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

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. 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