Published by Eric Bogatin on 30 Nov 2010
In my Essential Principles of Signal Integrity class, I discuss the fundamental behavior of signals on a transmission line and the origin of reflections whenever the instantaneous impedance the signal sees changes. (For more details on these topics, check out the other articles on our web site at www.beTheSignal.com )
Of course, this is the most common root cause of signal quality problems- reflections from multiple discontinuities either due to poor terminations, a branched routing topology or non uniform interconnect structure. Understanding this principle makes it obvious how to avoid these problems.
The impact from impedance changes at any boundary for a signal traveling from an impedance Z1 into an impedance Z2 can be described by a reflection coefficient and a transmission coefficient, given by:
rho = ratio of the reflected voltage divided by the incident voltage
t = the ratio of the transmitted voltage to the incident voltage
The most common question I get asked is about a seeming paradox when we try to apply this relationship to the case of the simple circuit of a driver with an output impedance of 10 Ohms, driving a 50 Ohm transmission line, as shown below.
What is the voltage that is transmitted into the transmission line and measured at the terminating resistor, V_received?
Without invoking incident, reflected and transmitted signals, we would have answered this question by saying we have a simple voltage divider. The impedance we see looking into the transmission line is 50 Ohms, so the circuit really looks like this:
Of course, in this circuit, the voltage received is the voltage divider result:
The transmission coefficient in this example would be 5/6. And, this is the correct answer.
But, if we were to apply the transmission coefficient above to this problem, we would have naively expected the transmitted signal to be
This analysis says we will have more than the source voltage transmitted into the transmission line. In addition to this result being different from the first analysis, it just doesn’t seem reasonable. This is the paradox.
Why doesn’t the approach of using the transmission coefficient give the right answer? What’s going on?
I like this paradox and others like it because it really helps exercise some of our fundamental understanding of signals and transmission lines. Spend a little time thinking about what the answer is- why does this seem like a paradox and how is it resolved? When you are ready to see the answer, read on.
When we apply the transmission line analysis to derive the reflection and transmission coefficient, we are using the model of signals propagating on a transmission line, seeing some instantaneous impedance. It is instructive to redraw the circuit model slightly to illustrate how to think of this problem in a transmission line view as dynamic signals propagating.
In fact, the signal does not just come out of a point like lumped resistor. We can think about the source impedance being a 10 Ohm source resistor and a short length of 10 Ohm transmission line. Its length is irrelevant. It could be a micron, it could be 10 inches. There is no difference in the result. Here is what the equivalent circuit really is:
Adding the 10 Ohm transmission line is perfectly equivalent to the original schematic, and explicitly shows the path for the signal propagating from the source to the 50 Ohm transmission line.
In this new circuit, it’s clear that the incident signal to the interface between the two transmission lines that the signal sees is not going to be V_source. It is going to be ½ x V_source. The signal has to get through the voltage divided between the 10 Ohm source impedance and the 10 Ohm transmission line.
Based on this correct view of the problem, the transmission coefficient is still going to be the same, 10/6, but the incident voltage is ½ V_source, so we predict the received signal will be
Using the correct perspective- thinking about signals dynamically propagating on transmission lines, we see the transmission line view or the pure resistor voltage divider view, gives exactly the same answer.
For more information on this and other topics, check out www.beTheSignal.com.