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Return Loss: An Old Problem that Still Haunts
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At some point in his or her career, every engineer puts the wrong emphasis on return loss. Let’s look at what it is and when it is important and when it is not. We’ll take a look at the effects in cable systems, both in the RF and optical domains.

What is it?

Return loss is a measure of how closely the impedance of a source matches that of a load. The source might be a signal processor and the load a headend combining network. Or the source might be a receiving antenna and the load the coaxial cable feeding signal to the headend. Return loss is one of several methods of measuring the degree of match between the two. Another, used by ham operators and others who look at power transfer, is voltage standing wave ratio (VSWR). Return loss and VSWR are simply two ways to look at the same thing.

To get an idea of what return loss is about, think of ripples of water in a bucket. If you drop a stone in the middle of a bucket of water, you know that waves will ripple out from the stone, and when they hit the side, they will bounce back again. This is exactly what is going on with return loss: when a signal encounters an impedance discontinuity, a portion of it is reflected back to the source. A coaxial cable (or any other medium) has what is known as a "characteristic impedance" determined by the geometry and the dielectric constant of any insulators involved. This characteristic impedance is the impedance that must be connected to each end to eliminate any reflections. If the source and load impedance are exactly equal to the characteristic impedance, then a wave entering or leaving the cable does not "see" any discontinuity, and it is not reflected. If the source and terminating impedances are not equal to the characteristic impedance, then some power will be reflected. The percentage of the incident power reflected is proportional to the ratio of the source or load resistance to the characteristic impedance.

The effect on power transfer

There is some fairly simple math you work in college circuit theory courses that says you must have the source and load impedances conjugately matched in order to maximize power transferred from the source to the load. The term "conjugate match" means that the resistive components of the source and load must be the same. If one has a certain capacitive reactance, the other must have the same reactance, but it must be inductive. To keep it simple, we usually assume that the source and load are pure resistances.

Figure 1 plots the return loss against the transmission loss. The figure in the graph shows the model. It defines return loss (the amount of incident power reflected back to the source, RS) and transmission loss (the amount of incident power delivered to the load, RL). The math is a simple conversion of the reflected and transmitted power, from decibels to fractions of the incident power. Add them up and force the result to equal the incident power. Finally, solve for the transmission loss as a function of the return loss. This analysis assumes that any power reflected back to the source is absorbed in the source resistance, RS. Often this is a good assumption.

Notice that if you have any kind of a reasonable return loss, you really are not losing much transmitted power. A return loss of only 7 dB means that you are suffering 1 dB loss in transmitted power. If your return loss is 15 dB, you are only losing 0.14 dB of transmitted power. Striving for a good return loss will not get you much in the way of increased power. However, if you monitor the return loss of an antenna for example, you can get an early warning if something is going wrong. A change in return loss will be noticeable long before you experience any significant loss of transferred power.

Return loss affects frequency response

Often of much more importance than the power transfer, is the effect of a mismatch (low return loss) on the frequency response of a broadband system. Figure 2 illustrates three cases that show the range of what you might encounter.

We illustrate two different situations shown by the two diagrams at the top of the figure. The left diagram is an example of what we might encounter in a headend, except that we have assumed worse return loss than you should expect in real life. We model connecting two pieces of equipment with a coaxial cable. The two pieces are shown as a combining network and an optical transmitter, but they could be any two pieces. We show two cases for illustration. The first is interconnecting the two with a 6.2 meter cable having a round-trip delay of about 55 ns. The second example is a 3.4 meter cable having about 30 ns of round-trip propagation time.

The signal travels from the combiner to the optical transmitter. If the transmitter does not have a perfect (infinite) return loss, some of the signal is reflected (the first echo) back to the combiner. The output of the combiner has some finite return loss, so some of the first echo is reflected back toward the transmitter as the second echo. The echo magnitude is the sum of the return loss encountered at the two ends of the cable. When the second echo reaches the transmitter, its voltage adds with the incident, or first trip, signal. However, the second echo has traveled the length of the cable two times more than has the incident signal. Thus, the echo is delayed by the round-trip propagation delay of the cable, compared with the incident signal. At some frequencies, the second echo and the incident signals are in phase, so the two voltages add together. At other frequencies the two voltages are out-of-phase and they subtract. And at yet other frequencies, something in between happens. The result is the frequency response shown in the curve labeled 10 dB echo, 55 ns delay.

We show a second case, of a 17 dB echo with a 30 ns delay (the cable is only 3.4 m). Notice that because of the shorter delay, the ripple frequency is lower. Because the assumed echo magnitude is also lower (17 dB as opposed to 10 dB), the amplitude of the ripple is also lower. The curves are not quite symmetrical about the 0 dB relative amplitude axis. The asymmetry is particularly noticeable in the 10 dB echo curve, which is correct. If you were to look at a 0 dB net return loss, you would find that the amplitude peaked at 6 dB when the incident and reflected voltages add together. When they subtract, though, the relative amplitude goes to zero or minus infinite decibels. The shape of the curve is scalloped.

The third example is quite interesting. We have made it a bit more realistic. What we are modeling here is shown at the top right of Figure 2. We model a drop from tap to home. The drop is 30.5 meters long, with a propagation delay of 271.5 ns. Since the drop is so long, we have included the effect of cable loss in the computation. We assume that the total return loss of the two ends is 20 dB. This is a reasonable assumption: a television has very low return loss and the tap is typically in the low 20s. We said before that the return loss affecting the frequency response is the sum of the return loss at the two ends, but to compute the effect, you also must add twice the insertion loss of the cable. The reason is that the echo will experience this additional loss on each of its passages through the cable. Since cable exhibits more loss at high frequencies than at low, the frequency response ripple decreases as you go up in frequency.

The effect return loss has is a function of how much signal is reflected at each end of the cable. If either the source or the load had a perfect (infinite) return loss, we really wouldn’t care what the other end had. There would be no signal reflection to cause problems with the frequency response.

Jim Farmer is chief technical officer of Wave7 Optics. He may be reached at .

Next month: How return loss creates echos in a digital transmission system, and the effects of return loss in optical systems.

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