On Sun, 06 Jun 2004 23:18:37 -0500, Cecil Moore wrote:

Walter Maxwell wrote:
Cecil, I know V2 is the re-reflected voltage, but what I'm trying to persuade
you is that they do NOT superpose to form the forward voltage--they superpose
only to form the standing wave. You've go to accept that the standing wave
voltage is NOT the forward voltage. If you can't come to realize this is the key
to the problem I'm going to have to give up.

I'm sorry, Walt,
Your belief that V2 is a reflected wave is the root of the misunderstanding.
V2 is a re-reflected wave and is therefore forward-traveling toward the load.
V2 is equal to the reflected wave voltage multiplied by the reflection
coefficient. V1 and V2 are traveling in the same direction, toward the load.

Cecil, please read me in the first paragraph. By Steve's own words he says the
re-reflected wave must equal the reflected wave. This means the system is
matched in his account. Therefore, V2, is not the root of any misunderstanding
in my part.

You are still not getting the picture concerning that V1 and V2 cannot be added
to establish the forward wave, as Steve incorrectly believes, they add only to
form the standing wave.

I'm sorry that I didn't think of this earlier that Steve copied his Eq 6 in Part
1 from Johnson, except he placed Vfwd, the forward voltage, instead of E for the
standing wave voltage. Look it up in your Johnson on Pages 99 and 100. He
derives Eq 4.23 (the Eq Steve misunderstands) on Page 99, and expresses it on
Page 100. However, note on Page 98, the beginning of Sec 4.2:

"The equations for E and I along the line can be expressed...."

So Cecil, please understand that this equation does NOT yield the forward
voltage, as Steve believes, which is the root of his misunderstanding throughout
his entire paper.

Concerning tau, I've seen it described in an HP App Note, which I didn't bring
to Michigan, but I've never used it. However, if the power transmission
coefficient is (1 - rho^2) the coefficient is 0.75 for rho = 0.5. Therefore, for
100 w forward only 75 w are delivered. This condition is shown valid
experimentally.

Now let's use tau = 1 + rho as the voltage transmission coefficient. I interpret
this to mean tau x input voltage = forward voltage arriving at a mismatched
load. For a 100 w source at 50 ohms with the same rho as above, we have 70.71 x
1.5 = 106.07 v. But we know that the forward voltage on a matched 50-ohm line
with rho = 0.5 is 81.65 v. Why the difference?

I should have been more aware of the explanation in the HP App Note--there must
be a reason shown there to explain the difference.

Walt