Richard Harrison wrote:
The newly minted RF and the twice reflected RF are similar, both having
their volts and amps in-phase. So, the similar RF constituents merge to
have a go at the reflection point.

Wonder why we have protection circuitry in transmitters?
The superposed forward voltage and reflected voltage can damage an
unprotected transmitter. The superposed forward current and reflected
current can cause over heating in an unprotected transmitter.

The transmitter sees whatever impedance it sees and that impedance
can be highly reactive. The superposed voltage can be high or low.
The superposed current can be high or low. The phase between the
superposed voltage and superposed current can have lots of values.

Just a for instance - assume the transmitter is putting out 70.7v
in phase with 1.4a at zero deg. The arriving reflected wave is 50v
at 90 deg and 1.0a at -90 deg. The load seen by the transmitter is
86.6v at 35 deg and 1.72a at -35 deg. Over voltage and over current
exist at the transmitter output. The forward power is 100w and the
reflected power is 50w. The net power being delivered to the reactive
"load" seen by the transmitter is 86.6*1.72*cos(70.4) = 50w.

The math model is trying to dictate reality. It is supposed to be
exactly the opposite. There is no magic barrier that automatically
rejects reflected energy from a transmitter. Reflected energy
arriving at the transmitter can drastically alter the impedance
away from the designed-for load impedance. The transmitter sees
one of the transformed impedances that exists on the SWR circle.

Note that the problem disappears in a matched system where reflected
energy is not allowed to reach the transmitter.
--
73, Cecil http://www.qsl.net/w5dxp


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Cecil, W5DXP wrote:
"The superposed forward voltage and reflected voltage can damage an
unprotected transmitter."

To do so, they would be in-phase and not out-of-phase. Entirely possible
if the transmission line is the right length.

If the reflected volts are in the same phase as the newly munted volts,
which are larger? With a reflection coefficient of 1, and a lossless
line, the open-circuit value of transmitter volts would face some lower
value of line volts on opposite ends of the internal impedance of the
transmitter. Which way does the current flow?

Theory is that it flows from the higher to the lower. That is, from the
transmitter to the line.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:
Cecil, W5DXP wrote:
"The superposed forward voltage and reflected voltage can damage an
unprotected transmitter."

To do so, they would be in-phase and not out-of-phase.

Nope, they don't have to be in-phase. The reflected voltage can
arrive with any phase with respect to the forward voltage. You
trimmed out the example I gave. I will repeat it so you can
study it closer.

Just a for instance - assume the transmitter is putting out 70.7v
in phase with 1.4a at zero deg. The arriving reflected wave is 50v
at 90 deg and 1.0a at -90 deg. The load seen by the transmitter is
86.6v at 35 deg and 1.72a at -35 deg. Over voltage and over current
exist at the transmitter output. The forward power is 100w and the
reflected power is 50w. The net power being delivered to the reactive
"load" seen by the transmitter is 86.6*1.72*cos(70.4) = 50w.

In the above example, the designed for output voltage is 70.7v and
the designed-for output current is 1.4a. The superposed voltage is
86.6v, higher than the designed-for voltage. The superposed current
is 1.72a, higher than the designed-for current. In this example,
we have both over-voltage and over-current occurring *at the same
time* even when the forward voltage and reflected voltage are 90
degrees out of phase. The superposed voltage is not smaller than
the forward voltage until the phase angle between them is in the
neighborhood of 120 degrees. Two superposed voltages with a 90
degree phase angle are *always* larger than either voltage component,
i.e. SQRT(V1^2+V2^2) is always larger than either V1 or V2.

The superposed voltage will be higher than the forward voltage for
any phase angle from zero to 90 degrees. At some phase angle higher
than 90 degrees, in the neighborhood of 120 degrees, the superposed
voltage starts to decrease.

If you sit down and draw these phasors on a piece of paper, you
will discover that you are mistaken. It appears that you are
thinking one-dimensionally instead of two-dimensional phasors.
--
73, Cecil http://www.qsl.net/w5dxp


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Cecil, W5DXP wrote:
"Nope, they don`t have to be in-phase."

Cecil is right. The third side of a triangle can be longer or shorter
than one of the sides.

I should have said, it is entirely possible that the superposed forward
and reflected voltages can damage an unprotected transmitter, The
reflected voltage can add to the forward voltage applied to the
transmitter.

The phase of the reflected voltage should be out of phase with the
transmitter source voltage to maximize the volts across the internal
impedance of the transmitter if we want to damage an unprotected
transmitter.

Best regards, Richard Harrisin, KB5WZI

Richard Harrison wrote:
Cecil, W5DXP wrote:
"Nope, they don`t have to be in-phase."

Cecil is right. The third side of a triangle can be longer or shorter
than one of the sides.

I should have said, it is entirely possible that the superposed forward
and reflected voltages can damage an unprotected transmitter, The
reflected voltage can add to the forward voltage applied to the
transmitter.

The phase of the reflected voltage should be out of phase with the
transmitter source voltage to maximize the volts across the internal
impedance of the transmitter if we want to damage an unprotected
transmitter.

Now assuming, as Richard H. logically did, that we don't want to damage
an unprotected transmitter, let's introduce the concept of an active
match Vs a passive match.

The match at the transmitter output is passive, not active. Consider
the following system:

XMTR---1/4WL 300 ohm feedline---1800 ohm load

What does it really mean when we say that the XMTR "sees" 50 ohms?

It means that an interference pattern is established at the XMTR
that re-reflects (re-directs) the reflected energy back toward the
load. The transmitter has absolutely nothing to do with that reflection.

We can reconfigure the system as follows:

XMTR---1WL 50 ohm feedline--+--1/4WL 300 ohm feedline---1800 ohm load

Now we can see that the match is passive and the XMTR has nothing to
do with the match.

Destructive interference on the XMTR side of '+' eliminates the reflections.
The resulting equal magnitude of constructive interference on the load side
of '+' re-reflects (re-directs) the reflected energy back toward the load.

Optics engineers understand interference. Seems RF engineers might not
fully understand interference. Interference is the cause of the
elimination of reflections in optic systems. Interference is the cause
of elimination of reflections in RF systems, including at antenna tuners.
--
73, Cecil http://www.qsl.net/w5dxp


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Cecil, W5DXP posed a problem in which a transmitter seeking a 50-ohm
load (I suppose) is attached to 1 WL of 50-ohm cable which is attached
to 1/4-wave of 300-ohm feedline which is attached to an 1800-ohm load.

Cecil wrote:
"Destructive interference on the XMTR side of "+" (where 50-ohm cable
meets the 300-ohm Q-matching section) eliminates the reflections."

It seems to me, the product of 50 (the cable Zo) and 1800 (the load Z)
is 90,000. The sq.rt. of 90,000 is 300 (the Q-section impedance). The
numbers are right, so in my opinion, the Q-section converts the
1800-ohms to 50-ohms. This is the 50-ohms likely prescribed for the
transmitter`s load. The 50-ohms presented by the Q-section to the cable
should result in a match and thus there should be no reflected energy in
the 50-ohm cable to cause interference. The reflections should all be in
the 300-ohm Q-section.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:
Cecil, W5DXP posed a problem in which a transmitter seeking a 50-ohm
load (I suppose) is attached to 1 WL of 50-ohm cable which is attached
to 1/4-wave of 300-ohm feedline which is attached to an 1800-ohm load.

Cecil wrote:
"Destructive interference on the XMTR side of "+" (where 50-ohm cable
meets the 300-ohm Q-matching section) eliminates the reflections."

It seems to me, the product of 50 (the cable Zo) and 1800 (the load Z)
is 90,000. The sq.rt. of 90,000 is 300 (the Q-section impedance). The
numbers are right, so in my opinion, the Q-section converts the
1800-ohms to 50-ohms. This is the 50-ohms likely prescribed for the
transmitter`s load. The 50-ohms presented by the Q-section to the cable
should result in a match and thus there should be no reflected energy in
the 50-ohm cable to cause interference. The reflections should all be in
the 300-ohm Q-section.

Exactly, thanks to destructive interference. In S-parameter terms:

b1 = a1*s11 + a2*s12

where b1 is the normalized reflected voltage back toward the source
and is equal to zero because complete destructive interference between
a1*s11 and a2*s12 causes their phasor sum to equal zero.

Those two voltage components cancel to zero toward the source. That
destructive interference is necessary and sufficient for a matched
system. The result is constructive interference toward the load.
--
73, Cecil http://www.qsl.net/w5dxp


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