On Mar 29, 7:18 pm, Roger Sparks wrote:
On Sat, 29 Mar 2008 12:45:48 -0700 (PDT)

Keith Dysart wrote:
On Mar 27, 2:06 am, Roger Sparks wrote:
Cecil Moore wrote:
Roger Sparks wrote:
You need to take a look at the spreadsheets.

Roger, in a nutshell, what is the bottom line?

The bottom line in a nutshell? I'll try.

First, I added a note to both spreadsheets indicating that zero degrees
is CURRENT zero degrees. This because the source turns out to be
reactive, with current peak 45 degrees from voltage peak.

http://www.fairpoint.net/~rsparks/Sm...Reflection.pdf

I was not able to discern the derivation of the various equations
though the data in the columns looked somewhat reasonable. Were
the equation really based on the opening paragraph statement of
100 Vrms, or is it scaled to a different source voltage. The sin
functions have an amplituted of 100, which suggests a source of
200 volts or Vrms of 141.4 volts.

I simplified and changed the phase for the equation in column C from '100sin(wt-90) + 100sin(wt-180)' to '100sin(wt+90) + 100sin(-wt)' so that the total voltage in column D better matches with the voltages from my formula in column G.

The equation for column B is just the sine wave for the base wave, showing the voltage resulting to the source resistor from the peak current. The equation in column C is the same (from current) but recognizes that current from the reflected wave will cancel the current from the forward wave coming through Rs. The two waves are traveling in opposite directions so one angle must be labeled with a negative sign so that it will rotate in the opposite direction. In column C, the first term is the reflected wave and the second term is the base wave.

My formula, displayed in column G, uses the applied voltage of 141.4 volts. I thought it would be easier to see how the spreadsheet was built using current to find the voltages. You can see how the results of column D match well with those from column G.

I am still having difficulty matching these equations with mine. Just
to make sure we are discussing the same problem....

The circuit is a voltage source
Vs(t) = 141.4 sin(wt-45)
driving a source resistor of 50 ohms and 45 degress of 50 ohm
transmission line that is shorted at the end.

My calculations suggest that
Vrs.total(t) = 100 sin(wt-90)
Irs.total(t) = 2 sin(wt-90)

which agrees with you column D but not the introduction which
says that the zero current is at 0 degrees. With -100 volts
across the source resistor at 0 degrees, the current should
be -2 amps through the source resistor at 0 degrees; in other
words, a current maximum.

Using superposition, I compute the the contribution of the
source to be
Vrs.source(t) = 70.7 sin(wt-45)
and the contribution from the reflected wave to be
Vrs.reflected(t) = -70.7 sin(wt+45)
which sum to
Vrs.total(t) = 100 sin(wt-90)
as expected.

So while I can construct an expression for Column D, the
contributing values do not agree with those you have
provided in Columns B and C.

Column E follows from Column D and my calculations agree.
And also with Column F.

The spreadsheet addresses the following issues:

Does the traveling wave carry power? Yes. The spreadsheet was built
assuming that power is carried by traveling waves. Because the
resulting wave form and powers seem correct, the underlaying assumption
seems correct.

It was not obvious which columns were used to draw this correlation.

Column E and column F display power. Column F recognizes that if 100 watts is applied continueously (on the average) over an entire 360 degree cycle, the final power applied over time would be 360 * 100 = 36000 watt-degrees. Part of the power comes from the reflection, part comes directly. Obviously, interference is very much at work in this example.

Column E is the power dissipated in the resistor, and Column F
is the integral of Column and represents the total energy which
has flowed in to the resistor over the cycle. It is also the
average energy per cycle.

If you were to extend your analysis to compute the energy
in each degree of the reflected wave and add it to the energy
in each degree of Vrs.source(t) and sum these, you would
find that the instantaneous energy from Vrs.source and
Vrs.reflected does not agree with the instantaneous energy
dissipated in the source resistor. It is this disagreement
that is the root of my argument that the power in the
reflected wave is a dubious concept.

Using averages, the computed powers support the hypothesis,
but when examined with finer granularity, they do not.

However, even if this experiment is consistent with the hypothesis
it only takes one experiment which is not to disprove the hypothesis.

True!

Is power conserved on the transmission line, meaning, can the energy
contained in power be conserved and located over time on the
transmission line? Yes, the spreadsheet was built assuming that power
could be conserved and traced over time so the underlaying assumption
seems correct.

Does interference occur in this example? The spreadsheet was built
assuming that voltage and currents from superpose in a manner consistent
with constructive and destructive interference, so the underlaying
assumption seems correct.

Is power stored in the reactive component for release in later in the
cycle or during the next half cycle? Yes, power is stored on the
transmission line during the time it takes for power to enter the line,
travel to the end and return. The time of wave travel on the
transmission line is related to the value of the reactive component.

Does the direction of wave travel affect the measurement of voltage and
the application of power to a device? Yes. A wave loses energy (and
therefore voltage) as it travels through a resistance. As a result,
power from the prime source is ALWAYS applied across the sum of the
resistance from the resistor AND transmission line.

I am not sure that I would describe this as the wave losing energy,
but rather as the voltage dividing between the two impedances.
If the source resistance was replaced by another transmission,
which could easily be set to provide a 50 ohm impedance, would
you still describe it as the wave losing energy?

It should be OK to think of the voltage dividing between two impedances. The important thing is to consider how the reflected voltage sums with the forward voltage where it is measured. Because the two waves are traveling in opposite directions, the measured voltage is not the voltage applied to either Rs or the transmission line. This is why columns B and C must be added to find the total voltage across Rs.

Whether one needs to add or subtract is more a matter of the
convention
being used for the signs of the values. When Vf and Vr are derived
using
Vtot = Vf + Vr; Itot = If + Ir
one would expect to have to add the negative of Vr to the contribution
from Vs to arrive at the total voltage across the source resistor.

The spreadsheet was
built using this assumption and seems correct. (At times during the
cycle, the forward and reflected waves oppose, resulting in very little
current through the resistor. During those times, the power applied to
the transmission line is much HIGHER because the reflected wave reflects
from the load and source, and merges/adds to the forward wave from the
source.)

I am not convinced. When there is very little current through the
resistor,
there is also very little current into the transmission line. This
suggests to me that the power applied to the transmission line is low.

I don't follow you here. Right, power into the transmission line is low when the current in is low, and it is high when the current is high. It is clear that peak current and peak voltage do not occur at the same time except when measured across the resistor in column D.

Power into the transmission line is low when either the voltage or
the current is low; when either is zero, the power is zero.

Since the highest voltage occurs with zero current and the highest
current occurs with zero voltage, maximum power into the transmission
line occurs when the voltage and current are both medium; more
precisely, when they are both at .707 of their maximum values.

This is just one example, but it seems like the power is accounted for here. We need another example where power can NOT be accounted for.

I suggest it is the same example, but the granularity of the
analysis needs to be increased.

....Keith