going current waves.
Let's consider the positive going current and
voltage waves, and plug them into Equation 5.
∂∂x
V
+
ⅇ-ⅈβx=-ⅈωL
I
+
ⅇ-ⅈβx
x
V
+
β
x
ω
L
I
+
β
x
(20)
The x-derivative brings down a
-ⅈβ
β
, the
ⅇ-ⅈβx
β
x
's cancel, and we have
V
+
=ⅈωLⅈβ
I
+
V
+
ω
L
β
I
+
(21)
But, since
β=ωLC
β
ω
L
C
we have
V
+
=LC
I
+
≡
Z
0
I
+
V
+
L
C
I
+
Z
0
I
+
(22)
as we had before.
So, what has changed? Not much from the case of
transients on a line. We will now assume we have a
steady state problem. This means we turned
on the generator a long time ago. We assume that it has been
connected to the line long enough so that all transient behavior
has died away, and that voltages and currents are not changing
any more (except oscillating at frequency
ωω, of course).
If the line is semi-infinite (or matched with a
load equal to
Z
0
Z
0
) Figure 1 then it is pretty obvious that
V
+
=
Z
0
Z
0
+
Z
g
V
g
V
+
Z
0
Z
0
Z
g
V
g
(23)
where
Z
g
Z
g
is the source impedance, and
V
g
V
g
is the source voltage phasor.
