On Jun 16, 2:49*pm, Cecil Moore wrote:
On Jun 16, 1:29*am, lu6etj wrote:

For not to work too much, calculations could be reduced to three
resistive Zls, and three lengths of TL. Let me suggest 25, 50 and 100
ohms RLs and 0.5, 0.25 and 0.125 of lambda (to honor Cecil and Roy
papers), Vg voltage = 100 V RMS, Rs 50 ohms, Zo obviously 50 ohms too
(lossless line). Zl=50 ohms and 1/2 lambda TL are for beginners, as
me ;D

Miguel, Owen's equation, Prs=(Vs/2-Vr)^2/Rs, and the following power
density equation are equivalent and yield the same answers for Prs.
However, the power density equation indicates the magnitude of
interference present at Rs and the sign of the interference indicates
whether the interference is destructive or constructive. This general
equation is how optical physicists track the power density in their
light and laser beams.

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.

For all three of the RL=50 ohm examples, Pref=0, the line is matched
and Prs = Prl = 50 watts. There is no question of where the reflected
power goes because Pref=0.

For all three of the 0.125WL examples, since A=90 deg, the
interference term in the power density equation is zero so for the 25
ohm and 100 ohm loads:

Prs = Pfor + Pref = 55.556 watts

When there is no interference, all the reflected power is obviously
dissipated in the source resistor. This matches my "zero interference"
article.

For all four of the other examples, the SWR is 2:1.
Pfor = 50w, Pref = 5.556w, Pload = 44.444w

If the reflected wave arrives with the electric field in phase with
the forward wave,
angle A = 0 degrees so cos(A) = +1.0
The positive sign tells us that the interference is constructive.

Prs = Pfor + Pref + 2*SQRT(Pfor*Pref)*cos(A)

Prs = 50 + 5.556 + 2*SQRT(50*5.556)

Prs = 55.556 + 33.333 = 88.89 watts

In addition to the sum of the forward power and reflected power being
dissipated in Rs, the source is forced to supply the additional power
contained in the 33.333 watt constructive interference term.

So Psource = 100w + 33.333w = 133.33w

Because of the constructive interference, the source not only must
supply the 100 watts that it supplies during matched line conditions,
but it must also supply the constructive interference power of 33.333
watts.

When the reflected wave arrives at Rs 180 degrees out of phase with
the forward wave,
cos(A) = cos(180) = -1.0. The negative sign indicates that the
interference is destructive.

Prs = 50 + 5.556 - 2*SQRT(50*5.556)

Prs = 55.556 - 33.333 = 22.223 watts

The source is forced to throttle back on its power output by an amount
equal to the destructive interference power of 33.33 watts. It is
supplying the 22.223w dissipated in Rs plus the difference in the
forward power and the reflected power.

Psource = Prs + Pfor - Pref = 66.667w

Note that the source is no longer supplying the reflected power. The
destructive interference has apparently redistributed the incident
reflected energy back toward the load as part of the forward wave.

In attempted chart form, here are the Prs values:

line length, 0.5WL, 0.25WL, 0.125WL

25 ohm load, 88.889w, 22.223w, 55.556w

50 ohm load, 50w, 50w, 50w

100 ohm load, 22.223w, 88.889w, 55.556w
--
73, Cecil, w5dxp.com

lets see one case off the top of my head... 100 ohms resistive at 1/4
wave, transformed back to the source results in 25 ohms, pure
resistive. 50 ohm source in series with 25 ohm transformed load gives
100v/75ohms = 1.333A from at the source terminals and 100v*25ohm/(25ohm
+50ohm) = 33.33V. power from the source has to equal power into the
load since the line is lossless so PL = 33.33v*1.333a = 44.44w.
power dissipated in the source resistor =1.333a*(100v-33.33v)=88.88w

no sines, no cosines, no square roots, no Pfor/Pref... so much simpler
using voltage or current and simple transformations.