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Dr. Slick wrote:
You're right, but we are talking about a one-port network, the antenna and transmission line. Hmmmmmm, I could have sworn that you were talking about the source impedance. :-) -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
wrote:
Let's apply these expressions to some simple examples. Connect a length of 50 Ohm transmission line to a 9 Volt battery and wait for the transient to die: Why do you need to obfuscate the discussion by resorting to DC? Why can't you explain things using RF? One previous poster refused to discuss a single source system for some reason. Since reactance disappears during DC steady-state, the relevance of a DC example is certainly questionable. Hint: The Z0 of a piece of coax at DC is NOT 50 ohms. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
W5DXP wrote:
wrote: Let's apply these expressions to some simple examples. Connect a length of 50 Ohm transmission line to a 9 Volt battery and wait for the transient to die: Why do you need to obfuscate the discussion by resorting to DC? Mostly because a sensible answer should be able to explain from DC to RF. If it only works at RF but collapses with the much simply DC, then the answer seems suspect. Why is there so much protestation about having the answer explain DC behaviour as well? Why can't you explain things using RF? I like my answers to explain the more complex (RF) as well as the simpler (DC). As well, applying the answer to the simpler system sometimes points out absurdities in the answer. One previous poster refused to discuss a single source system for some reason. I fully agree. If the explanation works for the more complex system, it should also work for the simpler. Since reactance disappears during DC steady-state, the relevance of a DC example is certainly questionable. Perhaps, but we are dealing with simple circuit theory here using ideal components. All seems to work. Hint: The Z0 of a piece of coax at DC is NOT 50 ohms. I've seen nothing in the derivations for ideal transmission lines that indicate frequency dependence. Care to elaborate? In any case, if you wish to stay away from DC, we could just use an arbitrarily low AC signal. How about 10**-100 Hertz? Indistinguishable from DC for any engineering purpose? Or should we go lower? Just pick the number. ....Keith |
W5DXP wrote:
wrote: W5DXP wrote: Unfortunately, the analogy is not a good one. In a transmission line, there must exist a discontinuity to cause a reversal of momentum of the waves. No such discontinuity exists so there is nothing to reverse the momentum of the forward and reflected waves. But then I was not talking waves, but charge. The waves are a manifestation of the changes in the charge distribution. Yes, I know that. When are you going to stop confusing the charge carriers with the waves? :-) The charges have exactly as much to do with the waves as water molecules have to do with water waves. Water molecules don't cross certain boundaries but that doesn't keep tsunamis from wiping out an island. Well, charge does seem to be key. Current is charge per time. Power is voltage times current. I fail to understand the opposition to using charge in the discussion. It is a basic element of circuit theory. Correct. I hope I have not given any other impression. For the most part in my discussions I mean instantaneous power, since conversion to average loses too much information to enable understanding. Instantaneous power is essentially useless to this discussion according to Hecht in _Optics_. Here's a quote: "Since the power arriving cannot be measured instantaneously, the detector must integrate the energy flux over some finite time." Irradiance is *average* power. This may be a challenge in optics but it is not at the lower AC frequencies especially in a circuit. My power supplier does it all the time. And sends me a bill to prove it. But for the most part, greater understanding will arise from sticking with instantaneous energy flow. I hope, for your mental health, that you don't really believe that. :-) If it doesn't work in optics, there is no reason to believe that it will work for other photonic waves, including RF. Actually, I suspect that optics has more to learn from RF than vice versa since, with RF, one has the luxury of measuring both voltage and current (and simultaneously at lower frequencies). Optics seems constrained to pure average power measurements. But at certain points (1/2 wavelength apart), the voltage IS always 0 so the energy flow IS always 0. Unless you are prepared to discard Pinst = Vinst * Iinst. The *NET* energy flow is zero. The component energy flow encounters absolutely nothing that can change its momentum. There are NO impedance discontinuities to cause reflections. It would take a Transmission Line God to do that. It appears that you are prepared to ignore Pinst = Vinst * Iinst. Could you expand on when Pinst IS equal to Vinst * Iinst and when it isn't? Simply convince me that Pinst is not always equal to Vinst * Iinst and my difficultes will disappear. ....Keith |
p(t) = v(t) * i(t). Period. No phase, no vectors, no cross products.
You can't calculate instantaneous power from phasor quantities, because phasors assume a frequency that's the same for all quantities, and the frequency of the power wave is twice that of the voltage and current waves. Of course, a lot of people mean "average power" when they say "power", so you have to be very careful when reading anything written by those folks. Roy Lewallen, W7EL W5DXP wrote: wrote: W5DXP wrote: wrote: Do not be afraid to admit that you have changed the definition of P = V x I and therefore do not accept the standard definition. Well, I was taking 'x' as a multiplication sign. Did you mean it as a cross product sign? No, simple multiply. Well, then your equation is wrong. It should contain phase. The power is V*I*cos(theta) which is known to be *AVERAGE* power since V and I are RMS values. When the total voltage is zero and the total current is 2*I, the phase angle is still 90 degrees for a lossless stub. . . . |
W5DXP wrote:
wrote: Let's apply these expressions to some simple examples. Connect a length of 50 Ohm transmission line to a 9 Volt battery and wait for the transient to die: Why do you need to obfuscate the discussion by resorting to DC? Why can't you explain things using RF? One previous poster refused to discuss a single source system for some reason. Since reactance disappears during DC steady-state, the relevance of a DC example is certainly questionable. Hint: The Z0 of a piece of coax at DC is NOT 50 ohms. . . . It is for the time it takes for the transient to die. Y'see, when you connect a battery to a transmission line, you don't have DC. You only have DC after you wait around forever or, for practical purposes, until everything settles to its final value within some small error bound. It's also wise to remember that, likewise, when you connect a source to a line, you're not dealing with a single frequency. The system settles down to a single frequency only after all transients have died out. That's why a time domain analysis is necessary, or at least highly preferable, to analyze transient conditions. But I do agree that, although it's instructive and can help understanding of transmission line phenomena, we have to be careful when extrapolating pulse or step results to sine wave situations -- especially steady state. Roy Lewallen, W7EL |
wrote:
I see your confusion and apologize for not being completely clear. When I say P = V * I, P, V and I are instantaneous values, the only ones worth exploring if an understanding is desired. Instantaneous power is useless if an understanding is desired. I agree with Hecht and all my other references. Kraus, Jasik, Balanis, Hecht, Ramo & Whinnery all agree that dealing with instantaneous power is a waste of time. Instantaneous power is essentially meaningless since the definition of power requires a length of time. There is simply no such thing as instantaneous power. Power is always the energy passing a point during a slot of time. Zero energy passes a point in dt of time as delta-t approaches zero, by definition. Well, if you do a bit of fudging you can always make it work. But I do not observe these fudge terms in the expressions related to forward and reverse power in transmission lines. Then you haven't read Dr. Best's QEX article. The interference is there and takes the form of 2*Sqrt(P1)*Sqrt(P2) for 100% constructive interference. Of circuit theory I have a reasonable grasp, optics I leave to others. Too bad. The field of optics has already solved the problems with which you are wrestling. Take a look at this web page to figure out how Z0- matching works. http://www.mellesgriot.com/products/optics/oc_2_1.htm As mentioned previously: stored in the line and moving back and forth between quarter wave boundaries. Mentioned, but not explained. How does the energy's momentum change direction in a constant Z0 environment? What causes the back and forth movement? Back in basic circuit theory, some years ago, it was permitted to short points of equal voltage and open conductors with no current. Has this changed in the intervening years? Ahhhh, I see your confusion. Circuit theory and transmission line theory are not the same thing. Many have tried to mix the two models and fallen on their faces. Transmission line theory is simply more complicated than basic circuit theory. If you don't believe it, replace a transmission line with ghosting with an equivalent circuit - the ghosting disappears. Are they really equivalent? Because the observed voltages and currents are the same. The circuit has been replaced by one which is indistinguishable from the first. BS! I can certainly distinguish between a short and a non-shorted transmission line. Please try again. You have gone from 100% transmission of waves to 100% reflection of waves. An intriguing assertion, but one which can not be demonstrated through any observations made on the circuit. Of course it can by observing ghosting in a TV signal. With the line shorted, there is no ghosting. With the line not shorted, there is ghosting. Why is that so difficult to understand? I am not sure why you are looking so hard for reflections. I am not attempting to claim that there are any. I simply claim that no energy can cross the boundaries because the current or voltage is always 0. In the real world, if no RF energy crosses a boundary, then that energy is reflected since it cannot stand still. Of course, the supernatural world is an entirely different matter. Basic electriciy. This says nothing about the presence or absence of reflections or whether there is a mechanism to prevent the energy crossing the boundary or it just happens through the dynamics of the distributed capacitance and inductance. Unfortunately, this is beyond the ability of basic electricity to handle. It takes the wave reflection model (or quantum physics) to handle it. Einstein said a model should be as simple as possible, but not too simple. Once again, you seem to be trying to force reality to obey your model instead of vice versa. Oh, I fully understand what a 'directional watt' meter will indicate. Then why are you so confused? :-) May I suggest that you perform the same experiment with a real instantaneous watt meter: This is obviously a diversionary tactic. Instantaneous power is essentially meaningless according to all my references and I have a bunch of them. Would you care to provide a reference that seriously deals with instantaneous power in transmission lines? I'm sorry, I do not grasp what you are attempting to say here. That's more than obvious. -- 73, Cecil http://www.qsl.net/w5dxp "One thing I have learned in a long life: that all our science, measured against reality, is primitive and childlike ..." Albert Einstein -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
wrote:
Well, charge does seem to be key. Sorry, charge is not the key. The charges are simply the medium that makes the waves possible. Actually, I suspect that optics has more to learn from RF than vice versa ... You have got to be kidding. Optics is at the leading edge. RF is at the trailing edge. Things that are common knowledge in optics are dismissed by this newsgroup (including you?). It appears that you are prepared to ignore Pinst = Vinst * Iinst. I will go with all my references - instantaneous power is essentially useless for any kind of serious analysis. Average power is the only thing that matters at RF and optical frequencies. Please provide a transmission line reference that extols the virtues of instantaneous RF power. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Roy Lewallen wrote:
p(t) = v(t) * i(t). Period. No phase, no vectors, no cross products. Millions of power engineers have been taught that V*I*cos(theta) is power. Are you disagreeing with that teaching? p(t) = v(t) * i(t) is virtually worthless to a power engineer since a generator is a giant heat sink. Most of my references say that instantaneous power is virtually meaningless for RF and optics work. Can you provide a reference that extols the virtues of instantaneous power applied to RF? Can you list any benefit for considering instantaneous RF power? -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Roy Lewallen wrote:
W5DXP wrote: wrote: Connect a length of 50 Ohm transmission line to a 9 Volt battery and wait for the transient to die: Hint: The Z0 of a piece of coax at DC is NOT 50 ohms. It is for the time it takes for the transient to die. Did you not see where he said "wait for the transient to die"? -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
W5DXP wrote:
Instantaneous power is useless if an understanding is desired. A number of texts, in presenting analyses that enhance understanding, use both instantaneous power and the instantaneous value of the Poynting vector as part of the explanation. Perhaps you mean that it adds unwelcome complexity to your simplified analysis? I agree with Hecht and all my other references. Kraus, Jasik, Balanis, Hecht, Ramo & Whinnery all agree that dealing with instantaneous power is a waste of time. Instantaneous power is essentially meaningless since the definition of power requires a length of time. There is simply no such thing as instantaneous power. Instantaneous power is defined in the first half dozen books I pulled from my bookshelf. They we Physics: Weidner & Sells Circuit analysis: Pearson & Maler Network analysis: Van Valkenburg ("Power, instantaneous" is even indexed) Electromagnetics: Johnk, Holt, Kraus You certainly have a lot to straighten us out on in your forthcoming paper! Power is always the energy passing a point during a slot of time. Zero energy passes a point in dt of time as delta-t approaches zero, by definition. By exactly the same reasoning, instantaneous current can't exist, since it has the same relationship to charge as power does to energy. By your argument, we should all through away that "useless" and "boring" analysis using time-varying currents, and deal only with average currents. The new math indeed. . . . Roy Lewallen, W7EL |
W5DXP wrote:
. . . This is obviously a diversionary tactic. Instantaneous power is essentially meaningless according to all my references and I have a bunch of them. Would you care to provide a reference that seriously deals with instantaneous power in transmission lines? It was just day before yesterday I posted just such a reference (Magid). But you seemingly rejected it out of hand as being probably wrong without making an effort to even look at it. So what's the point in providing you with references? Roy Lewallen, W7EL |
W5DXP wrote:
Roy Lewallen wrote: p(t) = v(t) * i(t). Period. No phase, no vectors, no cross products. Millions of power engineers have been taught that V*I*cos(theta) is power. Are you disagreeing with that teaching? Yes. p(t) = v(t) * i(t) is virtually worthless to a power engineer since a generator is a giant heat sink. Most of my references say that instantaneous power is virtually meaningless for RF and optics work. Can you provide a reference that extols the virtues of instantaneous power applied to RF? Can you list any benefit for considering instantaneous RF power? I've addressed the remainder of this in another posting, except for the last question. The last question has been the subject of many postings I've made over the past several months. If your memory needs refreshing, a Google search should locate them quite easily. Roy Lewallen, W7EL |
The issue of the max. value of the magnitude of the reflection
coefficient came up in our student mess. It was shown in lab that it must approach the square root of 5, which is close to 2.41. But what is all of the noise about reflection coefficient (or SWR), which measures what is happening down stream, having anything at all to do with what is happening up stream. Wow. I have never had a student who was that illogical. However, if I were to encounter such a person at this time of life, I would suggest a career in politics or the UN. My hat is off to those of you who are succeeding in educating the lurkers. I marvel at your restraint towards others. 73 Mac N8TT -- J. Mc Laughlin - Michigan USA Home: |
Roy Lewallen wrote:
. . . By your argument, we should all through away that "useless" . . . That's "throw", not "through", of course. I really do know better. Roy Lewallen, W7EL |
Ian'
I don't see from what what Reg said that this should have been unloaded on him in such a way. I sure hope the " establishment" or "cabal" as I call them don't follow you and pile it on . Seems to me that Reg is a giver not a taker as far as ham radio is concerned, and just as much as a asset to this group as those wish to fire off a shot at the first oportunity to give dutch courage to all those that wil follow. Now you can stick the label on me also, audacity was it? After all I did say "cabal" ? What really surprises me is that you have not noticed over the years how many people have been attacked so wickedly on this newsgroup by people that infer that "all is known" to which you whimsically referred to. The fact is that is if this were true a thread would never exceed five postings after a particular expert had posted. I can only imagine that Reg has not swallowed all you have said and thus has raised your ire such that you have lost your cool Art A real XG "Ian White, G3SEK" wrote in message ... Reg Edwards wrote: It should not be forgotten this newsgroup is primarily intended for exchange of information between radio amateurs. It is fairly clear there's a number of retired professional engineers who are also radio amateurs who frequent this newsgroup. Fine! I'm a long-retired engineer myself. There are also a sprinkling of engineers who are professionals in that they are paid for their services of some sort or another. There are some people, perhaps with some small historical justification, have the audacity to consider themselves to form an informal Establishment. Reg, this "Establishment" is something that you just invented. The audacity is yours. Yes, there are people on this newsgroup who agree on lots of technical points. And yes, they are often quite insistent about it. (At this point, I'll change to "we", because I am one of the people we're talking about. And despite his chosen role as maverick, provocateur and general gadly, so too is Reg ;-) Nobody knows everything, but each of us has been around for long enough to have covered most of the same territory, and to have made all the common mistakes. But we've also been around long enough - and been persistent enough - to have come out the other end with some correct answers. (Here comes Philosophy of Science in two paragraphs. Please don't give me a hard time over the things I'm going to leave out.) What's a correct answer? One that accurately describes external, physical, observable, measurable reality, as nearly as we can measure it or need to know. People who take that view are not too comfortable with approximations and half-truths, because sooner or later we're going to be caught out by a new situation where the "nearly correct" answers don't work any more. In the long run, it's simply easier to work out a correct answer that can be trusted. Each of the people Reg is talking about (in fact, especially Reg himself) worked out many of those answers for ourselves, way before we met on the Internet. But when we did meet, it wasn't at all surprising to find ourselves agreeing on so much. That's because all correct answers are describing the same physical reality, so they *have* to agree. The only allowable difference is to choose different viewpoints upon the same reality; but if they are correct, all those alternative viewpoints still *must* be consistent with one another - they must *not* contradict in even the smallest detail (unless it's acknowledged to involve an approximation). Those rules are rigid, but nobody made them - reality did. The only available choice is whether to accept them and stick by them, or to sometimes let them slide. I choose to accept them, as much as I possibly can, because it's actually easier in the long run. If someone else's better logic forces me to re-evaluate, that's really not a problem because I'm better-off than before. Yes, people like that can be rather insistent! But it's not because we want our personal faiths or beliefs to prevail - it's not about us at all. It's because we want external, physical, observable, measurable reality to prevail. Cue for somebody to pipe up about quantum physics, Heisenberg, uncertainty and all that. There are certain areas of science and engineering where these apply and even dominate, and mean that certain things are fundamentally un-knowable. Antennas, transmission lines and electronic engineering are *not* one of those areas. It's an excellent approximation to say that everything in these areas of physics and engineering is fixed and "real", and - if we work hard enough at it - knowable. The exceptions, such as noise and the internal physics of semiconductors, can all be treated by engineers as macro-scale observables. A good example is the tunnel diode, which relies on a probabilistic quantum phenomenon that just can't happen according to classical science... but at the engineering level you can still rely on the device data. You don't see "Probably" stamped on each page. Cue now for someone else to pipe up "But we don't know everything yet." No, we don't... but for every genuine new discovery, there are billions of mistakes, experimental and logical errors. When straightened out, these simply confirm that what we already knew is correct. You'd better learn to trust that body of existing knowledge much more that you trust yourself! I certainly do, and again life is easier as a result. The good thing about the Internet is that we can all learn much faster than before. I certainly have. The bad thing is that we no longer make our mistakes in private. Stuff that used to end up as crumpled balls of paper, to be thought about more clearly tomorrow, is now being published every night to be read by thousands of people and archived forevermore. It's becoming harder and harder to identify the good information among all the rest. To anyone who has already made those same mistakes, it seems such a waste to see people wandering off and doing it all over again. It's very hard to stand by and let it happen... |
"J. McLaughlin" wrote
The issue of the max. value of the magnitude of the reflection coefficient came up in our student mess. It was shown in lab that it must approach the square root of 5, which is close to 2.41. ================================ You don't need a lab. All you need is a pencil and the back of a cigarette packet. Theoretical Max possible value = 1+Sqrt(2) exactly = 2.4142136 . . . . . Can't imagine where you get 5 from. It occurs when line Zo = Ro - jXo has an angle of -45 degrees, ie., when Xo = -Ro, and when the line is terminated in an inductive reactance of +jXo. ---- Reg. |
On Tue, 19 Aug 2003 03:06:15 -0700, Roy Lewallen
wrote: Thanks for the analogy. One can mathematically and conceptually conceive two opposite-traveling waves that add up to the observed standing wave, and that's fine. The problem comes with assigning power or energy to the waves. Then you run into the problem of how one wave got the energy over the barrier into the pocket and the other wave took the same amount back out, without transfering any air molecules across the barrier in the process. The average power analysis looks to me something like this. Suppose you have two batteries each with exactly 2 volts potential and zero internal resistance, with a 2 ohm resistor connected between their positive terminals. The negative terminals are connected together. You replace the battery on the left with a short (turning it off), and observe that the current through the resistor is one amp to the left. Then you hook the left hand battery back up and turn the right hand battery "off" by replacing it with a short. You observe that there's one amp now flowing through the resistor to the right. Finally, turn both batteries "on" by putting them both in place. You can use superposition to conclude, correctly, that there's zero current flowing in the resistor. But it's silly to insist that there's a forward two watt "power wave" flowing to the right, and another two watt wave flowing to the left. You subtract one from the other and, sure enough, get zero. But are the "power waves" real? Studying and analyzing these imaginary waves is surely a lot more interesting than simply looking at the circuit and noting that the "boring" (as Cecil calls it) net power is zero. But aren't you studying ghosts? Interesting question, isn't it? It seems to me that this is a class of problem that is succeptable to solution by several approaches, classical network theory, or transmission line theory and perhaps others. Identifying the solution that best matches physically observable phenominia isn't always obvious. Applying Ockham's Razor (spell it any of the alternative ways you wish) one might prefer the simpilist theory that produces correct results. But, to some extent the artifical conditions necessary in such a statement are responsible for some of difficulty. I = E/R but as as you make R zero, you have introduced a singularity into Ohm's law, so in some sense I isn't determinable. There is also a finite propagation time for current flow across the resistor, (and through the battery, for that matter) and that might argue for a transmission line solution. It also reminds me a bit of the question posed in electromagnetic theory class. You've probably seen it before as well. You have two wires going through the room you are in. With a clamp-on ammeter, you measure 1 amp in each wire, in opposite directions. With a volt meter, you measure 1,000 volts between the two wires. When asked to calculate the power flow, you answer 1,000 watts, being 1KV * 1A. However, the actual circuit is 4 wires, and you only see two in the room. Two loops of 1 volt batteries with a 1 ohm series resistance, with the two loops bridged via a 1KV source. No current flow from the 1 KV battery and the total power is 2 watts, being 1 watt in each loop. As I recall, the only way to get the correct answer is to properly determine the Poynting vector, but it's damn near 40 years since I looked at it. 1 ohm A B 1 ohm +---WWW--------+ +---WWW--------+ | | | | | | | | + + + = 1 volt battery | = | + | | | | +--------------+--+||---+--------------+ A B 1KV battery You can only see the wires at A-A and B-B. Jack K8ZOA Even more risky is adding the things. This time hook two one volt batteries in series with the 2 ohm resistor and energize one at a time. With the upper one on and the lower one "off" (replaced with a short) you get 1/2 amp. You've got a "power wave" of I^2 * R = 1/2 watt. Turn the lower one on and the upper one "off", and you get another "power wave" of 1/2 watt, in the same direction. Turn them both on, and you have a power flow of, um, 2 watts. Welcome to the new math. Roy Lewallen, W7EL Sounds like a variation on the old two identical caps with equal charge are shorted together. Where does the excess energy go? Jack Smith wrote: Roy: Interesting point and I don't recall reading or hearing it elsewhere. The following is dashed off without fully thinking it through, so no warranty on its accuracy. If you think of a sound wave (longitudinal transmission, of course) in a lossless acoustic transmission line terminated with a short, the individual air molecules within each 1/4 wave section are likewise trapped since at the 1/4 wave points there is zero sound pressure. This may be a useful analogy for the electromagnetic transverse propagating T-line. Jack K8ZOA |
W5DXP wrote:
wrote: I see your confusion and apologize for not being completely clear. When I say P = V * I, P, V and I are instantaneous values, the only ones worth exploring if an understanding is desired. Instantaneous power is useless if an understanding is desired. I agree with Hecht and all my other references. Kraus, Jasik, Balanis, Hecht, Ramo & Whinnery all agree that dealing with instantaneous power is a waste of time. Instantaneous power is essentially meaningless since the definition of power requires a length of time. There is simply no such thing as instantaneous power. May I recommend differentiation: In the limit, as t approaches 0.... Very useful results can be obtained with this technique. Power is always the energy passing a point during a slot of time. Zero energy passes a point in dt of time as delta-t approaches zero, by definition. Well, if you do a bit of fudging you can always make it work. But I do not observe these fudge terms in the expressions related to forward and reverse power in transmission lines. Then you haven't read Dr. Best's QEX article. The interference is there and takes the form of 2*Sqrt(P1)*Sqrt(P2) for 100% constructive interference. Of circuit theory I have a reasonable grasp, optics I leave to others. Too bad. The field of optics has already solved the problems with which you are wrestling. Take a look at this web page to figure out how Z0- matching works. http://www.mellesgriot.com/products/optics/oc_2_1.htm As mentioned previously: stored in the line and moving back and forth between quarter wave boundaries. Mentioned, but not explained. How does the energy's momentum change direction in a constant Z0 environment? What causes the back and forth movement? Back in basic circuit theory, some years ago, it was permitted to short points of equal voltage and open conductors with no current. Has this changed in the intervening years? Ahhhh, I see your confusion. Circuit theory and transmission line theory are not the same thing. Many have tried to mix the two models and fallen on their faces. Transmission line theory is simply more complicated than basic circuit theory. Well, slightly more. The capacitances and inductances are spatially distributed. Should make the math a bit more complicated but hasn't introduced anything new. If you don't believe it, replace a transmission line with ghosting with an equivalent circuit - the ghosting disappears. Are they really equivalent? Well if the ghosting disappears, it was not the 'equivalent circuit'. No surprises there. Because the observed voltages and currents are the same. The circuit has been replaced by one which is indistinguishable from the first. BS! I can certainly distinguish between a short and a non-shorted transmission line. Please try again. Just to recap, the experiment was: a matched source connected to an open transmission line. After the source is turned on (and a brief wait for settling), measure any voltage or current you desire at any point on the line. I short the points of zero voltage and cut the points with zero current. You can now remeasure any current or voltage at any points you choose to tell me whether the shorts or cuts are in or out. Since you can not tell the difference, the circuit with cuts and opens is identical to the one without. You have gone from 100% transmission of waves to 100% reflection of waves. An intriguing assertion, but one which can not be demonstrated through any observations made on the circuit. Of course it can by observing ghosting in a TV signal. With the line shorted, there is no ghosting. With the line not shorted, there is ghosting. Why is that so difficult to understand? We were, I thought, discussing an open transmission line, not one carrying a TV signal. You have changed the experiment. I am not sure why you are looking so hard for reflections. I am not attempting to claim that there are any. I simply claim that no energy can cross the boundaries because the current or voltage is always 0. In the real world, if no RF energy crosses a boundary, then that energy is reflected since it cannot stand still. Of course, the supernatural world is an entirely different matter. Basic electriciy. This says nothing about the presence or absence of reflections or whether there is a mechanism to prevent the energy crossing the boundary or it just happens through the dynamics of the distributed capacitance and inductance. Unfortunately, this is beyond the ability of basic electricity to handle. It takes the wave reflection model (or quantum physics) Oh, I don't think we need quantum physics to understand energy distributions in ideal transmission lines. Let's not scare off the neophytes. to handle it. Einstein said a model should be as simple as possible, but not too simple. Once again, you seem to be trying to force reality to obey your model instead of vice versa. Oh, I fully understand what a 'directional watt' meter will indicate. Then why are you so confused? :-) May I suggest that you perform the same experiment with a real instantaneous watt meter: This is obviously a diversionary tactic. Not at all. A much better understanding will be had if both the indications of real wattmeters and 'directional watt' meters are understood. Limiting yourself to only one instrument limits the opportunity for full understanding. Instantaneous power is essentially meaningless according to all my references and I have a bunch of them. Would you care to provide a reference that seriously deals with instantaneous power in transmission lines? It's not so hard that you need a text book. V, I, cos(theta) and a little thinking will suffice. Draw a sine wave for the voltage and one for the current shifted by 90 degrees. Compute the power (which will be a new sine wave at double the frequency). Observe that the power goes positive and negative representing forward and reverse energy flow. At any instance (at a single point) the power is either positive or negative. Energy is either flowing in one direction or the other at any given instance. And remember that at any point where the voltage or current is always 0, no energy is ever flowing. ....Keith |
Dear Reg:
Lab equipment was not used. The lab was just a convenient gathering place with a blackboard. (No cigarettes in our lab or anywhere inside buildings.) I suspect that we made an arithmetic error. Need to revisit what we did - much too late at night right now. Thanks for the poke. 73 Mac N8TT -- J. Mc Laughlin - Michigan USA Home: "Reg Edwards" wrote in message ... "J. McLaughlin" wrote The issue of the max. value of the magnitude of the reflection coefficient came up in our student mess. It was shown in lab that it must approach the square root of 5, which is close to 2.41. ================================ You don't need a lab. All you need is a pencil and the back of a cigarette packet. Theoretical Max possible value = 1+Sqrt(2) exactly = 2.4142136 . . .. . . Can't imagine where you get 5 from. It occurs when line Zo = Ro - jXo has an angle of -45 degrees, ie., when Xo = -Ro, and when the line is terminated in an inductive reactance of +jXo. ---- Reg. |
Roy Lewallen wrote:
p(t) = v(t) * i(t). What is 100 volts at zero degrees multiplied by 2 amps at 90 degrees? You really want us to believe there is not a cos(90deg) in there somewhere? :-) -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
wrote:
It does go back to 'double think' unless you can explain how energy can flow when the power is zero. Already explained. The forward power flow vector is equal to the reflected power flow vector so the *NET* power is zero just as explained in Ramo & Whinnery. So I will repeat the challenge: "It appears that you are prepared to ignore Pinst = Vinst * Iinst. Could you expand on when Pinst IS equal to Vinst * Iinst and when it isn't? Sorry, this appears to be just a diversion from a subject you don't want to discuss. According to Hecht, Pinst brings nothing of value to the discussion. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
wrote:
Why the resistance to examining this case? I don't chase irrelevant logical diversions. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
wrote:
You can now remeasure any current or voltage at any points you choose to tell me whether the shorts or cuts are in or out. That's easy. The line is shorted or cut where the signals disappear. Since you can not tell the difference, the circuit with cuts and opens is identical to the one without. I can't tell the difference when someone cuts my transmission line? Can I have a hit off of whatever you are smokin'? -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Goodness, didn't you do this at Texas A & M? Or did they just give you
the formula for average power and tell you to forget the rest? But the calculation that follows requires only high school level trigonometry, not engineering mathematics. v = 100 cos(wt) i = 2 cos(wt + 90) v * i = 200 cos(wt)cos(wt + 90) = 100 cos(-90) + 100 cos(2wt + 90) = 100 cos(2wt + 90). What you have here is a sinusoidal waveform of radian frequency 2w, centered about zero. Cos(90) is zero. The result above, the power, or rate of energy transfer, is not zero. It shows that energy moves back and forth at twice the rate of v and i, and that the same amount that moves one way moves back on the alternating cycle. Roy Lewallen, W7EL W5DXP wrote: Roy Lewallen wrote: p(t) = v(t) * i(t). What is 100 volts at zero degrees multiplied by 2 amps at 90 degrees? You really want us to believe there is not a cos(90deg) in there somewhere? :-) |
Roy Lewallen wrote:
Goodness, didn't you do this at Texas A & M? = 100 cos(2wt + 90). I thought you said there wasn't a cos(90) in there anywhere. If t=0, there's a cos(90) term. What you have here is a sinusoidal waveform of radian frequency 2w, centered about zero. Cos(90) is zero. The result above, the power, or rate of energy transfer, is not zero. It shows that energy moves back and forth at twice the rate of v and i, and that the same amount that moves one way moves back on the alternating cycle. And exactly why is that information important? -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Art Unwin KB9MZ wrote:
I don't see from what what Reg said that this should have been unloaded on him in such a way. I sure hope the " establishment" or "cabal" as I call them don't follow you and pile it on . Seems to me that Reg is a giver not a taker as far as ham radio is concerned, and just as much as a asset to this group as those wish to fire off a shot at the first oportunity to give dutch courage to all those that wil follow. Now you can stick the label on me also, audacity was it? After all I did say "cabal" ? What really surprises me is that you have not noticed over the years how many people have been attacked so wickedly on this newsgroup by people that infer that "all is known" to which you whimsically referred to. The fact is that is if this were true a thread would never exceed five postings after a particular expert had posted. I can only imagine that Reg has not swallowed all you have said and thus has raised your ire such that you have lost your cool If you still think there is a cabal, then you missed the whole point. If you thought it was a personal attack against Reg, then you completely misunderstood my motives in writing all that. (And anyway, Reg probably regards "audacity" as the highest of compliments :-) And if you thought any of it was written in anger, then you never understood a single word. -- 73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB) Editor, 'The VHF/UHF DX Book' http://www.ifwtech.co.uk/g3sek |
W5DXP wrote:
wrote: You can now remeasure any current or voltage at any points you choose to tell me whether the shorts or cuts are in or out. That's easy. The line is shorted or cut where the signals disappear. You must have misread the question. Shorts are only applied at points with zero volts and only wires with zero current are cut. Since there is no 'signal' present at these points, the 'signal' will not disappear. For your convenience, I have provided the experiment again, below. "Just to recap, the experiment was: a matched source connected to an open transmission line. After the source is turned on (and a brief wait for settling), measure any voltage or current you desire at any point on the line. I short the points of zero voltage and cut the points with zero current. You can now remeasure any current or voltage at any points you choose to tell me whether the shorts or cuts are in or out. Since you can not tell the difference, the circuit with cuts and opens is identical to the one without." ....Keith |
Keith wrote:
"Since you can not tell the difference, the circuit with cuts and opens is identical to one without." There he goes again! If you short an r-f transmission line where SWR has produced zero volts, you make a difference. You change the reflection point to the short you impose from its former location. Best regards, Richard Harrison, KB5WZI |
On Wed, 20 Aug 2003 09:36:52 +0100, "Ian White, G3SEK"
wrote: The fact is that is if this were true[, then] a thread would never exceed five postings after a particular expert had posted. bracketed [ ] inclusions made by me. Hi Ian, This posting (#5) is simply to confirm suspicions. :-) 73's Richard Clark, KB7QHC |
wrote:
W5DXP wrote: You must have misread the question. Shorts are only applied at points with zero volts and only wires with zero current are cut. Since there is no 'signal' present at these points, the 'signal' will not disappear. Let me see if I have this straight. You cut off 1/4WL of feedline and current and voltage still exist in the cut off parts? Have you patented that perpetual motion machine? All I have to do to prove you wrong is measure the current and voltage in the part that was isolated from the source. In the previous example: 1/4WL coax source---------coax--------------------+---wattmeter-----short When you cut the line at '+', the wattmeter will go to zero. It's a no brainer. What do you mean I can't tell the difference? You can now remeasure any current or voltage at any points you choose to tell me whether the shorts or cuts are in or out. Since you can not tell the difference, the circuit with cuts and opens is identical to the one without." But I can tell the difference because I choose to measure the current and voltage in the parts of the feedline that have been cut off and isolated from the source. It's a no brainer. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
My mistake. I forgot to make clear that the transmission line was
ideal for this thought experiment; that is, R=G=0; no loss. The signal does indeed continue to circulate for ever. Please re-evaluate your answer after this clarification. ....Keith W5DXP wrote: wrote: W5DXP wrote: You must have misread the question. Shorts are only applied at points with zero volts and only wires with zero current are cut. Since there is no 'signal' present at these points, the 'signal' will not disappear. Let me see if I have this straight. You cut off 1/4WL of feedline and current and voltage still exist in the cut off parts? Have you patented that perpetual motion machine? All I have to do to prove you wrong is measure the current and voltage in the part that was isolated from the source. In the previous example: 1/4WL coax source---------coax--------------------+---wattmeter-----short When you cut the line at '+', the wattmeter will go to zero. It's a no brainer. What do you mean I can't tell the difference? You can now remeasure any current or voltage at any points you choose to tell me whether the shorts or cuts are in or out. Since you can not tell the difference, the circuit with cuts and opens is identical to the one without." But I can tell the difference because I choose to measure the current and voltage in the parts of the feedline that have been cut off and isolated from the source. It's a no brainer. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
|
"Reg Edwards" wrote in message ...
Dear Dr Slick, it's very easy. Take a real, long telephone line with Zo = 300 - j250 ohms at 1000 Hz. Load it with a real resistor of 10 ohms in series with a real inductance of 40 millihenrys. The inductance has a reactance of 250 ohms at 1000 Hz. If you agree with the following formula, Magnitude of Reflection Coefficient of the load, ZL, relative to line impedance = ( ZL - Zo ) / ( ZL + Zo ) = 1.865 which exceeds unity, and has an angle of -59.9 degrees. The resulting standing waves may also be calculated. Are you happy now ? --- Reg, G4FGQ Nope. Re-do your calculations, and you will get something like 0.935: Actually, my first posting was right all along, if Zo is always real. From Les Besser's Applied RF Techniques: "For passive circuits, 0=[rho]=1, And strictly speaking: Reflection Coefficient = (Zl-Zo*)/(Zl-Zo) Where * indicates conjugate. But most of the literature assumes that Zo is real, therefore Zo*=Zo." And then i looked at the trusty ARRL handbook, 1993, page 16-2, and lo and behold, the reflection coefficient equation doesn't have a term for line reactance, so both this book and Pozar have indeed assumed that the Zo will be purely real. That doesn't mean Zl cannot have reactance (be complex). Try your calculation again, and you will see that you can never have a [rho] (magnitude of R.C.)greater than 1 for a passive network. How could you get more power reflected than what you put in? If you guys can tell us, we could fix our power problems in CA! But thanks for checking my work, and this is a subtle detail that is good to know. Slick |
"Reg Edwards" wrote in message ...
You don't need a lab. All you need is a pencil and the back of a cigarette packet. Theoretical Max possible value = 1+Sqrt(2) exactly = 2.4142136 . . . . . Can't imagine where you get 5 from. It occurs when line Zo = Ro - jXo has an angle of -45 degrees, ie., when Xo = -Ro, and when the line is terminated in an inductive reactance of +jXo. ---- Reg. Actually, my first posting was right all along, if Zo is always real. From Les Besser's Applied RF Techniques: "For passive circuits, 0=[rho]=1, And strictly speaking: Reflection Coefficient = (Zl-Zo*)/(Zl-Zo) Where * indicates conjugate. But most of the literature assumes that Zo is real, therefore Zo*=Zo." And then i looked at the trusty ARRL handbook, 1993, page 16-2, and lo and behold, the reflection coefficient equation doesn't have a term for line reactance, so both this book and Pozar have indeed assumed that the Zo will be purely real. That doesn't mean Zl cannot have reactance (be complex). Try your calculation again, and you will see that you can never have a [rho] (magnitude of R.C.)greater than 1 for a passive network. How could you get more power reflected than what you put in? If you guys can tell us, we could fix our power problems in CA! But thanks for checking my work, and this is a subtle detail that is good to know. Slick |
wrote:
My mistake. I forgot to make clear that the transmission line was ideal for this thought experiment; that is, R=G=0; no loss. The signal does indeed continue to circulate for ever. So exactly how many angels can dance on the head of a pin? -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
wrote:
Given an ideal transmission line (necessary to achieve 0 volts), there are no measurements you can make which will indicate whether the short is present or not. Given a Transmission Line God, there are no measurements you can make which will indicate whether the short is present or not. How about we limit these technical discussions to the real world? -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
"J. McLaughlin" wrote -
Dear Reg: Lab equipment was not used. The lab was just a convenient gathering place with a blackboard. (No cigarettes in our lab or anywhere inside buildings.) I suspect that we made an arithmetic error. Need to revisit what we did - much too late at night right now. Thanks for the poke. 73 Mac N8TT =================================== Mac, you didn't make an arithmetical error. I think you used the wrong value of terminating reactance Xt. Just a Sherlock Holmes deduction. ;o) The value of terminating reactance which gives greatest possible magitutude of reflection coefficient is equal to the magnitude of Zo which is equal to |Zo| = Sqrt( Sqr( Ro) + Sqr( Xo) ). As you have already deduced, maximum value occurs when the angle of Zo is equal to -45 degrees at which Ro = -Xo. You incorrectly set the terminating inductive reactance XL equal to -Xo whereas XL should have been Sqrt( 2 ) times greater. Now if XL is incorrectly made -Xo then the reflection coefficient calculates to Sqrt( 5). Which is where your 5 comes from. Somewhere in these threads I made the exactly same error myself in a non-calculating context where it was not likely to be noticed. The pair of errors, yours and mine, were probably just coincidental. I do hope I have not just introduced another. ;o) You appear to be in an educational establishment. There are two programs available from the website below which you may find useful. Download in a few seconds and run immediately programs COAXPAIR and RJELINE3. They have been written according to classical transmission line formulae, generally accurate to all figures displayed and may be used to check other work which uses only engineering approximations. There are others also of educational value. ---- ======================= Regards from Reg, G4FGQ For Free Radio Design Software go to http://www.g4fgq.com ======================= |
Ahhhhh, finally. The not unexpected descent into non-sequitors.
Much safer than actually thinking about questions which might cause you to change your world view. ....Keith W5DXP wrote: wrote: My mistake. I forgot to make clear that the transmission line was ideal for this thought experiment; that is, R=G=0; no loss. The signal does indeed continue to circulate for ever. So exactly how many angels can dance on the head of a pin? -- 73, Cecil http://www.qsl.net/w5dxp |
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