On Sun, 30 Mar 2008 07:43:59 -0700 (PDT)
Keith Dysart wrote:
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.
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http://www.fairpoint.net/~rsparks/Sm...Reflection.pdf
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I am still having difficulty matching these equations with mine. Just
to make sure we are discussing the same problem....
As I tried to understand your problem, I finally realized that it was really my problem. As a result, I redid the spreadsheet. I now label "my formula" as being erroneous, this because it does relate to the current, not the voltage, and must be displaced by 45 degrees.
My formula (labeled erroneous) was the result of the formulas of columns B and C added with trig identities. What I did not realize was that the adding, while correct, rotates the phase. The formula is wrong because it does not recognize the phase shift that had been assumed for line one.
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.
I think the energy adds correctly now. Of course it depends upon how we measure the energy because energy is stored in the transmission line at all times, but the stored energy is only measured across either Rs or the transmission line. We do not measure the energy within the transmission line in this example, which is the sum of the applied power for 90 degrees.
Using averages, the computed powers support the hypothesis,
but when examined with finer granularity, they do not.
I think each degree has the correct power now. I apologize for the errors, and hope that I have them all removed.
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
I hope I have the spread sheet displaying correct values now. Thank you for your careful analysis.
I doubt that this will satisfy your power location concerns because the spread sheet shows more power being delivered to the resistor than is present in the voltage. This is because the impedance of the power equation has changed due to the contribution of the current component. Consider that for columns B and C, the same current flows whether the voltage in B is applied or the voltage in C is applied. This can only happen if the impedance seen by each respective voltage is different. This is interference at work
--
73, Roger, W7WKB