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Old June 16th 10, 07:48 AM posted to rec.radio.amateur.antenna
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Default Where does it go? (mismatched power)

lu6etj wrote in
:

....
For not to work too much, calculations could be reduced to three
resistive Zls, and three lengths of TL. Let me suggest 25, 50 and 100
ohms RLs and 0.5, 0.25 and 0.125 of lambda (to honor Cecil and Roy
papers), Vg voltage = 100 V RMS, Rs 50 ohms, Zo obviously 50 ohms too
(lossless line). Zl=50 ohms and 1/2 lambda TL are for beginners, as
me ;D


Miguel, I gave an expression for power in your simple source circuit in a
recent post, did you see it. Well here it is again:

Prs=(Vs/2-Vr)^2/Rs where Vs is the o/c source voltage, Vr is the complex
reflected wave voltage equivalent, Rs is the source resistance.

If the reflected wave was always, and entirely dissipated in Rs in that
type of source, then Prs=Vr^2/Zo=Vr^2/Rs... but it isn't, see the above
expression.

You can work one case, three cases, three x three cases, but they are not
as complete as the expression above that shows that in the steady state,
Prs is not simply equal to the 'reflected power'.

But if you want, work the cases and report the results here, it is a
trivial exercise. You will accept the results more readily if you work it
our yourself.

Owen

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Old June 16th 10, 08:28 AM posted to rec.radio.amateur.antenna
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Default Where does it go? (mismatched power)

On 16 jun, 03:48, Owen Duffy wrote:
lu6etj wrote :

...

For not to work too much, calculations could be reduced to three
resistive Zls, and three lengths of TL. Let me suggest 25, 50 and 100
ohms RLs and 0.5, 0.25 and 0.125 of lambda (to honor Cecil and Roy
papers), Vg voltage = 100 V RMS, Rs 50 ohms, Zo obviously 50 ohms too
(lossless line). Zl=50 ohms and 1/2 lambda TL are for beginners, as
me ;D


Miguel, I gave an expression for power in your simple source circuit in a
recent post, did you see it. Well here it is again:

Prs=(Vs/2-Vr)^2/Rs where Vs is the o/c source voltage, Vr is the complex
reflected wave voltage equivalent, Rs is the source resistance.

If the reflected wave was always, and entirely dissipated in Rs in that
type of source, then Prs=Vr^2/Zo=Vr^2/Rs... but it isn't, see the above
expression.

You can work one case, three cases, three x three cases, but they are not
as complete as the expression above that shows that in the steady state,
Prs is not simply equal to the 'reflected power'.

But if you want, work the cases and report the results here, it is a
trivial exercise. You will accept the results more readily if you work it
our yourself.

Owen


Do not argue with me Owen... If you do not want put your numbers you
are free, Roy played theirs, Cecil too, it is a simple and loving
exercise of numeric agreement. No arguments, no algebra, no calculus,
no photons, no Quantum, no Maxwell, no clever and slippery words...
simple and crude numbers... Not more than five minutes. :)

73. Here 04:26 I am go to dream with the little angels... Thanks,
Miguel LU6ETJ
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Old June 16th 10, 01:42 PM posted to rec.radio.amateur.antenna
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Default Where does it go? (mismatched power)

On Jun 16, 1:48*am, Owen Duffy wrote:
Prs is not simply equal to the 'reflected power'.


I don't remember anyone saying that Prs is equal to the reflected
power and of course it is not. What you guys are missing are the
effects accompanying interference from superposition. There is another
mechanism besides the reflection mechanism that can redistribute the
reflected power. The power density equation includes an interference
term that indicates what happens to the energy.

Prs = Pfor + Pref + 2*SQRT(Pfor*Pref)*cos(A)

This equation gives the same answer as your equation but it also
indicates what happens to the energy components. A is the phase angle
between Vfor and Vref. The last term is the *interference* term.

If the interference term is zero, there is no interference and all of
the reflected power is dissipated in Rs.

If the interference term is negative, there exists destructive
interference at Rs and power is redistributed toward the load as
constructive interference. This is technically not a reflection
although the results are the same as a reflection.

If the interference term is positive, there exists constructive
interference at Rs and excess power is dissipated in Rs.

Let's look at your equation, Prs=(Vs/2-Vr)^2/Rs

Vfor = Vs/2, so Prs = (Vfor+Vref)^2/Rs (phasor addition)

Prs = Vfor^2/Rs + Vref^2/Rs + 2*Vfor*Vref*cos(A)/Rs

Prs = Pfor + Pref + 2*SQRT(Pfor*Pref)*cos(A)

I just derived the power density equation (with its interference term)
from your equation. The power density equation reveals the
interference term which tells us exactly where the reflected power
goes.

I learned about destructive and constructive interference at Texas A&M
in the 1950s.
--
73, Cecil, w5dxp.com
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Old June 16th 10, 06:09 PM posted to rec.radio.amateur.antenna
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Default Where does it go? (mismatched power)

Owen Duffy wrote in
:

....
Miguel, I gave an expression for power in your simple source circuit
in a recent post, did you see it. Well here it is again:

Prs=(Vs/2-Vr)^2/Rs where Vs is the o/c source voltage, Vr is the
complex reflected wave voltage equivalent, Rs is the source
resistance.

If the reflected wave was always, and entirely dissipated in Rs in
that type of source, then Prs=Vr^2/Zo=Vr^2/Rs... but it isn't, see the
above expression.


Thinking about this some more, the last paragraph should say:

If the reflected wave was always, and entirely dissipated in Rs in that
type of source, then Prs=a+Vr^2/Zo=a+Vr^2/Rs (where a is some constant
value)... but it isn't, see the above expression.

Owen

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Old June 16th 10, 06:25 PM posted to rec.radio.amateur.antenna
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Posts: 1,169
Default Where does it go? (mismatched power)

Owen Duffy wrote in news:Xns9D9A202A37637nonenowhere@
61.9.191.5:

Owen Duffy wrote in
:

...
Miguel, I gave an expression for power in your simple source circuit
in a recent post, did you see it. Well here it is again:

Prs=(Vs/2-Vr)^2/Rs where Vs is the o/c source voltage, Vr is the
complex reflected wave voltage equivalent, Rs is the source
resistance.

If the reflected wave was always, and entirely dissipated in Rs in
that type of source, then Prs=Vr^2/Zo=Vr^2/Rs... but it isn't, see the
above expression.


Thinking about this some more, the last paragraph should say:

If the reflected wave was always, and entirely dissipated in Rs in that
type of source, then Prs=a+Vr^2/Zo=a+Vr^2/Rs (where a is some constant
value)... but it isn't, see the above expression.


To deal properly with the fact that Vr has magnitude and phase, the above
should we written as:

Prs=|(Vs/2-Vr)|^2/Rs where Vs is the o/c source voltage, Vr is the
complex reflected wave voltage equivalent, Rs is the source resistance.

If the reflected wave was always, and entirely dissipated in Rs in that
type of source, then Prs=a+|Vr|^2/Zo=a+|Vr|^2/Rs (where a is some
constant value)... but it isn't, see the above expression.

Owen



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Old June 16th 10, 07:10 PM posted to rec.radio.amateur.antenna
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Default Where does it go? (mismatched power)

On Jun 16, 12:09*pm, Owen Duffy wrote:
If the reflected wave was always, and entirely dissipated in Rs in that
type of source, then Prs=a+Vr^2/Zo=a+Vr^2/Rs (where a is some constant
value)... but it isn't, see the above expression.


Again, a straw man. To the best of my knowledge, nobody has said that
reflected power is "entirely dissipated in Rs" (except for the special
case when interference doesn't exist.)
--
73, Cecil, w5dxp.com
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