Roger wrote:
. . .

This does raise the question of the description of the traveling wave
used in an earlier posting.

The example was the open ended 1/2 wavelength transmission line, Zo = 50
ohms, with 1v p-p applied at the source end. The term wt is the phase
reference. At the center of the line, (using the source end as a
reference), you gave vr(t) =-.05*sin(wt-90 deg.) and ir(t) =
0.01*sin(wt-90 deg.)

By using the time (wt-90), I think you mean that the peak occurs 90
degrees behind the leading edge.

The leading edge and peak of what? The function sin(wt -90) looks
exactly like the function sin(wt) except that it's delayed 90 degrees in
phase. So the forward voltage wave is delayed by 90 degrees relative to
the source voltage. This is due to the propagation time down 90
electrical degrees of transmission line.

This posting was certainly correct if we consider only the first
reflected wave. However, I think we should consider that TWO reflected
waves may exist on the line under final stable conditions. This might
happen because the leading edge of the reflected wave will not reach the
source until the entire second half of the initial exciting wave has
been delivered. Thus we have a full wave delivered to the 1/2
wavelength line before the source ever "knows" that the transmission
line is not infinitely long. We need to consider the entire wave period
from (wt-0) to (wt-360.

If these things occur, then at the center of the line, final stable
vr(t) and ir(t) are composed of two parts, vr(wt-90) and vr(wt-270) and
corresponding ir(wt-90) and ir(wt-270). We should be describing vr(t)
as vrt(t) where vrt is the summed voltage of the two reflected waves.

Sorry, it's much worse than this.

Unless you have a perfect termination at the source or the load, there
will be an *infinite number*, not just one or two, sets of forward and
reflected waves beginning from the time the source is first turned on.
You can try to keep track of them separately if you want, but you'll
have an infinite number to deal with. After the first reflected wave
reaches the source, its reflection becomes a new forward wave and it
adds to the already present forward and reflected waves. The general
approach to dealing with the infinity of following waves is to note that
exactly the same fraction of the new forward wave will be reflected as
of the first forward wave. So the second set of forward and reflected
waves have exactly the same relationship as the first set. This is true
of each set in turn. Superposition holds, so we can sum the forward and
reverse waves into any number of groups we want and solve problems
separately for each group. Commonly, all the forward waves are added
together into a total forward wave, and the reverse waves into a total
reverse wave. These total waves have exactly the same relationship to
each other that the first forward and reflected waves did -- the only
result of all the reflections which followed the first is that the
magnitude and phase of the total forward and total reverse waves are
different from the first pair. But they've been changed by exactly the
same factor.

It's not terribly difficult to do a fundamental analysis of what happens
at each reflection, then sum the infinite series to get the total
forward and total reverse waves. When you do, you'll get the values used
in transmission line equations. I've gone through this exercise a number
of times, and I recommend it to anyone wanting a deeper understanding of
wave phenomena. Again, the results using this analysis method are
identical to a direct steady state solution assuming that all
reflections have already occurred.

The infinite number of waves could, of course, be combined into two or
more sets instead of just one, with analysis done on each. If done
correctly, you should get exactly the same result but with considerably
more work.

I do want to add one caution, however. The analysis of a line from
startup and including all reflections doesn't work well in some
theoretical but physically unrealizable cases. One such case happens to
be the one recently under discussion, where a line has a zero loss
termination at both ends (in that case, a perfect voltage source at one
end and an open circuit at the other. In those situations, infinite
currents or voltages occur during runup, and the re-reflections continue
to occur forever, so convergence is never reached. Other approaches are
more productive to solving that class of theoretical circuits.

. . .

which is the total power (rate of energy delivery) contained in the
standing wave at the points 45 degrees each side of center. If we want
to find the total energy contained in the standing wave, we would
integrate over the entire time period of 180 degrees.

So think I.

I haven't gone through your analysis, because it doesn't look like
you're including the infinity of forward and reverse waves into your two.

. . .

Roy Lewallen, W7EL

Roy Lewallen wrote:
Roger wrote:
. . .

This does raise the question of the description of the traveling wave
used in an earlier posting.

The example was the open ended 1/2 wavelength transmission line, Zo = 50
ohms, with 1v p-p applied at the source end. The term wt is the phase
reference. At the center of the line, (using the source end as a
reference), you gave vr(t) =-.05*sin(wt-90 deg.) and ir(t) =
0.01*sin(wt-90 deg.)

By using the time (wt-90), I think you mean that the peak occurs 90
degrees behind the leading edge.

The leading edge and peak of what? The function sin(wt -90) looks
exactly like the function sin(wt) except that it's delayed 90 degrees in
phase. So the forward voltage wave is delayed by 90 degrees relative to
the source voltage. This is due to the propagation time down 90
electrical degrees of transmission line.

I think we are in sync here, but something is missing. When I think of
a traveling sine wave, it must have a beginning as a point of beginning
discussion. I pick a point which is the zero voltage point between wave
halves. It follows that the maximum voltage point will be 90 degrees
later. I think you are doing the same thing, but maybe not.

Next I imagine the whole wave moving down the transmission line as an
intact physical object, with the peak always 90 degrees behind the
leading edge. In our example 1/2 wave line, the leading edge would
reach the open end 180 degrees in time after entering the example. We
can see then, that the current peak will be at the center of the
transmission line when the leading edge reaches the end.


This posting was certainly correct if we consider only the first
reflected wave. However, I think we should consider that TWO reflected
waves may exist on the line under final stable conditions. This might
happen because the leading edge of the reflected wave will not reach the
source until the entire second half of the initial exciting wave has
been delivered. Thus we have a full wave delivered to the 1/2
wavelength line before the source ever "knows" that the transmission
line is not infinitely long. We need to consider the entire wave period
from (wt-0) to (wt-360.

If these things occur, then at the center of the line, final stable
vr(t) and ir(t) are composed of two parts, vr(wt-90) and vr(wt-270) and
corresponding ir(wt-90) and ir(wt-270). We should be describing vr(t)
as vrt(t) where vrt is the summed voltage of the two reflected waves.

Sorry, it's much worse than this.

Unless you have a perfect termination at the source or the load, there
will be an *infinite number*, not just one or two, sets of forward and
reflected waves beginning from the time the source is first turned on.
You can try to keep track of them separately if you want, but you'll
have an infinite number to deal with. After the first reflected wave
reaches the source, its reflection becomes a new forward wave and it
adds to the already present forward and reflected waves. The general
approach to dealing with the infinity of following waves is to note that
exactly the same fraction of the new forward wave will be reflected as
of the first forward wave. So the second set of forward and reflected
waves have exactly the same relationship as the first set. This is true
of each set in turn. Superposition holds, so we can sum the forward and
reverse waves into any number of groups we want and solve problems
separately for each group. Commonly, all the forward waves are added
together into a total forward wave, and the reverse waves into a total
reverse wave. These total waves have exactly the same relationship to
each other that the first forward and reflected waves did -- the only
result of all the reflections which followed the first is that the
magnitude and phase of the total forward and total reverse waves are
different from the first pair. But they've been changed by exactly the
same factor.

It's not terribly difficult to do a fundamental analysis of what happens
at each reflection, then sum the infinite series to get the total
forward and total reverse waves. When you do, you'll get the values used
in transmission line equations. I've gone through this exercise a number
of times, and I recommend it to anyone wanting a deeper understanding of
wave phenomena. Again, the results using this analysis method are
identical to a direct steady state solution assuming that all
reflections have already occurred.

The infinite number of waves could, of course, be combined into two or
more sets instead of just one, with analysis done on each. If done
correctly, you should get exactly the same result but with considerably
more work.

I do want to add one caution, however. The analysis of a line from
startup and including all reflections doesn't work well in some
theoretical but physically unrealizable cases. One such case happens to
be the one recently under discussion, where a line has a zero loss
termination at both ends (in that case, a perfect voltage source at one
end and an open circuit at the other. In those situations, infinite
currents or voltages occur during runup, and the re-reflections continue
to occur forever, so convergence is never reached. Other approaches are
more productive to solving that class of theoretical circuits.

Again, we seem to be in complete agreement except for the statement "In
those situations, infinite currents or voltages occur during runup".
For many years I thought that "initial current into a transmission line
at startup" would be very high, limited only by the inductive
characteristics of the line. With this understanding, I thought that
voltage would lead current at runup. It was not until I saw the formula
Zo = 1/cC that I realized that a transmission line presents a true
resistive load at startup. Current and voltage are always in phase at
startup.

If we agree that voltage and current are always in phase in the
traveling wave, then we should find that in our example, the system
comes to complete stability after one whole wave (two half cycles) is
applied to the system.
. . .

which is the total power (rate of energy delivery) contained in the
standing wave at the points 45 degrees each side of center. If we want
to find the total energy contained in the standing wave, we would
integrate over the entire time period of 180 degrees.

So think I.

I haven't gone through your analysis, because it doesn't look like
you're including the infinity of forward and reverse waves into your two.

. . .

Roy Lewallen, W7EL

73, Roger, W7WKB

Roger wrote:
If we agree that voltage and current are always in phase in the
traveling wave, then we should find that in our example, the system
comes to complete stability after one whole wave (two half cycles) is
applied to the system.

Assume a constant power source and you will get the results
that Roy is talking about.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Roger wrote:
If we agree that voltage and current are always in phase in the
traveling wave, then we should find that in our example, the system
comes to complete stability after one whole wave (two half cycles) is
applied to the system.

Assume a constant power source and you will get the results
that Roy is talking about.
No, because you would find two waves of equal voltage and current
traveling in opposite directions, always arriving at exactly out of
phase, at the source.

Now the question here is "Do the waves bounce off one another?" which
would result in a doubling of observed voltage, or "Do the wave pass
through one another?" which would allow a condition of energy entering
the system equal to energy leaving the system.

73, Roger, W7WKB

Roger wrote:
Cecil Moore wrote:
Assume a constant power source and you will get the results
that Roy is talking about.

No, because you would find two waves of equal voltage and current
traveling in opposite directions, always arriving at exactly out of
phase, at the source.

No, because a *constant power source* is pumping joules/second
into the system no matter what voltage or current it requires
to move those joules/second into the system. It's like the
power source in "Forbidden Planet".

Now the question here is "Do the waves bounce off one another?"

Waves do NOT "bounce" off one another. At a physical
impedance discontinuity, the component waves can
superpose in such a way as to redistribute their energy
contents in a different direction. (Redistribution of energy
in a different direction in a transmission line implies
reflections.) In the absence of a physical impedance
discontinuity, waves just pass through each other.

www.mellesgriot.com/products/optics/oc_2_1.htm

micro.magnet.fsu.edu/primer/java/scienceopticsu/interference/waveinteractions/index.html
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Roger wrote:
Cecil Moore wrote:
Assume a constant power source and you will get the results
that Roy is talking about.

No, because you would find two waves of equal voltage and current
traveling in opposite directions, always arriving at exactly out of
phase, at the source.

No, because a *constant power source* is pumping joules/second
into the system no matter what voltage or current it requires
to move those joules/second into the system. It's like the
power source in "Forbidden Planet".

Now the question here is "Do the waves bounce off one another?"

Waves do NOT "bounce" off one another. At a physical
impedance discontinuity, the component waves can
superpose in such a way as to redistribute their energy
contents in a different direction. (Redistribution of energy
in a different direction in a transmission line implies
reflections.) In the absence of a physical impedance
discontinuity, waves just pass through each other.

www.mellesgriot.com/products/optics/oc_2_1.htm

micro.magnet.fsu.edu/primer/java/scienceopticsu/interference/waveinteractions/index.html

There is a third possibility. The interaction of the two waves can
establish a very high resistance, so high that no current flows-zero.

Does this place us at a logical impasse, with current reversing and
voltage doubling at in one argument (at the open end), but not doubling
at the source end? No, the voltage will double at the source end when
stability is reached after one full cycle (in the 1/2 wave example).

Logically then, we must recognize that our source voltage WILL NOT
remain constant following the arrival of the reflected wave. Certainly
this is what we find when we retune our transmitters after changing
frequency.

What would be the logic of insisting that the input voltage be held
constant to the 1/2 wave example after it is shown that the reflected
wave must interact with the incoming wave give a very high impedance at
the source?

73, Roger, W7WKB

Roger wrote:
There is a third possibility. The interaction of the two waves can
establish a very high resistance, so high that no current flows-zero.

This is a common confusion of cause and effect. Take a
short-circuit 1/4WL stub for instance. A very high
resistance is established at the mouth of the stub but
that high resistance has zero effect on the forward current
which keeps on flowing into the stub. Without the forward
current flowing uninhibited into the stub, the very high
resistance could not be maintained. In fact, the current
is a maximum at the shorted end of a 1/4WL stub. If you
don't believe it, measure it. Where did that current come
from if current cannot flow into the stub?

The stub impedance is the result of the ratio of voltage to
current. It is a virtual impedance and since it is an effect,
it cannot be the cause of anything. The people who say that
a virtual impedance is the same thing as an impedor have not
read the definitions of those things in the IEEE Dictionary.
--
73, Cecil http://www.w5dxp.com

Roger wrote:

I think we are in sync here, but something is missing. When I think of
a traveling sine wave, it must have a beginning as a point of beginning
discussion. I pick a point which is the zero voltage point between wave
halves. It follows that the maximum voltage point will be 90 degrees
later. I think you are doing the same thing, but maybe not.

Next I imagine the whole wave moving down the transmission line as an
intact physical object, with the peak always 90 degrees behind the
leading edge. In our example 1/2 wave line, the leading edge would
reach the open end 180 degrees in time after entering the example. We
can see then, that the current peak will be at the center of the
transmission line when the leading edge reaches the end.

Yes, you're describing some of the properties of a sinusoidal traveling
wave. I generally describe them mathematically.

Again, we seem to be in complete agreement except for the statement "In
those situations, infinite currents or voltages occur during runup". For
many years I thought that "initial current into a transmission line at
startup" would be very high, limited only by the inductive
characteristics of the line. With this understanding, I thought that
voltage would lead current at runup. It was not until I saw the formula
Zo = 1/cC that I realized that a transmission line presents a true
resistive load at startup. Current and voltage are always in phase at
startup.

They are provided that Z0 is purely resistive. That follows from the
simplifying assumption that loss is zero or in the special case of a
distortionless line, and it's often a reasonable approximation. But it's
generally not strictly true.

But that doesn't have anything to do with my statement, which deals with
theoretical cases where neither end of the line has loss. For example,
look at a half wavelength short circuited line driven by a voltage
source. Everything is fine until the initial traveling wave reaches the
end and returns to the source end.

If we agree that voltage and current are always in phase in the
traveling wave, then we should find that in our example, the system
comes to complete stability after one whole wave (two half cycles) is
applied to the system.

Assuming you're talking about the half wavelength open circuited line
driven by a voltage source -- please do the math and show the magnitude
and phase of the initial forward wave, the reflected wave, the wave
re-reflected from the source, and so forth for a few cycles, to show
that what you say is true. My calculations show it is not. I'd do it,
but I find that the effort of showing anything mathematically is pretty
much a waste of effort here, since it's generally ignored. It appears
that the general reader isn't comfortable with high school level
trigonometry and basic complex arithmetic, which is a good explanation
of why this is such fertile ground for pseudo-science. But I promise
I'll read your mathematical analysis of the transmission line run-up.

Roy Lewallen, W7EL

Roy Lewallen wrote:
Roger wrote:

I think we are in sync here, but something is missing. When I think
of a traveling sine wave, it must have a beginning as a point of
beginning discussion. I pick a point which is the zero voltage point
between wave halves. It follows that the maximum voltage point will
be 90 degrees later. I think you are doing the same thing, but maybe
not.

Next I imagine the whole wave moving down the transmission line as an
intact physical object, with the peak always 90 degrees behind the
leading edge. In our example 1/2 wave line, the leading edge would
reach the open end 180 degrees in time after entering the example. We
can see then, that the current peak will be at the center of the
transmission line when the leading edge reaches the end.

Yes, you're describing some of the properties of a sinusoidal traveling
wave. I generally describe them mathematically.

Again, we seem to be in complete agreement except for the statement
"In those situations, infinite currents or voltages occur during
runup". For many years I thought that "initial current into a
transmission line at startup" would be very high, limited only by the
inductive characteristics of the line. With this understanding, I
thought that voltage would lead current at runup. It was not until I
saw the formula Zo = 1/cC that I realized that a transmission line
presents a true resistive load at startup. Current and voltage are
always in phase at startup.

They are provided that Z0 is purely resistive. That follows from the
simplifying assumption that loss is zero or in the special case of a
distortionless line, and it's often a reasonable approximation. But it's
generally not strictly true.

But that doesn't have anything to do with my statement, which deals with
theoretical cases where neither end of the line has loss. For example,
look at a half wavelength short circuited line driven by a voltage
source. Everything is fine until the initial traveling wave reaches the
end and returns to the source end.

If we agree that voltage and current are always in phase in the
traveling wave, then we should find that in our example, the system
comes to complete stability after one whole wave (two half cycles) is
applied to the system.

Assuming you're talking about the half wavelength open circuited line
driven by a voltage source -- please do the math and show the magnitude
and phase of the initial forward wave, the reflected wave, the wave
re-reflected from the source, and so forth for a few cycles, to show
that what you say is true. My calculations show it is not. I'd do it,
but I find that the effort of showing anything mathematically is pretty
much a waste of effort here, since it's generally ignored. It appears
that the general reader isn't comfortable with high school level
trigonometry and basic complex arithmetic, which is a good explanation
of why this is such fertile ground for pseudo-science. But I promise
I'll read your mathematical analysis of the transmission line run-up.

Roy Lewallen, W7EL

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.

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)

where 1v p-p has been originally applied and vf(t) = vr(t) and vt(t) is
total voltage at any time point. Notice that the total voltage is now
2vr(t) = 2vf(t). This doubling continues as the wave moves forward, with

vt(t) = vr(wt-2) + vf(wt-362) ......... = 2vf(t) = 2vr(t)

The current is similar except, very important, the current was reversed
when reflected from the open end.

it(t) = vf(wt-362) - vr(wt-2) ...... = vf(t) - vr(t) = 0


The effect on the old source point is to make the impedance infinitely
high for all ongoing wave forward motion, which is not stopping the
wave, only indicating that power no longer moves past this point.

At time 720 degrees, the reflected wave (wt-0) reaches our revised
source point where it matches with wt-720 and begins raising the source
impedance, stopping power movement into the system. From this time on,
no further power enters the revised system because of the high impedance.

We should notice here that no power leaves the system after this time as
well. The high impedance works both ways, for forward and reflected
wave. There is no need for additional reflection analysis because both
source and full wave systems are stable and self contained after this
720 degree point. The source is effectively "turned off", and the full
wave system isolated.

If the source was parallel with a 50 ohm resistor (assuming a 50 ohm
transmission line), then the reflected wave would be matched with the
resistor and absorbed. Power would be continually moved through the
transmission line, giving hot spots on the line at 90 and 270 degree
points. We would still see the doubled voltage points at the end (360
degrees) and 180 degree points.

If we move to the shorted transmission line case, the math is identical
except that the voltages are reversed but currents add. The result at
the source is a low impedance where power can no longer be applied
because the voltage is always zero. Could we make current flow through
a resistor in parallel to the transmission line at the source in this
case? I think not.

So I think.

Thanks for taking time to consider these words. It takes real time to
carefully consider the arguments (and to present them).

73, Roger, W7WKB

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