Transmission line engineers came up with bounce diagrams to keep track of all the waves bouncing back and forth on the line.
Now this
new
will head back towards the load and ...Hmmm... things are going
to get kind of messy and complicated. Fortunately for us,transmission line engineers came up with a scheme for keeping
track of all of the waves bouncing back and forth on theline. The scheme is called a
bounce diagram . A
bounce diagram consists of a horizontal distance line, whichrepresents distance along the transmission line, and a vertical
time axis, which represents time since the battery was firstconnected to the line. Just to keep things conceptually clear,
we usually first start out by showing the line, the battery, theload and a switch, S, which is used to connect the source to the
line. It doesnt hurt to make a little sketch like
, and write down the length of the line,
and
,
along with the source and load resistances. Now we draw thebounce diagram, which is shown in
Transient problem
A "bounce diagram"
Normally, you would not put the formula for
and
by 0 and
in the diagram, but
rather their values. This will become clear when we do anexample. The next thing we do is calculate
and draw a straight line on the bounce diagram (nominally at a
slope of
)
which will represent the initial signal going down the line. Wemark a
on the vertical axis to show how long it takes for the wave to
reach the end of the line
.
Diagram with first wave
Once the initial wave hits the load, a second, reflected wave
is sent back the other way. So we add it to the bounce
diagram. This is shown in
. Since all of
the waves move with the same phase velocity, we should becareful to draw all of the lines with the same slope. Note that
the time when the reflected wave hits the generator end is atotal round trip time of
. (This simple concept is one which students often
forget come test time, so be forewarned!)
Adding the first reflected wave
We saw that the next thing that happens is that another wave is
reflected from the generator, so we add that to the bouncediagram as well. This is shown in
.
The third wave
Finally, one last wave, as we are almost bounced right off the
diagram, as shown in
!
And the fourth
OK, so we've got a bounce diagram, so what? Having the diagram
is only part of the solution. We still have to see what goodthey are. Let's do a numerical example, as it is maybe a little
more illustrative, and certainly will be easier to write outthan all these ratios all the time. We will just pick some
typical numbers, and then work out the answers. Let's let
,
,
and
.
The line will be 100m long, and
.
A numerical example
First we calculate the reflection coefficients
and
The initial voltage signal
is
and the propagation time is
So we draw the bounce diagrams seen in
.
The bounce diagram
Now, here's how we use a bounce diagram, once we have
it. Suppose we want to know what
, the voltage as a function of time, would look like
half-way down the line. We draw a vertical line at the place weare interested in (the dotted line in
) and
then just go up along the line, adding voltage to whatever wehad before whenever we cross one of the "bouncing"
signal lines. Thus for the line as shown we would have for
what we see in
.
V(t) at 50m down the line
For the first 0.25μs we have no voltage, because
has not reached the half-way point yet. The voltage then jumps
to +10V when
comes by. It stays like that until the -5V
comes by 0.5μs later. The voltage then remains constant at 5V
until the -2.5V
comes along to drop the total voltage down to only 2.5
volts. When
comes along, it has been switched back to a positive voltage
wave by the negative load reflection coefficient, and so now thevoltage jumps back up to 3.75V. It will keep oscillating back
and forth until it finally settles down to some asymptoticvalue.
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
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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?
