On Mon, 03 Nov 2008 04:46:57 GMT, Owen Duffy wrote:

Jeff Liebermann wrote in
:

...
In the real world, the power that a transmitter delivers to a non
ideal load is not so simply predicted, and it is entirely possible
that it delivers more power to the mismatched load.

Owen

I beg to differ somewhat. In order for the reflected power to
contribute to the incident power, the reflected power would first be
attenuated by the coax loss. It would then require a substantial

Are you proposing vector addition of power?

Nope. No need to complexicate things. I can safely assume a real 75
ohm load. I can cheat a bit and assume some multiple of 1/2 wave coax
cables, thus eliminating any imaginary contributions from the coax.

Please note that my purpose was to demonstrate that 75 ohm antennas
and 75 ohm coax will work adequately in a 50 ohm system. I think I've
done most of that. Including complex impedances to the calculations
will yield a more accurate result, but the resultant reflected power
that will be added to the forward delivered power, will be LESS than
the results produced by my calculations using only the real part of
the impedances.

mismatch at the transmitter, which is unlikely. However, assuming
there is a mismatch at the source, some of the reflected power will be
sent back to the load (antenna), after getting attenuated by the coax
for a 2nd time. There may be some contribution, but it will very very
very very small.

Let's try some more or less real numbers. I kinda prefer doing
everything in dBm but hams have this thing about using watts...

Of course it doesn't matter, which unit system you use, but if you start
adding 'forward' and 'reflected' power in dBm because it is real
convenient, you have peformed a flawed vector addition of power.
Is that valid?

It's only valid for the level of accuracy with which you are working.
For convenience, perhaps we can just assume that the re-reflected
contribution to the forward power is in phase, thus yielding the
maximum delivered power. Any phase shifts between the two signals
will result in LESS delivered power than the in phase simplistic
calculation. I'm sure the accuracy might be useful for academic
purposes, but my example demonstrated that only 35 mw was added to 33
watts, an error of 0.1%. Of course, that's ridiculous because the
initial measurement of the originating 50 watts is probably only
accurate to 2 significant figures.

Start with a 50 watt xmitter and 20 meters of LMR-240 coax at
0.09dB/meter for an attenuation of 1.8dB.

The power delivered to the antenna is:
50 watts / ((1.8/10)^10) = 50 / 1.5 = 33 watts
The 1.5:1 VSWR reflects 4% of 33 watts for 1.3 watts reflected.

The 1.3 watts is again attenuated by the 1.8dB coax loss resulting in:
1.3 watts / (1.8/10)^10) = 1.3 / 1.5 = 0.87 watts

You start with a limited view of the mismatch, VSWR conveys only one
dimension of a two dimensional mismatch.

Sure. It's good enough for a back of the envelope estimate of how
much power the re-reflected signal can possibly add to the forward
power.

Your treatment of the forward wave and reflected waves as independently
attenuated is an approximation that will lead to significant errors in
some cases.

True. However, as long as I assume a 1.8dB coax cable loss, the
reflected and re-reflected powers will be sufficiently low to be
considered negligible. Including the necessary vector arithmetic to
include the possibility of random coax cable lengths will improve
accuracy, but not affect the result very much.

For example, what percentage of the power at the source end of the line
is lost as heat in 1m of LMR400 at 1MHz with a) a 5+j0 ohm load, and b) a
500+j0 ohm load. The VSWR is approximatly the same in both cases but the
answers are very different, one is almost 100 times the other.

I'll work out the exact numbers tomorrow, but I see your point.
However, please note that I made an effort to use a REALISTIC example
of a typical 2m radio, coax, and coaxial antenna arrangement. Of
course, you can conjure a set of numbers that will result in a
substantially increased calculation error. I can do the same thing if
I take my example and simply reduce the coax cable length to the point
where coax attenuation is dramatically smaller. A 100:1 load
impedance change is not the same as a 1.5:1 impedance change (from 50
to 75 ohms)

Doesn't it stand to reason that as the length of the transmission line
approaches zero, that the power lost transmission in this type of line in
the high voltage low current load scenario is lower than the low voltage
high current load scenario.

Ummmm... you lost me there. I've got a headache tonite. I'll see it
makes more sense tomorrow morning.

Another issue is that the V/I characteristics of a transmitter output
stage is not necessarily (or usually for most ham transmitters) a
straight line, in other words it does not exibit a constant Thevenin
equivalent source impedance with varying loads and the application of
some linear circuit analysis techniques to the output stage are
inappropriate.

I really don't know if that's true for a 2m FM transmitter. I'm not
sure it even matters. The V/I characteristic (slope) is just the
source impedance of the output stage. Whether it's 50 or 75 ohms is
close enough for my simplistic calculation to be accurate without
throwing in non-linearities. The source impedance may be different
for a 50 watt radio, running at 1 watt, but not enough to make a big
difference. If the source impedance were magically 10 times as high,
the 35 milliwatts of re-reflected RF would become 350 mw and still be
a fairly negligible contribution to the delivered 33 watts.

--
Jeff Liebermann
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558