On Feb 19, 3:30*pm, Cecil Moore wrote:
Keith Dysart wrote:
* w5dxp wrote:
Pa(R0) = fPa + rPa + 2*(fPa*rPa)cos(180-GA)

Opps, sorry - a typo. That equation should be:

Pa(R0) = fPa + rPa + 2*SQRT(fPa*rPa)cos(180-GA)

Could you expand on why the expression on the right is equal to
the average power dissipated in R0(Rs)? How was the expression
derived?

This is essentially the same as the irradiance-interference
equation from optical physics. It's derivation is covered
in detail in "Optics" by Hecht, 4th edition, pages 383-388.
It is also the same as the power equation explained in detail
by Dr. Steven Best in his QEX article, "Wave Mechanics of
Transmission Lines, Part 3", QEX, Nov/Dec 2001, Eq 12. It
can also be derived independently by squaring the s-parameter
equation: *b1^2 = (s11*a1 + s12*a2)^2

Steven Best's article does have an equation of similar form:

PFTOTAL = P1 + P2 + 2 * sqrt(P1) * sqrt(P2) * cosè

Starting with superposition of two voltages, he derived an
equation based on powers for computing the total power.

So that made sense. But why would adding in the forward
power in the line be useful for computing the power in the
source resistor? And the answer: Pure happenstance. The
value of the source resistor is the same as the value
of the line impedance, so identical currents produce
identical powers.

Perhaps you could carry out the calculations for a slightly
different experiment. Use a 25 ohm source impedance and a
voltage oF 70.7 RMS.

Forward and reverse power for the short circuited load will
still be 100 W, but only 200 W will be dissipated in the
source resistor. It is not obvious to me how to compute the
interference term for this case.

As well, what would be the equivalent expression for the following
example?

Pa(R0) = fPa + rPa + 2*SQRT(fPa*rPa)cos(GA)

Note the 180 degree phase difference between the two
examples. Why that is so is explained below.

* * * *+------+-------------+----------------------+
* * * *| * * * | * * * * * * * * * * * * * * * * * *|
* * * *^ * * * | * Rs * * * * * * * * * * * * * * * |
* * * Is * * * +-/\/\/-+ * * * *1/2 wavelength * *ZLoad
* * 2.828A * * *50 ohm | * * * * 50 ohm line * * * *|
* * * *| * * * * * * * | * * * * * * * * * * * * * *|
* * * *+---------------+-----+----------------------+

The forward power is the same, the source impedance is the same,
but the conditions which cause maximum dissipation in the source
resistor are completely different.

If by "completely different", you mean 180 degrees different,
you are absolutely correct. The 1/2WL short-circuit and open-
circuit results are reversed when going from a voltage source
to a current source.

Why is it not the same expression as previous since the conditions
on the line are the same?

We are dealing with interference patterns between the forward
wave and the reflected wave. In the voltage source example,
the forward wave and reflected wave are flowing in opposite
directions through the resistor. In the current source example,
the forward wave and reflected wave are flowing in the same
direction through the resistor.

I see that now. It works because you were effectively using
superposition to compute the voltage across the source resistor
and then converting this to power using the expression derived
by Steven Best.

But it was happenstance that the voltage across the resistor
was the same as the forward voltage on the line. A very misleading
coincidence.

That results in a 180 degree
difference in the cosine term above. I believe all your other
excellent questions are answered above.
--
73, Cecil *http://www.w5dxp.com