On Jun 16, 4:44*pm, Roy Lewallen wrote:
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.

I would really like to see you prove that assertion.
If Rs = 50 ohms and Z0 = 50 ohms, why would the reflected wave change
the forward power since the reflected wave sees a matched 50 ohm
"load"? Hint: The forward power won't change unless there is a re-
reflection at the source. You have deliberately designed your food-for-
thought example to avoid re-reflections at the source. Lumped-circuit
models seem to have atrophied a lot of brains.

Those are the kind of strange concepts that result from over-
simplification of the math models. When one denies the ExH energy
content of an EM wave (even a reflection) one chooses to follow
oneself down a primrose path. Optical physicists can track their light
wave energy down to the last photon. Why are some RF engineers so
ignorant in the tracking of energy that they assert there is no energy
in a reflected wave?
--
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, 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, 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, 11:10*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, 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 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 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?".

Owen Duffy wrote:
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

I stand corrected. In the special case of Rs = Z0, the steady state
forward voltage and therefore "forward power" become independent of the
transmission line length and load impedance, just as you said and show
in your analysis.

But this isn't true in the general case where Rs isn't equal to Z0. In
the general case, the forward voltage is

Vf = Vfi * exp(-j*theta) / (1 - Gs*Gl*exp(-j*2*thetal))

where Gs = reflection coefficient at the source looking back from the
line = (Zs - Z0)/(Zs + Z0)
Gl = reflection coefficient at the load = (Zl - Z0)/(Zl + Z0)
theta = distance along the line from the source end
thetal = length of the line in radians
Vfi = initial forward voltage = Vs * Z0 / (Zs + Z0)

At the input end of the line, theta = 0 so

Vf1 = Vfi / (1 - Gs*Gl*exp(-j*2*thetal))

and the reverse voltage at the input is

Vr1 = Gl * Vf1 * exp(-j*2*thetal)

Note that although Vf is a function of Gs, reflected voltage Vr is also
a function of Gs and their ratio is not. So VSWR and Pf/Pr don't depend
on Gs. But the general equation for Vf does include line length and Z0
as well as both source and load Z.

In the example I posted earlier (Transmitter power 100 watts, load Z =
50 + j0, line length 1/2 wavelength), no mention was made of the source
impedance, only the power being delivered by the transmitter, which was
the same for both cases since the impedance seen by the transmitter
didn't change. In the first case, where Z0 = Zl = 50 + j0, the forward
power on the line was 100 watts and reverse power was zero. In the
second case, the line Z0 was changed to 200 ohms. This caused an
increase in forward power to 156.25 watts (calculated from VSWR). From
these we can infer that the forward voltage increased by a factor of 2.5
from sqrt(100 * 50) = 70.71 Vrms to sqrt(156.25 * 200) = 176.8 Vrms.
Let's see if this agrees with the equations.

When the line is a half wavelength long, thetal = pi, so the equation
for Vf1 simplifies to:

Vf1 (half wavelength line) = Vfi / (1 - Gs*Gl)

Expanding Vfi, Gs, and Gl results in

Vf1 (half wavelength line) = Vs * (Z0 + Zl)/(2 * (Zs + Zl))

Various combinations of Vs and Zs can give us the 100 assumed watts, but
we don't need to specify any specific values. Leaving Vs and Zs as
unknown constants and Zl as fixed, we can find the ratio for two
different values of Z0:

Vf1 = Vs * (Z01 + Zl)/(2 * (Zs + Zl))
Vf2 = Vs * (Z02 + Zl)/(2 * (Zs + Zl))

Vf2/Vf1 = (Z02 + Zl)/(Z01 + Zl)

In the example, case 1 had Z0 = Z01 = 50; case 2 had Z0 = Z02 = 200, and
Zl was 50 for both. So Vf2/Vf1 = (200 + 50)/(50 + 50) = 2.5, which
agrees with the calculation based on VSWR.

Once again, the general equations for Vf and Vr and therefore Pf and Pr
include Zs, Zl, line length, and line Z0. While one or more of these
terms might disappear in special cases or ratios, they all have to
appear in any equation of general applicability.

Roy Lewallen, W7EL

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