Load

greater than its own, but smaller than an open circuit-just to get a figure, let's say one three times as large as the line impedance the result will be similar to that when we have an open, but not as extreme. The reflection coefiicient in this case would be xh instead of 1, which would make the reflected wave half as strong as the incident wave. When a cancellation would occur with an open circuit, only half of the incident wave would be cancelled in this case and the other halt would remain. When a reinforcement would occur, the resulting sum would be a full-strength incident wave and a half-strength reflected wave, or 144 times the "matched" value. Both voltage and current would vary from half normal to \-Vz times normal, for a 3-to-l variation.

At the quarter-wave point, the current would be I-V2 times normal and the voltage would be only half normal. This higher-current/lower-voltage condition defines an impedance LOWER than that of the line although the load impedance in this case is greater. We would expect this because we have already seen that a quarter-wave line transforms an open circuit into a short. The value of this lower impedance, though, bears the same ratio to the line impedance as the line impedance (joes to the load; in each case it's 3 to 1. That's necessary in order to keep the reflection coefficient constant all along the line. We can step up as well as down this way, and it's widely used.

What is a Standing Wave Ratio? We have just seen how any reflection upon a transmission line produces an interaction between the incident and reilected waves to create a pattern of high and low voltage points (loops and nodes) from that reflection back to the generator, and that this pattern does not vary with the instantaneous voltage or current of the outgoing energy.

Such a stationary pattern of voltage or current distribution is called a "standing wave" since it is not travelling. The amount of standing wave present on any line is measured by the "Standing Wave Ratio" or SWR, which is the ratio of maximum voltage to minimum voltage along the line, or alternatively the ratio of maximum to minimum current. This is very similar to the definition of reflection coefficient, and in fact SWR and reflection coefficient are related so closely that either can be converted to the other by a little arithmetic. The major difference is that reflection coefficient compares values in the forward wave with those in the reflected wave, while

SWR compares actual maximum value on the line with actual minimum value and is thus somewhat easier to measure directly. Reflection coefficient, however, is easier to measure indirectly, and most of our "SWR meters" actually measure reflection coefficient instead, making the conversion to SWR by their scale calibration.

Some authorities speak of the 'Voltage SWR" or VSWR when they mean the ratio we have just defined, and use the bare term SWR to mean "power SWR". Power SWR is the square of voltage SWR, since the power in any resistor is proportional to the square of the voltage (or the current) applied to that resistor.

Fig. 3 shows a waveform view of what we've been talking about for so long, now that we have most of the words. While the illustration is based upon voltage waveforms, the same sort of thing is true of current waveforms as we saw a while back. Fig. 4 compares voltage and current waveforms in the SWR patterns created by several types of loads on the same type of line.

We saw last month that the presence of a standing wave upon an antenna structure made it easier for radiation to occur, and that also holds true of transmission lines. A high SWR will increase the likelihood of radiation from the line-which is usually an unwanted situation, since radiation is one major source of Une losses.

A high SWR will also increase both leakage and resistive losses within the line, since it will increase current at the high-current points along the line ("current losses") and will increase voltage at the high-voltage points as well. Higher current means more power loss in a given resistance (P=I~R), and higher voltage also means more power loss.

Most operators who take pride in a "good" operation, for these reasons, strive to achieve the lowest possible SWR in their transmission lines.

As we have indicated, SWR can be measured by actually measuring either voltage or current all along the line and then dete nni ning the maximu m-t o-mi ni m u m ratio, or by measuring the reflection coefficient with a "reflectometer" 01 "directional coupler* and converting to SWR, It can also be calculated by a much simpler rule: if the load impedance is known, and is a pure resistance without reactance, then the VSWR is equal to the ratio of load impedance to line impedance (or line impedance to load impedance, whichever yields the highest figure). Our 3-to-l example a few paragraphs back shows this, and how it comes about.

The VSWR, incidentally, can never be better than l-to-l5 since this says that the load and line have identical impedances and under this condition the line is perfectly matched. An SWR figure less than 1.0 simply says that you have a pretty bad SWR and have calculated it the wrong way: divide 1 by the figure you have to get the real SWR.

The VSWR of an open or a short circuit is infinite—but you can never measure it to be such a high value, because of li? = e losses. The more loss in the line, the more the reflected wave will be reduced. The more the reflected wave is reduced, the less effect it can have upon actual net voltage or current an any point, and the smaller effect will produce a lower SWR. A long, lossy line is one of the best dummy loads available, because it cun present a near-perfect SWR at its input end even when the load end is open circuited. For the same reason, SWR measurements should be made as close to the load end of the line as possible to escape this "swamp-ing-out7' effect of line losses.

Let s take as an example to calculate most of the things about SWR which the FCC wants you to know the case of a 52-ohm line terminated in a 104-ohm resistive load. We know immediately that the VSWR is 1 04/52, or 2 to 1. We also know t at if 10 amperes are flowing in the antenna terminals, either twice as much or half as much will be flowing at a point % wave back from the load because the VSWR is 2 to 1, Since the antenna's impedance is higher than that of the line, the !£-wave point will have an impedance lower because of the transformer action—and in a lower impedance, more current must flow. This tells us that twice as much current will be flowing, and also that the point % wave back is a current loop. Therefore 20 amps will be flowing at the loops, and 10 amps at the nodes.

Had the load resistance been less than that of the line but with the same VSWR (which would have required a 26-ohm load), the current at the Vi-wave point would have been only half that at the load. This would mean, in turn, that the load represented the highest current point or loop, and the other loops would have been at the '/2-wave intervals back along the line.

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