On Nov 17, 4:03 pm, Cecil Moore wrote:
K7ITM wrote:
Of course it was obvious from the base posting in this thread that the
"waves" must be on a TEM transmission line. But you still haven't
said anything about HOW you launched two distinct waves but got them
to combine into one.

The waves are launched by the external reflection from a
Z0-match and the internal reflection from the load.
--
73, Cecil http://www.w5dxp.com

So the waves are going opposite directions along the line??

K7ITM wrote:
On Nov 17, 4:03 pm, Cecil Moore wrote:
The waves are launched by the external reflection from a
Z0-match and the internal reflection from the load.

So the waves are going opposite directions along the line??

No, all reflections travel toward the source and
therefore, are traveling in the same direction.
Their Poynting vectors are all toward the source.
Given the following Z0-match impedance discontinuity
in a transmission line with the source to the left
and the load to the right:

Z0-match
------Z01---+---Z02------
Pfor1-- Pfor2--
--Pref1=0 --Pref2

The power reflection coefficient is
rho^2 = [(Z02-Z01)/(Z02+Z01)]^2

Pref1 is a combination of two reflected waves

1. P1 = Pfor1(rho^2) "the external reflection from the
Z0-match"

2. P2 = Pref2(1-rho^2) "the internal reflection from
the load"

Pref1 = P1 + P2 + 2*SQRT(P1*P2)cos(A)

Pref1 equals zero at a Z0-match so P1+P2 and A=180 deg.
--
73, Cecil http://www.w5dxp.com

On Nov 18, 7:22 am, Cecil Moore wrote:
K7ITM wrote:
On Nov 17, 4:03 pm, Cecil Moore wrote:
The waves are launched by the external reflection from a
Z0-match and the internal reflection from the load.

So the waves are going opposite directions along the line??

No, all reflections travel toward the source and
therefore, are traveling in the same direction.
Their Poynting vectors are all toward the source.
Given the following Z0-match impedance discontinuity
in a transmission line with the source to the left
and the load to the right:

Z0-match
------Z01---+---Z02------
Pfor1-- Pfor2--
--Pref1=0 --Pref2

The power reflection coefficient is
rho^2 = [(Z02-Z01)/(Z02+Z01)]^2

Pref1 is a combination of two reflected waves

1. P1 = Pfor1(rho^2) "the external reflection from the
Z0-match"

2. P2 = Pref2(1-rho^2) "the internal reflection from
the load"

Pref1 = P1 + P2 + 2*SQRT(P1*P2)cos(A)

Pref1 equals zero at a Z0-match so P1+P2 and A=180 deg.
--
73, Cecil http://www.w5dxp.com

Ah, finally you get around to telling us the setup.

So to get to the conditions in the original posting, we must have a
total power coming into that Z01:Z02 junction exactly equal to the
power leaving it. For example, if Z01 = 50 ohms as implied by your
numbers, and Z02 = 100 ohms, barring stupid math errors, I make out
that the left-to-right power on the Z01 line is 450 watts, and the
right-to-left power on the Z02 line is 56.25 watts, for a total of
506.25 watts. Since you've only accounted for 171 watts, the
remainder must be going off to the right from that junction. Change
the phases, and the power will split differently.

This seems to all agree with standard superpostion. So what the heck
was the point of the original posting in this thread?

Or, why do I even bother reading these things in the first place,
since they all turn out to be pretty boring?

Once again, we see that everything interesting going on in the system
is happening right at the discontinuity where waves arrive and are
reflected. Same in a Wilkinson combiner, same in a "magic T" (which I
suppose the Wilkinson is, if you look at it the right way), same as in
a resistive combining network (if you account for power dissipated in
the resistors), ...

K7ITM wrote:
This seems to all agree with standard superpostion. So what the heck
was the point of the original posting in this thread?

Some posters deny that destructive interference is associated
with zero reflected energy toward the source when a Z0-match
exists. If you are not one of those posters, the thread was
not aimed at you.
--
73, Cecil http://www.w5dxp.com