Walter Maxwell wrote:
Cecil, let's take it one step further and include current, forward current, that
is, Ifwd.

Ifwd = Isouce x sqrt[1/(1 - rho^2)]

This equation for voltage was wrong. Perhaps, we should resolve that
problem first.

Forward current can be calculated from the square root of (forward
power divided by Z0). So IF2 = SQRT(133.33W/150ohms) = 0.943 amps

Funny thing is that 141.4v*0.943a = 133.33 watts, the known forward
power on the 150 ohm line.
--
73, Cecil http://www.qsl.net/w5dxp



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Walter, W2DU wrote:
"I`ll give you two ways to determine Vfwd."

Walter makes it too easy by using the power relationships, but that`s
the way it is. The forward (incident) wave is opposed by Ro in a
practical line. The reward (reflected) wave is also opposed by Ro (the
surge impedance) of the line.

Power from the transmitter is nearly the same as that delivered to the
load as loss is small and no significant room exists in the line to
store RF.

Power to the load is the difference between forward power and reflected
power.

The voltage at any point on a mismatched line is the sum of the forward
and reflected waves at that point but is merely a manifestatoion of SWR
and has little practical value. At a current or a voltage loop
(maximum), the forward and reflected amps or volts are in-phase. So, if
a line is opened at a loop point, the impedance looking toward the load
is a pure resistance regardless of the nature of the load (see page 37
of "Transmission Lines, Antennas, and Wave Guides" by King, Mimno, and
Wing).

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:
The voltage at any point on a mismatched line is the sum of the forward
and reflected waves at that point but is merely a manifestatoion of SWR
and has little practical value.

All true and completely irrelevant to Dr. Best's article since he was
*never* talking about the net voltage. He is talking about the forward
voltage. Any notion that Dr. Best was talking about net voltage is a
mistaken notion. (I can't believe I'm defending Dr. Best but he said
what he said.)
--
73, Cecil http://www.qsl.net/w5dxp



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On Tue, 08 Jun 2004 13:29:02 -0500, Cecil Moore wrote:

Walter Maxwell wrote:
I'll make it a little easier for you, Cecil. I' ll give you two ways to
determine Vfwd.

OK, here's the example:

100W XMTR---50 ohm line---x----1/2WL 150 ohm line---50 ohm load
PF1=100W-- PF2=133.33W--
--PR1=0W --PR2=33.33W

Way 1. Vfwd = sqrt(Power fwd x Zo)

So Vfwd = sqrt(133.33W x 150ohms) = 141.4 volts (that's what I said)

Way 2. Vfwd = Vsource x sqrt[1/(1- rho^2)]
Now plug rho = 0.5 into the expression for Way 2 and see what comes up.

OK, Vfwd = 70.7V x sqrt[1/0.75] = 81.6 volts (something wrong there)
We know Vfwd has to be 141.4V as calculated above using Ohm's law.

Walt, your two ways don't yield the same value. Perhaps that is the
source of your confusion about what Dr. Best said. Your Way 2 appears
to be a calculation of the voltage delivered to a mismatched load.

Cecil, you've hit the problem on the nose. Until a few posts past I didn't
realize you were using a 150-ohm line. I've been using 50 ohms for the line Zo
terminated with 150 ohms, and you'll find my numbers are correct for that Zo.
Vfwd = 81.6 v is correct for 100 w on a matched 50 ohm line with 3:1 mismatched
load. You'll also get 81.6 using my Way 1.

However, on a 50-ohm line with a 3:1 mismatch and Vfwd = 81.6v (actually it's
81.65 v), the reflected wave is 40.83 v. When the system is matched the phase of
the re-reflected wave is 0° and is also 40.83 v. Therefore, V1 = 81.65 v and V2
= 40.83,v, which means V1 + V2 cannot equal Vfwd. Yet, in the paragraph
preceding Steve's Eq 13 he says that when the phase is 0° Eq. 13 derives
PFtotal. It does not.

Using his Eq 7 V1 yields P1 = 133.33 w, and using his Eq 8 V2 P2 = = 33.33 w.
Using these values of P1 and P2 in his Eq. 12 yields PFtotal = 238 watts, which
is also what we get when plugging V1 and V2 into his Eq 9. We know 238 watts is
incorrect, because it need to be 133.33 w. It makes no sense to plug P1 from Eq
7 into Eq 12 to find PFtotal, because P1 is already PF total.

This clearly proves that Eq 12 is invalid, because Eq 7 and 8 are valid.

I am studying your case with the 150-ohm line, but I haven't yet corralled it.

Walt




re-reflected voltage is also 40.83 v.

On Tue, 8 Jun 2004 13:56:06 -0500, (Richard Harrison)
wrote:

Walter, W2DU wrote:
"I`ll give you two ways to determine Vfwd."

Walter makes it too easy by using the power relationships, but that`s
the way it is. The forward (incident) wave is opposed by Ro in a
practical line. The reward (reflected) wave is also opposed by Ro (the
surge impedance) of the line.

Power from the transmitter is nearly the same as that delivered to the
load as loss is small and no significant room exists in the line to
store RF.

Power to the load is the difference between forward power and reflected
power.

The voltage at any point on a mismatched line is the sum of the forward
and reflected waves at that point but is merely a manifestatoion of SWR
and has little practical value. At a current or a voltage loop
(maximum), the forward and reflected amps or volts are in-phase. So, if
a line is opened at a loop point, the impedance looking toward the load
is a pure resistance regardless of the nature of the load (see page 37
of "Transmission Lines, Antennas, and Wave Guides" by King, Mimno, and
Wing).

Best regards, Richard Harrison, KB5WZI

Thanks, Richard, what you say is true, but problem is in determining the method
to find the forward voltage when the line is mismatched.

Walt

On Tue, 08 Jun 2004 12:58:17 -0500, Cecil Moore wrote:

Walter Maxwell wrote:
On Mon, 07 Jun 2004 22:22:49 -0500, Cecil Moore wrote:
snip.
Please consider the following matched system with a 1/2 second long lossless
transmission line. The physical rho is 0.5. Assume VF1 is a constant 70.7V.

100W XMTR---50 ohm line---x---1/2 second long lossless 150 ohm line---50 ohm load
VF1=70.7V-- VF2--
--VR1 --VR2

V1 is equal to 70.7(1.5) = 106.06V and is a constant value
V2 starts out at zero and builds up to 35.35V
VF2=V1+V2 assuming they are in phase

Here is (conceptually simplified) how VF2 builds up to its steady-state value.
VR2(t+1) is 1/2 of VF2(t) and V2(t+1) is 1/4 of VF2(t)

Time in
Seconds V1 VR2 V2 VF2
0 106.06 0 0 106.06
1 106.06 53.03 26.5 132.56
2 106.06 66.26 33.13 139.19
3 106.06 69.58 34.79 140.85
4 106.06 70.4 35.2 141.26
5 106.06 70.63 35.32 141.38
6 106.06 70.69 35.34 141.39
7 106.06 70.7 35.35 141.4 very close to steady-state

My rounding is a little off but above is a close approximation to what happens.
Note that VF2 is equal to V1 + V2 and converges on the proper value of 141.4V.

Sorry to spoil your fun, Cecil, but 141.4 v is not the forward voltage, it's the
max voltage of the standing wave. This is the same mistake that Steve made. I
asked you to rethink what the forward voltage and you have come up wrong.

OK Cecil, I now understand why 141.4 v is the forward voltage. But there is more
below.

Nope, I haven't, Walt. The maximum voltage of the standing wave equals the
maximum forward voltage plus the maximum reflected voltage. The forward
voltage is 141.4V. The reflected voltage is 70.7V. The maximum standing-wave
voltage is 141.4V + 70.7V = 212.1V. The minimum standing-wave voltage is
141.4V - 70.7V = 70.7V. The VSWR = 212.1V/70.7V = 3:1. Rho = 0.5
SWR = (1+Rho)/(1-Rho) = 1.5/0.5 = 3:1 Everything is perfectly consistent.

So the maximum standing-wave voltage is 212.1V, not 141.4V.

And yes, Cecil I now concur that the standing wave voltage is 212.1 v. There is
still more.

Please reconsider - here's the steady-state powers for the above
matched example:

100W XMTR-----50 ohm line---x---150 ohm line---50 ohm load
PF1=100W-- PF2=133.33W--
--PR1=0 --PR2=33.33W

The forward power is 133.33W and Z0=150 ohms. Last time I looked,
the square root of [133.33W(150 ohms)] was 141.4V.

141.4V^2/150 ohms equals 133.33W. The forward voltage is indeed 141.4V.

Yep, I agree, but as I said above, there is still more.

Until you discover your mental block, whatever it is, this discussion is
not going to progress. The above example is exceptionally simple so your
mistake must also be simple. You, yourself, have used the above example
and always agreed that 100W + 33.33W = 133.33W, i.e. forward power equals
generated power plus reflected power.

I assure you, Cecil, there is no mental block.

Returning to Steve's paper, and as I said in the preceding post, Steve precedes
his Eq 13 stating this equation works under the conditions where V1 and V2 are
in phase. This means that the system is matched and thus the re-reflected
voltage equals the reflected voltage. With forward voltage V1 = 141.4 and
reflected voltage V2 = 70.71, and using these values in his Eqs 7 and 8 and then
plugging the resulting values of P1 and P2 in Eq 13, we get PFtotal = 300 w.
This again proves that Eq 13 is invalid, because P1 from Eq 7 already derived
the total forward power, yet he uses P2 in Eq 13 to obtain the total forward
power.

Do you not see what's wrong here?

Walt

Walter Maxwell wrote:
Cecil, you've hit the problem on the nose. Until a few posts past I didn't
realize you were using a 150-ohm line.

I've only explained and presented it about ten times now.

The impedances are irrelevant, Walt. The equations have to
work no matter what the impedances.

I have presented ASCII schematics of what I was discussing.
Would you be so kind as to do likewise? I am getting sick
and tired of this run-around.
--
73, Cecil http://www.qsl.net/w5dxp



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Walter Maxwell wrote:
This means that the system is matched and thus the re-reflected
voltage equals the reflected voltage.

No! NO! NO! The voltage re-reflected from any physical impedance
discontinuity equals the reflected voltage times the physical reflection
coefficient. That is the S-parameter term s22(a2). s22 is the physical
reflection coefficient. a2 is the reflected voltage incident upon port 2.

If the reflected voltage is 70.7V and the physical voltage reflection
coefficient is 0.5, then the re-reflected voltage is 35.35V. This is
all explained in HP's AN 95-1 on S-parameter analysis.
--
73, Cecil http://www.qsl.net/w5dxp



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On Tue, 08 Jun 2004 16:43:04 -0500, Cecil Moore
wrote:

Walter Maxwell wrote:
This means that the system is matched and thus the re-reflected
voltage equals the reflected voltage.

No! NO! NO! The voltage re-reflected from any physical impedance
discontinuity equals the reflected voltage times the physical reflection
coefficient. That is the S-parameter term s22(a2). s22 is the physical
reflection coefficient. a2 is the reflected voltage incident upon port 2.

If the reflected voltage is 70.7V and the physical voltage reflection
coefficient is 0.5, then the re-reflected voltage is 35.35V. This is
all explained in HP's AN 95-1 on S-parameter analysis.

Cecil, when a mismatched line is matched at the input with a tuner the
reflection coefficient at the matching point in the tuner is 1.0 when the tuner
is matched, thus the re-reflected voltage wave must be equal to that of the
reflected wave. My statements are all based in these conditions.

These conditions are in effect when using an antenna tuner, so any equations
that work in general must work with the tuner setup. When the tuner is matched
the re-reflected voltage is in phase with the source wave. Then, according to
the conditions Steve set forth to use Eq 13 this equation should be viable for
use with conditions set up with the tuner, but it doesn't. You say this equation
is valid, so can you come up with the correct answers using it with a tuner
matching the mismatched line?

Unfortunately, my S-parameter texts are behind in Florida, and I've forgotten
the significance of the 'a' and 'b' terms.

Walter Maxwell wrote:
Cecil, when a mismatched line is matched at the input with a tuner the
reflection coefficient at the matching point in the tuner is 1.0 when the tuner
is matched, thus the re-reflected voltage wave must be equal to that of the
reflected wave. My statements are all based in these conditions.

That is NOT true of an S-parameter analysis, Walt, which is essentially
what Dr. Best is using in his article. In a system with reflections the
S-parameter reflection coefficients are NEVER equal to 1.0. How can I make
you understand that you and Dr. Best are using entirely different systems
of analysis and yours has no bearing on his.

Here's your arguments:

Steve: "It's a plant."

Walt: "No, it's a tree."

Steve: "No, no, it's a plant."

Walt: "No, no, it's a tree."

In the following matched system, the reflection coefficient in Dr. Best's
system of analysis is ALWAYS |0.5| and is NEVER 1.0. This is also true for
an S-parameter analysis. The only difference between Dr. Best's analysis
and an S-parameter analysis is that he didn't normalize his voltages to
SQRT(Z0). I presented that fact a couple of postings ago.

100W XMTR---50 ohm line---x---1/2WL 150 ohm line---50 ohm load.

Nowhere at no time is Dr. Best's reflection coefficient equal to anything
except |0.5|. It is an absolute constant whether the XMTR is off, in the
transient state, or in the steady-state. Please read this paragraph over
and over until it soaks in. In Dr. Best's system of analysis, the only
time that the reflection coefficient is 1.0 is in a system with a short,
open, or pure reactance. In his system of analysis, a reflection coefficient
of 1.0 NEVER occurs at a matched or mismatched impedance discontinuity if
the impedances are not zero or infinite (or purely reactive).

Unfortunately, my S-parameter texts are behind in Florida, and I've forgotten
the significance of the 'a' and 'b' terms.

The HP AN 95-1 ap note is available at:

http://www.sss-mag.com/pdf/hpan95-1.pdf

The S-parameter reflection coefficients are constant and do NOT change as the
reflected power changes. They are the same with no signal applied, or during
the transient state, or during the steady-state. The S-parameter reflection
coefficient is *NEVER* 1.0 when there exists an impedance discontinuity.

You and Dr. Best are NOT USING THE SAME REFLECTION COEFFICIENTS!!!!
You and Dr. Best are NOT USING THE SAME SYSTEM OF ANALYSIS!!!!
Nothing you say about your system of analysis is relevant to the
system of analysis that Dr. Best used.
--
73, Cecil http://www.qsl.net/w5dxp



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