# The Dark Side of the Conjugate Match

Discussion in 'General Technical Questions and Answers' started by KL7AJ, Mar 12, 2010.

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1. ### W5DXPHam MemberQRZ Page

Not necessarily. A non-dissipative resistance definition is covered in The IEEE Dictionary under "(B) the real part of impedance", i.e. no dissipation required. An example is the Z0 of a transmission line which is mostly resistive yet non-dissipative.

The conjugate matching theorem applies only to lossless systems and therefore an ideal conjugate match cannot exist in reality. But it is easy to check for a near-conjugate match in a low loss system.

Adjust your antenna tuner until your transmitter sees 50 ohms, i.e. an SWR of 1:1. Disconnect the transmitter cable and put a 50 ohm non-inductive resistor across the tuner input. Disconnect the tuner output and measure the impedance looking up the transmission line toward the antenna, e.g. R+jX. Measure the impedance looking back into the output terminal of the tuner. If that impedance is close to R-jX, the system is tuned to as close to a conjugate match as is possible in a real-world system with losses. In the real world, we are usually forced to settle for a Z0-match which, in a lossless system, would guarantee a conjugate match.

You see, one of the premises of the conjugate matching theorem is that if there is a conjugate match at one point, there is a conjugate match at all points. Unfortunately, that can only happen in a lossless system. But if in a low-loss system we can prove that the impedance measured in one direction is nearly the conjugate of the impedance measured in the other direction, we are as close to a conjugate match as we are going to get in the real world.

Last edited: Mar 13, 2010
2. ### W5DXPHam MemberQRZ Page

For that to be true, the transmitter would have to be re-reflecting 100% of the reflected wave energy incident upon the transmitter. But that would prove a Z0-match exists and we know it almost always doesn't exist. What you have stated is an imperfect assumed solution to an age-old conundrum.

It is a "by definition only" rule because it is very difficult to determine how much reflected energy is being dissipated in the source. A signal generator equipped with a circulator-load driving a long transmission line can be proven to violate that concept. The measured energy being dissipated in the circulator-load at any instant can be proven by modulation to have made a round trip to the load and back. I have observed such using a TV signal where the signal driving the circulator-load has been delayed by the transmission line.

What actually happens inside a transmitter is a combination of reflection (depending upon the impedance discontinuity) and wave cancellation (depending upon the phase of the forward and reflected waves). Depending on whether the internal interference between the forward wave and the reflected wave is constructive or destructive, any percentage of the reflected wave energy between 0% and 100% can be dissipated inside the transmitter. I can prove it for an ideal voltage source and series source resistor.

3. ### AF6LJHam MemberQRZ Page

This is good stuff, I am just lapping it up like a cat to tuna juice.

4. ### W5DXPHam MemberQRZ Page

A few more comments: Sometimes RF math models and shortcuts run afoul of the laws of physics. In particular, parts of the standing wave math model fall into those categories. Even though the math may work, it sometimes diverges from reality. Here are a few random points.

The energy in an RF EM wave is, like visible light, photonic in nature and must obey the known laws of physics. Photons must move at the speed of light in the medium, taking the velocity factor into account. This is a fact for light waves and RF waves, whether traveling or "standing". Photons cannot stand still.

It is obvious that an RF standing wave is not really "standing" since photons cannot stand still. There only exists an illusion of "standing" and many people have been fooled by that illusion. There is always forward wave energy moving in one direction at the speed of light and reverse wave energy moving in the other direction at the speed of light. Unless those two waves encounter an impedance discontinuity, they do not interact within a fixed Z0 transmission line and have (almost) no effect on each other. When we say that a standing wave contains no (available) energy, we are talking about energy available to be delivered to a load. There is indeed no energy delivered to the load when an ideal standing wave exists. But energy continues to exist in the forward traveling wave and in the reverse traveling wave. That's why the losses in one wavelength of feedline go up as the SWR goes up.

Here is a purely conceptual example that will illustrate the above point. Assume a 100 watt constant power source feeding a one second long lossless transmission line with a power reflection coefficient of 0.5 (SWR=5.82:1). The forward power is 200 watts and the reflected power is 100 watts. From the key-down transient state to steady-state, it can be shown that 300 joules of energy has been generated that has not yet reached the load. Note that is pure RF photonic energy, not watts. During steady-state, there exists 200 joules of RF photonic energy in the forward wave and 100 joules of RF photonic energy in the reflected wave for a total of 300 joules of RF photonic energy existing in the transmission line. All of that energy is moving at the speed of light in the transmission line medium. It is impossible for it to be standing still in the standing waves.

Here's a couple of quotes from two textbooks:

Quoting one of my college textbooks, Electrical Communication, by Albert:

"Such a plot of voltage is usually referred to as a voltage standing wave or as a stationary wave. Neither of these terms is particularly descriptive of the phenomenon. A plot of effective values of voltage, appearing as in Fig. 6(e), is not a wave in the usual sense. However, the term 'standing wave' is in widespread use."

From "College Physics", by Bueche and Hecht:

"These ... patterns are called standing waves, as compared to the propagating (traveling) waves considered above. They might better not be called waves at all, since they do not transport energy and momentum."

If someone told you that there is no traffic on the Golden Gate Bridge because the northbound traffic equals the southbound traffic, would you believe him?

Last edited: Mar 13, 2010
5. ### N8CPAHam MemberQRZ Page

In a way, you're right. If you misconjugate the match, you might copulate your whole system.

6. ### N0SYAHam MemberQRZ Page

isnt a conjugate match what the prisoners get?

7. ### K8ERVHam MemberQRZ Page

Please! They are NOT prisoners. They are Detainees. Where u been?

TOM K8ERV Montrose Colo

8. ### VK2TDNHam MemberQRZ Page

Hi Eric,

Only just discovered this thread whilst searching about the meaning of "a conjugate match" after a question I posed in the microwaves101 forum.

thanks so much

Dave
VK2TDN

9. ### AB9LZHam MemberQRZ Page

Bu.. bu.. but none of this 'splains how to find the G5RV menu setting on my new rig.

73 m/4

10. ### WB2WIKPlatinum SubscriberPlatinum SubscriberQRZ Page

I think I have a conjugal match

Is that pretty much the same thing?

If not, I'll need a tuner for the XYL, and she's not going to like it.