This module looks at terminated transmission lines.
If, on the other hand, we have a finite line,
terminated with some load impedance, we have a somewhat morecomplicated problem to deal with
.
A finite terminated transmission line
There are several things we should note
before we head off into equation-land
again. First of all, unlike the transient problems we looked atin a
previous chapter , there can
be no more than
two voltage and current
signals on the line, just
and
, (and
and
). We no longer have the luxury of having
,
,
etc. , because we are talking now about a
steady state system . All of the transient
solutions which built up when the generator was first connectedto 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
and a total current function
Note also that until we have solved for
and
, we do not know
or
anywhere on the line. In particular, we do not know
and
which would tell us what the apparent impedance is
looking into the line.
Until we know what kind of impedance the generator is seeing, we
can not figure out how much of the generator's voltage will becoupled 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 usedpreviously, find a relationship between
and
, 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 (
) to (
), and from this, and knowledge of properties of the
generator and its impedance, determine what the actual voltagesand currents are.
Let's take a look at the load. We again use KVL
and KCL (
) to match voltages and currents
in the line and voltages and currents in the load:
and
Doing kirchoff at the end of the line
Change variables! Now, we
could substitute
for the two currents on the line and
for
, and then try to solve for
in terms of
using
and
but we can be a little clever at the outset, and make our
(complex) algebra a good bit cleaner
. Let's make a change of variable and let
S=0 at the load and so the exponentials go away!
This then gives us for the voltage on the line (using
)
Usually, we just fold the (constant) phase terms
terms in with the
and
and so we have:
Note that when we do this, we now have a
positive exponential in the first term
associated with
and a
negative exponential
associated with the
term. Of course, we also get for
:
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
the value for
at the load
(
), the equations simplify to:
and
which we then re-write as
This is beginning to look almost exactly like a
previous chapter . As a reminder, we
solve
for
and substitute for
in
From which we then solve for the reflection coefficient
, the ratio of
to
.
Note that since, in general,
will be complex, we can expect that
will also be a complex number with both a magnitude
and a phase angle
. Also, as with the case when we were looking at
transients,
.
Since we now know
in terms of
, we can now write an expression for
the voltage anywhere on the line.
Note again the change in signs in the two exponentials. Since
our spatial variable
is going in
the opposite direction from
, the
phasor now goes as
and the
phasor now goes as
.
We now substitute in
for
in
, and for reasons that will
become apparent soon, factor out an
.
We could have also written down an equation for
, the current along the line. It will be a good test
of your understanding of the basic equations we are developinghere to show yourself that indeed
Questions & Answers
A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
Adjei
please, I'm a physics student and I need help in physics
Adjanou
chemistry could also be understood like the sexual attraction/repulsion of the male and female elements. the reaction varies depending on the energy differences of each given gender. + masculine -female.
Pedro
A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
