Transmission line problems.


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The Antenna-discussion list at http://www.antennex.com/ produced the following transmission line problems of general interest.


Bill Miller wrote:

[We have a coax cable which connects a d.c. battery to a load and there is a switch at the battery to connect it to the line.] About the coax cable. Since David Jefferies posed a conundrum, I will venture forth with a koan.

Let us look at the coax, battery and load in a different fashion. Let us assume that there is no overt action to precisely match the source and load impedances to the coax Zo. When the switch is closed between the battery and the coax, a pulse, whose amplitude is defined by the source impedance, [the battery voltage, and] the coax's Zo will zip down the line at the speed of light on the line. At the [load] resistor, unless it perfectly matches the coax, a portion of the pulse will be "reflected" back to the source. (Yes, I know there is not a trampoline at the end of the coax. I'm using "reflect" to describe the effect of the process.) At the source end, the (now-diminished) pulse will encounter a source mismatch and a portion will be reflected back to the resistor. Meanwhile, current is merrily pouring into the coax from the source, (ad infinitem.)

After a while, when one places a DC voltmeter anywhere on the coax, one reads the [battery] same voltage.

Questions: Where did all the pulses go? Why? What tests could we structure to verify this?


Bill,

Yes, a nice one. I use this as an example in microwave engineering classes.

If there is an imperfect reflection at either end, and at time zero you throw a switch to connect the battery to the line, an edge equal in amplitude to the battery voltage progresses down the line to the load where is is reflected with a reflection coefficient Gamma(load) less than magnitude 1. An edge of smaller amplitude propagates back to the source where similarly it is reflected with a reflection coefficient Gamma(source) which is also less than magnitude 1. You can see that the edges progressively diminish as time passes, getting exponentially smaller until all ac activity on the line has died out. You are then left with a DC voltage on the line, and across the load, equal to the battery voltage. This supplies energy and current to the load.

This is analogous to the transient behaviour in a damped oscillatory circuit, RLC, where if you apply an edge at time zero the oscillations die out over a time comparable to 1/Q.

If Gamma(load)=Gamma(source)=1. This is like having an undamped circuit and the oscillations persist for all time. But this is only possible in a thought experiment. In that case, the time average voltage on the line equals the battery voltage V and the peak amplitude on the line equals 2V for a duty cycle of 1/2

regards

David.


Bill Miller wrote:

Here is a continuation of the coax koan. Let us modify the coax and resistor by putting a switch at an arbitrary distance between the battery and the load. That distance is such that it takes T seconds for a pulse or RF signal to traverse it. Let us make the load R equal to the Zo of the line. Open the switch and charge the line. Close the switch and measure the width of the pulse at the load.

Shouldn't it be T? Was it? Why not?

Cheers,

Bill


Bill,

That took me a bit of time to figure out. I reckon that after you close the switch, the current flows for 2T and the voltage across the resistor is (1/2)V. I have replaced the combination of line Zo terminated by Zo by a simple resistor Zo without loss of generality.

This supposes that the generator end is open circuit after you have charged the line. The energy equations all add up as does the transmission line theory. If you'd like me to expound further I'll write it up and post it at

http://www.ee.surrey.ac.uk/Personal/D.Jefferies/figures/koan.html

David.

Postscript: Your way of looking at a charged length of transmission line as containing a Fourier superposition of propagating waves in each direction is interesting.

What we can say is that stored energy on a transmission line must be represented by photons or waves propagating at the wave velocity on the line. Even in the case of a static charged line, then, this gives rise to your scenario where the energy near the switch has to propagate back to the open circuit, and then return to the switch after time 2T after reflection at the open.


Hi David...

You have the correct answer. I'd love to see the math. I have an alternate explanation, but want to see your derivation, since you are the first person to whom I have posed the question that got the "right" answer.

73, Bill


Bill,

To save time I'm drawing this freehand and posting it as three inline .jpg files...


A note of explanation. Before the switch is closed, the current everywhere is zero. After the switch is closed, current begins to flow in the load. This has to be supplied from the charged section of line, which requires the setting up of an edge travelling along the charged line towards the generator. To determine the size of this edge we have to equate voltage either side of the switch. A little thought shows that the only possibility is that the voltage at the switch is V/2. That leads to current V/2Zo in the load, and voltage V-(V/2Zo)Zo on the charged section of line.



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Copyright © D.Jefferies 2003.


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11th July 2003.