On Jun 25, 7:07*pm, Keith Dysart wrote:
Using superposition, when you add Vrev2(tau) to Vfor1(rho) you get
zero. With zero voltage comes 0 energy transfer.

Completing your above sentence: With zero voltage comes 0 energy
transfer *in the direction of travel of the original waves that were
superposed*. Assuming that you believe in the conservation of energy
principle, what happened to the energy in the two component voltage
waves necessary for their existence before they cancel each other? If
they didn't contain any energy, they would be zero but we know they
are not zero, i.e. they are 35.7 volts each. That original wave energy
is redistributed and *transfered* in the opposite direction, the only
other direction available in a transmission line.

One cannot argue with a forked tongue that the superposed waves never
existed in the first place because that would violate the laws of
physics and superposition.

Do you really need rho^2 to understand what goes on in a transmission
line?

Not using rho^2 is why you are so confused. If you actually cared
where the energy goes, you would be forced to use rho^2 or at least
multiply the superposition component voltages and currents to obtain
the power in the superposition component wavefronts.

In the earlier example, the impedance discontinuity has a physical
voltage reflection coefficient of 0.7143 and a physical power
reflection coefficient of 0.51. If you consider the steady-state power
conditions, you will calculate a virtual power reflection coefficient
of 0.0 and a virtual voltage reflection coefficient of 0.0. Which
reflection coefficient is correct? Obviously, physical trumps virtual
every time.

The 50v source voltage reflected at the 0.7143 reflection coefficient
is 35.7 volts and it exists in a 50 ohm environment. Simple math
yields the power = (35.7)^2/50 = 25.5 watts. Where did the energy in
that 25.5 watt EM wave go? One can obtain the same value by
calculating the current: 1a(0.7143) = 0.7143. Power = 35.7(0.7143) =
25.5 watts.

So you can get by without using rho^2 but to determine where the
energy is going, one needs to at least multiply the EM traveling-wave
voltage by the EM traveling-wave current (or calculate the ExH
Poynting vectors).

In fact, this would be a good application for your instantaneous power
calculations. Where is the energy going that is in the instantaneous
power being reflected by the impedance discontinuity?
--
73, Cecil, w5dxp.com