On Jun 28, 7:20 pm, K1TTT wrote:
On Jun 28, 10:50 pm, Keith Dysart wrote:
On Jun 28, 5:11 pm, K1TTT wrote:

On Jun 27, 11:59 pm, Keith Dysart wrote:
I suppose, but then you have to give up on P(t)=V(t)*I(t), generally
considered to be a rather fundamental equation.

very fundamental, and very restricted. only good for one point in
space at one time, and for one pair of voltage and current
measurements... can not be applied to separate waves that are
superimposed, only to the final total voltage and current at the
measurement point at that instant.

True, but others reject it completely.

In your example, the RF energy does seem to disappear and re-appear,
when tracked on a moment by moment basis.

when doing conservation of energy you must include the WHOLE system!
it doesn't work on one section of a transmission line any more than it
works for the infamous undergraduate teaser:

take a refrigerator, put it in a perfectly insulated room, and then
open the doors... what happens to the temperature in the room?

The teaser is amusing, but hardly relevant. In my example, all of
the energy is tracked. Or, I invite you to point out that which was
overlooked. Cecil has not found any and would rather prattle on about
the difference between energy and power than actually understand.

of course its relevant... so what happens to the temperature?

Any energy that might raise the temperature is accounted for in the
source and load resistors, so I still suggest that no energy sources
or sinks have been missed.

Well, it would help if you could actually find and articulate a flaw
inhttp://sites.google.com/site/keithdysart/radio6.

...Keith

that site is rather worthless... you say Vs can be used to get the
time reference for the other signals, but time is a variable, as is
space. you seem to have a snapshot of a bunch of sine waves on an
angular scale, but is that scale time or distance?

Time, of course. I agree, though, it is not as clear as it could
have been. It helps a bit if you look at Cecil’s schematic.

Still, it is complicated and will probably take some effort to
understand.

It would probably be better to start with the step wave example
offered previously in another post and copied below for
convenience:

example
I am not sure where you think there is an error. Perhaps you can
point them out in the following example:
Generator:
- 100V step in to an open circuit
- 50 ohm source impedance
Line:
- 50 ohm
- open circuit
Generator is commanded to produce a step.
This will produce 50 V and 1 A at the line input which will
propagate down the line.
The open end of the line has a reflection co-efficient of 1.0.
Just before the 50 V step reaches the end of the line, the
whole line will be at 50 V and 1 A will be flowing everywhere.
The 50 V step hits the end and is reflected, producing a 50 V
step (on top of the 50V already there) which propagates back
to the generator. In front of the 50 V step, the current is
still 1 A (which provides the charge necessary to produce
the reverse propagating 50 V step. Behind the step, the
current is 0.
When the reverse 50 V step (which is actually a step from
50V to 100V) reaches the generator, the source impedance
matches the line impedance so there is no further reflection.
The line state is now 100V and 0A all along its length.
The settling time was one round-trip.
The generator is still producing the step, so the forward
step voltage wave is still 'flowing' and being reflected so
there is still a reflected step voltage wave, each of 50 V.
Since the generator open circuit voltage is 100 V and the
line voltage is now 100 V, current is no longer flowing
from the generator to the line.
Does this agree with your understanding?
/example

...Keith

this is different than what you claimed before. on june 17th you
claimed:
If(t) = 50/50 = 1a
Ir(t) = 50/50 = 1a
now you say current is no longer flowing.

i will object to you saying the voltage 'wave' is still flowing. The
line is at a constant 100v, there is no current, there can be no em
wave without current AND voltage, therefore voltage can not be
flowing.

I think I detect a bit of a problem originating with definitions.

Start by considering that old stalwart from the text books: the
infinite transmission line. For convenience, the characteristic
impedance is 50 ohms and at the left end is attached a generator
with a 50 ohm source impedance. The transmission line extends
to the right forever, and this time the generator produces a
sinusoid.

Turn on the generator and it starts to produce a sinusoidal wave.
For greater certainty in the description let us say that the
function which describes the wave is:
V(t) = 100 cos(wt)
where w is the ascii excuse for omega, the angular frequency.

Because the line is accepting this signal and transporting it
away from the generator, the voltage distribution on the line
is also a sinusoid, but spatially distributed along the line.

We can visualise this spatially distributed sine wave as
moving along the line and thus we have the term travelling wave.
If we run fast enough along the line we can get to a point
where the signal has not yet arrived and the line is at 0V,
and if we wait, the sine wave will eventually arrive and
we will observe the same sinusoidal spatially distributed
voltage pattern moving along the line.

We can run the experiment with the generator producing other
functions of time and we would observe the same function
spatially distributed along the line and moving.

A generator producing a square wave will produce a square
wave voltage distribution moving along the line.

A generator producing pulses will result in pulses moving
along the line.

A voice signal will result in a complicated voltage pattern
moving along the line.

This holds for any wave, V(t), produced by the generator.

But most lines are not infinite. What happens when the line
is terminated? Most texts will explain that if a section
of line is terminated in its characteristic impedance, that
section will behave as if it was connected to a line that
continued forever. That is, the observations on that
section will be no different than if the line went on and
on and on.

Thus in that section we would observe the moving sine,
square, pulse, or whatever wave, but, we can no longer
run far down the line to find a place where the wave
has not yet arrived, since the line has been truncated.

Now what happens if our section of line is open-circuited?
The wave reflection model allows us to compute that the
reflection coefficient (rc) will be 1. This means that if
a forward travelling voltage wave, Vf(t), encounters the
open circuit, it will be reflected as a reverse wave,
Vr(t).

Mathematically
Vr(t) = rc * Vf(t)

Going back to our waveforms, this is most readily
visualised with pulses where the repetition rate is low
enough that only one pulse is on the line at a time.
The pulse can be seen to travel down the line, hit the
end and travel in the reverse direction back to the
generator. What happens when the pulse gets back to
the generator? Well the generator was specified to
have a source impedance that was the same as the line
characteristic impedance, so the behaviour on the
section of line will be exactly the same as if the
line continued forever. There is no reflection, and
we can forget about the pulse.

The wave reflection model also tells us that the voltage
at any point on the line will be
V(t) = Vf(t) + Vr(t)
where Vf(t) is the voltage from the forward wave at that
point on the line and Vr(t) is the voltage from the
reverse wave at that point on the line.

For pulses, where there is only one pulse on the line
at a time, this is easy to visualise, but for continuous
signals, the voltage distribution on the line is now
the sum of the forward and reverse travelling waves and
for the most part, the forward and reverse travelling
waves are no longer easily visible. For example, with
a sine wave, the resulting voltage distribution on the
line is the time varying, so called, standing wave.

It is important to note that even though the original
travelling waves are no longer visible in the spatial
distribution of the voltage on the line, the travelling
wave continues to be provided by the generator, it
travels to the end of the line where it is reflected and
this reverse wave travels back to the generator where it
behaves as if the line continued forever because the
line is terminated by its characteristic impedance.

Recall that all of this holds for any Vf(t) produced by
the generator. So what happens with that original
sticking point, the step function?

Consider the infinite line again. The step is moving
along the line. If we run far enough ahead, we can
watch the step pass, but behind the step the voltage
is constant. This makes it hard to visualise as a
moving wave because there are no peaks and valleys
to observe, but it is still moving. If in doubt,
just run further down the line to observe the step
pass again.

Now if we terminate the line, we can no longer run
on ahead to find the step, but the wave is still
moving; we just can’t see it because of the constant
nature of the wave after the step.

With the open circuited line, the step function
is reflected and travels as a reverse wave back to
the generator. Once the step reaches the generator,
just as happens for any other V(t) wave, the wave
continues to move; it just is not visible. And the
voltage at any point on the line can be correctly
computed from
V(t) = Vf(t) + Vr(t)

Another way to think of this is to produce the step
wave in a different manner. Start with a pulse wave.
Then increase the pulse width until it approaches
the pulse period. The variation in voltage can still
be observed on the line. Then slowly increase the
width until it is equal to the period. There is now
no longer any variation in the wave to be observed,
but the wave is still moving just as it was when
the width was infinitesimally shorter than the
period.

Another view is to start with a square wave, then
increase the period until it is much longer than
the duration of the experiment. Over the duration
of the experiment, no change in voltage will be
observed, but it is a square wave. One could
perform a Fourier transform on the square wave,
treat all the constituent sine waves as travelling
waves, work out the reflections for each, and sum
them back. Exactly the same results will be
obtained as will be for treating the step as a
moving wave, even though there are no peaks and
valleys to aid the visualisation.

So I hope that I have persuaded you that it is
completely valid to view a step function as a
wave that continues to move (though it is hard
to see) even after the step has passed.

If not, re-read the above to locate the flaw in
my argument.

If you have been convinced, then we can get back
to
If(t) = 50/50 = 1a
Ir(t) = 50/50 = 1a

Consider a generator, 50 ohms, that produces a
50 V step in to 50 ohms. For this
Vf(t) = 50V
The line has a 50 ohm impedance, so
If(t) = 50/50 = 1A

This is the voltage and current that would be
observed on the infinite or terminated line
after the step passes.

With the open circuit line, where the reflection
coefficient is 1, the voltage will be reflected
so that
Vr(t) = 50V
after the reverse step passes and
Ir(t) = 50/50 = 1A

Recalling that at any point on the line
V(t) = Vf(t) + Vr(t)
and
I(t) = If(t) – Ir(t)

we have, after the reverse step has returned to the
generator, everywhere on the line:
V(t) = 50 + 50 = 100V
I(t) = 1 – 1 = 0A

This is exactly as expected for a generator with
an open circuit voltage of 100V and an open circuit
line that has no current flowing.

So the wave reflection model, with continuously
forward and reverse waves, accurately describes
the resulting observations on the line, even for
a step wave.

....Keith