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tuner - feedline - antenna question ?
Jim Kelley wrote:
Cecil Moore wrote: Your example is the same as putting a load resistor on an open transmission line, measuring the dissipated power, and then claiming the same thing happens without the load resistor there. No, it is more akin to presenting all the evidence. Your approach is akin to rolling dice in the dark where you are the only one allowed to report the results. :-) Are you willing to assert that the power being dissipated in the circulator resistor didn't make a round trip to the load and back even though the actual delay is easy to measure? Do the reflected waves that you see when looking at yourself in the mirror contain any joules/sec? -- 73, Cecil http://www.w5dxp.com |
tuner - feedline - antenna question ?
Jeff wrote:
Change the source impedance to 100ohms and the picture changes to 5.5:1 at the load and 4.75:1 at the source end. Here's the equation for rho at the load. rho = (Z0-Zload)/(Z0+Zload) SWR = (1+rho)/(1-rho) I don't see the source impedance in those equations. -- 73, Cecil http://www.w5dxp.com |
tuner - feedline - antenna question ?
Here's the equation for rho at the load.
rho = (Z0-Zload)/(Z0+Zload) SWR = (1+rho)/(1-rho) I don't see the source impedance in those equations. Your analysis is fine if the source is matched to the coax, but you are neglecting the mismatch at the source to coax interface. If you used a TDR, for example, to look at the set-up you would see 2 points of discontinuity, firstly at the 100 ohm source to 50 ohm cable interface, and secondly at the cable to 200 ohm load. BOTH of these discontinuities add to the overall mismatch as seen by the 100 ohm load. Your application of the above equations neglects the first discontinuity. 73 Jeff |
tuner - feedline - antenna question ?
" The solution is 4.9 dB. snip If anyone wants to challenge the 4.9 dB solution, they can impeach my reference "Reference Data for Radio Engineers," (various editions). I can supply other references that have been named in this group too, but I would suggest with tackling one authority at a time. 73's Richard Clark, KB7QHC I have to admit to an error in my analysis of the problem, I made a mistake with the attenuation of the line. Re analysing with the correct values gives me the following: Loss 4.78dB S11 -5.25dB Vswr as seen by the source 3.41:1 I think these values are close enough to Richard's answers to make little difference. I have not got Reference Data for Radio Engineers to hand, but it may be a graphical solution that could account for the slight discrepancy. My figures were obtained using Ansoft Designer RF Cad package. The error crept in because I did the analysis by using 5.35m of coax at 300m to get the 5.35 wavelengths, and forgot to scale the attenuation value for the coax. 73 Jeff |
tuner - feedline - antenna question ?
"Jeff" wrote in
.com: If you used a TDR, for example, to look at the set-up you would see 2 points of discontinuity, firstly at the 100 ohm source to 50 ohm cable interface, and secondly at the cable to 200 ohm load. BOTH of these discontinuities add to the overall mismatch as seen by the 100 ohm load. Your TDR does not work in the steady state frequency domain space, and is misleading you. In the steady state, the (complex) ratio of forward voltage to reflected voltage is determined solely by the load impedance and characteristic impedance of the line. In crude terms, during establishement of steady state, you can view that a load end reflected wave which is then partially reflected at a mismatched source end, will reach the load end and be reflected in the same ratio as the earlier passes. The subsequent round trips as steady state is approached do not change the (complex) ratio of forward voltage to reflected voltage in the steady state. I know you have support here for the assertion that source end mismatch affects VSWR in the steady state, but you won't find it in reputable text books. Owen |
tuner - feedline - antenna question ?
Richard Clark wrote in
: On Wed, 28 Feb 2007 08:11:30 GMT, Owen Duffy wrote: Now, are you prepared to post your solution? Hi Owen, Your quick computation of 3.3 dB is suitably close to my reference's first pass solution (3.27 dB), but it neglects the contribution of the source's resistance. The solution is 4.9 dB. Reminding you that your question was "What is the loss in the line?", check your own post. Well, you posted an answer, not a solution. It wouldn't have been your solution anyway, because it looks like it is copied straight out of a book. Looking at Reference Data for Engineers, Sixth Edition, p24-12, Example 3 (which is the same as the problem you posed), they give the answer as 3.27dB. I am happy that my answer rounded to 3.3dB is correct. The source resistance has no influence over the line loss at all. You posed this problem as difficult and one that no one has ever got right. No wonder, you have a different answer to the the book! Owen |
tuner - feedline - antenna question ?
Your TDR does not work in the steady state frequency domain space, and is misleading you. In the steady state, the (complex) ratio of forward voltage to reflected voltage is determined solely by the load impedance and characteristic impedance of the line. What utter rot. The TDR indicates a discontinuity that is IN ADDITION to the one that gives you your result. Ignoring that discontinuity will certainly give you the wrong answer, regardless if which domain you are working in!!! Jeff |
tuner - feedline - antenna question ?
I am happy that my answer rounded to 3.3dB is correct.
The source resistance has no influence over the line loss at all. Well it certainly does not agree with Ansoft Designer, the result it gives is very close to Richards, and it shows a marked effect when you change the source resistance. With a 50 ohm source and 200 ohm load the loss is calculated as is 3.93dB and VSWR is 3.98:1 Changing to a 100 ohm source the loss is calculated as is 4.78dB and VSWR is 3.41:1 So a very well respected CAD package agrees with Richard at least!! 73 Jeff |
tuner - feedline - antenna question ?
Jeff wrote:
rho = (Z0-Zload)/(Z0+Zload) SWR = (1+rho)/(1-rho) Your application of the above equations neglects the first discontinuity. The first discontinuity (inside the source) doesn't have any effect on the SWR on the transmission line. Tuners at the source present another discontinuity and have no effect on transmission line SWR. -- 73, Cecil http://www.w5dxp.com |
tuner - feedline - antenna question ?
Your application of the above equations neglects the first discontinuity.
The first discontinuity (inside the source) doesn't have any effect on the SWR on the transmission line. The discontinuity is NOT "inside the source", it is at the source to coax interface, and as such effects the VSWR that the source sees. Tuners at the source present another discontinuity and have no effect on transmission line SWR. There are no tuners involved in the current example. Jeff |
tuner - feedline - antenna question ?
Jeff wrote:
What utter rot. The TDR indicates a discontinuity that is IN ADDITION to the one that gives you your result. Ignoring that discontinuity will certainly give you the wrong answer, regardless if which domain you are working in!!! When line loss is given in watts, the discontinuity at the source certainly has an effect. When line loss is given in dB (10x log of a ratio), the discontinuity at the source has NO effect. The discontinuity at the source has NO effect on the RATIO of two powers. -- 73, Cecil http://www.w5dxp.com |
tuner - feedline - antenna question ?
Jeff wrote:
With a 50 ohm source and 200 ohm load the loss is calculated as is 3.93dB and VSWR is 3.98:1 Changing to a 100 ohm source the loss is calculated as is 4.78dB and VSWR is 3.41:1 Looks like your losses and VSWR are taken on the wrong side of the source resistor. source R source V----x--/\/\/\/\/\/\--y----T-line---load Loss and VSWR should be calculated at 'y', not at 'x'. We are interested in the VSWR and losses *on the T-line*. Please change your reference point from 'x' to 'y'. -- 73, Cecil http://www.w5dxp.com |
tuner - feedline - antenna question ?
Jeff wrote:
Your application of the above equations neglects the first discontinuity. The first discontinuity (inside the source) doesn't have any effect on the SWR on the transmission line. The discontinuity is NOT "inside the source", it is at the source to coax interface, and as such effects the VSWR that the source sees. There is usually a piece of coax running from the source connector back to a filter. I would suggest that the discontinuity that you are talking about is indeed some distance "inside the source". But we users don't care or measure what VSWR the source sees. Only the PA designer worries about such. We users only care and measure the VSWR *ON* the transmission line. -- 73, Cecil http://www.w5dxp.com |
tuner - feedline - antenna question ?
Owen Duffy wrote:
"Jeff" wrote in e.com: If you used a TDR, for example, to look at the set-up you would see 2 points of discontinuity, firstly at the 100 ohm source to 50 ohm cable interface, and secondly at the cable to 200 ohm load. BOTH of these discontinuities add to the overall mismatch as seen by the 100 ohm load. Your TDR does not work in the steady state frequency domain space, and is misleading you. In the steady state, the (complex) ratio of forward voltage to reflected voltage is determined solely by the load impedance and characteristic impedance of the line. In crude terms, during establishement of steady state, you can view that a load end reflected wave which is then partially reflected at a mismatched source end, will reach the load end and be reflected in the same ratio as the earlier passes. The subsequent round trips as steady state is approached do not change the (complex) ratio of forward voltage to reflected voltage in the steady state. I know you have support here for the assertion that source end mismatch affects VSWR in the steady state, but you won't find it in reputable text books. Owen and Cecil are right: the source (transmitter) has no effect whatever on the VSWR on the line. That isn't just an assertion - it is part of the bedrock transmission line theory. Owen referred to "reputable textbooks", one of which would surely be 'Theory and Problems of Transmission Lines' by R A Chipman [1]. This book gains a lot of its reputation from its very complete mathematical development of the theory, showing all the detailed working. Chipman treats standing wave patterns in two different ways: first by assuming the final steady-state conditions, and then in much more detail by considering multiple reflections between the load and the source. Given a sufficient number of reflections, the multiple-reflection model converges on exactly the same results as the steady-state analysis - just as it does in the physical world. VSWR on the line is determined by the ratio |Vmax|/|Vmin|. The complex impedance that the source sees at the input terminals of the line is the ratio V/I at that point (where V and I are both vector quantities which include phase information). An alternative way of calculating either VSWR or Zin is through the ratio Vforward/Vreflected (again vector quantities). All of these approaches are alternative pathways through the same body of theory. They are all consistent with one another, and there is no contradiction between any of them. You will notice that all these standing wave relationships involve ratios. Chipman's detailed analysis confirms that these ratios are determined EXCLUSIVELY by the properties of the line and the load - never the source. The source properties do determine the magnitudes of all of the individual voltages and currents - but when you change the source properties (output voltage and/or impedance) all the individual voltages and currents on the line and at the load are changed by the same factor. So when you take the ratio, the source properties cancel right out again. All this confirms that, if you sweat out the math in all the different levels of detail that Chipman did, the source (transmitter) still has no effect whatever on the VSWR on the line. [1] Out of print, but well worth searching for: ISBN 0-07-010747-5. The web bookstores currently have eight copies on offer, at a range of prices. -- 73 from Ian GM3SEK |
tuner - feedline - antenna question ?
Owen and Cecil are right: the source (transmitter) has no effect whatever
on the VSWR on the line. That isn't just an assertion - it is part of the bedrock transmission line theory. Owen referred to "reputable textbooks", one of which would surely be 'Theory and Problems of Transmission Lines' by R A Chipman [1]. This book gains a lot of its reputation from its very complete mathematical development of the theory, showing all the detailed working. I am sorry but you are not correct, I have not read Chipman so I cannot comment on his analysis or your interpretation of his results, but my understanding , practical experiments and CAD analysis would lead me to disagree. If we take the situation where the source is matched (50ohms) to the 5.35 wavelength transmission line (lossless to simplify things) with a 100ohm load, I agree that the vswr is 4:1, unchanging with frequency. Plotted on a Smith Chart when swept against frequency this gives a circle centred on 1 (50ohms) with a radius of 4. i.e. on a constant VSWR circle. Now if we change the source impedance to 100ohms and repeat the same sweep and re-plot, keeping the chart normalized to 50 ohms, the circle moves on the resistance axis, still with a radius of 4 and now passing though 2 (100 ohms) resistive. The centre moves to about 0.6 (30ohms). It then becomes obvious that the locus of the circle is NOT a constant VSWR against frequency. You will come to the same conclusion if you normalize the chart to 100 ohms, the new source impedance and re-plot. The coax is acting as an impedance transformer, causing a shift along the resistance axis. Looking at it another way, the vswr changes sinusoidally with frequency, in our example, between 2:1 and 8:1. (The same as the Smith chart plot with a circle of radius 4 centred at about 0.6). 73 Jeff |
tuner - feedline - antenna question ?
Cecil Moore wrote: Jim Kelley wrote: Most engineers equate the units of power to power, i.e. joules/sec = watts and so does the IEEE dictionary. I can't speak for most engineers, but I think the first time I saw it was in high school physics, and of course later in engineering school. That was about 35 years ago. I think of it a fundamental concept - one that I happen to understand very well. Not unlike the relationship between Joules and electron-volts. But I am content to assert that the joules in the joules per second of a reflected wave is real energy. Do you disagree? I don't agree that the terms power and energy become interchangeable by virtue of the fact that their units can both be expressed with the word Joule in them. One can find himself making unrealistic predictions if he is not precise in his application of the ideas which underlie these terms. 73, ac6xg |
tuner - feedline - antenna question ?
Jeff wrote:
You will come to the same conclusion if you normalize the chart to 100 ohms, the new source impedance and re-plot. The Z0 of the transmission line has not changed to 100 ohms so normalizing the chart to 100 ohms is not valid. -- 73, Cecil, http://www.qsl.net/w5dxp |
tuner - feedline - antenna question ?
" You will come to the same conclusion if you normalize the chart to 100 ohms, the new source impedance and re-plot. The Z0 of the transmission line has not changed to 100 ohms so normalizing the chart to 100 ohms is not valid. -- 73, Cecil, http://www.qsl.net/w5dxp It is just as valid as using 50 ohms, and the result is the same, a changing vswr. I see you have not commented on the main point of my post, that being that the smith chart shows a changing vswr when you change the source impedance. Hint: transmission line transformers would not work if the vswr did not change. 73 Jeff |
tuner - feedline - antenna question ?
On Thu, 01 Mar 2007 09:49:52 GMT, Owen Duffy wrote:
Reminding you that your question was "What is the loss in the line?", check your own post. Hi Owen, Can you offer why I should? Well, I suppose not or you would have. However, I am one to never turn aside a suggestion and I did review everything (except my own quote - I've repeated it enough, haven't I?) and I will respond to that review within the body of this text. Well, you posted an answer, not a solution. It wouldn't have been your solution anyway, because it looks like it is copied straight out of a book. Does it being someone else's solution make any difference to the outcome? Owen, your comment reveals a prejudice by implication. Copying it right out has removed any issue of authority has it not? It has also removed any issue of accuracy too - if you accept that authority. Ultimately, having copied it out makes for the best resolution. Having copied it out, and offering the citation, gives us both access to the chain of evidence. Did I withhold or otherwise linger with the citation? You asked for my solution and I immediately offered both. Ironically, does your suggestion that It wouldn't have been your solution anyway mean you would suspect I would have come up with a different answer? That is, ascribing to me the quality of being able to get it right instead? That would be generous, thank you. However, it appears I fell short of that mark (and may have been the intent of your elliptical pat on the back). Looking at Reference Data for Engineers, Sixth Edition, p24-12, Example 3 (which is the same as the problem you posed), they give the answer as 3.27dB. A simple review of example four distinctly reveals the details to the problem I posed; example three contains only some of them. Example 3 is a subset of example 4 (as that example dwells on at great length). However, example 4 does have one notable difference, it asks: "What is mismatch loss between the generator and the line?" for which the answer is: "1.62 dB" Ah, the devil is in the details. Continuing from example 4: "The transducer loss is found by using the results of 3 and 4 in (4). This is 1.27 + 2.00 + 1.62 = 4.9 decibels" I am happy that my answer rounded to 3.3dB is correct. Congratulations. You may note in my earlier correspondence I allowed exactly that. The source resistance has no influence over the line loss at all. Upon review of my own reference (not the recommendation you offer above) I must concur. I was trapped by what I have already described as being the classic confusion between systems of match and loss. My solution was not for the loss in the line, but for transducer loss, and specifically for the inclusion of mismatch loss within the transducer loss. All caloric, but mis-ascribed to the line loss. You posed this problem as difficult and one that no one has ever got right. No wonder, you have a different answer to the the book! Well, in fact my answer conforms exactly to the book. The problem is not one of inexactitude, it is of poor referencing. As to the matter of no one else having ever got it right, no one even consulted the book - even partially. You can count yourself among a population of one and hashing it through served us well. 73's Richard Clark, KB7QHC |
tuner - feedline - antenna question ?
Richard Clark wrote:
On Wed, 28 Feb 2007 13:55:47 -0800, Jim Kelley wrote: What I meant was, in what way were you able to attribute and apportion this heat to its various sources? What evidence were you able to obtain to show reflected energy re-entering the source output? What component in the system in fact dissipated the reflected energy? How were you able to determine the exact source and amount of energy at any given location within the source? Or did you just presume that you understood the underlying mechanisms? Hi Jim, This knowledge arrived by many avenues. But primarily, it seems, by speculation. I know how to measure heat, Richard. What I am asking, and what you have thus far been unable to answer (which is as I suspected), is how is it that you were able to ascertain that this heat energy was caused by energy that was reflected from the load rather than having come directly from the power supply within the source? How is it that this electromagnetic energy is so easily reflected from a load, but is utterly immune to reflection when it encounters the output of a source? I think it's been fairly well established that the output impedance of these things is far from 50 ohms. Why should reflected energy not be, at least in some part, re-reflected back toward the load? Someone who alleges to be so familiar with load lines should be able to contend with an increase in dissipation against a mismatched load without having to explain it as 're-absorbed' reflected energy. Inductive logic leads us to this junction as the principle target of reflected power (the signal path is symmetric, after all). Speculation could also lead to that juction. Experience has supported this logic. It could be experience coupled with misattributed fact. Possible? 73, Jim AC6XG |
tuner - feedline - antenna question ?
Cecil Moore wrote:
Are you willing to assert that the power being dissipated in the circulator resistor didn't make a round trip to the load and back even though the actual delay is easy to measure? How does inserting a circulator load into a circuit that doesn't have one illustrate anything about energy flow other than the behavior of a circuit with a circulator load in it? Do the reflected waves that you see when looking at yourself in the mirror contain any joules/sec? You seem to be laboring under a misconception about my point of view, Cecil. But it can't be for a lack of attempts on my part to persuade you of it. 73, Jim AC6XG |
tuner - feedline - antenna question ?
Jim Kelley wrote:
I don't agree that the terms power and energy become interchangeable by virtue of the fact that their units can both be expressed with the word Joule in them. That's why I am willing to switch from the words "Reflected Power" to "Reflected Energy" and measure that energy flow past a point on a transmission line in joules/second. Thus "power" and "watts" are dropped from the discussion along with any semantic disagreements over the definitions of those words. So the question is: With a forward RF energy flow of 200 joules/sec and a reverse RF energy flow of 100 joules/sec, would you agree that there is 300 joules of energy existing in a lossless one-second long transmission line? i.e. exactly the amount of energy required to support the forward RF energy wave and the reflected RF energy wave. Or if the above transmission line is one microsecond long, that 300 microjoules of energy exists in the line, i.e. exactly the amount of energy required to support the forward RF energy wave and the reflected RF energy wave. -- 73, Cecil, http://www.qsl.net/w5dxp |
tuner - feedline - antenna question ?
On Thu, 1 Mar 2007 11:00:30 -0000, "Jeff" wrote:
So a very well respected CAD package agrees with Richard at least!! Hi Jeff, Thanx for the flowers, and sorry for having been myopic in my correspondence and restricting my comments to response to Owen. HOWEVER, those flowers may wilt in my hands. That congruent result you found with my writing and what you have found in your CAD work, sadly, does not conform to the question put. This has been an object lesson in the difference between mismatch loss and line loss. Such issues are frequently polluted through the course of discussion, and by polluted I mean that two or more concepts are combined as though they were one. This thread has revealed just such mixture, other threads often violate the separation of Conjugate Matching and Impedance Matching. On the other hand, there is an upside, to you as an innocent bystander who has participated through your own analysis. Before they wilt, I hand that bouquet back to you for your discovery of how much impact source Z has upon the SYSTEM. Ultimately, that was my point all along. Having arrived there after deviating from the straight and narrow and having plunged into the slough of despond, redeems me (Owen, that was copied, or rather lifted, too, from "The Pilgrim's Progress from This World to That Which Is to Come," Bunyon, John, 1678 However, I only include this citation so that Art can sneer at his heritage in association to spitting on me. Any originality that I can claim is in my sense of irony.). I have other wallows in that slough to be revealed, as soon as Owen gets to them in his "power explanation." 73's Richard Clark, KB7QHC |
tuner - feedline - antenna question ?
Cecil Moore wrote:
Jim Kelley wrote: I don't agree that the terms power and energy become interchangeable by virtue of the fact that their units can both be expressed with the word Joule in them. That's why I am willing to switch from the words "Reflected Power" to "Reflected Energy" and measure that energy flow past a point on a transmission line in joules/second. SNIPPED Energy FLOW implies a unit of time. Flow indicates a flow rate. In the absence of a flow rate you are discussing static conditions. Conclusion: all the illogical rational in the world does not change the understanding of energy flow energy/unit time = power = watts = joules/second pass a point or dissipated. Cecil, as an engineer you should stick with standard vocabulary. |
tuner - feedline - antenna question ?
Jeff wrote:
w5dxp wrote: The Z0 of the transmission line has not changed to 100 ohms so normalizing the chart to 100 ohms is not valid. It is just as valid as using 50 ohms, and the result is the same, a changing vswr. No, the center of the Smith Chart is the Z0 of the transmission line (when used on a transmission line). One cannot willy nilly change the reference Z0. The confusion from doing such is obvious. I see you have not commented on the main point of my post, that being that the smith chart shows a changing vswr when you change the source impedance. I think I see the problem. It is an *error* to change the Smith Chart reference point when the source impedance changes while the T-line Z0 and load remain the same. Hint: transmission line transformers would not work if the vswr did not change. Hint: A lossless series-section transmission line transformer has a *constant SWR*. It is the *constant SWR circle* that causes the impedance transformation. A fixed-constant SWR on 300 ohm line looks like it changes when measured with a 50 ohm SWR meter but that is an illusion. The SWR meter *must* be calibrated to the Z0 of the transmission line in order to obtain a valid SWR reading. The impedance is indeed being transformed all around the constant SWR circle. With your software, you are conceptually doing the same thing as using a 50 ohm SWR meter on a 300 ohm transmission line. The meter reading is invalid when taken at face value. The meter reading does NOT indicate a valid SWR on the 300 ohm feedline and neither does your software. -- 73, Cecil, http://www.qsl.net/w5dxp |
tuner - feedline - antenna question ?
Jim Kelley wrote:
How does inserting a circulator load into a circuit that doesn't have one illustrate anything about energy flow other than the behavior of a circuit with a circulator load in it? It proves that the reflected energy made a round trip to the load and back. If there is no such thing as reflected energy, how is that possible? If it is possible in a system with a circulator load, why is it not possible when the circulator load is removed? You seem to be laboring under a misconception about my point of view, Cecil. But it can't be for a lack of attempts on my part to persuade you of it. This question of yours from another posting gives insight into what you are trying to say. how is it that you were able to ascertain that this heat energy was caused by energy that was reflected from the load rather than having come directly from the power supply within the source? How is it that you are able to ascertain that your reflection in the mirror was caused by reflections rather than having come directly from your face? -- 73, Cecil, http://www.qsl.net/w5dxp |
tuner - feedline - antenna question ?
Cecil Moore wrote:
So the question is: With a forward RF energy flow of 200 joules/sec and a reverse RF energy flow of 100 joules/sec, would you agree that there is 300 joules of energy existing in a lossless one-second long transmission line? i.e. exactly the amount of energy required to support the forward RF energy wave and the reflected RF energy wave. I think it depends on how long the energy has been flowing. But in the steady state it's rather like posing this question: With a forward speed of 200 knots, and with a headwind speed 100 knots, would you agree that the apparent airspeed of the aircraft is 300 knots? Or if the above transmission line is one microsecond long, that 300 microjoules of energy exists in the line, i.e. exactly the amount of energy required to support the forward RF energy wave and the reflected RF energy wave. Or even if the forward energy is 200 microjoules/sec, reverse of 100 microjoules/sec through a 1 second transmission line. It's kind of a boring problem though. Personally, I think it's more interesting and enlightening to consider what goes on prior to steady state. 73, ac6xg |
tuner - feedline - antenna question ?
On Thu, 01 Mar 2007 09:58:02 -0800, Jim Kelley
wrote: , is how is it that you were able to ascertain that this heat energy was caused by energy that was reflected from the load rather than having come directly from the power supply within the source? In the theological sense, this predicates that power never becomes dissociated from "the source." That is ambiguous, isn't it? Is that to include the batteries behind the collector supply? The power supply charging the batteries? The power grid feeding the power supply? The generator driving the grid? The Coal firing the steam spinning the generator? The sun through photosynthesis growing plants to provide the coal? The previous supernova that seeded the cosmos by which coalescence formed the sun? ...and into an infinite regression to that previous supernova? The energy dissipated is computed from the Galactic Load Line. I think it's been fairly well established that the output impedance of these things is far from 50 ohms. Can you offer what that complex number is? :-0 73's Richard Clark, KB7QHC |
tuner - feedline - antenna question ?
Dave wrote:
Cecil, as an engineer you should stick with standard vocabulary. Just trying to appease the physicists, Dave. They are arguing that it is not power until work is done. They say that since reflected energy is not doing any work, it cannot be reflected power. Therefore, reflected power doesn't exist. It's purely semantics. The very essence of an EM wave is its energy content. So the real question is: Since standing waves obviously exist and just as obviously cannot exist without two coherent waves traveling in opposite directions, does reflected energy exist? (That question seems to cause their skivvies to get all bunched up.) I will just be happy when they admit that reflected EM waves possess a certain amount of energy that cannot stand still and according to the theory of relativity must necessarily travel at the speed of light. -- 73, Cecil, http://www.qsl.net/w5dxp |
tuner - feedline - antenna question ?
Dave wrote: Cecil Moore wrote: Jim Kelley wrote: I don't agree that the terms power and energy become interchangeable by virtue of the fact that their units can both be expressed with the word Joule in them. That's why I am willing to switch from the words "Reflected Power" to "Reflected Energy" and measure that energy flow past a point on a transmission line in joules/second. SNIPPED Energy FLOW implies a unit of time. Flow indicates a flow rate. In the absence of a flow rate you are discussing static conditions. Conclusion: all the illogical rational in the world does not change the understanding of energy flow energy/unit time = power = watts = joules/second pass a point or dissipated. Cecil, as an engineer you should stick with standard vocabulary. Dave, I couldn't agree more. Energy flow is correct. Power flow is a bit more controversial. In some cases the notion can lead to power being reflected, algebraically summed, and it can ultimately interfere constructively and destructively. It can even, by making exactly the right misinterpretations, end up changing direction without the aid of a reflecting surface. That's why it can sometimes be important to make sure the hairs are properly split. :-) 73, Jim AC6XG |
tuner - feedline - antenna question ?
"Jeff" wrote in
.com: Owen and Cecil are right: the source (transmitter) has no effect whatever on the VSWR on the line. That isn't just an assertion - it is part of the bedrock transmission line theory. Owen referred to "reputable textbooks", one of which would surely be 'Theory and Problems of Transmission Lines' by R A Chipman [1]. This book gains a lot of its reputation from its very complete mathematical development of the theory, showing all the detailed working. I am sorry but you are not correct, I have not read Chipman so I cannot comment on his analysis or your interpretation of his results, but my understanding , practical experiments and CAD analysis would lead me to disagree. If we take the situation where the source is matched (50ohms) to the 5.35 wavelength transmission line (lossless to simplify things) with a 100ohm load, I agree that the vswr is 4:1, unchanging with frequency. Plotted on a Smith Chart when swept against frequency this gives a circle centred on 1 (50ohms) with a radius of 4. i.e. on a constant VSWR circle. Now if we change the source impedance to 100ohms and repeat the same sweep and re-plot, keeping the chart normalized to 50 ohms, the circle moves on the resistance axis, still with a radius of 4 and now passing though 2 (100 ohms) resistive. The centre moves to about 0.6 (30ohms). It then becomes obvious that the locus of the circle is NOT a constant VSWR against frequency. You will come to the same conclusion if you normalize the chart to 100 ohms, the new source impedance and re-plot. The coax is acting as an impedance transformer, causing a shift along the resistance axis. Looking at it another way, the vswr changes sinusoidally with frequency, in our example, between 2:1 and 8:1. (The same as the Smith chart plot with a circle of radius 4 centred at about 0.6). If you are asserting that VSWR on a real or even theoretical line varies sinudoidally with displacement, it is time to go back to basics. You need some time with a reputable text book. Owen |
tuner - feedline - antenna question ?
Richard Clark wrote: On Thu, 01 Mar 2007 09:58:02 -0800, Jim Kelley wrote: , is how is it that you were able to ascertain that this heat energy was caused by energy that was reflected from the load rather than having come directly from the power supply within the source? In the theological sense, this predicates that power never becomes dissociated from "the source." That is ambiguous, isn't it? Is that to include the batteries behind the collector supply? The power supply charging the batteries? The power grid feeding the power supply? The generator driving the grid? The Coal firing the steam spinning the generator? The sun through photosynthesis growing plants to provide the coal? The previous supernova that seeded the cosmos by which coalescence formed the sun? ...and into an infinite regression to that previous supernova? The energy dissipated is computed from the Galactic Load Line. Sarcasm clearly noted, and surprisingly uncalled for. I'll try asking one more time. It is a simple metrology question: How were you able to directly ascertain that the heat being dissipated in the source was produced by energy being reflected from the load? Thanks, Jim, AC6XG |
tuner - feedline - antenna question ?
Jim Kelley wrote:
Cecil Moore wrote: I think it depends on how long the energy has been flowing. But in the steady state it's rather like posing this question: With a forward speed of 200 knots, and with a headwind speed 100 knots, would you agree that the apparent airspeed of the aircraft is 300 knots? Of course, I was talking about steady-state conditions. And I can prove that there is exactly the amount of energy stored in the transmission line necessary to support the forward joules/sec and reflected joules/sec readings. Why violate the laws of physics to try to come up with some other more esoteric explanation than forward and reflected EM traveling wave energy moving at the speed of light? It's kind of a boring problem though. Personally, I think it's more interesting and enlightening to consider what goes on prior to steady state. That's exactly what I have done. During the initial transient build-up period, *exactly* enough energy is supplied by the source to the transmission line to support the steady-state forward and reflected joules/sec - no more and no less. If the source is disconnected during steady-state, the energy stored in a lossless transmission line is dissipated in the load during that final transient decay period. -- 73, Cecil, http://www.qsl.net/w5dxp |
tuner - feedline - antenna question ?
Owen Duffy wrote:
"Jeff" wrote in e.com: Owen and Cecil are right: the source (transmitter) has no effect whatever on the VSWR on the line. That isn't just an assertion - it is part of the bedrock transmission line theory. Owen referred to "reputable textbooks", one of which would surely be 'Theory and Problems of Transmission Lines' by R A Chipman [1]. This book gains a lot of its reputation from its very complete mathematical development of the theory, showing all the detailed working. I am sorry but you are not correct, I have not read Chipman so I cannot comment on his analysis or your interpretation of his results, but my understanding , practical experiments and CAD analysis would lead me to disagree. [...] Looking at it another way, the vswr changes sinusoidally with frequency, in our example, between 2:1 and 8:1. (The same as the Smith chart plot with a circle of radius 4 centred at about 0.6). If you are asserting that VSWR on a real or even theoretical line varies sinudoidally with displacement, it is time to go back to basics. You need some time with a reputable text book. Agreed, but make that a textbook that specifically deals with the subject in enough detail. Chipman was highly recommended by contributors to earlier rounds of this debate. It isn't an easy read, but it's certainly thorough. I ordered the book from the other side of the world because I wanted to be very sure of my answers next time around. We don't know where you are, Jeff, but it would probably be easier and cheaper for you to do the same. -- 73 from Ian GM3SEK |
tuner - feedline - antenna question ?
"Cecil Moore" wrote in message . .. Jeff wrote: w5dxp wrote: The Z0 of the transmission line has not changed to 100 ohms so normalizing the chart to 100 ohms is not valid. It is just as valid as using 50 ohms, and the result is the same, a changing vswr. No, the center of the Smith Chart is the Z0 of the transmission line (when used on a transmission line). One cannot willy nilly change the reference Z0. The confusion from doing such is obvious. You are mis-representing what I said; which was that you can plot the problem using with the chart normalized to EITHER 50 or 100 ohms (the impedance of the generator or that of the line) and the net result will be the same answer. The chart does not necessarily be normalized to the impedance of a transmission line that you are trying add, otherwise you would never be able to include a series line of an impedance other than that of the chart in a matching network. Jeff |
tuner - feedline - antenna question ?
" If you are asserting that VSWR on a real or even theoretical line varies sinudoidally with displacement, it is time to go back to basics. You need some time with a reputable text book. Owen No, I am saying that a fixed length line, mismatched at the far end, and of a different impedance to that of the generator, will not present a constant vswr to the generator with varying frequency. That variation being sinusoidal between a max and min value. Jeff |
tuner - feedline - antenna question ?
On Thu, 01 Mar 2007 11:27:36 -0800, Jim Kelley
wrote: Sarcasm clearly noted, and surprisingly uncalled for. Hi Jim, I responded in kind is all, you revealed a trap and I jumped into it with both feet. If that broke it, return it to the vendor for a refund. Is power/energy separable from its source? If this question is obnoxious, why did you raise the prospect? When it is generally accepted that our sources do not exhibit 50 Ohms source resistance/impedance, what resistance/impedance do they exhibit? If you don't have an answer, what was the purpose of this uninforming assertion? If these two questions have the trappings of sacrasm, I did not originate their discussion. And putting your mock-shock aside, they are part of the chain of denial you are adding links to, aren't they? 73's Richard Clark, KB7QHC |
tuner - feedline - antenna question ?
On Thu, 01 Mar 2007 19:21:43 GMT, Owen Duffy wrote:
If you are asserting that VSWR on a real or even theoretical line varies sinudoidally with displacement, it is time to go back to basics. You need some time with a reputable text book. Hi Owen, Under the right circumstances (and they have been presumed in some discussion here), then the power terms (as expressed by a power meter inserted into the line) will vary sinusoidally with displacement, even if the SWR does not. 73's Richard Clark, KB7QHC |
tuner - feedline - antenna question ?
Jeff wrote:
The chart does not necessarily be normalized to the impedance of a transmission line that you are trying add, otherwise you would never be able to include a series line of an impedance other than that of the chart in a matching network. The point is that it is best to use one Smith Chart for each Z0. Trying to plot multiple Z0's on the same chart leads to the present confusion. The fact that a piece of transmission line with an SWR1 transforms impedances is NOT proof that the SWR is changing. It is *only* proof the the impedance is changing and that happens with constant SWR. -- 73, Cecil, http://www.qsl.net/w5dxp |
tuner - feedline - antenna question ?
Jeff wrote:
No, I am saying that a fixed length line, mismatched at the far end, and of a different impedance to that of the generator, will not present a constant vswr to the generator with varying frequency. That variation being sinusoidal between a max and min value. Changing the length of the ladder-line changes the 50 ohm SWR seen by the source. That is a well known fact. But that doesn't mean the SWR on the ladder- line is changing. It simply means that the impedance is being transformed by the constant SWR circle. -- 73, Cecil, http://www.qsl.net/w5dxp |
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