Roy Lewallen wrote:
Roger wrote:
Roy Lewallen wrote:
Roger wrote:
Roy Lewallen wrote:

I wonder why we can not get the same results?

If you mean for the calculation of the voltage at 100 ns and 20
degrees down the line, it's because of my error. It should be sin(36
+ 20 degrees) ~ 0.83.

From the equation vf(t,x) = sin(wt-x), I am getting
vf(36 degrees, 20 degrees) = sin(36-20) = sin(16) = 0.276v.

I should quit before I get farther behind! In my haste I made a second
error by adding rather than subtracting the 20 degrees. Again, you're
correct and I apologize for the error.

I'll take extra care to try and avoid goofing up again.

Could we look at five points on the example? (The example has
frequency of 1 MHz, entered the transmission line 100 ns prior to the
time of interest, and traveled 36 degrees into the line)

Ok.

Using zero voltage on the leading edge as a reference point on the
sine wave, and the input
point on the transmission line as the second reference point, find the
voltage at (36, -20), (36, 0), (36, 20), (36, 36)and (36, 40). All
points are defined in degrees.

I'm a little confused by your dual reference points, both of which seem
to refer to physical positions. In my analysis, I give an equation for a
voltage as a function of both time and physical position. The time (t)
reference point is the time at which the source is connected. At that
instant, the sine wave source voltage is zero. The reference point for
position (x) is the input end of the line. The voltage at any time and
any position can be determined from the equation, provided that the wave
being referred to has reached the position on the line at or before the
time being evaluated (and provided that I don't make a stupid arithmetic
error -- or two). You'll note that in my analysis, I usually give a time
at which the equation is valid -- this is to insure that the wave has
reached any point of interest on the line.

Not on the line yet, at -20 degrees, sin(36+20) = 0.83.

The equation vf = sin(wt - x) isn't valid until the wave has reached
point x on the line. It's also not valid at any point not on the line.
So it can't be used under those conditions.

At line input, at 0 degrees, sin(36+0) = 0.59v.

100 ns after turn-on, the wave has propagated 36 degrees down the line,
so it's present at the input end (x = 0) and the equation is valid. The
voltage at the line input is as you calculated, 0.59 v. at that time.

On the line, at +20 degrees, sin(36-20) = 0.276v.

Correct.

On the line, at +36 degrees, sin(36-36) = 0v.

Correct.

On the line, at +40 degrees, sin(undefined, -40) = (wave has not arrived)

Correct.

Each of us must be using a different reference point because we are
getting different results.

Without my careless errors, we get the same result except for the first
case.

I make errors as well, so no criticism from here. I am just glad that
we are on the same page on how we use the equation.

As for the (36, -20) point, I was indeed specifying a point that had not
been supplied to the transmission line. It was a prediction, as if the
wave existed physically and was moving toward our experiment. We could
not actually observe it until time = 56 degrees.

As I mentioned, time domain analysis of transmission lines is
relatively rare. But there's a very good treatment in Johnk,
_Engineering Electromagnetic Fields and Waves_. Another good
reference is Johnson, _Electric Transmission Lines_. Near the end of
Chapter 14, he actually has an example with a zero-impedance source:
"Let us assume that the generator is without impedance, so that any
wave arriving at the transmitting end of the line is totally
reflected with reversal of voltage; the reflection factor at the
sending end is thus -1." And he goes through a brief version of
essentially the same thing I did to arrive at the steady state.

I wonder if it is possible that Johnson, _Electric Transmission
Lines_, presented an incorrect example? That occasionally happens,
but rarely.

Certainly it happens, but both he and I get the same, correct result
using the same method. And it seems unlikely that both he and Johnk made
the same error.

Or maybe some subtle condition assumption is different making the
example unrelated to our experiment?

I'm not sure why you see this as necessary. Do you have an analysis
which correctly predicts the voltage at all times on the line after
startup but which uses some different interaction of waves returning to
the source? If so, please present it. I've presented mine and shown that
I arrived at the correct result.

clip.......
The "perfect voltage source" controls or better, overwhelms any
effect that might be caused by the reflected wave. It completely
defeats any argument or description about reflected waves.

If you'd like, I can pretty easily modify my analysis to include a
finite source resistance of your choice. It will do is modify the
source reflection coefficient and allow the system to converge to
steady state. Would you like me to? There's no reason the choice of a
perfect voltage source should interfere with the understanding of
what's happening -- none of the phenomena require it.

Just for conversation, we could place a 50 ohm resistor in parallel
with the full wave 50 ohm transmission line, which is open ended. At
startup, the "perfect voltage source" would see a load of 50/2 = 25 ohms.

No, at startup, the source always sees Z0, regardless of the load
termination. The load termination has no effect at the source end until
the reflected wave returns. And if the line is terminated in Z0, then
there is no reflected wave and the source sees Z0 forever. To the load,
a terminated line looks exactly like a plain resistor of value Z0. (I'm
assuming a lossless line as we have been all along.)

At steady state, the "perfect voltage source" would see a load of 50
ohms in parallel with "something" from the transmission line. The
power output from the "perfect voltage source" would be reduced below
the startup output. We would arrive at that conclusion using
traveling waves, tracing the waves as they move toward stability.

With the line terminated in 50 ohms, steady state is reached as soon as
the initial forward wave reaches the load.

If the termination is some value other than Z0, the steady state
impedance seen by the transmission line will be the load impedance (for
our example one wavelength line or any line an integral number of half
wavelengths long). When starting, the initial impedance is always Z0 for
one round trip, for any line length. Then it will jump or down. With
each successive round trip, the impedance will either monotonically step
in the direction of the final value, or oscillate around it but with the
excursion decreasing each time. Which behavior you'll see and the rate
at which it converges are dictated by the relationship among source,
load, and characteristic impedances.

Would it be acceptable to use a perfect CURRENT source, along with a
parallel resistor. Then CURRENT would remain constant, but voltage
would vary. Again, power into the test circuit would vary.

The transmission line will react to a perfect current source in parallel
with a resistance exactly the same as for a perfect voltage source with
a series resistance. Put them in boxes and there's no test you can
devise which can distinguish them by tests of the terminal
characteristics. So yes, that's fine, or just a black box about which
the contents are completely unknown except for any two of the open
circuit voltage, short circuit current, or impedance. (It's assumed that
the box contains some linear circuit that doesn't change during the
analysis.) The results will be identical for any of the three choices.

When we define both the source voltage and the source impedance, we
also define the source power. Two of the three variables in the power
equation are defined, so power is defined.

We define the maximum amount of power which can be extracted from the
box, but not the power that it's delivering. If we open circuit or short
circuit the box or terminate it with a pure reactance, it's delivering
no power at all. If we terminate it with the complex conjugate of its
impedance, it's delivering V^2 / (2 * Rp) where Rp is the parallel
equivalent load resistance and V is the open circuit voltage. With any
other termination it's delivering less than that but more than zero.

Now if the we use such a source, a reflection would bring additional
power back to the input.

This presumes that there are power waves bouncing around on the line.

We would need to begin the analysis all over
again as if we were restarting the experiment, this time with two
voltages applied (the source and reflected voltages). Then we would
have the question: Should the two voltages should be added in series,
or in parallel?

That's not a question I can answer, since it's a consequence of a
premise I don't believe. If your premise has merit, you should be able
to express it mathematically and arrive at the answer to the question.
Feel free to do an analysis using multiple sources and traveling waves
of power. If you get the correct answer for how a line actually behaves,
we'll try it out with a number of different conditions and see if it
holds up.

Your answer has been to use a reflection factor of -1, which would be
to reverse the polarity. This presents a dilemma because when the
reflected voltage is equal to the forward voltage, the sum of either
the parallel or series addition is zero.

That's correct but no dilemma.

You can see what that does to our analysis. Power just disappears so
long as the reflective wave is returning, as if we are turning off the
experiment during the time the reflective wave returns.

By "our" analysis do you mean my analysis or the one you propose?
There's no rule saying power can't disappear -- power is not conserved.
In fact, look at any point along the open circuited transmission line
and you'll find that the power "disappears" -- goes to zero -- twice
each V or I cycle. When you first turn on the source, the source
produces average power. In steady state, the average power is zero. It
"disappeared", without any dissipative elements in the circuit. Energy
*is* conserved, however, and there is no energy disappearing in my
analysis. Feel free to try and conserve power in your analysis if you
want. But energy had better be conserved in the process.

Would a "perfect CURRENT source" without any restrictions about
impedance work as an initial point for you? That would be a source
that supplied one amp but rate of power delivery could vary.

Sure. The only difference to the analysis procedure would be to change
the source reflection coefficient from -1 to +1, since a perfect current
source has an infinite impedance. I don't know how it would affect the
final outcome without going through the steps. Of course, the reflection
coefficient for the current wave would go from +1 to -1, so I imagine
you'd run into the same conceptual problem if you tried doing an
analysis of the current waves.

Roy Lewallen, W7EL

I agree that we would have the same problem with a perfect current
source which had an infinite impedance.

How about using a perfect POWER source, that could not absorb power.
Output power would be limited by the external impedance.
Mathematically, the perfect POWER source would be described by

Ps = (Vp^2)/Zvp = Zip * Ip^2 where Ps = maximum power output, Vp =
voltage from perfect voltage source, Zvp = 0, Zip = infinity, and Ip =
current from a perfect current source.

The power output could be infinite, but power could never be absorbed by
the source.

Just as for both perfect voltage and perfect current sources, the actual
power output would be limited by external loads.

The impedance of the perfect POWER source would be Vp/Ip, both
controlled by external loads. To me that means that the output power
from the perfect POWER source would follow the impedance presented by
the load, but power going into the source would be defined as being
zero.

Would the "perfect power source" be acceptable to you?

73, Roger, W7WKB