Cecil Moore wrote:
Jim Kelley wrote:

Your expectation of obtaining 400 watts from two 100 watt radios hints
at a possible error in your understanding regarding simultaneously
superposing voltages and currents. Just a thought.


Put two 70.7 vdc batteries in series with a 50 ohm load.
Switch in one battery at a time. Each battery will
supply 100 watts. Now switch in both batteries. They
will supply a combined 400 watts. The batteries are
required to supply the extra constructive interference
energy assuming each battery maintains its constant
70.7 vdc voltage output.

This is what has to happen when two coherent RF voltages
are superposed in phase. I just don't think an IC-706
will do that, i.e. it is not a linear device, thus
violating the requirements for superposition. In phase
superposition would require that:

Vtot = V1/_0 deg + V2/_0 deg = |V1|+|V2|

Ptot = P1 + P2 + 2*SQRT(P1*P2)cos(0)

Ptot = 100w + 100w + 2*SQRT(100w*100w) = 400w

Presupposition of linearity of any amateur radio
transmitter without proof is foolish as is the
presupposition that a ten cent resistor or a $10
resistor will gobble up all the reflected energy.

Holy cow. Happy Good Friday by the way.

A 100 watt source is something which ostensibly produces a maximum of
70.7 volts and a maximum 1.414 amps into a 50 ohm load. There are two
ways to superpose the outputs of 100 watts sources such as this. One
way is to superpose voltages by arranging them in series. The total
output voltage available would then 141.4 volts. But doing so does
not allow the source to produce current beyond some limit. Maximum
current would still ostensibly be 1.414 amps. To increase the
available current one could arrange the sources in parallel and
basically 'superpose' the currents. However this does not increase
the available voltage. That is still ostensibly 70.7. volts. Total
current from the two sources in parallel into a 50 ohm load would now
be 2.828 amps.

A car battery does not have a 100 watt limit, or in particular, a
1.414 amp limit. When you put two batteries in series you are more of
less superposing the voltages, but you aren't superposing currents.
Since the battery doesn't have a 1.414 amp current limit, the
batteries can each now produce 141.4 volts divided by 50 ohms worth of
current. Since they are arranged in series, each battery will produce
2.828 amps of current. It isn't a superposition thing. Put the
batteries in parallel as a check on your notion of superposition.

Your numbers highlight the problem using irradiance equations with
power terms in them. You're getting the wrong answer because power
doesn't interfere, so it's pretty ridiculous to put it into an
interference equation in the first place. Look at the risk you run
doing it. There is nothing wrong with the proper use of superposition
and interference. Given that, what do you think the problem might be?

73, Jim AC6XG