On 17 jun, 11:30, Cecil Moore wrote:
On Jun 16, 11:47*pm, lu6etj wrote:

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) *:)

If, instead of a voltage analysis, one does an energy analysis, what
happens to the reflected energy becomes obvious. I posted an earlier
example that nobody solved. So let me repeat it here.

------Z01------+------Z02------

The power reflection coefficient at point '+' is
rho^2 = 0.5. The power transmission coefficient
is (1-rho^2) = 0.5
Pfor1 on the Z01 line is 100w
Pref1 on the Z01 line is 0w
What are Pfor2 and Pref2 on the Z02 line?
What is the SWR on the Z02 line?

The power reflected back toward the source at point '+' is

Pfor1(rho^2) = 100(0.5) = 50w

We know that Pref1 is zero, so

Pref2(1-rho^2) = 50w

Since 1-rho^2 is 0.5, Pref2 must be 100w and

Pref2(rho^2) = 50w

We also know that Pfor1(1-rho^2) = 50w

So we have two 50w waves trying to flow toward the source and two 50w
waves trying to flow toward the load. Do we have 100 watts flowing in
both directions? No, that would be superposition of power which is a
no-no. Using the power density equation indicates what is happening.

Pref1 = 0 = Pfor1(rho^2) + Pref2(1-rho^2) minus total destructive
interference at the Z0-match

Pref1 = 50 + 50 - 2*SQRT(50*50) = 0

Pfor2 = Pfor1(1-rho^2) + Pref2(rho^2) plus total constructive
interference

Pfor2 = 50 + 50 + 2*SQRT(50*50) = 200w

Since Pref2/Pfor2 = 0.5, the SWR on the Z02 line is 5.83:1

Note that we have solved the problem without knowing the value of
Vfor1, Vfor2, and Vref2. But let's assume Z01 is 50 ohms which makes
Z02 equal to 291.5 ohms. Now we can calculate the voltages assuming
Vfor1 at '+' is our phase reference:

Vfor1 = 70.7 volts at zero degrees at '+'
Vref1 = 0
Vfor2 = 241.4 volts at zero degrees at '+'
Vref2 = 170.7 volts at 180 deg at '+'

Vfor1(rho1) = 50v at zero deg
Vref2(tau2) = 50v at 180 deg

Vfor1(tau1) = 120.7v at zero deg
Vref2(rho2) = 120.7v at zero deg

There's the interference, voltage style. Two voltages superpose to
zero toward the source while engaged in total destructive
interference. What happens to the energy components that are
supporting those two voltages? There's only one thing that can happen.
They are obviously not flowing in their original directions so they
must necessarily flow in the only other direction possible. Total
destructive interference toward the source results in total
constructive interference toward the load.

Someone earlier remarked about the fact that when two 50w waves
combine, the result is one 200w wave. That is total constructive
interference in action and that extra energy had to come from
somewhere.
--
73, Cecil, w5dxp.com

Good day:

wave as 'reflected power' leads to some of the misconception.

Ugh! The slippery word again...! Please Owen, remember me what power
definition are you using here and expand the sentence idea.

EM waves cannot exist without ExH energy per unit time per
unit area.

This is what teachers taught me, but when the EM TL waves reachs a
"target" seems the issue arise in the newsgroup.
Puzzle to me why we can (want? :) ) not reconcile those physics
concepts with usual electricity concepts if today they arise from the
same place ultimately?
Certainly in electromagnetism we deal with vectors E,D,B,H,S and in
electricity with scalars V(t), I(t) or phasors, this is for
convenience and simplicity, but I accept we should be able to
understand them in both forms without contradiction.

73 - Miguel Ghezzi - LU6ETJ