Here's an analysis of the input power and energy of the system similar
to the one described in my "Second analysis: +0.5 input reflection
coefficient" Please refer to that posting when reading this. I'm going
to make just one change to the system in that analysis: I'll use the 5
wavelength line of the SPICE analysis. This has absolutely no effect on
any of the values shown in the analysis except where I gave periods of
time when various equations were valid (e.g. t = 2*pi/w) -- those will
be five times greater. The effect of this change is to allow use of the
SPICE output as a visual aid. When speaking of the SPICE plot, I'll be
referring to http://eznec.com/images/TL2_input.gif, which is the voltage
at the input end of the line.

For this analysis I'm going to let f = 1 Hz, so w = 2*pi Hz.

From the voltage analysis and the SPICE plot, the initial voltage at
the input of the line is sin(wt). So the voltage across the input
resistor is 3 * sin(wt) (+ toward the source), and the current flowing
into the line is (3 * sin(wt)) / 150 = 20 * sin(wt) mA. The average
power being delivered to the line is Vin(rms) * Iin(rms) (since the
voltage and current are in phase) = (0.7071 v. * 14.14 mA) = 10 mW.
Since the line initially presents an impedance of Z0, this should also
be Vin(rms)^2 / Z0 or Iin(rms)^2 / Z0. Let's see: Vin(rms) = 0.7071, so
Vin(rms)^2 / Z0 = 0.5 / 50 = 10 mW; Iin(rms) = 0.01414 A, so Iin(rms)^2
* Z0 = 0.0002 * 50 W = 10 mW. Ok.

The input voltage remains unchanged until the first reflected wave
returns at t = 10 sec (remember, we're using a 5 wavelength line and
frequency of 1 Hz). So during that time we put 10 mW * 10 sec = 0.1
joule of energy into the line. Note that I could have calculated the
instantaneous power for each instant, then integrated it over the first
10 seconds to get the total energy. The result would be identical.

During that first 10 seconds, the resistor is dissipating an average of
Vr(rms) * Ir(rms) = 2.121 * 14.14 = 30 mW. The source is producing
Vs(rms) * Is(rms) = 2.828 * 14.14 = 40 mW. So here's the story so far:

For the first 10 seconds (one round trip):
The source produces 40 mW, for a total of 400 mj (millijoules)
The resistance dissipates 30 mW, for a total of 300 mj
The line gets 10 mW, for a total of 100 mj

All the power and energy is totally accounted for, so far. The sum of
power and energy delivered to the resistor and line equals the power and
energy produced by the source.

As you can see from the SPICE output or the voltage analysis, the input
voltage jumps to 2.5 volts at t = 10 seconds and stays there for the
next 10 seconds. During that time, the voltage across the resistor is (4
- 2.5) * sin(wt), so the current drops to 1.5 sin(wt) / 150 = 10 sin(wt)
mA which has an RMS value of about 7.071 mA. Using the same methods as
before,

For the next 10 seconds:
The source produces 20 mW, for a total of 200 mj
The resistance dissipates 7.50 mW, for a total of 75 mj
The line gets 12.5 mW, for a total of 125 mj

We've again accounted for all the power and energy, with no need to
invoke any kind of power waves.

Next the input voltage goes to 3.25 volts, so the current drops to 5 *
sin(wt) mA and

For the next 10 seconds,
The source produces 10 mW, for a total of 100 mj
The resistance dissipates 1.875 mW, for a total of 18.75 mj
The line gets 8.125 mW, for a total of 81.25 mj

There's something interesting about this energy flow. Notice that the
amount of power produced by the source decreases by a factor of two each
cycle, and the amount of power dissipated by the resistor decreases by a
factor of four each cycle. These are to be expected, since the line
input voltage is increasing each time. But look at the power supplied to
the line -- it actually increases during the second cycle relative to
the first, then drops back. What's happening? Well, the line has an
apparent impedance of Vin/Iin at any given time. Let's see what it is
(the peak values have the same ratio as the RMS values, so I'll use those):

For t = 0 to 10, Vin/Iin = 1/0.020 = 50 ohms
For t = 10 to 20, Vin/Iin = 2.5/0.010 = 250 ohms
For t = 20 to 30, Vin/Iin = 3.75/0.005 = 750 ohms

The input impedance will, of course, approach an infinite value as time
goes on, Vin approaches Vs, and Iin approaches zero. But during the time
interval of 10 to 20 seconds, the apparent line impedance provided a
better "impedance match" for the 150 ohm source than at other times,
which increased the line power input during that time.

Only because we have no energy leaving the line can be calculate a total
energy delivered by the source and dissipated by the resistance over an
arbitrarily long time period. That is, we can find the total energy
delivered if the line were connected and the source left on forever. We
can use the same formula for summing an infinite series as used in the
voltage analysis to find that:

The source produces a total of 400 / (1 - 0.5) = 800 mj
The resistor dissipates a total of 300 / (1 - 0.25) = 400 mj

from which we see that a total of 400 mj has been supplied to the line.
If we were to quickly replace the source and resistor with a 50 ohm
resistor across the line and no source, what should happen? Well, we
have forward and reverse waves of 2 volts peak traveling on the line, so
the end voltage should immediately drop to 2 * sin(wt) as the reverse
traveling wave exits with no further reflections from the input end of
the line, and the forward wave moves toward the far end. So the 50 ohm
resistor would have an RMS voltage of 1.414 volts across it, and a
dissipation of (1.414)^2 / 50 = 40 mW. This constant dissipation should
continue until the line is completely discharged, which will take one
round trip time or 10 sec. The total energy removed from the line and
dissipated during this time is 40 mW * 10 sec = 400 mj, which is exactly
the amount we put into the line during the charging process. All the
energy produced by the source is accounted for. I think I can set this
maneuver up with SPICE, and I'll do so if anyone who might be skeptical
would be convinced by the SPICE output.

A caution is in order. This analysis was greatly simplified by the lack
of any energy storing devices other than the transmission line. That is,
there was no reactance at either end of the line. An equally accurate
analysis could be done with reactances present, but calculation of
energy from power would have to account for the temporary energy storage
in the reactive components. Also, if a resistance-containing load is
connected to the output, it will also dissipate power and energy, so
that would also have to be accounted for. The bottom line is that care
should be taken in applying this method to more complex circuits without
suitable modification.

I've completely accounted for the power and energy leaving the voltage
source, being dissipated in the resistor, and entering the line, at all
times from startup to steady state, and done it quantitatively with
numerical results. And I did this without any mention of propagating
waves of power or energy. It was done very simply using Ohm's law and
elementary power relationships, with the only variable being the line
input voltage calculated from the voltage wave analysis. I offer a
challenge to those who embrace the notion of traveling waves of average
power or other alternative theories to do the same using mathematics and
premises consistent with the alternative theory. Until such an analysis
is produced, I remain unconvinced that any such theory is valid. *Any*
analysis has to produce results consistent with the law of the
conservation of energy, as this one has.

Roy Lewallen, W7EL