What will that asymptotic value be? One approach
is to write down the following equation.
Which we can re-write as
Now, remembering the infinite sum relationship:
for
(which is
always the case for a reflection
coefficient). We can substitute
for the
terms inside the parentheses in
and we get
We will leave it as an exercise to the reader to show that if we
substitute
,
and finally
into
we
will eventually get:
Look back at
and see if
makes any sense. It should. If we wait long
enough, it is reasonable to expect that any "transmission line"effects should go away, and we would be back to the same
situation we would have if the line was just some wireconnecting the source to the load. In this case, the load
resistor and the source resistor would form a voltage divider,and we would expect the voltage across the load to be determined
by the voltage divider equation. That's all
is saying!
What do we do if we want, say, the voltage
across the load with time? To do this we move up the RHS of thebounce diagram, and count voltage waves as we move across
them. We start out at zero, of course, and do not see anythinguntil we get to 0.5ms. Then we cross the 10V
wave
and we cross the -5V
wave at the same time, so the voltage only goes up to
+5V. Likewise, another 1ms later, we cross both the -2.5V
and the +1.25V
wave, and so the voltage ends up at the 3.75V position
.
V(t) across the load
We can also use the bounce diagram to find the voltage as a
function of position, for some fixed time,
.
Finding v(x) at t=0.75ΜS
To do this, we draw a horizontal line at the time we are
interested in, say 0.75μs. Now, for each position
, we go from the bottom of the
diagram, up to the horizontal line, adding up voltage as wego. Thus for the example: we get what we see in
. For the first half of the line, we cross the
+10V
, but that's it. For the second half of the line we
cross
both the +10V line as well as -5V
wave, and so the voltage drops down to 5V.
V(x) at t = 0.75ΜS
Of particular interest to many of you will be the way in which a
pulse moves down a line and is reflected
etc . This is also quite easy to do with a
reflection diagram, if we simply break the pulse into two waves,one which has a
positive swing at
and another which is a
negative going wave
at
,
where
is the pulse width of the pulse being generated. The way we dothis is suggested in
. We replace the
pulse generator with two battery/switch combinations. The firstcircuit is just like we have seen so far, with a battery equal
to the open circuit pulse height of the generator, and a switchwhich closes at
. The second circuit has a battery with an amplitude of
minus the pulse height, and a switch which
closes at
, the pulse width of the pulse itself.
Simulating a pulse with two batteries and two
switches
By superposition, we can just add these two generators, one
after the other, and see how the pulse goes down theline. Suppose
is 10 volts,
,
,
and
. With the numbers, we find that
.
and
.
Let's assume that the propagation time on the line is still0.5μs to get from one end of the line to the other.
We draw the
bounce
diagram , and launch
two waves , one
which leaves at
has an amplitude of
. The second wave leaves at a time
, later, and has an amplitude of -5V.
Pulse bounce diagram
Now when we want to see what the voltage as a function of time
looks like, we again draw a line up the middle, and add voltagesas we cross them. Here we see, again, no voltage until we cross
the first wave at 0.25μs, which pops us up to +5V. At a
time 0.25μs later however, the -5V wave comes along, and
we go back down to zero. At
, the reflected -1.67V pulse comes along, and so we see
that. Since the source is matched to the line,
and so this is the end of the story
.
V(t) half-way down the line
You can get somewhat more interesting waveforms if you go
someplace where the two pulses at least partially overlap. Let'slook at say,
.
Here is the bounce
diagram.
Finding v(t) near the load
And
here is the voltage waveform we
get.
V(t) near the load
This time the 1.67V pulse gets to us before the +5V pulse has
completely passed, and so we drop from 5V to 3.33V. Then, whenthe -5V wave goes by, we drop down to -1.67V for a little while,
until the +1.67V wave comes along to bring us back tozero.
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?
