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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. KR2D

    KR2D Ham Member QRZ Page

    OK, I think I misunderstood where the re-radiated signal was coming from. I thought it was from the received energy reflected at the feedpoint. I knew, but had forgotten, that antennas will re-radiate a portion of a received signal. What I did not know was that half of the signal was re-radiated. Is it always one half, or can that vary?
     
  2. WA0LYK

    WA0LYK Ham Member QRZ Page

    What you are dealing with is the 'maximum power transfer theorem'. You can google it and find several explanations.

    In a matched system, the "source" (antenna) resistance would be 50 ohms and the feedline would also be 50 ohms. Under this theorem, the source resistance absorbs half of the power and the load the other half. So with an antenna, it will re-radiate half the power and transfer half the power to the feedline.

    If the antenna is not matched to the feedline, you will get reflections back to the antenna to be re-radiated. Consequently, less than the maximum power will be sent down the feedline.

    Jim
    WA0LYK
     
  3. K5MC

    K5MC Ham Member QRZ Page

    Tom,

    1. I apologize for my technical "gurus" comment in my previous post.

    2. You and I will apparently continue to disagree about your spectrum plots. It is clear to me that your spectrum plots are either bogus or your two tested transmitters are most unusual, if not unique. Since you also post that you've "measured dozens of radios and they all follow this trend" (http://www.w8ji.com/occupied_bw_of_cw.htm), it's pretty obvious to me that you are not measuring the 99% occupied bandwidth properly. Last year I tested four CW transmitters (Ten Tec Orion II, Ten Tec Corsair, Kenwood 940, and Kenwood 2000) using an Agilent N9020A signal analyzer; for each of these transmitters the measured 99% occupied bandwidth did vary with the keying speed in a manner consistent with my Fourier analysis. (I also observed the RF output signal from each transmitter on an oscilloscope to be sure that the rise/fall characteristics were independent of the sending speeds I used during my tests.)

    3. I'm sorry that you aren't willing to share the details that I requested in post #380 concerning your 220-foot tower. I would think you could pass along the pertinent information in just a few sentences. In addition, I think practically all of us here now agree that the line efficiency can be higher for a line with reflections than the same line with no reflections. The difference in line efficiency appears to be quite small in the examples I've studied, however.

    Concerning Yaesu and some of the other manufacturers of ham gear, I think you are selling their design engineers somewhat short. I'm betting the reason some of the gear had (or has) terrible key clicks is because those engineers simply didn't attach much importance to the CW performance of their rigs, not because they didn't understand the basic relationship between key clicks and the rise/fall characteristics of the keying waveform. By the way, the keying waveforms on my four transmitters vary from pretty good (Corsair, 940, and 2000) to excellent (Orion II).

    I would challenge anyone who says that the occupied/power bandwidth of a CW signal varies only with the sending speed; it most definitely is also a function of the rise/fall characteristics just as I carefully demonstrated in my eHam article last year. The importance of proper waveshaping in minimizing key clicks cannot be overemphasized. However, once the rise/fall characteristics of the keying waveform are fixed for a given transmitter, the occupied bandwidth as defined by the FCC will vary with the keying speed. There's simply no way around that basic fact.

    Perhaps you and I can now "bury the hatchet" on this particular subject.

    73, K5MC
     
  4. W5DXP

    W5DXP Ham Member QRZ Page

    The maximum transfer of 50% power from the antenna to the receiver occurs when the antenna is conjugately matched. That has very little to do with the Z0 of the feedline. It has everything to do with the impedance looking back down the feedline from the antenna. The following lossless system is conjugately matched no matter what the characteristic impedance of the feedline.

    XCVR----1/2WL feedline----50 ohm antenna

    In an ideal lossless system, if the antenna system is conjugately matched during transmit, it is automatically conjugately matched during receive.
     
    Last edited: Jul 1, 2008
  5. WA0LYK

    WA0LYK Ham Member QRZ Page

    Here is a short answer to your question. Once the signal leaves the feedline, you no longer have "waves" per se, you are dealing with a simple conductor that is not long enough to support varying electric fields in space. You no longer have a rotating vector for the E field normal to the conductor, it is both parallel to and within the conductor and points in the direction of the current (or opposite depending on your definition of charge flow). The H field surrounds the conductor in the direction defined by the right hand rule.

    You have a situation that looks like this.

    Jfor=(Ifor sin(wt))/a --> ================ <-- Jrefl=(Irefl(-sin(wt))/a)

    J = current density (amperes/m^2)
    a = cross-sectional area of conductor (m^2)
    I = current (amperes)

    Jtotal/a = Jfor/a + (-Jrefl)/a
    multiply each side by 'a' and,

    Jtotal=Jfor + (-Jrefl)

    Jtotal = Jfor - Jrefl

    The same applies to the E field.

    Efor = Jfor/sigma Erefl = -Jrefl/sigma

    Etotal = Jfor/sigma + (-Jrefl/sigma)

    You can analyze this either with linear electron motion and inelastic collisions or with drift currents. In either case the resultant direction is in the direction of the higher E field and higher current density.

    My biggest problem with using "wave" explanations in conjunction with a lumped constant tuner is using a model that is correct for conditions. A simple conductor that is <<wavelength simply doesn't support waves with any significance in space. Sure you can start out with Maxwell's equations, but by the time you collapse all the relations that don't apply (like the space components) you end up with the basic circuit models. When a 6 inch piece of wire is not even one degree long, Bx and -Bx are the same for all practical purposes. As a consequence you end up with fields that only vary with time as above and that are well defined in circuit theory.

    Basically, the best way to settle some of this is to perform measurements. I am trying to sneak some money to buy new coax, ladder line, non-inductive power resistors, a temperature meter, etc. We'll see what happens.

    Jim
    WA0LYK
     
  6. WA0LYK

    WA0LYK Ham Member QRZ Page

    But you've already said that a conjugate match doesn't guarantee no reflections and that a conjugate match isn't 100% reflective. Therefore, there will be an SWR and something will be reflected back to the antenna from the xcvr end of the feedline. Since a conjugate match isn't 100% reflective, some of the reflected wave will continue on to the antenna. Unless the antenna is 100% reflective, it will absorb and re-radiate that part of the reflected wave from the xcvr end that makes it to the antenna..

    In essence, then you have a converging series that ends up with less signal to the xcvr than with a feedline whose Z0 matches the antenna.

    Jim
    WA0LYK
     
  7. W5DXP

    W5DXP Ham Member QRZ Page

    Here's one conceptual technique for alleviating those mental blocks by turning the problem into a true distributed network problem where the lumped circuit is completely useless. Given a typical CLC T-tuner, mentally install 1 wavelength of lossless 50 ohm coax between each of the tuner components and make transmission line calculations on each section. Conditions will still be the same during steady-state but the technical facts will become obvious. In the following diagram sC means series capacitor and pL means parallel shunt inductor.

    100w---tuner input--1WL--sC1--1WL--pL--1WL--sC2--1WL--tuner output---100-j100 ohm load

    XC1 = -j138.8 ohms, XL = j94.5 ohms, XC2 = -j83 ohms

    The tuner performs the same matching function as before during steady-state but now the nature of the forward waves and reflected waves are exposed. The forward and reflected power can be measured on each 1WL of transmission line.

    What forward and reflected power will a Bird directional wattmeter measure in each of the 1 WL sections?

    Between the source and C1: Pfor=100w, Pref=0w, SWR=1:1

    Between C1 and L: Pfor=292w, Pref=192w, SWR=9.6:1

    Between L and C2: Pfor=280w, Pref=180w, SWR=9.1:1

    Between C2 and the load: Pfor=163w, Pref=63w, SWR=4.3:1

    Seems this proves that the 50 ohm Z0-match point is at a tuner's input. Seems this also proves that reflections originate at every junction point inside the tuner. The distributed network model works perfectly inside a tuner. Now if the 1WL sections are replaced by 1ft sections of real-world 50 ohm coax, the results will be very similar at HF frequencies.

    100w---tuner input--1ft--sC1--1ft--pL--1ft--sC2--1ft--tuner output---100-j100 ohm load

    A real-world Bird directional wattmeter will read very close to the values that are posted above.

    Most of the misconceptions and confusion regarding the subject of this thread is the result of some people using the flawed lumped circuit model to try to solve distributed network problems. When the only tool one has is a hammer, every problem looks like a nail.
     
  8. W5DXP

    W5DXP Ham Member QRZ Page

    That is an absolutely false statement that you have been uttering from the beginning. If we accomplish nothing else, let's prove your statement is false by a number of lossless examples. Assume the antenna can deliver 1 microwatt. Pfor is toward the RCVR which is the 50 ohm load. Pref is toward the antenna which is the 1 uw 50 ohm source.

    1. 50 ohm RCVR---1/2WL 50 ohm feedline---50 ohm antenna (1 uw)

    1 uw is delivered to the RCVR. Pfor=1uw, Pref=0uw, SWR=1:1

    2. 50 ohm RCVR---1/2WL 75 ohm feedline---50 ohm antenna (1 uw)

    1 uw is delivered to the RCVR. Pfor=1.04uw, Pref=0.04uw, SWR=1.5:1

    3. 50 ohm RCVR---1/2WL 300 ohm feedline---50 ohm antenna (1 uw)

    1 uw is delivered to the RCVR. Pfor=2.04uw, Pref=1.04uw, SWR=6:1

    4. 50 ohm RCVR---1/2WL 450 ohm feedline---50 ohm antenna (1 uw)

    1 uw is delivered to the RCVR. Pfor=2.78uw, Pref=1.78uw, SWR=9:1

    5. 50 ohm RCVR--1/2WL 600 ohm feedline---50 ohm antenna (1 uw)

    1 uw is delivered to the RCVR. Pfor=6.50uw, Pref=5.50uw, SWR=12:1

    Does a pattern appear to have emerged?
    In ALL cases, ALL of the power available from the antenna (1 uw) is delivered to the RCVR.

    It appears what you are missing is the fact that, in the above examples, the voltage reflection coefficient at the source is the complement of the voltage reflection coefficient at the load. That makes the power reflection coefficient equal at both ends of the system resulting in a conjugate match and thus maximum power transfer.

    Can we please put this particular misconception (myth?) to rest now?
     
    Last edited: Jul 1, 2008
  9. WA0LYK

    WA0LYK Ham Member QRZ Page

    You are beginning to really operate like a troll. You ask for answers to a question, with math, and then ignore the answer and ask a totally irrelevant question.

    Please address the question you asked and the answer I gave to you about charge density and conductors. You simply can't ignore it because it won't go away. I explained to you why waves don't exist on short conductors. There is simply no variations of voltage and current based upon space when using a simple conductor. The current flowing past any cross-sectional area in the conductor is the same as in any other cross-sectional area. The E fields are contained within the conductor and are tangential to it. The only variations of voltage and current in the conductor are based upon time.

    Instead of addressing the issue you orignally asked about, you come up with another inane circuit that introduces something to carry waves within a regular lumped constant tuner. Analyzing this really has nothing to do with addressing the behavior of simple conductors. By the way, if there are reflections at intermediate points, why are the reflections not seen back at the source? Is there some magic at work that cancels the reflections in the direction of the source? Could it be the lumped components not propagating waves?

    Since you wanted to have the issue addressed, please respond the original question you asked, don't bring up something else and ignore both the question and answer. That is the hallmark of a troll, i.e. not really wanting responses.

    Jim
    WA0LYK
     
  10. WA0LYK

    WA0LYK Ham Member QRZ Page

    I am out of town for a day and I'll address this in more detail when I return. However, here is a hint. How much of the 1st reflected wave at the receiver end passes through to the antenna when it gets back to the antenna end? Does the antenna re-reflect 100% of this back into the line so it goes to the receiver or does it get radiated and subtracted from the system? This is important because each time a receiver end reflection gets back to the antenna end a portion of it is passed through to the antenna. If the antenna radiates, then there is a subtraction of net power from the system and then you can't possibly end up with as much going to the receiver as a perfect match with no reflections will provide.

    Jim
    WA0LYK
     
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