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Old June 11th 10, 05:18 PM posted to rec.radio.amateur.antenna
Cecil Moore Cecil Moore is offline
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First recorded activity by RadioBanter: Aug 2006
Posts: 572
Default Where does it go? (mismatched power)

On Jun 11, 10:36*am, K1TTT wrote:
i wouldn't call it a 'virtual' impedance, it is a very real impedance.


Yes, it meets the (B) non-dissipative definition of impedance from
"The IEEE Dictionary". It doesn't meet the (A) dissipative definition
and it is not an impedor as it only exists as a V/I ratio. That's why
I call it a virtual impedance. There is no resistor, there is no
inductor, and there is no capacitor. There is only a V/I ratio caused
by something else.

it is the steady state impedance seen by the transmitter at its output
terminals. *


And it is a V/I ratio caused by the superposition of the forward wave
and reflected wave, i.e. it would NOT be the same impedance without
the reflected wave. That fact of physics is undeniable.

When we have an actual impedor, i.e. a resistor plus an inductor or a
capacitor, the voltage/current ratio is caused by the impedor. When we
have a virtual impedance, the cause/effect procedure is reversed and
the impedance is caused by the voltage/current ratio which may (or may
not) contain both forward and reflected values.

For the two definitions of impedance, (A) and (B), given in "The IEEE
Dictionary", cause and effect are reversed.

Let's take an example. Assume a 50 ohm load resistor fed with 1/2WL of
300 ohm lossless line. If we assume a 100 watt (50 ohm) source, the
forward power will be 204 watts and the reflected power will be 104
watts.

100w Source----1/2WL 300 ohm feedline----50 ohm load

In the feedline, a forward power of 204 watts is a forward voltage of
247.3 volts and a forward current of 0.825 amps.

In the feedline, a reflected power of 104 watts is a reflected voltage
of 176.6 volts and a reflected current of 0.589 amps.

The reflected voltage is 180 degrees out of phase with the forward
voltage at the transmitter, so the superposed voltage at the
transmitter is 70.7 volts. The reflected current is in phase with the
forward current at the transmitter so the superposed current is 1.414
amps. Since everything is either in phase or 180 degrees out of phase,
phasor addition is not needed. So I repeat, the impedance seen by the
transmitter is:

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

Z = (247.3 - 176.6)/(0.825 + 0.589)

Z = 70.7/1.414 = 50 ohms, NON-DISSIPATIVE!

It is clear that the superposition of the forward wave with the
reflected wave is the CAUSE of the 50 ohm impedance. That particular
impedance cannot exist without the effects of the reflected wave.

Using a math model for answers to problems for so long that one comes
to believe that the model dictates reality is a major contributor to
the myths and old wives' tales that exist in amateur radio today. It's
time to get back to the basics even if it causes a few old rusty
brains to be put in gear for the first time in decades.
--
73, Cecil, w5dxp.com