Re: Standing-Wave Current vs Traveling-Wave Current

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


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