Roy Lewallen wrote:
Roger wrote:
Roy Lewallen wrote:

No interpretation necessary. Plug in a time t and distance x from the
input end of the line in degrees, and the result is the value of the
forward voltage wave at that time and place. For example, if the
frequency is 1 MHz, w = 2.828 X 10^6/sec. So 100 ns from the time you
connected the source (or 100 ns from the beginning of any source
cycle, since the signal is periodic), the value of the forward
voltage wave 20 degrees from the input of the cable is sin(2.828 X
10^6 * 100 X 10^-9 radians + 20 degrees) = sin(36.2 degrees) ~ 0.59
volts.

Sorry, I can not follow the numbers. For frequency of 1 MHz, I would
expect w = 2*pi* 10^6 = approximately 6.28 * 10^6 radians per second.
100 ns would be about 0.628 radians = 36 degrees.

The wave would have moved 36 degrees.

I apologize. I made an error and you're correct.

A point in 20 degrees from the input end would be found 16 degrees
behind the leading edge of the wave. The voltage should be sin(16) =
0.276v.

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.

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) 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.

Not on the line yet, at -20 degrees, sin(36+20) = 0.83.
At line input, at 0 degrees, sin(36+0) = 0.59v.
On the line, at +20 degrees, sin(36-20) = 0.276v.
On the line, at +36 degrees, sin(36-36) = 0v.
On the line, at +40 degrees, sin(undefined, -40) = (wave has not arrived)

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


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.

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

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. 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.

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.

When you use a "perfect voltage source" with a -1 reflection factor,
you are saying that a perfect polarity reversing plane (or
discontinuity) exists which reflects and reverses the reflected wave.

Yes, it behaves exactly like a short circuit to arriving waves.

However, the reflecting plane (or discontinuity) is one way because
it does allow passage of the forward wave. This is equivalent to
passing the forward wave through a rectifier. Is it fair to our
discussion to insist on using a voltage source that passes through a
rectifier?

The voltage source isn't acting like a rectifier, but a perfect source.
It resists as strongly as it can any change to what it's putting out.
I'm using superposition to separate what it's putting out from the
effects of other waves.

We could use a "black box" wave source. The only thing we would know
for sure about this source is that when it fed a Zo resistor through a
Zo feed line, there would be no swr on the feed line. Would that be
an acceptable voltage source for our discussions?

I have no problem with a "black box" source for which we know only the
voltage (as a function of time, e.g. Vs * sin(wt) where Vs is a
constant) and source impedance. The voltage is the open terminal voltage
of the black box, and the source impedance is that voltage divided by
the short circuit terminal current. One circuit (of an infinite number
of possibilities) having these characteristics is the Thevenin
equivalent, which is a perfect voltage source like I've been using in
the example, in series with the specified impedance. But for the
analysis I won't need to know what's in the box, just its voltage and
impedance. Analysis with any finite source (or load) resistance is
easier than the no-loss case because it allows convergence to steady state.

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.

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

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. 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.

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.

73, Roger, W7WKB