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Bart Rowlett wrote: Jim Kelley wrote: I think the source of part of the confusion here is that some people apparently interpret the 'forward power' reading on their meter to mean the power into their transmission line. It's not. I sincerely hope you don't think I was implying that YOU were confused by such a thing, Bart. ;-) 73, Jim AC6XG |
Robert Lay W9DMK wrote:
On Thu, 02 Dec 2004 09:47:54 -0600, Cecil Moore wrote: Consider an earlier example made up of lossless lines: 100W XMTR---50 ohm---+---one second long 291.5 ohm---50 ohm load The voltage reflection coefficient at the load is 0.707. The power reflection at the load is 0.5, i.e. half the power is reflected. After steady-state has been reached, the XMTR has output 300 more joules than the load has accepted. A smaller real-world experiment will easily verify that it is a fact that all energy sourced that has not reached the load must necessarily be confined to circulating energy or losses in the transmission line. Question: In the above example, where are those 300 joules of energy located and what is happening to them? We know that 300 joules is wave energy and RF waves always move at the speed of light, i.e. they cannot stand still. So please determine how much energy is moving and in which of only two possible directions. Dear Cecil, Here's my guesses - The 300 joules of energy decays at a particular rate - i.e., in a certain interval of time, 63.2% of it will have been converted to and become dissipated as heat. During that same time interval,Ti, there will be an equal amount of energy introduced to replenish the amount lost. In other words, there will be a continuum of energy transferring into the transmission line to exactly make up for that lost in any given period of time. Yep, the point is that the *net* energy in the transmission line will remain at 300 joules during steady-state. The transmission line is one second long and the source is sourcing 100 joules/sec. That means that 200 joules of "extra" energy is being stored in the transmission line. The forward power is 200 joules/sec, 100 joules/sec higher than the source power. The reflected power is 100 joules/sec. There is exactly where the 200 joules of energy are stored - half in the forward wave and half in the reflected wave whose superposition results in the standing wave. Eventually, someone will disconnect the source generator at time T2. However, the load will continue to receive energy for a length of time, Ti, at which point roughly 63.2% of 300 joules of energy will have been dissipated in the load. If we wait long enough, 99.9% of the 300 joules will have been dissipated, but it will take forever for the last little bit to disappear. It kind of makes you think in terms of "life everlasting", doesn't it? It probably gets close enough to zero for government work. Did you see my question on another posting? If one hooks up a 1/4WL stub to a signal generator, what is the dB loss in the stub? It might have something to do with your original posting. -- 73, Cecil http://www.qsl.net/w5dxp |
Instead of messing about calculating the additional loss due to SWR and then
adding it to the matched loss, I've just had a wonderful idea. Why not calculate the actual line loss directly and solve all your problems at one fell swoop. --- Reg. |
On Fri, 3 Dec 2004 19:00:14 +0000 (UTC), "Reg Edwards"
wrote: Glass half empty theory: Instead of messing about calculating the additional loss due to SWR and then adding it to the matched loss, I've just had a wonderful idea. Glass half full theory: Why not calculate the actual line loss directly and solve all your problems at one fell swoop. |
Jim Kelley wrote:
It's accurate to say that power is something which itself doesn't propagate in any fashion, at any wavelength. I know that's the physicists view, but some engineering views are slightly different because of a differing definition of "power". Let's say we have a one second EM pulse containing one joule of energy traveling along a transmission line with three joules/sec measuring points. For one second periods in succession, we will measure one watt at the measuring points. In each case, the one watt is the same one watt, displaced in time, being transferred to the load. That one watt appears at the first measuring point, then at the second measuring point, then at the third measuring point, and finally heats up the load. It certainly appears to be associated with the pulse moving from the source to the load. That's the way a lot of engineers deal with power, including all the power company engineers that I know, and that's the way the IEEE Dictionary deals with power. Your "tree falling in the forest making no sound" concept may not be the best approach for the real world. -- 73, Cecil http://www.qsl.net/w5dxp |
Reg Edwards wrote:
Instead of messing about calculating the additional loss due to SWR and then adding it to the matched loss, I've just had a wonderful idea. Why not calculate the actual line loss directly and solve all your problems at one fell swoop. What is the formula for the total dB loss? -- 73, Cecil http://www.qsl.net/w5dxp |
Cecil,
I am glad to see you have come around to the correct view that the "extra" energy is stored in a standing wave. Now if only we could convince you that mathematically, physically, and in "reality" there is not the slightest bit of difference between individual wave components and their superposition result then there would be real progress. Oh, by the way, is there some reason you highlight *net* energy? Is there any other kind? Energy is a scalar, so numerically it can be only positive or negative, with no phase angle or other vector properties. Unless part of the energy in your transmission line is negative the net energy would be the same as the gross energy. (Not including sales tax and shipping charges.) I am well acquainted with the concept of negative energy when describing atomic structure, for example, but I have not heard of negative energy in transmission lines. Can you explain the distinction you appear to make? 73, Gene W4SZ Cecil Moore wrote: Yep, the point is that the *net* energy in the transmission line will remain at 300 joules during steady-state. The transmission line is one second long and the source is sourcing 100 joules/sec. That means that 200 joules of "extra" energy is being stored in the transmission line. The forward power is 200 joules/sec, 100 joules/sec higher than the source power. The reflected power is 100 joules/sec. There is exactly where the 200 joules of energy are stored - half in the forward wave and half in the reflected wave whose superposition results in the standing wave. |
Jim Kelley wrote:
It's accurate to say that power is something which itself doesn't propagate in any fashion, at any wavelength. Will someone please try to explain to Cecil that engineering is an application of fundamental physical properties, and that the disciplines of physics and engineering are not at odds with one another? thanks, AC6XG |
Gene Fuller wrote:
Cecil, I am glad to see you have come around to the correct view that the "extra" energy is stored in a standing wave. Never have I ever budged from the view that the "extra" energy is stored in the standing wave components, i.e. in the forward wave and the reflected wave. What I object to is your merging of the forward wave and reflected wave as if they were some sort of homoginized energy. You can prove your beliefs by providing an example of a standing wave not composed of the superposition of a forward wave and a reflected wave. I have offered that challenge before and nobody has stepped up to the plate. Perhaps, you will be the first. I await with abated breath. Now if only we could convince you that mathematically, physically, and in "reality" there is not the slightest bit of difference between individual wave components and their superposition result then there would be real progress. What you are arguing is that there is no difference between a husband and a wife. They are just two components of a family who merge into a single entity so either or both may be male or female. Sorry, Gene, that's not my view of the world. IMHO, it is the *components* that are important. IMO, you liberals are trying to destroy science and society all at the same time. Oh, by the way, is there some reason you highlight *net* energy? Is there any other kind? Of course, there is another kind of energy besides net energy. Is there any difference between a wife and a husband? They are both parents, so you liberals would argue that there is no difference between them. I simply don't agree with you. I am well acquainted with the concept of negative energy when describing atomic structure, for example, but I have not heard of negative energy in transmission lines. Can you explain the distinction you appear to make? It's a convention, Gene. I'm surprised you have never heard of it. ExH traveling in one direction is positive. ExH traveling in the opposite direction is negative. The Z0 of a transmission line forces the E-field to H-field ratio to be Z0, i.e. forces the forward voltage to forward current ratio to be Z0 - and the same for reflected current. One of my references shows a Poynting Vector for the forward power and a separate Poynting Vector for the reflected power, i.e. husbands and wives are not interchangable! -- 73, Cecil http://www.qsl.net/w5dxp |
Jim Kelley wrote:
Will someone please try to explain to Cecil that engineering is an application of fundamental physical properties, and that the disciplines of physics and engineering are not at odds with one another? On the contrary, many of the pure-physics-oriented posters to this newsgroup have alleged that the IEEE Definitions are Bull$hit and are to be ignored. Check Google if you don't believe it. The IEEE Definitions are what engineers abide by. Of course, the disciplines of physics and engineering are certainly at odds with one another. For instance, engineers have an altogether different view of what "work" is, compared to the physicists. Physicists say no work is done if the starting line and the finish line are the same for a marathon. Engineers will say: "Then why am I so friggin' tired?" The POWER engineers at a POWER generating station operated by a POWER company simply don't buy into your pure physics BS. They assume that POWER (joules/sec) leaving a generating station will, after subtracting losses, make money for the company. -- 73, Cecil http://www.qsl.net/w5dxp |
On Fri, 03 Dec 2004 16:13:44 -0600, Cecil Moore
wrote: ExH traveling in one direction is positive. Radiation ExH traveling in the opposite direction is negative. Rearadiation You got transmitters in Texas that suck? |
Richard Clark wrote:
Cecil Moore wrote: ExH traveling in one direction is positive. Radiation ExH traveling in the opposite direction is negative. Rearadiation You got transmitters in Texas that suck? Suggest that you take time to understand the difference between an unterminated Rhombic and a terminated Rhombic and get back to us. Hint: The forward wave radiates in the forward direction. The reflected wave radiates in the reverse direction. The termination eliminates the reflected wave thus eliminating the reverse radiation. I'm surprised that you don't know that. -- 73, Cecil http://www.qsl.net/w5dxp |
Richard Clark wrote:
If that is too hard, how many candela total intensity did we begin with? You first, Richard. What is the difference between a duck? |
Richard Clark wrote: On Fri, 03 Dec 2004 16:13:44 -0600, Cecil Moore wrote: ExH traveling in one direction is positive. Radiation ExH traveling in the opposite direction is negative. Rearadiation You got transmitters in Texas that suck? Maybe so. Texas certainly does seem to have its own unique set of physical laws. ;-) "Physicists say no work is done if the starting line and the finish line are the same for a marathon." :-) No physicist I know would ever say something like that. Only a gross misapplication of Newtonian mechanics and/or thermodynamics could lead someone to such a belief. For example, such a person might think that a ball thrown up into the air has had zero work performed on it after it returns to the hand which threw it. The fact is, it requires as much work to return it to the earth as it does to throw it into the air. (There are some interesting physics demonstrations on how this doubling of work can be used to interesting advantage.) But since in mechanics there is no such thing as negative work, total work is accumulative. Potential and kinetic energies are of course restored to initial conditions, but the conversion from one to the other does not ordinarily occur without some form of external 'help'. The misunderstanding is at least consistent with some similar misunderstandings that have been expressed with regard to the physics of power and energy. ac6xg |
On Fri, 03 Dec 2004 17:57:32 -0600, Cecil Moore
wrote: You got transmitters in Texas that suck? Suggest that you take time to understand That's whining liberal talk - Up or Down, is it yes boy? |
On Fri, 03 Dec 2004 18:00:59 -0600, Cecil Moore
wrote: Richard Clark wrote: If that is too hard, how many candela total intensity did we begin with? You first, Richard. What is the difference between a duck? Your question, put in optical terms, and you can't answer it, that is the difference, duck. :-) |
On Fri, 03 Dec 2004 16:13:44 -0600, Cecil Moore
wrote: [snip] |I await with abated breath. If only it was true [g] |
"Cecil Moore" wrote Reg Edwards wrote: Instead of messing about calculating the additional loss due to SWR and then adding it to the matched loss, I've just had a wonderful idea. Why not calculate the actual line loss directly and solve all your problems at one fell swoop. What is the formula for the total dB loss? -- ================================= I wonder why I ever bothered to introduce Chipman to this newsgroup. Try him. ---- Reg. |
On Fri, 03 Dec 2004 17:54:33 -0800, Jim Kelley
wrote: I can if the light source is a sodium discharge lamp. :-) What sort of light source are you assuming? Hi Jim, I bet you can! Edison/Mazda style tungsten filament light bulb. Available at K-Mart or Walton's or any of a million retail outlets, even in Texas. If not in Texas, then heat a brandin' arn to incandescence (bet he can't tell us what temperature for 555nM tho'). Please, folks, Optics is for the professionals. Don't try this at home! ;-) 73's Richard Clark, KB7QHC |
"Reg Edwards" wrote in message ... "Cecil Moore" wrote Reg Edwards wrote: SNIP I wonder why I ever bothered to introduce Chipman to this newsgroup. Try him. ---- Reg. Reg, This thread is purely a platform for snide remarks or for the pursuit of appearing clever to readers. It is not for technical education but instead it is a duking out of smarmy comments so as to produce a suedo pecking order for onlookers to assuage who has the most accumen with respect antenna education. It also helps in ascertaining the personal ethics of one compared to others. Thus the introduction of Chipman would serve no real purpose other than to deflect the daggers or missiles thrown between individuals. I am amazed that Cecil is willing to stand there whether he is right or wrong when many (not all ) contributors have no interest in keeping to the subject other than to provide utterances that cannot be understood apparently or to throw a stone and then hide. Cecil, you just have nothing to gain by conversing with those whose only intent is to taunt you and not to provide true closure of posting discussion. Respond to those who have something to offer and let the others drown in their own saliva when left alone. Cheers and beers Art |
Cecil Moore wrote:
. . . The IEEE Definitions are what engineers abide by. . . If you believe that, you haven't had much contact with real, working engineers. In my experience, the IEEE definitions are often way out of step with common usage by working engineers. Nearly none in my acquaintance look to it as an authoritative source. A useful guideline, perhaps, at most. I can easily see three causes for the deficiency: 1. The IEEE Dictionary covers an extremely wide variety of rapidly evolving specialties, including power, digital, fields, control systems, fiber optics, electronics, EMC, and on and on. It would be extremely difficult to cover all these disparate specialties accurately and in depth without a huge amount of input from working engineers in each specialty. 2. As far as I can tell, the Dictionary is put together by volunteers, which limits the time and effort which can applied to it. 3. The active membership of the IEEE largely comprises academics rather than working engineers. Academics are a poor source of information about common usage by working engineers. And, working engineers don't tend to "abide by" the dictates of academics, in my experience. I don't have a recent copy of the IEEE Dictionary, but think and hope it's improved over the years. But I'm certain it hasn't come anywhere close to the point at which it's something "engineers" "abide by". Roy Lewallen, W7EL |
Cecil Moore wrote:
Of course, the disciplines of physics and engineering are certainly at odds with one another. That is nothing but an excuse for your own sloppy thinking. When it comes down to fundamentals, physics and engineering must always agree exactly - because they are both working with the same physical reality. That is a bedrock principle, known and shared by all competent physicists and all competent engineers. There's a reason why they call these subjects "disciplines", you know. Reality sets hard rules that you have to follow - or else you'll get it wrong. The only differences between physics and engineering are the acknowledged and clearly understood approximations that each side has to apply in order to follow its own particular interests. Physics is most interested in knowing things, while engineering is most interested in doing things - but neither to the exclusion of the other. If your ideas cannot make the physics and engineering approaches agree, it means that your ideas are wrong. That is a simple and completely reliable test. And it's strictly *your* problem. -- 73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB) http://www.ifwtech.co.uk/g3sek |
Roy Lewallen wrote:
Cecil Moore wrote: The IEEE Definitions are what engineers abide by. . . If you believe that, you haven't had much contact with real, working engineers. What dictionary do "real, working engineers" use? A language without a dictionary is a disaster waiting to happen. -- 73, Cecil http://www.qsl.net/w5dxp |
Ian White, G3SEK wrote:
If your ideas cannot make the physics and engineering approaches agree, it means that your ideas are wrong. If you believe physicists and engineers agree on everything, you are living in never-never land. It's difficult to find two physicists who agree on everything - or two engineers. Witness the arguments on this newsgroup. Ian, there are 13 definitions of "efficiency" in the IEEE Dictionary, each associated with a different engineering discipline. And that's not counting the definitions of "efficiency" that exist in the world of pure physics. Anyone who thinks a word has one and only one definition that everyone agrees upon and encompasses all subjects and all fields is clearly out of touch with reality. For instance, I point to the definition of "power" in the IEEE dictionary. Some posters on this newsgroup disagree with that definition and that's from people who had the same textbook as I did in college. Since that's the case, then of course, some of the ideas of engineering and physics will disagree. "Power" for a power engineer working at a power generating plant measuring megajoules/sec in a transmission line simply does not have the same definition as "power" for a physics professor. -- 73, Cecil http://www.qsl.net/w5dxp |
Cecil A. Moore wrote:
"Power" for a power engineer working at a power generating plant measuring megajoules/sec in a transmission line simply does not have the same definition as "power" for a physics professor. "Power" is just a single word, so it certainly does have to carry several different shades of usage. My point is that the competent engineer and the competent physics professor understand that their different usages are still completely consistent at a fundamental level. I don't believe you understand the discipline that that need for consistency imposes. -- 73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB) http://www.ifwtech.co.uk/g3sek |
"Ian White, G3SEK" wrote in message ... Cecil A. Moore wrote: "Power" for a power engineer working at a power generating plant measuring megajoules/sec in a transmission line simply does not have the same definition as "power" for a physics professor. "Power" is just a single word, so it certainly does have to carry several different shades of usage. My point is that the competent engineer and the competent physics professor understand that their different usages are still completely consistent at a fundamental level. I don't believe you understand the discipline that that need for consistency imposes. Cecil's back at it, I see. -- 73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB) http://www.ifwtech.co.uk/g3sek |
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On Sat, 04 Dec 2004 18:40:20 GMT, Richard Clark
wrote: At 30 MHz, this 1.5pF capacitance represents a reactance of 3.5K Ohm. This rather sweeps aside your specification for the load and replaces it with 4700 -j3537 Ohms. Actually 1700 -j2258 Ohms As you can see, this is the reason why open stubs are avoided. "Knowing" the parasitic capacitance is problematic. The fringing effect at the end is difficult to manage whereas a short is much simpler to define and implement. 73's Richard Clark, KB7QHC |
(PS : You can't deal with reflections without involving Distance, Velocity, Place and Time.) ====================================== Just an addition - You can't deal with Reflections without involving Distance, Velocity, ECHO's, Place, and, above all, TIME. Attempted steady-state analyses reduce to non-sense without involving TIME. But it is a simple matter to relate power levels, one to another, dB-wise or otherwise, at different PLACES and TIME. All you guru's, old wives and experts on forward and reflected powers, real or imaginary, should bear this basic, elementary stuff in mind. Whatever happened to your primay education? Power, work and energy are intimately related to TIME. The self-educated kids living in South American underground Rio sewers, subjected to culling by armed police, reduced to making a precarious living be selling laced cigarettes, can do better. But don't worry about it. Just continue taking your viagra. As for me, tonight I'm on white, dry, Bordeaux, best consumed within six months of purchase. And the French know what they are talking about. ---- Reg. |
On Sat, 04 Dec 2004 21:51:09 GMT, Richard Clark
wrote: On Sat, 04 Dec 2004 18:40:20 GMT, Richard Clark wrote: At 30 MHz, this 1.5pF capacitance represents a reactance of 3.5K Ohm. This rather sweeps aside your specification for the load and replaces it with 4700 -j3537 Ohms. Actually 1700 -j2258 Ohms As you can see, this is the reason why open stubs are avoided. "Knowing" the parasitic capacitance is problematic. The fringing effect at the end is difficult to manage whereas a short is much simpler to define and implement. 73's Richard Clark, KB7QHC Dear Richard, Each and every one of your concerns is deeply appreciated. I had seriously about measuring that 4700 ohm resistor by itself for just the reasons that you mention. The reason that I did not is quite simple. The Resistance dial on the General Radio Model 1606B RF Bridge only goes to 1000 ohms. C'est la Vie! I could have chosen a much lower resistance to start with, but then I would not have had the desired high SWR, which is critical to the experiment. You might well ask why I made the switch from a 1/4 wave open ended stub (at 10.6 Mhz) to the 30 MHz test using the same piece of line. That was to get a high SWR and still have reasonable confidence in the value without actually measuring it. Perhaps I should use a capacitor instead. That's still a possibility. I am thinking of doing it over with a measured 1 k resistor, and just accepting the SWR of only 20:1. I don't think I have any low ohm resistors that would give me something around 1 to 3 ohms and still be low inductance - not from my parts box. Actually, I doubt seriously that the capacitance effects are as bad as you are suggesting, but I won't argue that without some additional measurements. My reason for not going with a short circuit was even simpler. I wanted a concrete value for a non-zero load power. Regarding the artificial adjustment of the attenuation loss from around 0.9 up to about 1.72 did not bother me at the time that I did it, but I can see why it bothers an independant observer. I will have to do something about that, and perhaps doing it over with a 1 K resistor is the answer to all of that - so, here goes, and I will get back to you in a few hours. BTW - this is becoming more and more academic, since I think the real mystery has been resolved, but there's nothing like getting it wrapped up with a proper ribbon. Bob, W9DMK, Dahlgren, VA http://www.qsl.net/w9dmk |
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Cecil Moore wrote:
Reg Edwards wrote: Instead of messing about calculating the additional loss due to SWR and then adding it to the matched loss, I've just had a wonderful idea. Why not calculate the actual line loss directly and solve all your problems at one fell swoop. What is the formula for the total dB loss? I don't know of one, but here's a procedure which will get you the answer: 1) Determine the transmission line characteristics R, G, L & C at the frequency of interest. 2) Determine the transmission line termination impedance (or admittance) at the frequency of interest. 3) For convenience, define the excitation at the input to the transmission line. e.g. defined voltage, current, or generator with defined source impedance. 4) Solve the transmission line equation for both the voltage and current, as a function of time and position, subject to the imposed boundary conditions. Use any method you wish, though the usual approach is the so called method of reflections where the complete solution is taken as the summation of damped forward and reverse traveling waves. Calculation of Zo and Gamma require knowledge of R,G,L & C. The solution can be expressed algebraically in terms of hyperbolic functions and will be found in any good transmission line reference. Now known are v(x,t) and i(x,t). 5) Separately calculate the dissipative losses due to series resistance and shunt conductance (or dielectric power factor) as follows: a) The loss due to series resistance at any instant in time for some section of transmission line, is i(x,t)^2 * R * dx, integrated over the subject transmission line section. b) The loss due to shunt conductance at any instant in time for some section of transmission line, is v(x,t)^2 * G * dx, integrated over the subject transmission line section. c) The total dissipative power loss is the the sum of the time averaged series and shunt loss components, integrated over the entire transmission line length. Consider: The SWR (reflection coefficient at the termination) alone is not sufficient to determine the loss in general. This is an important concept to understand. For illustration, assume the limiting case where the shunt conductance can be disregarded. This is a useful approximation for many transmission line installations used at HF where the SWR is reasonably low (50). In such practical cases the losses will be dominated by series resistance. Consider a short (45 electrical degrees) transmission line terminated in a pure resistance which is much less than Ro. The standing wave pattern will be such that the current will be maximum, and the voltage minimum, at the termination. The losses can be calculated using the procedure given above. Now consider the same transmission line terminated in a pure resistance larger than Ro such that the SWR is the same as previously. The current will now be minimum at the termination and the total dissipative loss will be considerably less than when the termination resistance is low, even though the SWR is identical. Transmission line losses depend on both the actual load impedance and the transmission line parameters. Knowing matched attenuation (dB/length) and load SWR is not sufficient to calculate losses exactly. Any formulas using only SWR and matched transmission line loss are approximations and must be used carefully! Feeding electrically short antennas with 50 ohm coax as is commonly attempted by amateurs on 80 & 160m can result in transmission line losses much larger than predicted using the simple approximations given in the amateur literature. bart wb6hqk |
You are all floundering about like fish out of water. The ONLY way to predict the performance of a small resistor at UHF is to treat it as a transmission line - starting with the simple dimensions of length and diameter. L, C and R, and short-circuit input resistance versus frequency automatically drop into your laps with adequate ball-park accuracy. As so-called engineers, if you can't measure it, such obvious arithmetical techniques should already have been included in your elementary analytical educations. Spice (of which I have heard about only by reputation on newsgroups) is useless unless one already knows what it's all about beforehand. My criticism is not personal - only of the western world's educational systems and insistent dependence on weapons of mass-destruction. ---- Reg. |
Cecil Moore wrote:
Reg Edwards wrote: Cecil, when there are several different power levels at different places in a circuit, it is entirely up to you how you reference one to another in dB. Now Reg, that just cannot be true. It is true. Otherwise, there would exist no conventions. Multiple conventions exist in all the technical disciplines. A well versed practioneer is familiar with the various conventions commonly in use and will choose the one most appropriate to the problem at hand. General terms such as 'gain' and 'loss' always require the context be carefully established before the term can be used with any precision. For example, dissipative loss in a transmission line is different from transducer loss, but both are often expressed in decibels. One relates to an actual ratio of two physically meaningful power levels, and the other is notional. It's a simple question: When you tell me that the losses in a transmission line with reflections are 2 dB, exactly what power are you referencing those losses against? Reasonable question. Without knowing the context the statement is essentially meaningless. Reflected volts - yes! Reflected current - yes! Reflected power - NO! Reflected volts and reflected current existing without any associated joules/sec???? Reg didn't say any such thing. The wave equation is usually expressed in terms related to the two complementary field quantities associated with energy storage. In the case of TEM transmission lines, the two variables are voltage, and current. The method of reflections (images) used to facilitate solution of the wave equation, utilizes the concept of voltage and current reflection coefficient respectively. These concepts are used to find the v(x,t) and i(x,t) without any reference to to power or energy. Once v(x,t) and i(x,t) are known, the power, v(x,t) * i(x,t) can easily be calculated at any point x and time t. The energy storage density can be calculated as i(x,t)^2 * L/2 + v(x,t)^2 * C/2. As you might expect, the v(x,t)*i(x,t) is the derivative of the energy storage density. I've heard of waves without any trace of energy before, Reg, but I certainly didn't expect to hear miracle metaphysics from you. 'Solving' the wave equation for TEM transmission lines usually means determining the values of the v(x,t) and i(x,t) as a function of place and time. Energy distribution, and it's movement, is easily calculated from the knowledge of v(x,t) and i(x,t). Whatever happened to V*I*cos(theta) being power? Hasn't changed; works for DC out to upper microwave frequencies. The power companies would be surprised to learn that they are not transferring any joules/sec to their customers. Indeed they would. Who do you know who believes they are not transferring energy to their customers? bart wb6hqk |
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On Sun, 5 Dec 2004 17:30:23 +0000 (UTC), "Reg Edwards"
wrote: You are all floundering about like fish out of water. .... My criticism is not personal Of course not Reggie, But neither is it original, and if you had been following the technical correspondence instead of the fluff offerings, you would have observed neither of us is in danger of falling prey to all the ills you offer cautions against. Preach to those who need redemption and confine your comments along our technical interchanges to coin in kind. 73's Richard Clark, KB7QHC |
Bart Rowlett wrote:
A well versed practioneer is familiar with the various conventions commonly in use and will choose the one most appropriate to the problem at hand. My point exactly, Bart. It is not up to me to define a new standard. Indeed they would. Who do you know who believes they are not transferring energy to their customers? There are people who believe that energy being transferred past a point in the transmission line is not power. However, the IEEE dictionary says that power is the rate of generation, transfer, or consumption of energy. -- 73, Cecil http://www.qsl.net/w5dxp |
Bart, I don't doubt that your description of how to calculate line loss is correct. But it's more simple than that. The input and output currents (or voltages) of any line are very rigid and well defined. Regardless of generator internal impedance. Just as a transformer's input and output currents and voltages are defined. Their ratio is even more simply defined, albeit in complex terms. So when the line input RESISTANCE, regardless of input reactance, and the terminating RESISTANCE, regardless of termiating reactance, are known then the power input to power output ratio is easily directly calculated from I squared R. The hardest but still straightforward part of the calculation is for the input impedance, Rin+jXin, for given line attenuation and phase shift, and for given terminating impedance, Rt+jXt. The power dissipated in the line, although distributed in a highly complex manner along its length due to standing waves, is very simply Pin - Pout. Throughout the calculations, things like reflection coefficients and SWR do not appear, You cant use them as spin-off benefits. What practical use could you make of them anyway? The calculations have already been done, finished, ended, caput! ---- Reg. |
Preach to those who need redemption
and confine your comments along our technical interchanges to coin in kind. 73's Richard Clark, KB7QHC ============================== I am preaching to anybody who needs redemption. It is YOU who are taking it personally despite me explicitly letting you off the hook. But Rich, it's very unlike you to take cover in "off topic". ;o) --- Reg. |
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