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

On Jun 16, 1:29*am, 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, Owen's equation, Prs=(Vs/2-Vr)^2/Rs, and the following power
density equation are equivalent and yield the same answers for Prs.
However, the power density equation indicates the magnitude of
interference present at Rs and the sign of the interference indicates
whether the interference is destructive or constructive. This general
equation is how optical physicists track the power density in their
light and laser beams.

Ptot = P1 + P2 + 2*SQRT(P1*P2)*cos(A)

where A is the phase angle between the electric fields of wave1 and
wave2. In our case, it will be the phase angle between the forward
wave and the reflected wave at the source resistor.

For our purposes, based on your specifications, we can rewrite the
equation:

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

Pfor is fixed at 50 watts. For a 50 ohm load, Pref=0.
For a 25 ohm or 100 ohm load, Pref=5.556 watts.
Angle A is either 0, 90, or 180 degrees depending upon which example
is being examined.

For all three of the RL=50 ohm examples, Pref=0, the line is matched
and Prs = Prl = 50 watts. There is no question of where the reflected
power goes because Pref=0.

For all three of the 0.125WL examples, since A=90 deg, the
interference term in the power density equation is zero so for the 25
ohm and 100 ohm loads:

Prs = Pfor + Pref = 55.556 watts

When there is no interference, all the reflected power is obviously
dissipated in the source resistor. This matches my "zero interference"
article.

For all four of the other examples, the SWR is 2:1.
Pfor = 50w, Pref = 5.556w, Pload = 44.444w

If the reflected wave arrives with the electric field in phase with
the forward wave,
angle A = 0 degrees so cos(A) = +1.0
The positive sign tells us that the interference is constructive.

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

Prs = 50 + 5.556 + 2*SQRT(50*5.556)

Prs = 55.556 + 33.333 = 88.89 watts

In addition to the sum of the forward power and reflected power being
dissipated in Rs, the source is forced to supply the additional power
contained in the 33.333 watt constructive interference term.

So Psource = 100w + 33.333w = 133.33w

Because of the constructive interference, the source not only must
supply the 100 watts that it supplies during matched line conditions,
but it must also supply the constructive interference power of 33.333
watts.

When the reflected wave arrives at Rs 180 degrees out of phase with
the forward wave,
cos(A) = cos(180) = -1.0. The negative sign indicates that the
interference is destructive.

Prs = 50 + 5.556 - 2*SQRT(50*5.556)

Prs = 55.556 - 33.333 = 22.223 watts

The source is forced to throttle back on its power output by an amount
equal to the destructive interference power of 33.33 watts. It is
supplying the 22.223w dissipated in Rs plus the difference in the
forward power and the reflected power.

Psource = Prs + Pfor - Pref = 66.667w

Note that the source is no longer supplying the reflected power. The
destructive interference has apparently redistributed the incident
reflected energy back toward the load as part of the forward wave.

In attempted chart form, here are the Prs values:

line length, 0.5WL, 0.25WL, 0.125WL

25 ohm load, 88.889w, 22.223w, 55.556w

50 ohm load, 50w, 50w, 50w

100 ohm load, 22.223w, 88.889w, 55.556w
--
73, Cecil, w5dxp.com

Owen Duffy wrote in news:Xns9D995B6088AFDnonenowhere@
61.9.191.5:

....
Clearly, dissipation in Rs is related to Vr, but it is not simply
proportional to the square of Vr as believed by many who lack the basics
of linear circuit theory to come to a correct understanding.

Thinking about this some more, it should say:

Clearly, dissipation in Rs is related to Vr, but the contribution due to Vr
is not simply Vr^2/Rs as believed by many who lack the basics of linear
circuit theory to come to a correct understanding.

Owen

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

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

On Jun 16, 12:09*pm, Owen Duffy wrote:
Clearly, dissipation in Rs is related to Vr, but the contribution due to Vr
is not simply Vr^2/Rs as believed by many who lack the basics of linear
circuit theory to come to a correct understanding.

Come on, Owen, that's just a straw man. Exactly who said that Vref^2/
Rs is dissipated in Rs?
--
73, Cecil, w5dxp.com

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

Owen Duffy wrote:
. . .

I saw the challenge and note the lack of response.

Let me offer a steady state solution.

In the case of a simple source being an ideal AC voltage generator of Vs
and an ideal series resistance Rs of Ro, and that Zo=Ro, for any
arbitrary load, at the source terminals, Vf=Vs/2, Vl=Vf+Vr=Vs/2+Vr, and
the voltage difference across Rs is Vs/2-Vr (noting that Vr is a complex
quantity and can have a magnitude from 0 to Vs/2 at any phase angle),
therfore dissipation in Rs is given by:

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.

Clearly, dissipation in Rs is related to Vr, but it is not simply
proportional to the square of Vr as believed by many who lack the basics
of linear circuit theory to come to a correct understanding.

Roy, is that a solution?

Owen

Sorry, I don't think so. The very first equation, Vf=Vs/2, would be true
only if the load = Ro or for the time between system start and when the
first reflection returns. For the specified steady state and arbitrary
load, Vf would be a function of the impedance seen by the source which
is in turn a function of the line length and load impedance.

I'm not saying an equation relating "reflected power" and source
dissipation can't be written, but it does involve load impedance and
line length and Z0. (Equations written with Vf and Vr as variables tend
to conceal the inevitable dependence of these on line length and Z0 and
load impedance, but the dependence is still there.) And the equation
will show that there isn't any one-to-one correspondence between the two
quantities. For a given source voltage and resistance, the source
resistance dissipation depends only on the impedance seen by the source.
This in turn depends on the length and Z0 of the line and the load
impedance, and can be created by an infinite combination of loads and
lines having different "reflected powers". As I illustrated in an
earlier posting, a constant load with various lines having very
different "reflected powers" can present the same impedance to the
source and therefore result in the same source resistor dissipation.
Likewise, different lines having the same "reflected powers" can result
in different impedances seen by the source and therefore different
dissipations. So the two aren't really related, even though you could
write an equation which contains both terms.

Roy Lewallen, W7EL

Roy Lewallen wrote in
:

....

Sorry, I don't think so. The very first equation, Vf=Vs/2, would be true
only if the load = Ro or for the time between system start and when the
first reflection returns. For the specified steady state and arbitrary
load, Vf would be a function of the impedance seen by the source which
is in turn a function of the line length and load impedance.

Vf means the 'forward wave' voltage equivalent voltage at the source
terminals.

I provide the mathematical development of the case that Vf=Vs/2 where
Rs=Ro=Zo at http://vk1od.net/blog/?p=1028 .

Can you fault that development?

Owen

On Jun 16, 2:49*pm, Cecil Moore wrote:
On Jun 16, 1:29*am, 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, Owen's equation, Prs=(Vs/2-Vr)^2/Rs, and the following power
density equation are equivalent and yield the same answers for Prs.
However, the power density equation indicates the magnitude of
interference present at Rs and the sign of the interference indicates
whether the interference is destructive or constructive. This general
equation is how optical physicists track the power density in their
light and laser beams.

Ptot = P1 + P2 + 2*SQRT(P1*P2)*cos(A)

where A is the phase angle between the electric fields of wave1 and
wave2. In our case, it will be the phase angle between the forward
wave and the reflected wave at the source resistor.

For our purposes, based on your specifications, we can rewrite the
equation:

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

Pfor is fixed at 50 watts. For a 50 ohm load, Pref=0.
For a 25 ohm or 100 ohm load, Pref=5.556 watts.
Angle A is either 0, 90, or 180 degrees depending upon which example
is being examined.

For all three of the RL=50 ohm examples, Pref=0, the line is matched
and Prs = Prl = 50 watts. There is no question of where the reflected
power goes because Pref=0.

For all three of the 0.125WL examples, since A=90 deg, the
interference term in the power density equation is zero so for the 25
ohm and 100 ohm loads:

Prs = Pfor + Pref = 55.556 watts

When there is no interference, all the reflected power is obviously
dissipated in the source resistor. This matches my "zero interference"
article.

For all four of the other examples, the SWR is 2:1.
Pfor = 50w, Pref = 5.556w, Pload = 44.444w

If the reflected wave arrives with the electric field in phase with
the forward wave,
angle A = 0 degrees so cos(A) = +1.0
The positive sign tells us that the interference is constructive.

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

Prs = 50 + 5.556 + 2*SQRT(50*5.556)

Prs = 55.556 + 33.333 = 88.89 watts

In addition to the sum of the forward power and reflected power being
dissipated in Rs, the source is forced to supply the additional power
contained in the 33.333 watt constructive interference term.

So Psource = 100w + 33.333w = 133.33w

Because of the constructive interference, the source not only must
supply the 100 watts that it supplies during matched line conditions,
but it must also supply the constructive interference power of 33.333
watts.

When the reflected wave arrives at Rs 180 degrees out of phase with
the forward wave,
cos(A) = cos(180) = -1.0. The negative sign indicates that the
interference is destructive.

Prs = 50 + 5.556 - 2*SQRT(50*5.556)

Prs = 55.556 - 33.333 = 22.223 watts

The source is forced to throttle back on its power output by an amount
equal to the destructive interference power of 33.33 watts. It is
supplying the 22.223w dissipated in Rs plus the difference in the
forward power and the reflected power.

Psource = Prs + Pfor - Pref = 66.667w

Note that the source is no longer supplying the reflected power. The
destructive interference has apparently redistributed the incident
reflected energy back toward the load as part of the forward wave.

In attempted chart form, here are the Prs values:

line length, 0.5WL, 0.25WL, 0.125WL

25 ohm load, 88.889w, 22.223w, 55.556w

50 ohm load, 50w, 50w, 50w

100 ohm load, 22.223w, 88.889w, 55.556w
--
73, Cecil, w5dxp.com

lets see one case off the top of my head... 100 ohms resistive at 1/4
wave, transformed back to the source results in 25 ohms, pure
resistive. 50 ohm source in series with 25 ohm transformed load gives
100v/75ohms = 1.333A from at the source terminals and 100v*25ohm/(25ohm
+50ohm) = 33.33V. power from the source has to equal power into the
load since the line is lossless so PL = 33.33v*1.333a = 44.44w.
power dissipated in the source resistor =1.333a*(100v-33.33v)=88.88w

no sines, no cosines, no square roots, no Pfor/Pref... so much simpler
using voltage or current and simple transformations.