Let's take a look at the energy in pulses and sine waves. At the end of
the day, the energy all has to be accounted for, whether it superposes
or not. It's really not all that difficult to do the analysis, as long
as we're careful not to fall into the traps which seem to have tripped
up quite a few others.

First a rectangular pulse. The energy E it takes to launch a pulse of
voltage Vp and duration T (seconds) on a transmission line of
characteristic impedance (assumed purely real) Z0 is Pp * T = Vp * Ip *
T where

Pp is the constant power applied as the pulse as created
T is the length of time the power was applied
Vp and Ip are the voltage and current of the traveling pulse

The pulse is a traveling wave, so for a forward traveling pulse, Ip = Vp
/ Z0; consequently, E = Vp^2 * T / Z0 = Ip^2 * T * Z0. Note that it's
essential to assume a purely resistive Z0 for this simple time-domain
analysis, since a reactive Z0 would cause a distortion of the pulse shape.

Once launched onto the line, we don't have any guarantee that all the
energy will stay within the spatial boundaries of the pulse -- all we
know for sure is how much total energy we've put into the line. But we
can conceptually freeze the pulse at any instant and see where the
energy is. Let's do that.

The obvious way to determine the energy in the pulse is to integrate the
power, which we can easily calculate. This is, after all, what we did to
find the energy we put into the line in the first place. But we're
interested in the energy distribution as a function of physical position
at an instant of time, so we can't find it by integrating the power.
(This is a mistake that seems to be commonly made.) Why not? Well, first
of all, energy is the *time* integral of the power. If we integrate the
power over a time interval of zero, the result is zero. We could look at
a single position on the line and integrate the power during the time it
takes for the wave to move by, to get the amount of energy which went by
during the time interval. But that's an indirect way of seeing where the
energy is on the line at a given time, and can easily lead to invalid
results. There are at least two potential problems with integrating the
power over a period of time to get the energy which passes a point. The
first is that we assign a sign to power, negative when energy is
traveling one way and positive when traveling the other. Consequently,
the result of the integral can be positive or negative. Although the
concept of negative potential energy is a valid one, I don't believe it
really applies to this situation, so one would have to be very careful
in interpreting and dealing with the sign resulting from the
integration. The second potential problem is that an integral never
produces a unique answer, but only an answer that's correct to within a
constant which has to be separately determined. Careless evaluation of
the constant or ignoring it altogether can produce invalid results.

So what I'm going to do is to evaluate the stored energy *per unit
length* of the line at each position along the line. The meaning of this
is that if we were to choose some sufficiently short segment length, the
amount of energy stored on each segment will be proportional to the
energy per unit length evaluated at that segment. In other words, I'll
evaluate the energy density as a function of position, or the energy
distribution along the line. This tells us where the energy is at the
instant of evalulation. I'm going to use the convention that the stored
or potential energy of a discharged line (V and I = 0) is zero.

The energy per unit length stored in the electric field, or line
capacitance, is C'V^2/2, where C' is the capacitance per unit length and
V is the voltage on the segment of line being evaluated. V is assumed to
not vary significantly over the segment length. We can let the segment
length approach zero as a limit, and say that the energy per unit length
is this value at any particular point along the line, where V is the
voltage at that point.

Likewise, the energy per unit length stored in the magnetic field, or
line inductance, is L'I^2/2 where L' is the inductance per unit length.
The total energy stored per unit length at any point is

E' = (C'V^2 + L'I^2)/2

On our line with purely real Z0, Z0 = sqrt(L'/C'), so L' = Z0^2 * C' and

E' = C'(V^2 + (Z0*I)^2)/2

where
E' is the total stored energy per unit length (or energy density) at
some point
V is the voltage at that point
I is the current at that point
Z0 is the (purely real) line characteristic impedance
C' is the capacitance per unit length

Length units for E' and C' can be anything as long as they're the same
for both.

Now let's look at a traveling pulse. We'll freeze it at some instant
while it's traveling down the line.

At any point to the left or right of the pulse, V and I are zero, so the
energy density is zero except where the pulse is. Where the pulse is, V
is the pulse voltage Vp and I the pulse current Ip, so

E' = C'(Vp^2 + (Z0*Ip)^2)/2

For a traveling wave, I = V/Z0, so

E' = C'Vp^2

This energy density is constant over the whole length of the pulse,
since Vp and Ip are constant over that distance. The total energy in the
pulse is then

E = E' * len = C'Vp^2 * len where len is the length of the pulse in
the same length units as C' and E'.

Because the energy density beyond the pulse in both directions is zero,
this is also the total energy in the line, which must equal the amount
we put in originally. So

E = C'Vp^2 * len = Vp^2 * T / Z0

from which we can calculate C' = T / (Z0 * len). Some manipulation of
this gives

T / len = sqrt(L'C') which relates line delay to L' and C', a result
which can be derived by other means.

All the energy in the line is accounted for -- it's traveling along with
the pulse, confined to the width of the pulse as we'd expect.

Ok, now let's fire another pulse at it from the other end of the line,
and see what happens when they completely overlap. Call the pulse 1 and
2 voltages Vp1 and Vp2, and currents Ip1 and Ip2. Assume that both have
the same duration T and therefore the same length len.

Voltages and currents (or E and H fields) add in the overlap region, so
the total V and I are the sum of the individual pulses' V and I. The
energy density in the overlap region is then:

E' = C'((Vp1 + Vp2)^2 + (Z0*(Ip1 + Ip2))^2)/2 * len

= C'(Vp1^2 + Vp2^2 + 2*Vp1*Vp2 + Z0(Ip1^2 + Ip2^2 + 2*Ip1*Ip2))/2

But what's the simple sum of the energy densities of the two pulses?

E1' + E2' = C'(Vp1^2 + Vp2^2 + Z0*(Ip1^2 + Ip2^2))/2

Oops! The energy density of the sum of the two pulses isn't the same as
the sum of the energy densities of the two pulses! And Since the overlap
region length is the same as the single pulse length, the same holds
true for the total energy. The problem is the two additional terms in
the total energy density 2*Vp1*Vp2 and 2*Z0*Ip1*Ip2.

It turns out that we're saved -- For the forward traveling pulse, Ip1 =
Vp1/Z0. For the reverse traveling pulse, Ip2 = -Vp2/Z0. So when the
appropriate substitutions are made, we find that 2*Vp1*Vp2 +
2*Z0*Ip1*Ip2 = 0, so the energy in the sum of the pulses is equal to the
sum of the energies of the pulses. And this is true regardless of the
values of Vp1, Vp2, Ip1, and Ip2. That is, it's true for any two pulses,
for any overlap length. _Provided they're traveling in opposite directions._

What happens when one pulse is the inverse of the other, that is, one is
positive and the other negative? Don't they cancel?

No, they don't. In the overlap region, the voltage is indeed zero. But
the current is twice that of each original pulse. The energy is simply
all stored in the magnetic field (line inductance) during the overlap.
The above equations still hold.

Well, we ducked that bullet. But what if the two pulses are traveling in
the same direction? What then? The two troublesome terms don't cancel,
so some energy ends up getting created or destroyed. But before worrying
too much about that, try to imagine how you'd accomplish it. The
propagation speed is the same for all pulses, so there' no way one can
catch up with another if both are fired from the input. I believe you
can contrive a situation where two pulses can be generated, one from
each end which, if long enough, will partially overlap when going the
same direction, after reflection. But the overlap and energy
calculations will be different than for this example, and I'm sure the
energy of the summed pulses will equal the total energy in the line. I'd
appreciate seeing an analysis from anyone who thinks he can show
differently. Remember that this analysis assumed that no other pulse was
present at the input while the pulse was being generated. If one is, the
amount of energy going into the pulse will be different. It also assumed
a constant line Z0 and velocity factor (constant L' and C'), so a
different analysis would have to be used if that condition is violated.

The conclusion I reach is that yes, a specific amount of energy
accompanies a pulse on a transmission line having purely real Z0, and is
confined to the pulse width. Although it can swap between E and H
fields, the energy in the confines of the pulse stays constant in value,
and simply adding when pulses overlap.

Sine waves are another problem -- there, we can easily have overlapping
waves traveling in the same direction, so we'll run into trouble if
we're not careful. I haven't worked the problem yet, but when I do, the
energy will all be accounted for. Either the energy ends up spread out
beyond the overlap region, or the energy lost during reflections will
account for the apparent energy difference between the sum of the
energies and the energy of the sum. You can count on it!

As always, I appreciate any corrections to either the methodology or the
calculations.

Roy Lewallen, W7EL