# What effect does a tuner have at the antenna?

Discussion in 'General Technical Questions and Answers' started by W5DXP, Jun 4, 2008.

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1. ### WA0LYKHam MemberQRZ Page

This is not what w2du and you said in previous messages. (Color is added by me)

Which is it? Pure reflection at the tuner, or maximum power transfer to the source for waves traveling from the antenna to the source. You simply can't have it both ways.

Pardon me! Here is the quote again. "... when two waves of equal amplitude and wavelength that are 180-degrees ... out of phase with each other meet, they are not actually annihilated, ..." When two objects, people, or waves meet the usual interpretation is that they are traveling in opposite directions or at least have vectors with projections in opposite directions. When one object, wave, etc. overtakes another they must be traveling at different speeds or have at least passed the source and then their paths overlap. I've never heard meet used in this situation!

As to your assertion that two waves can interact such that energy is transferred from one to another please describe what happens to a photon when this occurs. Does the one giving up energy lose mass, lose speed, or change it's wavelength (color)?

You are equivocating again. See above. You simply can't have it both ways. Either a tuner reflects reflected waves back toward the load or it doesn't. Which is it?

Again, the question was that if a 100% percent reflection occurs in a tuner, how do you hear anything through it?

It is a simple question and deserves a simple answer. Most folks reading this could probably care less about the math and know less about how to obtain the answer. And I'll be honest, I don't see it as worthwhile until more is resolved. I'll make it simple for you to give an answer, pick one of the following that most closely resembles the answer.

A. All (or most) of a wave traveling from the load to the source is reflected back to the load and none (or little) passes through the tuner.

B. More of a wave traveling from the load to the source is reflected back to the load and less passes through the tuner.

C. Less of wave traveling from the load to the source is reflected back to the load and more passes through the tuner.

D. None (or little) of a wave traveling from the load to the source is reflected back to the load.

You are also ignoring the question I have asked about what happens to the reflected waves at the source end of a transmission line when they suddenly transition to a circuit based environment. The E and H fields are no longer transverse, the E field becomes parallel with the wire and in the direction of the current density (toward the source) and the H filed is a circular pattern around the wire in the direction shown by the right hand rule. The energy from the source is exactly opposite.

What happens?

This is important for knowing where reflections on the transmission line start and stop. It is important to know so one knows why reflected waves don't reach the source. It is important to know so one can calculate the effective impedance at that point. It is important to know why you don't have standing waves on connecting wires in a lumped constant tuner. On and on.

Please tell us what you envision happens.

Lastly, your insults regarding people's intelligence really undermine your arguments and explanations. This is the last gasp of people who have no intention of modifying their position in the face of overwhelming evidence otherwise. If you fully believe in your position, then simply state the facts as you understand them, and please refrain from the snide comments and personal attacks.

Jim
WA0LYK

2. ### W5DXPHam MemberQRZ Page

You're right, I was a little loose with words. Let me rephrase my statement:

The fact that zero reflected power is measured between the antenna tuner and the source proves that the reflected wave has been re-distributed back toward the load by the tuner. The re-distribution is a combination of reflections and wave cancellation as explained in detail in my Worldradio magazine article at: http://www.w5dxp.com/energy.htm. Have you read it yet?

I was trying to keep it simple and over-simplified the explanation. Thanks for keeping me honest. Let's define a couple of terms.

Reflection - what happens to a single wave at an impedance discontinuity.

Redistribution - what happens when two waves interact and interfere (wave cancellation).

The results of reflection or redistribution on the reflected energy are exactly the same. The reflected energy changes direction. It is common practice to lump reflection and redistribution together and call any reversal of direction/momentum of reflected waves a "re-reflection".

That may be the usual interpretation but that is not the meaning in this case. The two waves are indeed traveling in the same direction and wave cancellation takes place the instant that they meet but not before. So a meeting of two waves that cancel has somewhat of a special meaning in optical physics.

I am no quantum physicist but here is my understanding of what happens. Since the destructive interference can only occur in a medium at an impedance discontinuity (change in index of refraction), the interfering photons are absorbed by the electrons in the medium and re-emitted in the opposite direction in a transmission line. In a 3-D medium, it may not be the opposite direction, just a different direction. In a 1-D medium, like a transmission line, there is no doubt in what direction a redistribution must necessarily take place.

Already answered above. It is common practice to call the phenomena a 100% re-reflection when in actuality, it is a combination of reflection and wave cancellation. I apologize for not being more precise.

A moot point now that I have explained wave cancellation. Isn't it sufficient to know that, in an ideal lossless system, if a conjugate match exists for transmit, then it also automatically exists for receive. Why do you even need to perform a wave analysis? All that you need to know is that the impedance looking into the transmission line at the antenna is the conjugate of the antenna feedpoint impedance and we have proved that to be the case.

This is a question asked out of ignorance. You cannot willy-nilly switch math models. I have said that before. If you are using the distributed network model outside of the tuner, you MUST use the same model inside the tuner. Of course, the best approach is to use Maxwell's Equations for everything.

I have insulted no one's intelligence. Ignorance and intelligence are unrelated. A person of low intelligence can be quite knowledgeable about one thing and a person of high intelligence can be completely ignorant of another thing.

Ignorance seems to have developed into a bad word yet all of us are extremely ignorant about something. I do not use the word "ignorant" as an insult but as a factually descriptive term. I freely admit to being completely ignorant about a lot of subjects.

Only a person who considers himself to be omniscient would ever object to being called ignorant.

Last edited: Jun 30, 2008
3. ### W5DXPHam MemberQRZ Page

Walt, you covered that in section 4.3 of Reflections. You said: "Wave interference between these two complementary waves at the stub point causes a cancellation of energy flow in the direction toward the generator. This cancellation results from the difference between the two reflected waves. The wave interference also produces an energy maximum in the direction toward the load, resulting from the sum of the two reflected waves and the source wave."

The two reflected waves that cancel are flowing in the same direction toward the source. They are the s11*a1 and s12*a2 terms in the first s-parameter equation. When b1 = s11*a1 + s12*a2 = 0, no energy flows toward the source.

I offered the following example as the simplest configuration that I could think of. If someone wants to understand what is happening at a Z0-match, it is best to eliminate the antenna tuner and concentrate on the wave action at point '+', the Z0-match point, exposed for all the world to see.

100w XMTR---50 ohm coax---+---1/2 wavelength 300 ohm line---50 ohm load

The forward power on the coax is 100w. The reflected power on the coax is zero. The SWR on the coax is 1:1. The forward power on the 300 ohm line is 204 watts. The reflected power on the 300 ohm line is 104 watts. The SWR on the 300 ohm line is 6:1.

Reflected energy flows toward the XMTR on the 300 ohm line. What happens at point '+' to eliminate reflections on the coax? As long as one doesn't understand the answer to that question, it is futile to introduce an antenna tuner which just complicates things.

Following is a graphic from my Worldradio energy article at: http://www.w5dxp.com/energy.htm which should help people understand the reflections and wave cancellations that occur at a Z0-match.

4. ### K5MCHam MemberQRZ Page

97.307 also says (at the very beginning) that "No amateur station transmission shall OCCUPY more BANDWIDTH than NECESSARY for the INFORMATION RATE and emission type being transmitted, in accordance with good amateur practice." Part 2 of the FCC Rules is the place where both occupied bandwidth and necessary bandwidth are defined in a precise manner. The FCC gives a simple equation relating the necessary bandwidth to both the sending speed and the "hardness" of keying for simple on-off keying/telegraphy. (The hardness factor relates to the rise/fall characteristics of the keying waveform; "harder" keying implies shorter rise/fall times.) For a given hardness factor (equal to either 3 or 5), the necessary bandwidth as defined by the FCC is directly proportional to sending speed. The occupied bandwidth as defined in Part 2 is the frequency bandwidth such that, below its lower and above its upper frequency limits, the MEAN POWERS radiated are each equal to 0.5 percent of the total MEAN POWER radiated by a given emission. As I pointed out in my eHam article, this definition is equivalent to the 99% power bandwidth defined in the professional electrical engineering literature.

The only precise definition of "bandwidth" in Part 97 that I've found is in 97.3, where bandwidth is defined as the width of a frequency band outside of which the MEAN POWER of the transmitted signal is attenuated at least 26 dB below the MEAN POWER of the transmitted signal within the band. This definition is equivalent to the 99.75% power/occupied bandwidth. (Of course, the 99.75% occupied bandwidth is larger than the 99.0% occupied bandwidth.)

You do hams no service when you insist that only your definition of "bandwidth" has any meaning. Moreover, your comments and spectrum plots at http://www.w8ji.com/occupied_bw_of_cw.htm vividly illustrate that you do not understand the definition of occupied/power bandwidth. If you understood the definition of occupied bandwidth (and particularly the concept of MEAN/AVERAGE power), you would know that your spectrum plots are nonsense. (The definition of 99% occupied bandwidth used by the engineers at Hewlett-Packard/Agilent Technologies is, of course, exactly the same definition found in Part 2 of the FCC Rules and in the ITU Regulations.) As I stated before, the spectrum plots I obtained using an Agilent N9020A signal analyzer last year on four different CW transmitters clearly demonstrated that the occupied bandwidth is a function of keying speed just as my eHam article said, which, of course, is also what the professional literature says as well as the ARRL Handbook.

The fact is that there is no mathematically precise definition of "key-click" bandwidth. (Last year I challenged you to develop such a definition to submit to the FCC since you have such a hard time accepting the meaning of occupied bandwidth as defined by the FCC, the ITU, and the professional literature.) It is silly for you to say that the concept of occupied/power bandwidth applies to some communication systems but not to ham radio. The hams who pursue EME and QRSS modes, for example, will object to your narrow viewpoint. (See http://www.ussc.com/~turner/qrss1.html for a quick overview.)

The FCC understands, of course, that the occupied bandwidth definition isn't completely adequate to address the issue of key clicks. That's why the FCC has the additional language in 97.307(b) discussing keyclick interference that immediately follows the section I quoted in my first paragraph above.

It is obvious that my trying to help you understand that there is a place even in ham radio for the concept of occupied/power bandwidth is a waste of my time. However, I think there are many hams who do like to think for themselves and they will not be misled by your "teachings" as long as they are also exposed to this subject in a manner consistent with the professional electrical engineering literature.

73, K5MC

P.S. I apologize to W5DXP for the recent "derailment" of this thread by W8JI and myself over the subject of bandwidth. I also want W8JI to know that I'm still waiting to see from him the particular numerical details concerning the transmission line feeding his 220-foot tower that I requested back in post #380.

5. ### W8JIHam MemberQRZ Page

Mickey,

Sometimes we get into a mindset where we think something meaningless is suddenly all important even though no one can observe within reason what is being said. This is exactly what Warren Bruene did with the great conjugate match dispute that resulted in a very good book being removed from the ARRL.

This is why Doug Smith, Kevin Schmidth, and everyome else appears "stupid" to you on the keyclick issue. They understand the system, and apply bandwidth in a meaningful way to the system. You want to apply something that is technically correct but does not matter at all in the system we are talking about.

The receivers we use and the humans behind those receivers don't care that the power integrated over a long period of time because the humans and the receivers have to hear each individual dot and dash.

It isn't a scale where the weight of power in the sidebands averaged or integrated over seconds, weeks, months or years affects the user or the receiver.

I'll continue to talk about what the people using the system are affected by, and you will continue to talk about something that is correct in some other application.

Maybe someday you will think about what is meaningful to the operator or the system, and we will be able to have a useful conversation that will contribute to the hobby.

73 Tom

Last edited: Jun 30, 2008
6. ### W5DXPHam MemberQRZ Page

Before one can understand what happens with an antenna tuner, it is necessary for one to understand what happens without an antenna tuner. Given the following ideal lossless system during XMT:

50 ohm XMTR---1/2WL 300 ohm feedline---50 ohm antenna

and during RCV

50 ohm RCVR---1/2WL 300 ohm feedline---50 ohm antenna

During XMT and during RCV, the SWR on the 300 ohm feedline is 6:1, i.e. (reflected-power/forward-power)=0.51. If the XMTR is sourcing 100w, the forward power is 204 watts and the reflected power is 104 watts. During steady-state, what happens to the reflected wave on the 300 ohm feedline? How is the reflected energy reflected/redistributed/canceled such that zero reflected energy reaches the XMTR? i.e. Specifically, what reverses the direction and momentum of the reflected wave at the XMTR output terminal? Please show your math.

The system is conjugately matched during both XMT and RCV. Is maximum available power transferred during both XMT and RCV?

Until one comes to grips with the reflections on the 300 ohm feedline above, one will not understand what happens with an antenna tuner.

It is important to understand that "effective impedance" is a Z=(Vfor+Vref)/(Ifor+Iref) ratio, i.e. a virtual impedance, not an impedor. The IEEE Dictionary very carefully differentiates between the two kinds of impedances. Does the resistive portion of the effective impedance dissipate any power?

One does indeed have standing waves on connecting wires in a lumped constant tuner if the V/I ratio is different from the Z0 environment. But that is another discussion for another time. The distributed network needs to be first understood without the tuner. Lumped-circuit shortcuts have clouded the distributed network issues.

For anyone unfamiliar with a wave reflection analysis, the following graphic should help.

Last edited: Jun 30, 2008
7. ### K5MCHam MemberQRZ Page

And apparently you will continue to post misleading spectrum analyzer data that purports to be the 99% occupied bandwidth of two CW transmitters.

Among the main points of my eHam article were to let hams know that there are many definitions of "bandwidth" in the literature and that the "occupied" bandwidth does have a precise mathematical meaning. The occupied/power bandwidth is a useful concept in communication systems (yes, even in ham radio); my trying to make this fact known to the ham community does not mean that I am downplaying the importance of proper waveshaping to minimize the generation of key clicks. (I fully support the work of such folks as Doug Smith regarding the issue of key click interference.)

People in this hobby who try to portray themselves as technical "gurus" should use the proper terminology. In addition to using sloppy terminology, you continue to misrepresent my overall views on the subject of bandwidth.

73, K5MC

P.S. Please provide the details I requested in post #380 in this thread.

8. ### W8JIHam MemberQRZ Page

Mickey,

1.) There is no place in technical discussion to personally insult or attack another person.

2.) The 99% occupied BW I published is accurate for the mode I measured. The analyzer I used had the correct settings and is an accepted instrument.

3.) I don't take work assignments that will only prove something already proven in this thread. Anyone who understands transmission lines already knows for short transmission lines losses decrease when the mismatch is in a direction that reduces line current.

My point was the concept that lowest loss occurs with lowest SWR, or that a certain percentage power transfer to the load always occurs from a "reflected power %" is wrong. Even with 80% reflected power it is possible to transfer nearly 100% of the applied power. Most people should know that now, but the greatly mismatched line between my vertical and the matching network is an example... as are the transmission lines inside a high power PA I built.

I'm happy you agree with Doug Smith because I certainly do also. Doug's article was a big step away from the concept that CW bandwidth was independent of keying waveform and bandwidth was tied only to speed.

Yaesu (and others) had so many problems with radios causing terrible QRM because they thought the speed affected the bandwidth and not the rise and fall time or shape.

As long as we all know what caused a signal to be wide and we don't undo the good things, that's all that matters.

This is the same thing Walt Maxwell got into with Warren Bruene. Bruene undid all of the very good work Walt did instead of helping move things forward. We are still suffering from that today.

I hope we don't undo all the good work on CW bandwidth by focusing on something that just doesn't matter. A model that does NOT apply to the working system can be technically correct, but useless in real life. Power bandwidth is a useless model for CW, just as Bruene's Thevenin model was for discussing behavior of PA's.

If you ever want to work with me off line with this, I will as time permits. I figure I have about 20 years left to try and do some good and I don't want to spend even a month of that time seeing who can pee the furtherest with someone else. Think of the good Walt and Warren could have done if they didn't get in such a long brawl.

73 Tom

Last edited: Jun 30, 2008
9. ### WA0LYKHam MemberQRZ Page

Let me beg off some of the answers to you. There is a lot to absorb and cogitate on, at least a semester's worth. So, some of the questions I ask are the quick and dirty ones that appear to me to cause conflicts.

If I read your message right, in receive, there is an actual power loss. This means that you will not receive as well as transmit due to the absorption of reflections by the antenna, i.e. it is not a bilateral system.

Correct?

Jim
WA0LYK

10. ### KR2DHam MemberQRZ Page

You answered my next question before I asked it But here's another one:

If the tuner were at the antenna feedpoint instead of at the other end of the transmission line, would this reflection and re-radiation still occur?