This module continues derivation of Transmission Line equations. It also introduces characteristic impedance.
We need to solve the
telegrapher's
equations ,
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 inthe first and you end up with...well, let's try it and see.
Taking a derivative with respect to
of
Now we substitute in for
from
It should be
very easy for you to derive
Oh, I know you all
love differential
equations! Well, let's take a look at these and just
think for a minute. For
either
or
, we need to find a function that has some rather
stringent requirements. First of all, the function must be ofthe form such that no matter whether we take its second
derivative in space (
) or in time
(
), it must end up differing in
the way it behaves in
or
by no more than just a constant
(
).
In fact, we can be more specific than that. First
must have the same functional form for
both its
and
variation. At most, the two
derivatives must differ only by a constant. Let's try a "lucky"guess and let:
where
is the amplitude of the voltage, and
is some function, of a form yet
undetermined. Well
and
Note also, that
Now, let's take
,
, and
and substitute them into
:
Our "lucky" guess works as a solution as long as
So, what is this
? We don't know yet what its actual functional form
will be, but suppose at some point in time,
, the function looks like
.
F(x) at some point in time
at time
. At point
, the function takes on the value
. Now, let's advance to time
. We look at the function and we see
.
F(x) at a later point in time
at a later
. If
increases from
to
then
will have to increase from
to
in order for the argument of
to have the same value,
. Thus we find
which can be re-written as
where
is the velocity with which the function is moving alongthe 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
for our function, it should be pretty easy to see that
this would have given us a signal moving in the
minus
direction, instead of the plus
direction. Thus we shall denote
the
positive going voltage function and
which is the negative going voltage function. Notice that since
we are taking the
second derivative of
with respect to
, we are free to choose either a
or a
in front of the time argument inside
. Also note that these are our
only choices for a solution. As we know
from Differential Equations, a second order equation has, atmost, two independent solutions.
Since
has the
same differential
equation describing its behavior, the solutions for
must also be of the exact same
form. Thus we can let
represent the current function which goes in the positive
direction, and
represent the negative going current function.
Now, let's take
and
and substitute them into
:
This can be solved for
in terms of
.
where
is called the
characteristic impedance of the transmission line. We will leave it as an exercise to the
reader to ensure that indeed
has units of Ohms. For practice, and understanding
about just how these equations work, the reader should ensurehim/her self that
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
direction and the other moves in the
direction.
Both signals move at a constant velocity
given by
.
The voltage and current signals are related to one another by
the characteristic impedance
, with
