> But you left off the most important part of what I said here. Try this
> sometime with a 1/2 wave length of line. Take a piece of 75 ohm cable
> and terminate it with a 50 ohm load. Put your watt meter on it and you
> will see a perfect match. No reflected power. Forget about what the
> bird manual says. The meter will see a perfect flat 50 ohms. Also try
> it with other impedance lines as I mentioned here. You will get the
> same results.
There are at least two ways to look at the problem, both are correct
as long as you stay within the boundaries of the model. There
actually is a reflected wave with any mismatch, even in the system
with a 75 ohm line and 50 ohm load. You simply have moved the
meter to a point where the reflected wave **exactly** subtracts
from the forward wave and "compesses" the impedance to 50 ohms.
At that point, the voltage is exactly out-of-phase and current is in
phase. Because of that, reactance is zero and impedance is less
than 75 ohms (it happens to be exactly 50 ohms if the line is
lossless, 1/2 wave long, and the load is 50 ohms.
> It doesn't matter what the characteristic impedance of the line is.
> What matters is what the meter sees. I can take a 1/2 wavelength line
> that has a characteristic impedance of 100 ohms or 400 ohms or 900
> ohms. If I put a 50 ohm load on the end of that line and the other end
> connected to the watt meter the meter will see 50 ohms with no
> reflected power. Impedance transformation.
The other model that works is looking at it like a circuit (except it
doesn't explain why the impedance changes or repeats along the
length of the line), which is exactly what you are doing just above.
> Yes and the reason that I mentioned it is that you earlier stated that
> the reflected power coming from the antenna was re- reflected by the
> matching network in a tube radio but in a solid state radio there was
> no matching network. My point being that if you claim that the
> reflected power gets re- reflected by a matching network in a tube
> radio why would it not also be re- reflected by the matching network
> in the solid state radio.
The impedance of the source is often not correct for a perfect re-
reflection. The result is the PA output device can draw more current
than normal or develop more voltage than normal, or might not
change at all depending on the phase and amplitude of the
reflected wave.
It is incorrect folklore that all reflected power is absorbed in the PA
stage, even with a solid state PA. Under some conditions
dissipation increases with SWR, under some conditions it can
decrease. Of course it is also true that dissipation might not
change at all with high SWR.
We really can just treat it as a load impedance change at the
output device.
With the type of narrow-band systems we use, rise times of the
envelope are limited to many RF cycles. Because of that, we do
not have to worry about transient response of the feedline. We can
consider it an impedance mismatch, or a reflected wave equally
well.
A direction coupler meter that functions without a relatively long
transmission line actually is a bridge circuit that balances current
and voltage, after summing them at radio frequencies. It really
doesn't "know" about reflected waves, it simply detects phase and
level differences from an established ratio of current and voltage at
one point in the line. The ratio of current to voltage sets the
impedance of the directional coupler.
Since the coupler samples phase and level of both current and
voltage, it can read true power even with a mismatch if we simply
subtract reflected power from forward power indicated by the meter.
73, Tom W8JI
W8JI@contesting.com
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