Roger wrote:
OK. I think I should tweak the example just a little to clarify that
our source voltage will change when the reflected wave arrives back
at the source end. To do this, I suggest that we increase our
transmission line to one wavelength long. This so we can see what
happens to the source if we pretended that we had not moved it all.

We pick our lead edge at wt-0 and define it to be positive voltage.
The next positive leading edge will occur at wt-360. Of course, a
half cycle of positive voltage will follow for 180 degrees following
points wt-0 and wt-360.

Initiate the wave and let it travel 540 degrees down the transmission
line. At this point, the leading edge wt-0 has reflected and has
reached a point 180 degrees from the full wave source. This is the
point that was originally our source point on the 1/2 wave line.
Mathematically, wt-0 is parallel/matched with wt-360, but because the
wt-0 has been reflected, the current has been reversed but the
voltage has not been changed.

After the initial wave has been propagating 540 degrees along the one
wavelength line, it will be back at the input end of the line, not 180
degrees from the source. (I assume that by "full wave source" you mean
the source connected to the input end of the line.)

Lets move to wave point wt-1 and wt-361 so that we will have non-zero
voltage and current.

vt = vr(wt-1) + vf(wt-361) = 2*.5*sin(1)

I'm sorry, you've lost me already. Where exactly are these "wave
points"? By "wave point wt-1" do you mean 1 physical degree down the
line from the source, or 1 degree from the leading edge of the intial
wave? As it turns out, those two points would be the same after 540
degrees of propagation. But "wt-361" would be one degree beyond the end
of the line by the first interpretation, or 1 degree short of the end of
the line by the other. What's the significance of the sum of the
voltages at these two different points?

I'm increasingly lost from here. . .

Here's my analysis of what happens after the initial step is applied,
using your 360 degree line and notation I'm more familiar with:

If the source is sin(wt) (I've normalized to a peak voltage of 1 volt
for simplicity) and we turn it on at t = 0, a sine wave propagates down
the line, described by the function

vf(t, x) = sin(wt - x)

At time t = 2*pi/w (one period after t = 0), it arrives at the far end.
Just before the wave reaches the far end, we have:

vf(t, x) = sin(wt - x) evaluated at t = 2*pi/w, = sin(-x) = -sin(x)

at any point x, in degrees, along the line. This is also the total
voltage since there's no reflection yet.

Then the forward wave reaches the far end. The reflection coefficient of
an open circuit is +1, so the reflected voltage wave has the same
magnitude as the forward wave. It arrives at the source at t = 4*pi/w
(two periods after t = 0), where it's in phase with the forward wave.
The reverse wave (for the special case of a line an integral number of
half wavelengths long) is:

vr(t, x) = sin(wt + x)

So at any point x (in degrees) along the line,

v(t, x) = vf(t, x) + vr(t, x) = sin(wt - x) + sin(wt + x)

Using a trig identity,

v(t, x) = 2 * cos(x) * sin(wt)

Notice that a standing wave pattern has been formed -- the cos(x) term
describes the envelope of the voltage sine wave as a function of
position along the line. But notice that the peak amplitude of the total
voltage sine wave is 2 rather than 1 volt, except that it's now
modulated by the cos(x) position function. Also note that the time
function sin(wt) has no x term, which means that the voltage changes all
along the line at the same time.

At the moment the returning wave arrives at the source (t = 4*pi/w),
sin(wt) = 0, so

v(x) = 0

So the returning wave arrives at the source at the very moment that the
voltage is zero everywhere along the line. (For those interested in
energy, this means that the line's energy is stored entirely in the
magnetic field, or the equivalent line inductance, at this instant.)

Let's follow the returning wave as it hits the input end and
re-reflects. The reflection coefficient at the source is -1 due to the
zero-impedance ideal voltage source, so the re-reflected wave is

vf2(t, x) = -sin(wt - x)

and the total voltage anywhere along the transmission line just before
the re-reflection reaches the far end is vf(t, x) + vr(t, x) + vf2(t, x)
= sin(wt - x) + sin(wt + x) - sin(wt - x) = sin(wt + x). Now this is
interesting. When the second forward wave vf2 is added into the total,
the standing wave disappears and the total voltage is just a plain sine
wave with peak amplitude of 1. It's identical, in fact, to the original
forward voltage wave except reversed in phase.

And what happens when vf2 reaches the far end and reflects? Well,

vr2 = -sin(wt + x)

So just before it reaches the input end of the line, the total is now

vf + vr + vf2 + vr2 = sin(wt - x) + sin(wt + x) - sin(wt - x) - sin(wt
+ x) = 0

For the interval t = 6*pi/w to t = 8*pi/w, the total voltage on the line
is zero at all points along the line which the second reflected wave has
reached! Further analysis shows that the line continues to alternate among:

v(t, x) = sin(wt - x) [vf only] [Eq. 1]
v(t, x) = 2 * cos(x) * sin(wt) [vf + vr] [Eq. 2]
v(t, x) = sin(wt + x) [vf + vr + vf2] [Eq. 3]
v(t, x) = 0 [vf + vr + vf2 + vr2] [Eq. 4]

How about the value at the input end (x = 0) at various times?

When we had only the first forward wave, it was sin(wt). When we had the
original forward wave, the reflected wave, and the new forward wave, it
was also sin(wt). And as it turns out, it stays at sin(wt) at all times
as each wave returns and re-reflects. I was incorrect earlier in saying
that this example resulted in infinite currents. It doesn't, but a
shorted line, or open quarter wave line for example, would, when driven
by a perfect voltage source.

Just looking at the source makes it appear that we've reached
equilibrium. But we haven't. The total voltage along the line went from
a flat forward wave of peak amplitude of 1 to a standing wave
distribution with a peak amplitude of 2 when the reflection returned to
a flat distribution with peak amplitude of 1 when the first
re-reflection hit, then to zero when it returned. Maybe we'd better take
a look at what's happening at the far end of the line.

The voltage at the far end is of course zero from t = 0 to t = 2*pi/w,
when the initial wave reaches it. It will then become 2 * sin(wt), or
twice the source voltage where it will stay until the first re-reflected
wave (vf2) reaches it. The re-reflected forward wave vf2 will also
reflect off the end, making the total at the end:

vf + vr + vf2 + vr2 = sin(wt) + sin(wt) - sin(wt) - sin(wt) = 0

The voltage at the far end of the line will drop to zero and stay there
for the next round trip time of 4*pi/w! Then it will jump back to 2 *
sin(wt) for another period, then to zero, etc.

In real transmission line problems, some loss will always be present, so
reflections will become less and less over time, allowing the system to
reach an equilibrium state known as steady state. Our system doesn't
because it has no loss. The problem is analogous to exciting a resonant
circuit having infinite Q, with a lossless source. It turns out that
adding any non-zero series resistance at the source end, no matter how
small, will allow the system to converge to steady state. But not with
zero loss.

I set up a simple SPICE model to illustrate the line behavior I've just
described. I made the line five wavelengths long instead of one, to make
the display less confusing. The frequency is one Hz and the time to go
from one end to the other is five seconds. The perfect voltage source at
the input is sin(wt) (one volt peak).
http://eznec.com/images/TL_1_sec.gif is the total voltage at a point one
wavelength (one second) from the input end.
http://eznec.com/images/TL_5_sec.gif is the total voltage at the far end
of the line (five wavelengths, or five seconds from the input end).

First look at TL_1_sec.gif. During the time t = 0 to t = 1 sec., the
initial forward wave hasn't arrived at the one-wavelength sample point.
Then from t = 2 to t = 9, only the forward wave is present as in Eq. 1.
At t = 9 sec. the first reflected wave arrives, resulting in a total
voltage amplitude of 2 volts peak as predicted by Eq. 2. The
re-reflected wave arrives at t = 11 sec., at which time the amplitude
drops back to 1 volt peak as per Eq. 3. And at t = 19 seconds, vr2
arrives, dropping the voltage to 0 as predicted by Eq. 4. The pattern
then repeats, forever.

TL_5_sec.gif shows the total voltage at the open end of the line,
alternating between a 2 volt peak sine wave and zero as predicted.

A nearly identical analysis can be done for the line current.

Let me say once again that the introduction of any series source
resistance at all at the input will result in convergence rather than
the oscillating behavior shown here, so a steady state analysis of the
zero resistance case can be done as a limit as the source resistance
approaches zero. But any analysis of the start up conditions on the zero
source resistance line should produce the same results derived here
mathematically and confirmed by time domain modeling.

Roy Lewallen, W7EL