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Old February 26th 08, 01:17 PM posted to rec.radio.amateur.antenna
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First recorded activity by RadioBanter: May 2007
Posts: 492
Default Forward and Reverse Power

On Feb 21, 9:53*pm, Cecil Moore wrote:
Keith Dysart wrote:

Hey Keith, how about this one?

* * * * * * * *Rs * * * * * * Pfor=50w--
* * * * *+----/\/\/-----+----------------------+
* * * * *| * *50 ohm * * * * * *--Pref * * * *|
* * * * *| * * * * * * * * * * * * * * * * * * |
* * * * Vs * * * * * * * * * 45 degrees * * RLoad+j0
* * * 100v RMS * * * * * * * 50 ohm line * * * |
* * * * *| * * * * * * * * * * * * * * * * * * |
* * * * *| * * * * * * * * * * * * * * * * * * |
* * * * *+--------------+----------------------+

The dissipation in the source resistor is:

P(Rs) = 50w + Pref

How can anyone possibly argue that reflected power
is *never* dissipated in the source resistor? :-)


It is time for a more complete analysis. For the sake of an
example let us set Rload to 150 ohms. That makes a nice
voltage RC at the load of 0.5.

Vs = 100 V rms

At the generator end
Vf.g = 50 V rms
Pf.g = 50 W avg

Vr.g = 25 V rms
Pr.g = 12.5 W avg

At the source resistor, before the reflection returns, the
dissipation in the source resistor is
Prs.br = 50 W avg
and after the reflection returns it is
Prs.ar = 62.5 W avg

Yup, it sure looks like that reflected wave is dissipated in the
source resistor since 62.5 = 50 + 12.5, does it not?

But let us examine in more detail...

Vs = 100 V rms
Vs(t) = 141 cos(wt)

Vf.g = 50 V rms
Vf.g(t) = 70.7 cos(wt)

Pf.g = 50 W avg
Pf.g(t) = 50 cos(2wt) + 50

Vr.g = 25 V rms
Vr.t(t) = 35.35 cos(wt-90degrees)

Pr.g = 12.5 W avg
Pr.g(t) = 12.5 cos(2wt) + 12.5

The power in the source resistor, before the reflection returns is
Prs.br = 50 W avg
Prs.br(t) = 50 cos(2wt) + 50

and after the reflection returns
Prs.ar = 62.5 W avg
Prs.ar(t) = 62.5 cos(2wt-53.13degrees) + 62.5

So, while the average powers seem to sum nicely to support the
claim, when the actual power as a function of time is examined
it can be seen that the power in the source resistor is NOT the
sum of its dissipation pre-reflection plus the power from the
reflection.

This solidly disproves the claim that the reflected power is
dissipated in the source resistor.

You may find the spreadsheet at
http://keith.dysart.googlepages.com/...oad,reflection
useful in taking some of the tedium out of the calculations.
And yes, if your anti-virus tools do not like Excel macros, you may
have to invoke your authority over your tools. (They do know who is
boss, do they not?)

...Keith
 
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