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On Wed, 26 May 2004 00:05:40 GMT, "Henry Kolesnik"
wrote: I used 450 ohm ladder line and a balun along with a cheap SWR meter. Hi Hank, Classic! Thanx. 73's Richard Clark, KB7QHC |
On Tue, 25 May 2004 19:01:31 -0500, Tom Ring
wrote: Boy, I'm gonna have to try and find my old EE5xx Non-Linear Transistor Design notebook. I doubt I could translate it into words for this section of the thread, but it is incredibly applicable. Hi Tom, It's not that difficult. Edison corrupted the Thevenin model demanding that there be a RESISTANCE equal to the load to accomplish this equivalency (a case of being overly literal for a market advantage). The Thevenin model only requires an IMPEDANCE, hence these allusions to 50% maximum efficiency (an imposed condition much like debate here) were an invention by Edison to detract investment away from AC power transmission and attract it back to his DC dynamos. The real pro's new how to dance with numbers and concepts too. This resolves how the common Ham rig can present 50 Ohms Impedance and still offer more than 50% efficiency - although, a simple examination of the power in vs. the power out reveals it isn't nearly as good as 50% for ANY mode. Such is the gulf between reality and theory - some folk's mileage may vary. ;-) 73's Richard Clark, KB7QHC |
Richard Clark wrote:
Hi Tom, It's not that difficult. Edison corrupted the Thevenin model demanding that there be a RESISTANCE equal to the load to accomplish this equivalency (a case of being overly literal for a market advantage). The Thevenin model only requires an IMPEDANCE, hence these allusions to 50% maximum efficiency (an imposed condition much like debate here) were an invention by Edison to detract investment away from AC power transmission and attract it back to his DC dynamos. The real pro's new how to dance with numbers and concepts too. This resolves how the common Ham rig can present 50 Ohms Impedance and still offer more than 50% efficiency - although, a simple examination of the power in vs. the power out reveals it isn't nearly as good as 50% for ANY mode. Such is the gulf between reality and theory - some folk's mileage may vary. ;-) 73's Richard Clark, KB7QHC Understood. I still want to find it. Been about 30 years since I wrote it all down in 4 colors. I took serious notes. tom K0TAR |
Jim Kelley wrote:
Electromagnetic waves can cancel, but rocks can't. However, two EM waves have to exist before they can cancel. If they exist, they posses both energy and momentum. -- 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! =----- |
This is absolutely the best thread I've seen in years. It's
educational, thought provoking, entertaining, and revealing about the human psyche. The thread reminds of some of those that I monitor from time to time regarding aeronautics and fluid dynamics on the topic of how an airplane wing generates lift. Believe or not, there is still no consensus after 100+ years. It's interesting in that it parallels this thread in many ways -- attempting to interpret various abstract mathmatical definitions of a physical process in a way that it "makes sense." We're fortunate in that such things don't actually have to make sense to work. Al |
On Tue, 25 May 2004 19:43:42 GMT, "Henry Kolesnik"
wrote: They continued to work well and I always thought this was from overdriving but now I'm not so sure because it ran at the same high swr as the 30S-1. Hi Hank, I gave this some more thought from my Collins period. If I recall, overdriving an amplifier brings damage to the grid, which has its own dissipation limit. The usual failure mode is that it melts or sags and causes a grid-cathode short. This is especially true in lighthouse style tubes. 73's Richard Clark, KB7QHC |
Richard Clark wrote:
"The Thevenin model only requires an IMPEDANCE." Yes, but "To secure maximum power output from a generator whose emf and whose internal impedance are constant, the load must have an impedance equal to the conjugate of the generator`s internal impedance." (page 43, "Transmission Lines, Antennas, and Wave Guides", King, Mimno, and Wing) Best regards, Richard Harrison, KB5WZI |
"Richard Clark" wrote in message ... On Mon, 24 May 2004 13:19:41 -0500, Cecil Moore wrote: Well Richard, here's your chance. Please enlighten us on J.C. What was the original question? [accredited stock response] It's clear that R.C. loves tweeking Cecil. Steve |
"Dave Shrader" wrote in message news:D2usc.110697$iF6.9975401@attbi_s02... Cecil Moore wrote: snip For what it is worth, I believe that the first homo sapien originated about a quarter of a million years ago and was a female with dominant genes. -- 73, Cecil http://www.qsl.net/w5dxp My wife has me convinced that ALL women have the dominant genes!!!! Deacon Dave There's a guy claiming that the male gene (is it the X?) will dissapear in several hundred years because it is on the decline now... Steve |
"Tdonaly" wrote in message ... Richard Clark wrote, On Mon, 24 May 2004 15:28:40 -0500, Cecil Moore wrote: I am sure that our solar system and homo sapiens didn't exist when the original question was asked. So the question is which came first, the homo or the solar system? Homos before Helios, or Helios before homos? That's a profound question which I'll have to think about over my after-dinner port. On the surface it looks about as meaningful as "turkeys from Turlock," but first impressions are sometimes deceiving. 73, Tom Donaly, KA6RUH Hey! I know it HAD to be a PREGNANT Chicken that came first... |
Richard Harrison wrote:
Richard Clark wrote: "The Thevenin model only requires an IMPEDANCE." Yes, but "To secure maximum power output from a generator whose emf and whose internal impedance are constant, the load must have an impedance equal to the conjugate of the generator`s internal impedance." (page 43, "Transmission Lines, Antennas, and Wave Guides", King, Mimno, and Wing) Anything about dissipationless resistances or negative resistances? -- 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! =----- |
"Richard Clark" wrote in message ... On Mon, 24 May 2004 16:40:44 -0500, "Steve Nosko" wrote: This is an interesting twist, Tam. I think if this were the case, then there would be MORE power dissipated in the Tx Hi Steve, And why would that be? 73's Richard Clark, KB7QHC Reconstructing the long ago comment not having the original comment, Steve comments further: I think he had the power returning to the PA device and being dissipated there. Steve |
Yea! See. That's what I inferred and I said he was implying.
Man! This he said...he said...he said gets confusing who said what... Steve "Tam/WB2TT" wrote in message ... "Steve Nosko" wrote in message ... "Tam/WB2TT" wrote in message ... "Richard Fry" wrote in message ......................... ............................... Concept below However this is not an accurate model of a transmitter. For an example, take an old Heathkit DX-100 generating a measured 180 watts of CW RF into a matched 50 ohm load. To do this, it does NOT also dissipate 180 watts of RF into some "virtual" internal RF load in the DX-100. In fact, the PAs and power supply in the DX-100 could not produce a total RF output power of 360 watts without exceeding their ratings. The dissipation in the PA is essentially related only the DC to RF conversion efficiency of the PA, which in this case probably is about 75%, max (Class C). So a PA input power of about 240 watts DC is required to produce 180 watts of RF output power. The other 60 watts of plate input power is converted to heat by the PA tube anodes. The entire RF output generated by the PA stage is applied virtually 100% to the output connector. How much of that is absorbed by the load connected there is a function of load SWR and system losses. - RF There is a Motorola ap note that agrees with what Richard is saying. To paraphrase it, if the the DX100 had an output impedance of 50 Ohms, then the overall efficiency would be 37.5%. Unfortunately I can't read all the digressions in the thread. I skim by author... This is an interesting twist, Tam. I think if this were the case, then there would be MORE power dissipated in the Tx than Mr. Fry is saying - making the situation worse. By that, I mean, getting further from what is going on. I think this goes in the wrong direction. I believe the flaw is believing that the Rs=RL must exist for the transmitter. That is what I am saying. The efficiency goes from 75% to 37.5%; so, there is more power dissipated in the TX. Tam/WB2TT -- Steve N, K,9;d, c. i My email has no u's. |
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Richard Clark wrote:
(Richard Harrison) wrote: Yes, but "To secure maximum power output from a generator whose emf and whose internal impedance are constant, the load must have an impedance equal to the conjugate of the generator`s internal impedance." (page 43, "Transmission Lines, Antennas, and Wave Guides", King, Mimno, and Wing) Not if it is strictly resistive. Of course, it works for the purely resistive. R+j0 is the conjugate of R-j0 -- 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! =----- |
Ahhh! Finally some meat... SORRY,... LONG POST WITH MUCH techincal MEAT IN
IT... "Richard Clark" wrote in message ... On Tue, 25 May 2004 02:59:52 GMT, "Henry Kolesnik" wrote: Richard ...(what is) the "mechanism" in the transmitter (such that) it can't dissipate a reflection because there's some kind of one way ....checkvalve...reflects. tnx Hi Hank, ... It can also reflect (some or all of) what it does not absorb, ... is determined by the ratio of its Z to the line/load at the antenna terminal. (Of the transmitter, I believe you mean) Fine point: I think it must reflect ALL it does not absorb. Important point: However, I believe the case is that it absorbs little of this energy. In this case (bear with me), the source Z is NOW found at the load's mismatch, transformed through the line. This means matching works both ways (often the singlemost ignored aspect of obvious reciprocity in any of these arguments). Such circuit analysis is called superposition. Yes, no argument. Matching does, er um, CAN work both ways on the line, is true. And the TX is, indeed, a "load" for the reflected energy (or wave if you prefer). I say CAN because the match does work both ways, but that is not to imply that the impedances which exist on the line are matched - nor that one way IS matched while the other is NOT. If there is a mis-match in one direction, there will be another mismatch going the other way. This may be a difficult to understand considering the "this situation must be matched" mind-set we are in at this point, but I do not believe it is pixie dust, however choose to skip it for now. See my conclusion at the end.... I think "superposition" is ther wrong term for this. If I could spell reciprocity, I'd say that it is a better word. Digression, though minor: Superposition refers typically, to the application of two signals to a circuit and the resulting response being the sum of the individual responses. no? Now some meat I'd like to pursue... ...a 500 Ohm Source...you will never launch much power into a 50 Ohm system because it would immediately hit a reflective interface at the antenna connector. There's a question I see here as to wheather the power was "launched" in the first place and this may be part of the assumption (or set thereof) which (my gut feel says) leads the logic astray. This means that a 500 Ohm source, when confronted by power going towards it from a 50 Ohm system will reflect most of that power (but how did we get this power into the system in the first place - Karma?) This, by design, will never happen in any transistor ham rig built in the last quarter century. I think this is true...BUT let's ask this same question of a 5.0 ohm source?? How would you carry this through Richard C? I submit that THIS source CAN launch significant power down the line. ...snip non technical stuff The answer is the transmitter source Z is 50 Ohms at rated power. OK, so now I see you are of the "Zs=50 ohms" camp. If a watt of power is chooglin' down the line toward it, that 50 Ohms is going to dissipate into a watt worth of calories. This can be argued with wave mechanics, or lumped circuit equivalents - doesn't matter because it's all the same calories. All ok with me _IF_ Zs=50. I'm not convinced however. I have to think more about the modern broad-band transistor PA to have an strongly arguable opinion on this concept. Modern rigs can tolerate this...snip... === WARNING to the logic impaired, the following is a supposition === Now, lets simply accept those answers that require the magic pixie dust Uncalled for... I can make it work (so far though I am open) with the electrical/electronic principles I think are true and work everywhere else. of total reflection from the transmitter. Fine, the mismatched antenna reflects some power, this power returns to the transmitter, the transmitter simple routes it ALL back to the antenna, round-and-round until the antenna finally radiates it. If this were true, what do we need tuners for? I think the answer here is that if the transmitter can take the strange impedance it sees, then it isn't needed. Like Hanks 30L1 that, however, couldn't take it very well. Furthermore, there is an implicit assumption here in your reasoning which I believe is key and leading to some error. There is an unspoken assumption that the power which makes it to the antenna after all the " round-and-round until the antenna finally radiates it" is the FULL output and I believe it AIN'T. It is that little dribble that made it into the line in the first place. a.k.a. the Tx is not generating the full output. I think this assumption is causing much trouble. Unless, of course, I have been smoking the wrong brand. Comments R.C. ? Steve |
Cecil Moore wrote: Jim Kelley wrote: Electromagnetic waves can cancel, but rocks can't. However, two EM waves have to exist before they can cancel. And that makes rocks like waves? If they exist, they posses both energy and momentum. Bet ya can't prove it without first transfering it to something. 73, Jim AC6XG |
Cec and Co., I keep telling you it's a waste of bandwidth talking about
'conjugate matches'. In the transmitter/transmission line context conjugate matches do not exist because nobody knows what the internal impedance of the transmitter actually is. There's never any need to know. Of what possible use would it be if you knew it.? Except provide a topic of conversation for this newsgroup. ;o) |
Jim Kelley wrote:
Cecil Moore wrote: However, two EM waves have to exist before they can cancel. And that makes rocks like waves? That makes real waves tangible like real rocks. The wave particles are just smaller. OTOH, "People who live in glass houses shouldn't throw stones." is an intangible. If they exist, they posses both energy and momentum. Bet ya can't prove it without first transfering it to something. _Optics_, by Hecht is good enough for me. "It is possible to compute the resulting (momentum) force via Electromagnetic Theory, whereupon Newton's Second Law suggests that the *wave itself carries momentum*. (all emphasis his, not mine) ... As Maxwell showed, the *radiation pressure* equals the energy density of the EM wave. ... When the surface under illumination is perfectly reflecting, the beam that entered with a velocity of +c will emerge with a velocity of -c. This corresponds to twice the change in momentum that occurs on absorption, ..." It's obvious that the energy in the TV ghosting wave makes a round- trip to the match-point and back to the RCVR. That's obviously a change in the direction of momentum of the reflected wave. It is twice the change in momentum than if it encountered a circulator/ load and was dissipated. -- 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! =----- |
Cecil Moore wrote:
Tam/WB2TT wrote: You can get the same answer from knowing that the impedance looking into a 1/4 wave section of shorted transmission line is infinite. Ever measure the forward and reflected currents halfway into a shorted 1/4WL stub? How can currents be flowing unimpeded into and out of an infinite impedance? The question is a little misleading because the direction of the flow of current changes every half cycle and is transverse, or orthogonal to the direction of wave propagation. In a transmission line, the current flows through Z0, ostensibly, which is essentially the impedance from one conductor to the other at every point along the transmission line. Other than that, superposed forward and reflected waves behave just as you described, naturally. 73, Jim AC6XG |
Cecil Moore wrote: Jim Kelley wrote: Electromagnetic waves can cancel, but rocks can't. However, two EM waves have to exist before they can cancel. Yes, but what about the rock? :-) If they exist, they posses both energy and momentum. So if I "possess" an American Express card, do I "possess" money? No. I simply "possess" the potential to purchase something with it at a point of sale. 73, Jim AC6XG |
Cecil Moore wrote:
Jim Kelley wrote: Cecil Moore wrote: However, two EM waves have to exist before they can cancel. And that makes rocks like waves? That makes real waves tangible like real rocks. The wave particles are just smaller. OTOH, "People who live in glass houses shouldn't throw stones." is an intangible. If they exist, they posses both energy and momentum. Bet ya can't prove it without first transfering it to something. _Optics_, by Hecht is good enough for me. "It is possible to compute the resulting (momentum) force via Electromagnetic Theory, whereupon Newton's Second Law suggests that the *wave itself carries momentum*. (all emphasis his, not mine) ... As Excellent, now try and understand what is intended by _ALL_ of the words in the sentence. Maxwell showed, the *radiation pressure* equals the energy density of the EM wave. ... When the surface under illumination is perfectly reflecting, the beam that entered with a velocity of +c will emerge with a velocity of -c. This corresponds to twice the change in momentum that occurs on absorption, ..." Yes, I'm familiar with the subject. I've been familiar with it for a long time. However it is incorrect to infer that interactions between waves would be the same as interactions between waves and matter! It's obvious that the energy in the TV ghosting wave makes a round- trip to the match-point and back to the RCVR. It's obvious that the signal has taken multiple paths, at least. That's obviously a change in the direction of momentum of the reflected wave. Perhaps. You're describing back scattering, which should exhibit a Compton effect wavelength shift if true. But again, you're describing an interaction with matter. Photons don't interact with each other in the same way they interact with matter. 73, Jim AC6XG |
"Cecil Moore" wrote in message ... Tam/WB2TT wrote: "Cecil Moore" wrote: Richard, you know you are going against the conventional wisdom on this newsgroup. Ghosting cannot exist during steady-state so if ghosting exists it simply means that you are still in the transient state and the steady-state doesn't exist (yet). Looking for the smiley face. Is there a smiley face that means, "sad but true"? Many otherwise intelligent, knowledgeable, educated engineers have attempted to force their metaphysical "steady-state" agenda on uninitiated and unsuspecting victims. One is saddened by such an event and one wonders why. Does a steady-state religion or creed exist within amateur radio? If so, what are its purpose and goals? -- 73, Cecil http://www.qsl.net/w5dxp From this I infer that Cecil believes that once the transient portion of the response has concluded you are in the steady state and no more reflections are occurring. Is this a correct re-statement of your belief, Cecil? Steve N. |
"Tam/WB2TT" wrote in message ... "Henry Kolesnik" wrote in message .. . Richard ...I got out the book REFLECTIONS II by Walt Maxwell W2DU. I'm typing verbatim from page 2-2 and 23-1 I paraphrase / quote the quote: Power reflected from a mismatch back into the PA is not absorbed there. He writes: "when the pi-network tank is tuned to resonance, a virtual short circuit to rearward traveling waves is created at the input of the network. Consequently, instead of the reflected power reaching the tubes of the amplifier, it is totally re-reflected toward the load by the virtual short " Hank, I'm sorry, but I believe this is not correct as stated. The word "totally", I believe is misleading and contrary to what I think all will agree. Namely that there MUST be SOME real part (although the true amount is disputed here) to the Zout of the PA and therefore SOME power MUST be absorbed there. I believe the point of contention truely is just how much is absorbed and how much is reflected... ...Also what happens in a transistor final with no pi? AGAIN. Be careful that you DO NOT keep assuming that the full power is the incident power. Incident meaning forward power or that which the transmitter is sending toward the load. I have a real corker below. Be careful. Do not try this at home, I am a professional. (yea, I know this'll spawn all kinds of grief) Tam sez: Henry, Here is an example of what you just said. Take a sine wave source, and connect it to a 1/4 wave section of shorted transmission line through a series resistor R. The reflected wave will reach this resistor 1/2 cycle later, and will be in phase with the source. For a lossless transmission line, there will be *0 Volts across the resistor*. There will be 0 current through the resistor, and the reflected wave will be re reflected for all values of R, including R=Z0, because the reflected wave will not "know" what R is. You can get the same answer from knowing that the impedance looking into a 1/4 wave section of shorted transmission line is infinite. Tam/WB2TT Tam, Although I believe you have digressed somewhat, I will follow this path since it contains a closely related concept. You obviously have a pretty good handle on much of this (as others do, up to a point, and struggle to make their mental models fit the reality - or to make other's mental models fit theirs)...However, there is a paradox here which appears to be the root of this disagreement and your example hit it right on the head. You have some implicit assumptions here. 1) Zero volts across the resistor = (not said, but implied) the t-line acts like an open at the input end, therefore there is no current into the line which results in the zero V across R. V=I^2 * R. I think everybody will agree with this. (shorted 1/4 wave acts like an open and an open 1/4 wave looks like a short - mantra of all, no?). 2) Resistor current = zero therefore reverse traveling wave gets reflected toward the load end. The implication clearly is that with zero resistor current, no energy can be flowing out that way, so it must be reflected. HOWEVER... Been a while since I went to this depth and interesting to do so, though unnecessary, I will anyway... If the input to this stub acts like an open, there can be NO current, thus no power entering, therefore there can be no forward wave, no reflected wave and no summation of waves to make the open in the first place and no need to re-reflect the reflected wave from the resistor who (or is it whom) has no current. This appears to be the root cause of this problem. Now, we can say that (and I think it has been said) that it makes no difference how large these two waves (which cancel each other to form the open circuit at the stub input) can be any absolute magnitude. 1 amp 10 amps 100 amps doesn't matter - they cancel. So what are they really? There are two things to consider as you work out how you will resolve this paradox in your mental model. 1) I believe the most important -- If the Z looking in to the stub is high, how can you send a large amount of power down the line. If you say it is an infinite Z then you have a really big problem explaining how it got there in the first place. 2) There is a tendency to assume that the forward power is the same power as when the load is not a short, but Zo. Interesting puzzle, but I don;t need to go further. At some point you must 'believe' something and I can comfortably stop there...for now. Oh well... -- Steve N, K,9;d, c. i My email has no u's. |
"Henry Kolesnik" wrote in message . .. Before it came down my 20 meter antenna had a SWR approaching 20:1 and I was running a Collins 30S-1 and no tuner!. I did that for several years and would be still doing if the roofers hadn't destroyed ~130 ft. dipole. One of these days after my back improves I'll put another one up. Henry, I don't have a problem with the 20:1, obviously you were matching enough to get out, but I keep coming back to: Why did you keep doing it if it burned up your tubes? Steve N. |
"Henry Kolesnik" wrote in message .. . ...a cheap SWR meter. OOPS! Steve |
Cecil, W5DXP wrote:
"Anything about dissipationless resistances or negative resistances (in "Transmission Lines, Antennas, and Wave Guides")?" On page 73, the characteristic resistance of free-space is defined as the sq rt of the permeability of space devided by the dielectric constant of space. The units are henries/m and farads/m. The solution is a voltage to current ratio of 376.7 ohms, or 120 pi ohms. As free-space is a perfect (lossless) medium for radio waves, it is a dissipationless resistance. On page 13, the characteristic resistance (Ro) of a transmission line is given in formula (14.3) as the sq rt of L/C, but this is an approximation for low-loss lines as there are no perfect lines. Negative resistance is a gain instead of a loss. The authors of "Transmission Lines, Antennas, and Wave Guides" were writing for WW-2 officers being trained at Harvard University in radio and radar. The phone system was then using some "negative resistance repeaters" but neither these nor "active antennas" were big at that time. You can only make up the loss in a two-wire phone loop with amplification. Any more gain and the loop "sings" (breaks into oscillation). Best regards, Richard Harrison, KB5WZI |
"Cecil Moore" wrote in message ... Richard Clark wrote: wrote: You asked - I answered. Did you? What was the original question? I don't know. I wasn't around 7 billion years ago. Chuckle, chuckle. CUTE! Steve N. |
OH! NO! Vortex vs. Bernoulli
Steve N. -- Steve N, K,9;d, c. i My email has no u's. "alhearn" wrote in message om... This is absolutely the best thread I've seen in years. It's educational, thought provoking, entertaining, and revealing about the human psyche. The thread reminds of some of those that I monitor from time to time regarding aeronautics and fluid dynamics on the topic of how an airplane wing generates lift. Believe or not, there is still no consensus after 100+ years. It's interesting in that it parallels this thread in many ways -- attempting to interpret various abstract mathmatical definitions of a physical process in a way that it "makes sense." We're fortunate in that such things don't actually have to make sense to work. Al |
"Jim Kelley" wrote in message ... Cecil Moore wrote: Tam/WB2TT wrote: You can get the same answer from knowing that the impedance looking into a 1/4 wave section of shorted transmission line is infinite. Ever measure the forward and reflected currents halfway into a shorted 1/4WL stub? How can currents be flowing unimpeded into and out of an infinite impedance? The question is a little misleading because the direction of the flow of current changes every half cycle and is transverse, or orthogonal to the direction of wave propagation. In a transmission line, the current flows through Z0, ostensibly, which is essentially the impedance from one conductor to the other at every point along the transmission line. Other than that, superposed forward and reflected waves behave just as you described, naturally. 73, Jim AC6XG There is no current in the steady state. The steady state voltage is independent of source impedance, which affect only how long it takes to reach that. I ran a simulation on this, and you can see that as the voltage builds up, the current decreases Tam |
Tam/WB2TT wrote: "Jim Kelley" wrote in message ... Cecil Moore wrote: Tam/WB2TT wrote: You can get the same answer from knowing that the impedance looking into a 1/4 wave section of shorted transmission line is infinite. Ever measure the forward and reflected currents halfway into a shorted 1/4WL stub? How can currents be flowing unimpeded into and out of an infinite impedance? The question is a little misleading because the direction of the flow of current changes every half cycle and is transverse, or orthogonal to the direction of wave propagation. In a transmission line, the current flows through Z0, ostensibly, which is essentially the impedance from one conductor to the other at every point along the transmission line. Other than that, superposed forward and reflected waves behave just as you described, naturally. 73, Jim AC6XG There is no current in the steady state. The steady state voltage is independent of source impedance, which affect only how long it takes to reach that. I ran a simulation on this, and you can see that as the voltage builds up, the current decreases Tam Hi Tam, The simulation would be fun to play with. What do you use? 73 de jk |
On page 73, the characteristic resistance of free-space is defined as
the sq rt of the permeability of space devided by the dielectric ====================== Why must engineere resort to quoting from their worshipped Guru's who are hardly more likely to be correct than they are themslves. In all probability the recipient of the advice does't have a copy of the sacred text. If he did he wouldn't be asking the question anyway. And it must be extremely rare for anybody to spend weeks attempting to find a copy. He knows he will have lost interest long before he finds one. Do engineers these days, in their daily work, depend entirely on the 'gospel truths' to be found in text books on their extensive book shelves without much understanding of what its all about? It's a recipe for time-wasting errors! Or are they content to obtain money under false pretences? Sorry about the diversion. I'm in that sort of mood tonight. I'm on Pinot Noir 2001, red - a product of Romania, bottled before Romania became a EU member. But it's no worse than the Italian, Spanish, French, or even the Californian plonk! Never sampled a 6-shooter, John Wayne, Texan vintage. ---- Reg. |
Reg Edwards wrote:
Cec and Co., I keep telling you it's a waste of bandwidth talking about 'conjugate matches'. In the transmitter/transmission line context conjugate matches do not exist because nobody knows what the internal impedance of the transmitter actually is. If nobody knows what the internal impedance of a transmitter actually is, how do you know it's not conjugate matched? :-) -- 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! =----- |
Jim Kelley wrote:
So if I "possess" an American Express card, do I "possess" money? No. "money - 4. any article ... used as a means of payment." So the answer is yes, credit cards are a form of money. Sorry. -- 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! =----- |
Jim Kelley wrote:
Excellent, now try and understand what is intended by _ALL_ of the words in the sentence. I know, you have told me before, ad infinitum. All of the words intend exactly what Jim Kelly wants them to intend and absolutely nothing else. You are the only person in the entire universe capable of ascertaining the intent of words. So tell me something I don't already know. -- 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! =----- |
Richard Harrison wrote:
Negative resistance is a gain instead of a loss. Therefore, doesn't an amplifier have negative resistance? -- 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! =----- |
Tam/WB2TT wrote:
There is no current in the steady state. There is no *net* current in the steady state at the input of a shorted 1/4WL stub. What do you think the current at the short is? How did that large amount of current get there without flowing? -- 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! =----- |
If nobody knows what the internal impedance of a transmitter
actually is, how do you know it's not conjugate matched? :-) -------------------------------- Because nobody knows. And it wouldn't make slightest difference to anything if anyone did. And nobody knows why I keep answering your questions either. ;o) --- Yours, Reggie. |
Cecil Moore wrote: Jim Kelley wrote: So if I "possess" an American Express card, do I "possess" money? No. "money - 4. any article ... used as a means of payment." So the answer is yes, credit cards are a form of money. Sorry. A distinction is commonly drawn between plastic and cash, Cecil. I expected you wouldn't be too stubborn to acknowledge that. My mistake. 73, Jim AC6XG |
"Reg Edwards" wrote in message ... On page 73, the characteristic resistance of free-space is defined as snip- Sorry about the diversion. I'm in that sort of mood tonight. I'm on Pinot Noir 2001, red - a product of Romania, bottled before Romania became a EU member. But it's no worse than the Italian, Spanish, French, or even the Californian plonk! Never sampled a 6-shooter, John Wayne, Texan vintage. ---- Reg. Reg I shot a big rattlesnake just the other day with a 6-shooter. He was in the ham shack. Threatened my life. I used a Smith and Wesson model 19 in blued steel with a 4" barrel. It was loaded with rat shot. You're always welcome at my QTH and I'll give you a choice of 6-shooters, from a .32 S&W to a .44 magnum. Spirits are consumed only AFTER shooting, however. Tonight I'm looking forward to a Chateau St Michele merlot with my rare filet. 73 H. NQ5H |
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