Roy Lewallen wrote:
A high SWR doesn't mean there's "reflected energy" which is going to be
dissipated anywhere, least of all in your transmitter.

Most transmitters will reduce
output power if the SWR gets too high, which tells the transmitter that
the impedance it's seeing is beyond the range for which it's designed.

Not readily apparent is the contradiction between these
two statements above which needs to be resolved. Reflected
energy cannot exist without energy, i.e. without ExB watts.
It is the energy in the reflected waves that is the *cause*
of the impedance "seen" by the transmitter. There is *zero*
dissipation in that virtual impedance so it is NOT a
real resistor - it is a dissipationless resistance.

The impedance seen by the transmitter is not a resistor
or inductor or capacitor, but instead is a *virtual*
impedance *caused by* the magnitude and phase of the
reflected wave with respect to the magnitude and phase
of the forward wave. The impedance seen by the
transmitter is:

Z = (Vfor+Vref)/(Ifor+Iref)

where the voltages and currents are phasors, each with
a magnitude and associated phase. There is NO resistor!
There is NO inductor! There is NO capacitor! There is
no power dissipation! Virtual impedances cannot cause
anything.

It is interesting to note that the very people who
support the virtual impedance seen by a transmitter as
being the cause of the conditions there are the same
people who rail loud and long against a virtual short
being able to cause 100% re-reflection. Why does a
virtual impedance cause things to happen only at a
transmitter but nowhere else?

If there's no "reflected energy", the transmitter will
see the characteristic impedance of the transmission
line, e.g. 50 ohms. So the transmitter CANNOT see any
impedance other than Z0 unless reflected energy is the
cause of the deviation away from Z0.

Depending upon the phase of the reflected energy, all
or some or none of the reflected energy may make its
way into the transmitter. The exact magnitude of
joules/sec making its way into the transmitter is:

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

where 'A' is the phase angle between the E-fields of
EMWave1 and EMwave2 and P1 = E1xB1 and P2 = E2xB2
The last term is known as the "interference term".
If it is negative, it represents destructive
interference. If it is positive, it represents constructive
interference. It should be obvious that 'P' above, can
assume any value between zero and a maximum value so
the amount of reflected energy flowing into the transmitter
can be anything from zero to that maximum value. If the
transmitter is looking into an ideal shorted 1/4WL stub,
the reflected energy flowing into the transmitter will
be zero just as the food-for-thought example demonstrates,
i.e. none of the available power is dissipated in the
transmitter. If the transmitter is looking into an ideal
open-circuit 1/4WL stub, all of the reflected energy will
flow into the transmitter, i.e. all of the available power
will be dissipated in the transmitter. How much depends
upon the relative magnitudes and phases of the forward
and reflected waves.
--
73, Cecil http://www.w5dxp.com