## TDR and Ice Cube Trays

Dr Johnson, I've read your book and seen the impressive web site you maintain and wondered if you would be kind enough to throw some light on a problem of understanding I have.

I have been doing some TDR work here and was content with the standard explanations of what was happening. Helping in the work is an undergraduate student who kept asking me awkward questions. It was a bit like having your child asking why the sky is blue, all day long. It soon became obvious that while I might have fooled examiners over 20 years ago that I knew something about field theory, the intervening years have taken their toll. I went to the readable press, yours included but didn't find any one text that answered all my questions. I then tried to put together a picture for the mathematically challenged that made sense to me and offer it here (see below).

The problem of course is that while a story may sound good, it doesn't necessarily bear any relevance to the truth. In particular this story is in conflict with yours insomuch as you have simultaneous non-interacting waves deriving from a reflection and I don't. If we talk hydraulic analogy your model sounds like the undertow sliding under the incoming roller and mine is more like a blocked drain backing up to the sink.

The net current is the same but the dynamics of what is happening in the pipe is quite different. The problem of course is that since I can't visualize what TEM transmission 'really' looks like I just use the L C ladder model as you did in section 4.2.1 and carry it to its logical (to me) conclusion which is at variance with the received wisdom. That's why I'm asking for some expert guidance in this.

### TDR for the mathematically challenged

Consider a step function injected into a lossless coaxial transmission line. At the time of launch the wave has no knowledge of the end point. It is only constrained by the immediate characteristics of the transmission line it is entering. The wave enters a section, delta-x long of defined impedance Z made up of delta-c and delta-l whose values derive from the geometry of the cable (big D little d equation reference goes here).

The delta-c presents a short circuit, or at least a very low impedance, to the rising edge. The incoming current charges up the capacitor to the incoming waves potential. This provokes a flow of current into ground, which returns through the screen to the signal generator.

As the potential on the capacitor rises to meet that of the incoming wave front the capacitor becomes fully charged and current ceases to flow into it. By that time of course, the wave front has moved on to the next segment of length delta-x. As the wave front crosses the boundary from one small slice to the next, it experiences exactly the same boundary conditions of Z as it was leaving so v and i are conserved and the wave front propagates through the cable. Charge is brought to each slice of delta-c in turn and so there is a continuous generation of return current along the braid of the cable whose position of generation is always in phase with the wave front moving down the central conductor.

Consider now what happens if the cable terminates in a load of lower impedance than that of the cable itself. In the limit this load impedance could be zero, a short. For a lower impedance, the potential needed to transmit the actual current being supplied is much lower than that being presented, tending to zero for a short. This lower voltage must now propagate backwards to the source thus collapsing the charge that has been accumulated along the cable. This charge now flows forward into the load until electrostatic equilibrium is re- established while the collapsing potential propagates backwards towards the source.

This is described as a reflected wave of negative voltage, which subtracts from the incoming step function. If the mismatch was a complete short then the reflected negative wave has the same amplitude as the previous outgoing wave and will have fully canceled it out by the time it reaches the source.

Consider now what happens if the cable terminates in a load of higher impedance than that of the cable itself. In the limit this load impedance could be infinite, an open. In this case the current continues to arrive at the last slice and so the potential on the delta-c will start to rise. As it does so then there starts to develop a potential difference between the load and source ends of the cable. This in turn develops a wave front traveling back toward the source.

Equilibrium is reached when the potential at the load end reaches twice that at the source end. At that point the reflected current is equal to the incoming current and the potential no longer rises. The wave front propagates back to the source leaving the cable behind it in a static condition charged to twice the first incoming potential.

Thanks for your interest in high-speed digital design.

You've come up with a pretty good description. I especially admire the fact that you have properly perceived how current flows on the ground plane underneath the line. That is exactly what happens.

Here are a couple of ideas I use when I try to describe how lines work. First, I note that no analogy is going to be perfect, and that there is no substitute for just believing the mathematical model. The analogies are useful, but in the case of conflict between the analogy and the standard mathematical model, the mathematical model wins.

Anyway, one idea I use just to get across the idea of basic wave propagation is the "ice cube tray" model.

[You are in Europe, so I have to check a couple of facts before I go ahead here. In the U.S., we freeze ice cubes in shallow plastic (or metal) trays. The trays contain about a dozen little square compartments for the cubes, in two neat rows. The dividing walls between sections are not quite as high as the outer walls of the tray, so you can slosh the water about between sections to equalize cube size.]

There are two ways to fill an ice cube tray. Some people like to hold the tray horizontal under the faucet, moving it back and forth to fill all the individual compartments.

Other people (like me) like to hold the tray at a tilt, letting the water flow in at one end. We like to watch the water fill up one section, then spill into the next, and then fill the next, etc., until the whole tray is full.

The faucet provides water at a constant rate of flow, which propagates a step wave of water, moving further and further down the tray.

The front end of the tray develops a constant pressure head (the height of the first compartment). The ratio of pressure to flow (V to I) at the first compartment is constant).

This is precisely how the L-C model of a transmission line works.

When you impress a step of current into a distributed L-C model, the current initially charges the first unit element of capacitance, returning along the power and ground planes to its source. In the time the first capacitor takes to fill, the first unit element of inductance goes from a high impedance to a low impedance, allowing current to flow into the next section. As each successive section fills, the inductors switch the flow into the next. At any point in time, current is rushing down the line, filling up one little unit element of capacitance, and returning along the planes.

From the perspective of the front end of the line, you get a constant flow of current, and a constant pressure head (voltage).

In other words, the input impedance to a transmission line, in the presence of only one mode of propagation (nothing returning from the far end) is purely resistive (constant ratio of V/I).

This ratio is called Z0, the characteristic impedance of the cable.

OK, now what you do is use the principle of superposition to explain that with the transmission line, unlike with an ice-cube tray, I can superimpose two current sources at arbitrary positions and create two waves which pass each other like trains in the night.

At every position, the voltages add (and the currents, if they are going in different directions, subtract).

The easiest way to see this is to imagine an infinitely long line, and position your current sources at two intermediate points A (on the left) and B (on the right). The source at A creates two waves, spreading either direction away from A. At B, we get the same thing. Between the two points, we get two waves traveling in opposite directions.

I'm next going to change things around in order to show how to do the boundary calculations. Imagine that current source A puts out a single unit step.

Let's start out by asking a simple question: "What current would I have to inject at point B in order to make A think a resistive load Z0 was connected at point B instead of an infinite length of cable going to the right?" The answer is: "None". The cable to the right of B ALREADY looks like a resistor Z0.

This is the end-termination condition. When you terminate a line in a resistor of value Z0, nothing comes back in your face, regardless of what you send down the line.

Here's a more difficult question: "What current would I have to inject at point B in order to make A think a short circuit was connected at point B?" The conditions for a short circuit are that V=0 at point B. If we set the step current amplitude for B equal to negative one, and time the generation of B to correspond with the arrival of the step from A, then the voltage at B will be zero for all times. That's the same as a short circuit.

What we have learned is that having a short circuit at B is the same thing as having something at the end of the line which sends back towards us a wave of equal amplitude but opposite polarity.

The final question (and the one that causes the most confusion): "What current would I have to inject at point B in order to make A think an open circuit was connected at point B?" The conditions for an open circuit are that I=0 at point B. If we set the step current amplitude for B equal to positive one, and time the generation of B to correspond with the arrival of the step from A, then the current just to the left of position B will be zero for all times. That's the same as an open circuit. Note that the superposition of the wave from A, plus the wave from B generates a doubling of the apparent voltage at position B. (Don't worry about the current to the right, we are just interested in knowing what size wave we need going from B to A in order to make A think it hit an open circuit). [If you follow through with all the math you should know that the unit step currents produce step voltages of I*Z0/2, because each generated wave must split and go in two directions].

This is the general format for solving all boundary- condition problems in transmission lines: assume the line is truly infinite, and figure out what equivalent sources we would have to position at the boundary locations in order to cause equivalent actions within the perimeter of the model.

P.S. -- one little item I left out that may be of use to you -- I like to think about the POWER contained within the incoming waves. When an incoming step edge slams into the end of a transmission line, if the line is either shorted or open, there is nothing there to dissipate any power. Whatever power I shoved into the system must then reflect back in my face. In other words, the magnitude of the reflection coefficient has to be unity.

Best Regards,
Dr. Howard Johnson