Transmission lines: unravelling the mystery 2

RF transmission lines have always presented an element of mystery. However, understanding the behavior of transmission lines is important to those of us who routinely work with these conduits of RF energy. Call them feedlines, transmission lines or simply “coax”-the purpose is to get energy from a transmitter output to an antenna, or from an antenna to a receiver input.

Non-resonant lines When a transmission line is terminated in its characteristic impedance, all of the power delivered to the load by the incident (forward-traveling) wave is absorbed by the load. The line is said to be non-resonant in this situation. The impedance at each point along the line, at any point between the generator and the load, is the same. There is no reflected wave on the line. The impedance will be equal to the characteristic impedance of the line. The length of the line in this case is not important, except where loss is concerned. You might say that, compared to resonant lines, non-resonant lines are boring.

Resonant lines When a transmission line is terminated with an impedance that is not equal to the characteristic impedance of the line, things change. The impedance along the line between the generator and load varies but repeats at every half wavelength (lambda/2) from the load. The load impedance may be purely resistive, or it may contain an element of reactance, either inductive or capacitive.

In Figure 1A at the left, a 50ohm transmission line (in a 50ohm radio system) is open-circuited at the load end. This is a theoretically perfect transmission line. It exhibits no loss and has a velocity factor of 1. Let’s look down the line from the load end toward the generator, or source, in one-thirty-second-wavelength (l/32lambda) increments. We only need to cover lambda/2 of line because the situation starts repeating at lambda/2. The second column in Table 1 below shows what the impedance will be at each incremental point on the line. Remember, this is for an open-circuited line-a near-perfect open circuit.

Now, if we place a short circuit on the end of the transmission line shown in Figure 1B, the situation changes. The third column of Table 1 (for the short-circuited line) lists the impedance value at each location moving from the end of the line toward the generator. This is for a near-perfect short circuit.

Some interesting points can be gleaned by studying Table 1. First, for the open-circuited line, the impedance consists of infinite resistance and infinite reactance at the end of the line. As the distance increases down the line (toward the generator), the resistive component becomes 0ohm and the inductive reactive component decreases until it reaches 0ohm at lambda/4 from the end of the line. Thus, lambda/4 from the end of the line, the impedance has shifted from an open circuit to a short circuit-a complete impedance transformation.

Now, moving further down the line from the end, between lambda/4 and lambda/2, the resistive component remains at 0V while the reactive component switches from inductive to capacitive and increases up to infinity at the lambda/2 point. It is interesting to note that the reactive component is equal to the characteristic impedance of the line at both the lambda/8 and 3lambda/8 points. However, at the lambda/8 point, the reactance is capacitive, but it is inductive at the 3lambda/8 point. At the lambda/2 point, the impedance again becomes infinity (infinity +/- j(infinity)). Thus, the impedance repeats at lambda/2. This entire sequence would repeat for every lambda/2 section of the transmission line. So, it is only necessary to study lambda/2 section of line to understand what is happening on the entire line.

The third column of Table 1 summarizes what happens with a short-circuited transmission line. Of course, at the short-circuited end, the impedance is 0 +/-j0ohm. Between the end of the line and the lambda/4 point, the reactance is inductive. This is the reverse of the open-circuited line. Again, at the lambda/8 and 3lambda/8 points, the reactance is equal to the characteristic impedance of the line. At the lambda/4 point, the impedance becomes an open circuit (infinity +/- j(infinity)). Between the lambda/4 and lambda/2 points, the reactance switches to capacitive reactance. And, at the lambda/2 point, the short circuit at the end of the line is repeated and will do so at every multiple of lambda/2 down the line toward the generator.

From this it can be concluded that for an open- or short-circuited transmission line, the impedance at any point between the end of the line and the lambda/2 point will consist of 0ohm resistance and a reactive component that extends from 0 to infinity. Note that for an open-circuited line, the resistance component switches abruptly from 0ohm to infinity at the lambda/2 point and every multiple thereof referenced to the end of the line. Conversely, for a short-circuited line, the resistance component switches abruptly from 0ohm to infinity at the lambda/4 point and every odd multiple thereof referenced to the end of the line.

It can further be stated that if the load impedance is purely resistive and greater than the characteristic impedance of the line, the reactance will be capacitive at any point between the end of the line and lambda/4 from the end, and inductive between the lambda/4 and lambda /2 points. If the load impedance is purely resistive and less than the characteristic impedance of the line, the reactance will be inductive at any point between the end of the line and the lambda /4 point, and capacitive between the lambda/4 and lambda/2 points.

Figure 2 at the left illustrates a transmission line, with a characteristic impedance of 50ohm, terminated with a load that is a pure resistance of 25ohm (25 +/- j0). The VSWR will be 2:1. We can determine the impedance at lambda/4 from the end of the line by using the following formula:

Z(sub1) = Z^2/Z(subL)

where Z is the characteristic impedance of the line, Z(sub1) is the impedance at the lambda/4 point and Z(subL) is the load impedance. In Figure 2, the load impedance is a pure resistance of 25ohm and the characteristic impedance of the line is 50ohm. Now, plugging these values into the equation above we get:

Z(sub1) = 50^2/25 = 2,500/25 == 100

Thus, the impedance at the lambda/4 point is 100ohm. The VSWR will still be 2:1. Assuming a lossless line, the VSWR at any point on a line will be the same. If in Figure 2 the 50ohm transmission line was terminated by a load with a pure resistance of 100ohm, the VSWR would still be 2:1. The impedance at a point lambda 4 away from the end (according to the formula above) will be:

50^2/100 = 2,500/100 = 25

Suppose the line is terminated in a complex impedance, one that contains both resistive and reactive components. The formula still holds true, but now we must enter the load impedance in complex form. In Figure 3, a 50V transmission line is terminated in an impedance of 25 = j25 ohms. Plugging this into the formula we get:

Z = 50^2/(25 = j25) = 2,500/(25 + j25) = 2,500/25(1 = j) = 100/(1+j) = 100/(1 + j) X (1 – j)/(1-j) = 100(1 – j)/2 = 50 – j50

Thus, the impedance lambda/4 down the line is equal to 50 – j50ohms. The inductive reactance present in the load impedance has been transformed into capacitive reactance at the lambda/4 point down the line. To reiterate, although the impedance on the line changes, the VSWR does not change.

Next month’s column deals with practical applications of this information. So, until next time . stay tuned!