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

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.

On Jun 16, 5:58*pm, K1TTT wrote:
no sines, no cosines, no square roots, no Pfor/Pref... so much simpler
using voltage or current and simple transformations.

Yes, but determining where the reflected energy goes is the title of
this thread. My method shows where the reflected energy goes and yours
does not! That's the entire point.
--
73, Cecil, w5dxp.com

On Jun 16, 11:10*pm, Cecil Moore wrote:
On Jun 16, 5:58*pm, K1TTT wrote:

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

Yes, but determining where the reflected energy goes is the title of
this thread. My method shows where the reflected energy goes and yours
does not! That's the entire point.
--
73, Cecil, w5dxp.com

ah, but it does... it gives the exact same results but without all the
power manipulations. the point is, in sinusoidal steady state you can
calculate the impedance seen by the source and get all those same
results much simpler than trying to track forward and reflected
powers. and if you really want to know the source and load powers
they are easy enough to get as i showed.

On Jun 16, 6:47*pm, K1TTT wrote:
ah, but it does... it gives the exact same results but without all the
power manipulations.

Give us a break. If what you say is true, why didn't you tell us where
the reflected power goes a long time ago? :-)
--
73, Cecil, w5dxp.com

On Jun 16, 2:49*pm, Cecil Moore wrote:
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.

would you care to provide us with the general equation for A given a
complex load impedance and a line length other than a multiple of 90
degrees?

On Jun 16, 6:04*pm, K1TTT wrote:
would you care to provide us with the general equation for A given a
complex load impedance and a line length other than a multiple of 90
degrees?

It's simply the relative phase angle between the forward voltage and
reflected voltage at the source resistor. I trust that you can manage
that without my help.
--
73, Cecil, w5dxp.com

On Jun 16, 11:16*pm, Cecil Moore wrote:
On Jun 16, 6:04*pm, K1TTT wrote:

would you care to provide us with the general equation for A given a
complex load impedance and a line length other than a multiple of 90
degrees?

It's simply the relative phase angle between the forward voltage and
reflected voltage at the source resistor. I trust that you can manage
that without my help.
--
73, Cecil, w5dxp.com

ah, so you do have to calculate the voltages, which would requires two
transformations of the length of the coax plus calculating the
magnitude and angle of the complex reflection coefficient then adding
them all up. still sounds like more work than necessary.

On 16 jun, 20:50, K1TTT wrote:
On Jun 16, 11:16*pm, Cecil Moore wrote:

On Jun 16, 6:04*pm, K1TTT wrote:

would you care to provide us with the general equation for A given a
complex load impedance and a line length other than a multiple of 90
degrees?

It's simply the relative phase angle between the forward voltage and
reflected voltage at the source resistor. I trust that you can manage
that without my help.
--
73, Cecil, w5dxp.com

ah, so you do have to calculate the voltages, which would requires two
transformations of the length of the coax plus calculating the
magnitude and angle of the complex reflection coefficient then adding
them all up. *still sounds like more work than necessary.

Hello all

If I do not make any mistake, my numbers agree with yours Cecil, it is
a pleasure (my procedure was the old plain and simple standard) :)
I know, 1000 examples probe nothing, only one popperianists man can
bring a black swan at any time and falsify the induction :). but now
seems more clear to me you have a predictive model that render
identical numbers of my classical one, at least in basic tests, it is
a step forward to better considerate your efforts and with goodwill
begin to establish basic points of agreement.
I recognize it is a more simple approach the classic method (as said
K1TTT) to me too, but also I think not always one good method or model
it is more convenient to understand another useful view of phenomena.
Phasorial solutions are good, practical and likely complete electric
solutions but in my opinion they do not married very well with other
more general electromagnetic and physics models (wave guides perhaps?
I have not experience); unification has its advantages too (However my
favourite answer was definitely the Roy's one in the fourth post of
the thread, hi hi...)

73 - Miguel - LU6ETJ

PS: Owen do not be upset with me :) most of my available newsgroup
time it is spent in translate to english without flaws that may induce
to misinterpretations (all of you are very demanding with precise
wording and exact definitions), one mistake and I will need three or
four painly translations more to clarify :D

Today I tested your interesting formula with a Half wave 50 ohms TL,
loaded with 100 and 25 ohms respectively. Vs=100 Vrms, Rs=50 ohm give
2:1 VSWR. In both cases then Pf 50 W, Pr=5.5 W, Pnet=44.4 W. Giving
Vf=50 V and Vr=16,6 V aprox. for both loads.
I am using (Pf=Pnet / (1-Rho^2) and Pr=Pnet / ((1/1/Rho^2) -1) formula
which not gives different sign to Vr, in such case applying to
PRs=(((Vs/2)-Vr)^2)/Rs = ((50/2)-16.66)^2/50. I get PRs=22,2 with
both loads because the sign of Vf (always +) from simple Pr and Pf
formulas, changing the sign of Pr render Rs=88,8 W for Prs (OK).

I have in my disk a very descriptive, advisable and friendly article
downloaded from: http://www.ittc.ku.edu/~jstiles/622/...es_package.pdf

In page number 88 there is a agreement with your formula in "Is zo of
aa HF ham tx typycally 50+j0?".

lu6etj wrote in
:

....
PS: Owen do not be upset with me :) most of my available newsgroup
time it is spent in translate to english without flaws that may induce
to misinterpretations (all of you are very demanding with precise
wording and exact definitions), one mistake and I will need three or
four painly translations more to clarify :D

Today I tested your interesting formula with a Half wave 50 ohms TL,
loaded with 100 and 25 ohms respectively. Vs=100 Vrms, Rs=50 ohm give
2:1 VSWR. In both cases then Pf 50 W, Pr=5.5 W, Pnet=44.4 W. Giving
Vf=50 V and Vr=16,6 V aprox. for both loads.

I posted a correction to the formula to properly account for the fact
that Vr is a complex quantity.

The corrected expression is Prs=|(Vs/2-Vr)|^2/Rs . Apologies if you
missed it. Of course, the V quantities are RMS, it is a bit of a botch of
RMS with phase as we often do in thinking... but I see you using the same
shorthand in your calcs. Although we are using RMS values, don't overlook
that where they add (eg Vf+Vr) you must properly account for the phase.

For your cases:

For the 25+j0 load, Vr=-16.67+j0, so Prs=|50--16.67|^2/50=88.9W.

For the 100+j0 load, Vr=16.67+j0, so Prs=|50-16.67|^2/50=22.2W.

Of course, for a 50+j0 load, Vr=0, so Prs=|50-0|^2/50=50.0W.

As you can see, the first two cases have the same |Vr| (though different
phase), the same 'reflected power', and yet Prs is very different.

Consider an extreme case, Zl=1e6, VSWR is extreme, almost infinity, Vref=
50+j0, so Prs=|50-50|^2/50=0.0W. Here, your 'reflected power' is as large
as it gets, but the power dissipated in the source is zero.

The notion that reflected power is simply and always absorbed in the real
source resistance is quite wrong. Sure you can build special cases where
that might happen, but there is more to it. Thinking of the reflected
wave as 'reflected power' leads to some of the misconception.

Owen