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An antenna question--43 ft vertical
On 7/7/2015 2:32 PM, Ian Jackson wrote:
In message , Jerry Stuckle writes On 7/7/2015 6:25 AM, Ian Jackson wrote: In message , Jerry Stuckle writes Sure, there is ALWAYS VSWR. It may be 1:1, but it's always there. If there's no reflection, there can be no standing wave. So, being pedantic, there's no such thing as an SWR of 1:1! Wrong. An SWR of 1:1 indicates a perfect match, with no reflected power. It is recognized by all electronics texts and experts. My suggestion would be for you to learn some transmission line theory. Your statement here just showed you have no knowledge of it at all. Even when I took my novice test many years ago I had to understand SWR better than that. The point I'm trying to make is not technical. It's simply one of verbal logic. Without the presence of a standing wave, you can't possibly have something called a "standing wave ratio". But, like all RF engineers, an SWR of 1-to-1 is something I too strive to achieve! "Yesterday, upon the stair, I met a man who wasn't there. He wasn't there again today, I wish, I wish he'd go away..." It doesn't matter what your verbal logic is, Ian. The correct term is "Standing Wave Ratio", and an SWR of 1:1 means there is no reflected power and you have a perfect match. That is the technical definition of a technical term. -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
An antenna question--43 ft vertical
On 7/7/2015 11:09 AM, Ian Jackson wrote:
In message , rickman writes On 7/7/2015 6:25 AM, Ian Jackson wrote: In message , Jerry Stuckle writes Sure, there is ALWAYS VSWR. It may be 1:1, but it's always there. If there's no reflection, there can be no standing wave. So, being pedantic, there's no such thing as an SWR of 1:1! Why do you say that? If there is no reflection the voltage on the line is purely due to the forward signal and so the VSWR is 1:1. What's wrong with that? A standing wave is caused by a reflection. If there IS no reflection, there is NO standing wave. So while you can have an SWR of 1.00000000000001-to-1 (because a standing wave DOES exist), you can't really have one of 1-to-1 (because there IS no standing wave). ;o)) [Just a bit of pedantic, lateral thinking on my part. Don't worry too much about it. It has absolutely no bearing whatsoever on the current discussions.] Sounds great, but that is not how the VSWR is defined. :) -- Rick |
An antenna question--43 ft vertical
On 7/7/2015 11:02 AM, John S wrote:
On 7/7/2015 8:33 AM, rickman wrote: On 7/7/2015 8:52 AM, John S wrote: On 7/7/2015 7:19 AM, rickman wrote: On 7/7/2015 3:19 AM, John S wrote: On 7/6/2015 12:02 PM, rickman wrote: On 7/6/2015 12:50 PM, John S wrote: On 7/5/2015 11:39 PM, rickman wrote: On 7/5/2015 4:45 PM, Jerry Stuckle wrote: On 7/5/2015 3:58 PM, Roger Hayter wrote: Jerry Stuckle wrote: On 7/5/2015 9:23 AM, Ian Jackson wrote: In message , writes Wayne wrote: "Jeff Liebermann" wrote in message ... On Sat, 04 Jul 2015 19:04:01 -0400, Jerry Stuckle wrote: Think of it this way, without the math. On the transmitter side of the network, the match is 1:1, with nothing reflected back to the transmitter. So you have a signal coming back from the antenna. You have a perfect matching network, which means nothing is lost in the network. The feedline is perfect, so there is no loss in it. The only place for the signal to go is back to the antenna. Wikipedia says that if the source is matched to the line, any reflections that come back are absorbed, not reflected back to the antenna: https://en.wikipedia.org/wiki/Impedance_matching "If the source impedance matches the line, reflections from the load end will be absorbed at the source end. If the transmission line is not matched at both ends reflections from the load will be re-reflected at the source and re-re-reflected at the load end ad infinitum, losing energy on each transit of the transmission line." Well, I looked at that section of the writeup. And, I have no idea what the hell they are talking about. Looks like a good section for a knowledgeable person to edit. If the termination matches the line impedance, there is no reflection. Both the antenna and the source are terminations. This is a bit difficult to visualze with an RF transmitter, but is more easily seen with pulses. Being essentially a simple soul, that's how I sometimes try to work out what happening. You'll be better off if you killfile the troll. You'll get a lot less bad information and your life will become much easier. The wikipedia entry is correct as written. In the real world, the output of an amateur transmitter will seldom be exactly 50 Ohms unless there is an adjustable network of some sort. I've always understood that the resistive part of a TX output impedance was usually less than 50 ohms. If a transmitter output impedance WAS 50 ohms, I would have thought that the efficiency of the output stage could never exceed 50% (and aren't class-C PAs supposed to be around 66.%?). Also, as much power would be dissipated in the PA stage as in the load. Fixed output amateur transmitters are a nominal 50 ohms. It can vary, but that is due to normal variances in components, and the difference can be ignored in real life. But output impedance has little to do with efficiency. A Class C amplifier can run 90%+ efficiency. It's output may be anything, i.e. high with tubes but low with solid state. But the output impedance can be converted to 50 ohms or any other reasonable impedance through a matching network. A perfect matching network will have no loss, so everything the transmitter puts out will go through the matching network. Of course, nothing is perfect, so there will be some loss. But the amount of loss in a 1:1 match will not be significant. Also, the amplifier generates the power; in a perfect world, 100% of that power is transferred to the load. The transmitter doesn't dissipate 1/2 of the power and the load the other 1/2. It's not like having two resistors in a circuit where each will dissipate 1/2 of the power. BTW - the resistive part of the impedance is not the same as resistance. For a simple case - take a series circuit of a capacitor, an inductor and a 50 ohm resistor. At the resonant frequency, the impedance will be 50 +j0 (50 ohms from the resistor, capacitive and inductive reactances cancel). But the DC resistance is infinity. Again, a simple example, but it shows a point. The resistive part of the impedance is exactly the same as a resistance as far as the frequency you are using is concerned. And if the amplifier output impedance *and* the feeder input resistance were *both* matched to 50 ohms resistive then 50% of the power generated (after circuit losses due of inefficiency of generation) would be dissipated in the transmitter. It would, at the working frequency, be *exactly* like having two equal resistors in the circuit each taking half the generated power. So the amplifier has a much lower output impedance than 50ohms and no attempt is made to match it to 50 ohms. OK, so then please explain how I can have a Class C amplifier with 1KW DC input and a 50 ohm output, 50 ohm coax and a matching network at the antenna can show 900 watt (actually about 870 watts due to feedline loss)? According to your statement, that is impossible. I should not be able get more than 450W or so at the antenna. It is true that if the transmitter is thought of as a fixed voltage generator in series with, say, a 5 ohm resistor then the maximum power transfer in theory would occur with a 5 ohm load. But to achieve this the output power would have to be 100 times higher (ten times the voltage) and half of it would be dissipated in the PA. Not the best way to run things, it is better to have a voltage generator chosen to give the right power with the load being much bigger than its generator resistance. So why do all fixed-tuning amateur transmitters have a nominal 50 ohm output instead of 1 or two ohms? And why do commercial radio stations spend tens of thousands of dollars ensuring impedance is matched throughout the system? The maximum power transfer at equal impedance theorem only applies if you started with a *fixed* output voltage generator. We don't; we start with a load impedance (50 ohm resistive), then we decide what power output we want, and we choose the voltage to be generated accordingly. (Thank you for giving me the opportunity to think about this!) Actually, it doesn't matter if it's a fixed or a variable output voltage - maximum power transfer always occurs when there is an impedance match. How about we quit with the speculation and come up with some numbers? Here is a simulation of a 50 ohm load with a 50 ohm matched series output impedance and a voltage source of 200 VAC peak. Power into the load is 100 W. http://arius.com/sims/Matched%20Load%20Power.png Same exact circuit with the series impedance of just 1 ohm, power into the load is 385 W. http://arius.com/sims/UnMatched%20Load%20Power.png I'd say that is pretty clear evidence that matched loads are not the way to maximize power transfer when the load impedance is fixed and the output impedance is controllable. There can be a lossless resistive part of source impedance according to the IEEE (and most every other well educated EEs). After all, a transmission line has a resistance but it's loss resistance is much lower. Can you provide a reference to any of this? IEEE Standard Dictionary of Electrical and Electron Terms, IEEE Std 100-1972. For a complete treatment on the topic, see Reflections III by Walter Maxwell, W2DU, Appendix 10. Anything more accessible? I will not lead you by the hand. I have supplied the requested references. If you can't find the referenced material, then you need to learn how to research, buy books, go to the library, etc. It is your loss. Education is more than online chatter. Lol. You are a trip. I'm not going to spend $100 on a book just to see if you are right. I was intrigued by the idea that a wire could carry a signal without the resistance dissipating power according to P = I^2 R. I guess there is some communication failure. Yes, there is. I did NOT say that a wire was dissipationless. More communication failure. I never said *that*. Please reread the above. Ditto. I said "There can be a lossless resistive part of source impedance". What you said is clearly quoted above. Source impedance may or may not contain resistance which dissipates. A 50 ohm piece of coax has an impedance of approximately 50+j0 ohms. So, does the 50 ohms dissipate? Look at the specifications of the coax. How much loss? I've never seen a coax specified with a complex impedance. But I'll take your word that this is accurate. However, if you drive a 50 ohm cable, how much current flows? If it is unterminated, I believe the result is none because it all flows back from the reflection and is returned to the source. If you terminate it with a 50 ohm load it is all absorbed and dissipates the appropriate wattage. So I don't see how the 50 ohm impedance of the cable is in any way a useful analogy to say the real part of the impedance does not dissipate. A transmission line is not a load. If you are unwilling to go to the library or do research or buy a book on you own, then you are beyond help. What is your ignorance worth to you? If you don't want to discuss this, why do you type so many words? Maybe it would be useful to explain how an impedance matching network would provide a conjugate match. That is, show an example with real values for components. If this is too much work and you don't wish to do that, fine. I'm ok with it. But no need to insult me. Better to just not respond. -- Rick |
An antenna question--43 ft vertical
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An antenna question--43 ft vertical
On 7/7/2015 12:47 PM, Dave Platt wrote:
In article , rickman wrote: Lol. You are a trip. I'm not going to spend $100 on a book just to see if you are right. I was intrigued by the idea that a wire could carry a signal without the resistance dissipating power according to P = I^2 R. I guess there is some communication failure. Yah. It's a question of terminology. Unfortunately, one term has come to be used for two (related but different) concepts. There is "resistance", as in the E=I^2*R sort. If I recall correctly, Maxwell refers to this as "dissipative" impedance. If you put current through a dissipative resistance, a voltage drop develops across the resistance, and power is dissipated. There are plenty of examples of this, with which I'm sure you're familiar. There is also "resistance", as in "the 'real', non-reactive component of a complex impedance, in which current is in phase with voltage." This type of "resistance" is fundamentally non-dissipative - that is, you can run power through it without dissipating the power as heat. There are also good examples of this. One "textbook" example would be a perfectly-lossless transmission line... say, one made out of a wire and tube of a superconductor, cooled to below the superconducting temperature. You can (in principle) build such a superconducting coax to have almost any convenient impedance... 50 or 75 ohms, for example. Since we're theorizing, let's assume we can built one a few trillion miles long... so long that the far end is light-years away. If you hook a transmitter to one end of this and start transmitting, it will "look" to the transmitter like a 50-ohm dummy load. The transmitter itself won't be able to tell the difference. The transmitter puts out an RF voltage, and the line "takes current" exactly in phase with the voltage, in a ratio of one RF ampere per 50 RF volts. But, there's a fundamental difference between this "resistance" and that of a dummy load. A 50-ohm dummy load's resistance is dissipative... all of the power going into it turns into heat, and is dissipated in accordance with the fundamental laws of thermodynamics. *None* of the power being fed into the superconducting coax, is dissipated as heat in the coax. All of the power still exists, in its original RF form. It's being stored/propagated down the coax without loss. When it hits a load at the other end, it may be dissipated as heat there. Or, perhaps not. What if what's at the other end of the superconducting coax is a superconducting antenna, tweaked to present an impedance of exactly 50 ohms? The RF will be radiated into space. And, "free space" is another great example of a medium that has a well-defined "resistance" (in the non-dissipated sense). https://en.wikipedia.org/wiki/Impedance_of_free_space One of the fundamental jobs of an antenna, is to match the impedance of its feedline to the impedance of free space. Now, any coax you can buy at the store has *both* types of "resistance", of course. It has a dissipative component, and a non-dissipative component. Typically, the more you spend and the more you have to strain your back carrying it around, the lower the amount of dissipative resistance (which is only good for keeping the pigeons' feet warm) and the more predictable and precisely-defined the non-dissipative part. Excellent explanation, Dave. I don't have the ability to educate like you do. Thanks. John |
An antenna question--43 ft vertical
On 7/7/2015 1:58 PM, wrote:
John S wrote: On 7/5/2015 7:21 PM, wrote: John S wrote: On 7/5/2015 5:24 PM, wrote: Roger Hayter wrote: wrote: The output impedance of an amateur transmitter IS approximately 50 Ohms as is trivially shown by reading the specifications for the transmitter which was designed and manufactured to match a 50 Ohm load. Do you think all those manuals are lies? You are starting with a false premise which makes everything after that false. A quick google demonstrates dozens of specification sheets that say the transmitter is designed for a 50 ohm load, and none that mention its output impedance. If the source impedance were other than 50 Ohms, the SWR with 50 Ohm coax and a 50 Ohm antenna would be high. It is not. Where is the source impedance found on a Smith chart? Also, if you have EZNEC, you will not find a place to specify source impedance but it will show the SWR. A Smith chart is normalized to 1. EZNEC allows you to set the impedance to anything you want and assumes the transmission line matches the transmitter. The EZNEC help file is very comprehensive. Please find any reference to your assertion that there is an assumption of source impedance there and provide information for us to verify your assertion. Why don't you email the author and get his take on your assumptions? Why don't YOU? You are the one in need of knowledge. If I do it and report back here you will just doubt it or find something else to argue about. Better you should do it first-hand. |
An antenna question--43 ft vertical
On 7/7/2015 1:37 PM, wrote:
John S wrote: On 7/6/2015 1:03 PM, wrote: John S wrote: On 7/6/2015 11:01 AM, Jerry Stuckle wrote: On 7/6/2015 4:20 AM, Ian Jackson wrote: In message , rickman writes How about we quit with the speculation and come up with some numbers? Here is a simulation of a 50 ohm load with a 50 ohm matched series output impedance and a voltage source of 200 VAC peak. Power into the load is 100 W. http://arius.com/sims/Matched%20Load%20Power.png Same exact circuit with the series impedance of just 1 ohm, power into the load is 385 W. http://arius.com/sims/UnMatched%20Load%20Power.png I'd say that is pretty clear evidence that matched loads are not the way to maximize power transfer when the load impedance is fixed and the output impedance is controllable. Quite simply, if your prime objective is to get maximum power out of a power (energy?) source, the source having an internal resistance is a BAD THING. You don't design the source to have an internal resistance equal to its intended load resistance. No one designs lead-acid batteries that way (do they?), so why RF transmitters? While theoretically you can extract the maximum power available from the source when the load resistance equals the source resistance, you can only do so provided that the heat you generate in the source does not cause the source to malfunction (in the worst case, blow up). Because DC power transfer is not the same as AC power transfer. Why not? Does something happen to the laws of physics with AC? Yes, quite a lot, you get a whole new set of laws. If you apply 1vDC to a 1 ohm resistor, you get 1A of current. If you apply 1vAC RMS (at any frequency) to a 1 ohm resistor, you get 1A of current. How does the AC change the law? What part of "you get a whole new set of laws" was it you failed to understand? Here's a clue for you; at DC the reactive components of a length of wire are irrelevant but at AC they are not. So, at 1Hz the law has changed, eh? What new law do I need to use? |
An antenna question--43 ft vertical
On 7/7/2015 10:05 AM, Jerry Stuckle wrote:
On 7/7/2015 3:05 AM, John S wrote: On 7/6/2015 1:03 PM, wrote: John S wrote: On 7/6/2015 11:01 AM, Jerry Stuckle wrote: On 7/6/2015 4:20 AM, Ian Jackson wrote: In message , rickman writes How about we quit with the speculation and come up with some numbers? Here is a simulation of a 50 ohm load with a 50 ohm matched series output impedance and a voltage source of 200 VAC peak. Power into the load is 100 W. http://arius.com/sims/Matched%20Load%20Power.png Same exact circuit with the series impedance of just 1 ohm, power into the load is 385 W. http://arius.com/sims/UnMatched%20Load%20Power.png I'd say that is pretty clear evidence that matched loads are not the way to maximize power transfer when the load impedance is fixed and the output impedance is controllable. Quite simply, if your prime objective is to get maximum power out of a power (energy?) source, the source having an internal resistance is a BAD THING. You don't design the source to have an internal resistance equal to its intended load resistance. No one designs lead-acid batteries that way (do they?), so why RF transmitters? While theoretically you can extract the maximum power available from the source when the load resistance equals the source resistance, you can only do so provided that the heat you generate in the source does not cause the source to malfunction (in the worst case, blow up). Because DC power transfer is not the same as AC power transfer. Why not? Does something happen to the laws of physics with AC? Yes, quite a lot, you get a whole new set of laws. If you apply 1vDC to a 1 ohm resistor, you get 1A of current. If you apply 1vAC RMS (at any frequency) to a 1 ohm resistor, you get 1A of current. How does the AC change the law? You apply 1vdc to a 0.159 microfarad capacitor and you get 0 amps flowing (open circuit). You apply 1vac at 1MHz to that same capacitor and you get 1 amp flowing, with the current leading the voltage by 90 degrees. You apply 1vdc to a 0.159 microhenry inductor and you get infinite amps flowing (short circuit). You apply 1vdc at 1MHz to that same inductor, and you get 1 amp flowing with the voltage leading the current by 90 degrees. You place the capacitor and inductor in series. Fed with DC, you get 0 amps flowing (open circuit). Fed with 1MHz AC, you get infinite current flowing (short circuit). You place the capacitor and inductor in parallel. Fed with DC, you get infinite current flowing (short circuit). Fed with 1MHz AC you get 0 amps flowing (open circuit). There is a huge difference between ac and dc! Yes, but the LAWS have not changed. The components have changed. So, changing the components changes the laws of physics? Suppose you apply .01Hz AC RMS to the components you specified. What then? |
An antenna question--43 ft vertical
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An antenna question--43 ft vertical
On 7/8/2015 6:32 AM, Jeff wrote:
So, at 1Hz the law has changed, eh? What new law do I need to use? The laws are exactly the same for dc as ac, just at dc the frequency dependant parts tend to the point that they are of no consequence. Jeff Thanks, Jeff. I agree, of course. |
An antenna question--43 ft vertical
Jeff wrote:
So, at 1Hz the law has changed, eh? What new law do I need to use? The laws are exactly the same for dc as ac, just at dc the frequency dependant parts tend to the point that they are of no consequence. Jeff And, of course, they all do matter in the short time after switch-on when things are settling. -- Roger Hayter |
An antenna question--43 ft vertical
On 7/2/2015 1:38 PM, rickman wrote:
On 7/2/2015 1:56 PM, Ralph Mowery wrote: "Jerry Stuckle" wrote in message ... Try this - connect the output of an HF transmitter to an SWR bridge. Now connect a piece of 75 ohm coax such as RG-59 to the output of the SWR meter, and connect that to a 75 ohm resistive load. Do you think the SWR bridge will show a 1:1 SWR? Not a chance. It will be 1.5:1. What you have described is a case of using the wrong swr bridge. You are trying to use a 50 ohm bridge on a 75 ohm system. If a 75 ohm bridge is used it will show a 1:1 SWR. The real SWR is 1:1. With a 75 ohm line and 75 ohm load there is no reflected power. My knowledge of antenna systems is limited, but I do know that this is correct, there will be no reflection from the antenna. If there is no reflections from the antenna, how can there be a loss in the source end? There is NO power returned according to your own statement. If the transmitter output is 50 ohms there will be a loss in this matching that will result in less power being delivered to the feed line, but that will not result in reflections in the feed line. Why? What causes the loss? The transmitter output resistance? So that would mean that one can never achieve more that 50% efficiency at the transmitter's OUTPUT! And that would mean that a 1000W transmitter is dissipating 500 watts under the BEST circumstances. Good luck on getting that to work to your satisfaction. |
An antenna question--43 ft vertical
On 7/7/2015 1:47 PM, Dave Platt wrote:
In article , rickman wrote: Lol. You are a trip. I'm not going to spend $100 on a book just to see if you are right. I was intrigued by the idea that a wire could carry a signal without the resistance dissipating power according to P = I^2 R. I guess there is some communication failure. Yah. It's a question of terminology. Unfortunately, one term has come to be used for two (related but different) concepts. There is "resistance", as in the E=I^2*R sort. If I recall correctly, Maxwell refers to this as "dissipative" impedance. If you put current through a dissipative resistance, a voltage drop develops across the resistance, and power is dissipated. There are plenty of examples of this, with which I'm sure you're familiar. There is also "resistance", as in "the 'real', non-reactive component of a complex impedance, in which current is in phase with voltage." This type of "resistance" is fundamentally non-dissipative - that is, you can run power through it without dissipating the power as heat. There are also good examples of this. One "textbook" example would be a perfectly-lossless transmission line... say, one made out of a wire and tube of a superconductor, cooled to below the superconducting temperature. I have analyzed this previously. Unless your transmission line is infinitely long, eventually the wave reaches the other end and is either dissipatively absorbed or is reflected back to the source where it interacts. An infinitely long transmission line is not very interesting. Dealing with the reflection from a finite transmission line is what we are trying to analyze, so that model is not very useful. So, no, I am not familiar with a non-dissipative load with current in phase with the voltage. You can (in principle) build such a superconducting coax to have almost any convenient impedance... 50 or 75 ohms, for example. Since we're theorizing, let's assume we can built one a few trillion miles long... so long that the far end is light-years away. If you hook a transmitter to one end of this and start transmitting, it will "look" to the transmitter like a 50-ohm dummy load. The transmitter itself won't be able to tell the difference. The transmitter puts out an RF voltage, and the line "takes current" exactly in phase with the voltage, in a ratio of one RF ampere per 50 RF volts. But, there's a fundamental difference between this "resistance" and that of a dummy load. A 50-ohm dummy load's resistance is dissipative... all of the power going into it turns into heat, and is dissipated in accordance with the fundamental laws of thermodynamics. *None* of the power being fed into the superconducting coax, is dissipated as heat in the coax. All of the power still exists, in its original RF form. It's being stored/propagated down the coax without loss. When it hits a load at the other end, it may be dissipated as heat there. Or, perhaps not. What if what's at the other end of the superconducting coax is a superconducting antenna, tweaked to present an impedance of exactly 50 ohms? The RF will be radiated into space. And, "free space" is another great example of a medium that has a well-defined "resistance" (in the non-dissipated sense). I believe this radiation *is* dissipative in the sense that the power is removed from the system being analyzed. That is *exactly* why it is considered to be due to a radiation "resistance". Note they do not refer to it as a radiation "impedance". https://en.wikipedia.org/wiki/Impedance_of_free_space One of the fundamental jobs of an antenna, is to match the impedance of its feedline to the impedance of free space. Now, any coax you can buy at the store has *both* types of "resistance", of course. It has a dissipative component, and a non-dissipative component. Typically, the more you spend and the more you have to strain your back carrying it around, the lower the amount of dissipative resistance (which is only good for keeping the pigeons' feet warm) and the more predictable and precisely-defined the non-dissipative part. What you are calling "non-dissipative resistance" is only a way to characterize the AC behavior. It has nothing to do with what we are discussing and is in no way similar to resistance. Trying to analyze a transmission line without considering the reflections from the other end is only a transient solution which ignores the behavior of the transmission line. How about we construct an example circuit with a conjugate matching network rather than deal with abstractions that have nothing to do with the discussion? If you want to continue to discuss the transmission line, then we need to consider the reflection and find a steady state solution, not a transient one, right? -- Rick |
An antenna question--43 ft vertical
On 7/8/2015 10:09 AM, John S wrote:
On 7/2/2015 1:38 PM, rickman wrote: On 7/2/2015 1:56 PM, Ralph Mowery wrote: "Jerry Stuckle" wrote in message ... Try this - connect the output of an HF transmitter to an SWR bridge. Now connect a piece of 75 ohm coax such as RG-59 to the output of the SWR meter, and connect that to a 75 ohm resistive load. Do you think the SWR bridge will show a 1:1 SWR? Not a chance. It will be 1.5:1. What you have described is a case of using the wrong swr bridge. You are trying to use a 50 ohm bridge on a 75 ohm system. If a 75 ohm bridge is used it will show a 1:1 SWR. The real SWR is 1:1. With a 75 ohm line and 75 ohm load there is no reflected power. My knowledge of antenna systems is limited, but I do know that this is correct, there will be no reflection from the antenna. If there is no reflections from the antenna, how can there be a loss in the source end? There is NO power returned according to your own statement. I don't see any contradiction. The power comes from the source through the source impedance. The source impedance will create a loss, no? If the transmitter output is 50 ohms there will be a loss in this matching that will result in less power being delivered to the feed line, but that will not result in reflections in the feed line. Why? What causes the loss? The transmitter output resistance? So that would mean that one can never achieve more that 50% efficiency at the transmitter's OUTPUT! And that would mean that a 1000W transmitter is dissipating 500 watts under the BEST circumstances. Good luck on getting that to work to your satisfaction. Maybe "loss" isn't the right term then. The output of a 50 ohm source driving a 75 ohm load will deliver 4% less power into the load than when driving a 50 ohm load. That comes to -0.177 dB. Is there any part of that you disagree with? Nothing in this analysis addresses the theoretical maximum possible efficiency of an arbitrary transmitter and an arbitrary load. In particular I posted the results of a simulation that showed very clearly that the loss in the transmitter output impedance can be well below 50% of the total power drawn from the PSU. Just set the load impedance and make your output impedance as low as you would like. It is when you set the output impedance of the transmitter to a fixed value that a matched load impedance will draw the maximum power from the transmitter while the loss in the transmitter output will be 50%. -- Rick |
An antenna question--43 ft vertical
John S wrote:
So, at 1Hz the law has changed, eh? What new law do I need to use? To be pendatic, there is only one set of physical laws that govern electromagnetics. However for DC all the complex parts of those laws have no effect and all the equations can be simplified to remove the complex parts. In the real, practical world people look upon this as two sets of laws, one for AC and one for DC. A good example of this is the transmission line which does not exist at DC; at DC a transmission line is nothing more than two wires with some resistance that is totally and only due to the ohmic resistance of the material that makes up the wires. -- Jim Pennino |
An antenna question--43 ft vertical
Roger Hayter wrote:
Jeff wrote: So, at 1Hz the law has changed, eh? What new law do I need to use? The laws are exactly the same for dc as ac, just at dc the frequency dependant parts tend to the point that they are of no consequence. Jeff And, of course, they all do matter in the short time after switch-on when things are settling. To be pendatic again, there are 3 types of analysis: DC, AC, and transient. In the real world, there are 3 sets of "laws" or equations for each case, with DC being time invariant, AC in the frequency domain, and transient in the time domain. -- Jim Pennino |
An antenna question--43 ft vertical
Jeff wrote:
On 07/07/2015 19:44, wrote: Ian Jackson wrote: In message , Jerry Stuckle writes Sure, there is ALWAYS VSWR. It may be 1:1, but it's always there. If there's no reflection, there can be no standing wave. So, being pedantic, there's no such thing as an SWR of 1:1! Despite the name, VSWR is defined in terms of complex impedances and wavelengths, not "waves" of any kind. VSWR is defined as Vmax/Vmin on the transmission line and is independent of phase or wavelength. It can also be expressed in terms of the *magnitude* of the reflection co-efficient. 1+|p|/1-|p| The reflection coefficient can also be expresses as (Zl - Zo)/(Zl + Zo) where Zl is the complex load impedance and Zo is the complex source impedance. The complex impedances are functions of wavelengths, i.e. frequency. All complex numbers have a frequency dependant part. In the real world of transmission lines and antennas, the source impedance is usually 50 + j0 and thus the second part is ignored. or in terms of forward and reflected power 1+sqtr(Pr/Pf)/1-sqrt(Pr/Pf)) VSWR is the same regardless, of phase, when measured at any point on a lossless line. The phase of the reflection co-efficient will change but not its magnitude. Jeff -- Jim Pennino |
An antenna question--43 ft vertical
John S wrote:
On 7/7/2015 1:44 PM, wrote: Ian Jackson wrote: In message , Jerry Stuckle writes Sure, there is ALWAYS VSWR. It may be 1:1, but it's always there. If there's no reflection, there can be no standing wave. So, being pedantic, there's no such thing as an SWR of 1:1! Despite the name, VSWR is defined in terms of complex impedances and wavelengths, not "waves" of any kind. Actually, VSWR is defined as the ratio of Vmax/Vmin. Actually, VSWR can be defined several ways, one of which is: (1 + |r|)/(1 - |r|) Where r is the reflection coefficient which can be defined a: (Zl - Zo)/(Zl + Zo) Where Zl is the complex load impedance and Zo is the complex source impedance. Note that a complex impedance has a frequency dependant part. -- Jim Pennino |
An antenna question--43 ft vertical
John S wrote:
On 7/7/2015 1:52 PM, wrote: Brian Reay wrote: Do the experiment. Did it decades ago in electromagnetics lab with calibrated test equipmemnt, not with amateur radio equipment. Post the original lab notes, please. That way we cannot challenge the accuracy of your memory. Sorry, that was decades ago. If you are so convinced, do the experiments yourself and post the results. Or you could read an electromagnetics text on transmission lines and show me the errors of my statements. -- Jim Pennino |
An antenna question--43 ft vertical
Brian Reay wrote:
wrote: Brian Reay wrote: On 06/07/15 01:21, wrote: John S wrote: On 7/5/2015 5:24 PM, wrote: Roger Hayter wrote: wrote: The output impedance of an amateur transmitter IS approximately 50 Ohms as is trivially shown by reading the specifications for the transmitter which was designed and manufactured to match a 50 Ohm load. Do you think all those manuals are lies? You are starting with a false premise which makes everything after that false. A quick google demonstrates dozens of specification sheets that say the transmitter is designed for a 50 ohm load, and none that mention its output impedance. If the source impedance were other than 50 Ohms, the SWR with 50 Ohm coax and a 50 Ohm antenna would be high. It is not. Where is the source impedance found on a Smith chart? Also, if you have EZNEC, you will not find a place to specify source impedance but it will show the SWR. A Smith chart is normalized to 1. That is true but is doesn't address the point. There should still be somewhere to represent the source impedance, albeit normalised. The purpose of a Smith chart it to match a SOURCE to a LOAD. EZNEC allows you to set the impedance to anything you want and assumes the transmission line matches the transmitter. Likewise, that is a sweeping statement which evades the point. The main purpose of EZNEC is to design an antenna for amateur radio use and all commercial amateur radio transmitters have an output impedance of 50 Ohms. Neither of those responses address the points. It is clear you cannot support your assertions. EZNEC can set the source impedance to any value one desires, but the default is 50 Ohms as most people are interested in 50 Ohm systems as the majority of transmitters are designed for 50 Ohm loads. EZNEC calculates the SWR for an antenna presented to the SOURCE in the model. For most simulations, the SOURCE is placed at the antenna terminals, which represents what an attached transmission line will see. The transmission line most commonly used is 50 Ohm coax. EZNEC will also allow you to put a SOURCE at one end of a transmission line with the other end of the line at the antenna terminals. In this simulation, EZNEC simulates what the transmitter would see if it were connected to the transmission line/antenna system. SWR is defined in terms of SOURCE impedance and LOAD impedance. SWR = (1 + |r|)/(1 - |r|) Where r = reflection coefficient and r = (Zl - Zo)/(Zl + Zo) Where Zl = complex load impedance, Zo = complex source impedance. -- Jim Pennino |
An antenna question--43 ft vertical
John S wrote:
On 7/7/2015 1:58 PM, wrote: John S wrote: On 7/5/2015 7:21 PM, wrote: John S wrote: On 7/5/2015 5:24 PM, wrote: Roger Hayter wrote: wrote: The output impedance of an amateur transmitter IS approximately 50 Ohms as is trivially shown by reading the specifications for the transmitter which was designed and manufactured to match a 50 Ohm load. Do you think all those manuals are lies? You are starting with a false premise which makes everything after that false. A quick google demonstrates dozens of specification sheets that say the transmitter is designed for a 50 ohm load, and none that mention its output impedance. If the source impedance were other than 50 Ohms, the SWR with 50 Ohm coax and a 50 Ohm antenna would be high. It is not. Where is the source impedance found on a Smith chart? Also, if you have EZNEC, you will not find a place to specify source impedance but it will show the SWR. A Smith chart is normalized to 1. EZNEC allows you to set the impedance to anything you want and assumes the transmission line matches the transmitter. The EZNEC help file is very comprehensive. Please find any reference to your assertion that there is an assumption of source impedance there and provide information for us to verify your assertion. Why don't you email the author and get his take on your assumptions? Why don't YOU? You are the one in need of knowledge. If I do it and report back here you will just doubt it or find something else to argue about. Better you should do it first-hand. EZNEC calculates the SWR presented to the SOURCE which is usually placed at the antenna terminals. EZNEC also calculates the SWR presented to the SOURCE which can be modeled as a SOURCE at one end of a transmission line and the antenna at the other end. SWR is defined in terms of SOURCE impedance and LOAD impedance. I am tired of typing in the same equations over and over again. -- Jim Pennino |
An antenna question--43 ft vertical
On 7/8/2015 6:09 AM, John S wrote:
On 7/7/2015 10:05 AM, Jerry Stuckle wrote: On 7/7/2015 3:05 AM, John S wrote: On 7/6/2015 1:03 PM, wrote: John S wrote: On 7/6/2015 11:01 AM, Jerry Stuckle wrote: On 7/6/2015 4:20 AM, Ian Jackson wrote: In message , rickman writes How about we quit with the speculation and come up with some numbers? Here is a simulation of a 50 ohm load with a 50 ohm matched series output impedance and a voltage source of 200 VAC peak. Power into the load is 100 W. http://arius.com/sims/Matched%20Load%20Power.png Same exact circuit with the series impedance of just 1 ohm, power into the load is 385 W. http://arius.com/sims/UnMatched%20Load%20Power.png I'd say that is pretty clear evidence that matched loads are not the way to maximize power transfer when the load impedance is fixed and the output impedance is controllable. Quite simply, if your prime objective is to get maximum power out of a power (energy?) source, the source having an internal resistance is a BAD THING. You don't design the source to have an internal resistance equal to its intended load resistance. No one designs lead-acid batteries that way (do they?), so why RF transmitters? While theoretically you can extract the maximum power available from the source when the load resistance equals the source resistance, you can only do so provided that the heat you generate in the source does not cause the source to malfunction (in the worst case, blow up). Because DC power transfer is not the same as AC power transfer. Why not? Does something happen to the laws of physics with AC? Yes, quite a lot, you get a whole new set of laws. If you apply 1vDC to a 1 ohm resistor, you get 1A of current. If you apply 1vAC RMS (at any frequency) to a 1 ohm resistor, you get 1A of current. How does the AC change the law? You apply 1vdc to a 0.159 microfarad capacitor and you get 0 amps flowing (open circuit). You apply 1vac at 1MHz to that same capacitor and you get 1 amp flowing, with the current leading the voltage by 90 degrees. You apply 1vdc to a 0.159 microhenry inductor and you get infinite amps flowing (short circuit). You apply 1vdc at 1MHz to that same inductor, and you get 1 amp flowing with the voltage leading the current by 90 degrees. You place the capacitor and inductor in series. Fed with DC, you get 0 amps flowing (open circuit). Fed with 1MHz AC, you get infinite current flowing (short circuit). You place the capacitor and inductor in parallel. Fed with DC, you get infinite current flowing (short circuit). Fed with 1MHz AC you get 0 amps flowing (open circuit). There is a huge difference between ac and dc! Yes, but the LAWS have not changed. The components have changed. So, changing the components changes the laws of physics? Suppose you apply .01Hz AC RMS to the components you specified. What then? The point is - the rules for AC are different than the rules for DC. I'm not going to waste my time figuring out the calculations - you can do that. But the bottom line will be there will be some impedance in every case. It will be neither zero nor infinity, as it would be in a DC circuit. -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
An antenna question--43 ft vertical
In message , John S
writes On 7/8/2015 6:32 AM, Jeff wrote: So, at 1Hz the law has changed, eh? What new law do I need to use? The laws are exactly the same for dc as ac, just at dc the frequency dependant parts tend to the point that they are of no consequence. Jeff Thanks, Jeff. I agree, of course. The thinking that the laws of physics suddenly change when AC is so slow that it suddenly becomes DC is probably about as sound as my tongue-in-cheek suggestion that a SWR suddenly ceases to exist if there is no reflection. ;o) -- Ian |
An antenna question--43 ft vertical
On 08/07/15 19:36, wrote:
Brian Reay wrote: wrote: Brian Reay wrote: On 06/07/15 01:21, wrote: John S wrote: On 7/5/2015 5:24 PM, wrote: Roger Hayter wrote: wrote: The output impedance of an amateur transmitter IS approximately 50 Ohms as is trivially shown by reading the specifications for the transmitter which was designed and manufactured to match a 50 Ohm load. Do you think all those manuals are lies? You are starting with a false premise which makes everything after that false. A quick google demonstrates dozens of specification sheets that say the transmitter is designed for a 50 ohm load, and none that mention its output impedance. If the source impedance were other than 50 Ohms, the SWR with 50 Ohm coax and a 50 Ohm antenna would be high. It is not. Where is the source impedance found on a Smith chart? Also, if you have EZNEC, you will not find a place to specify source impedance but it will show the SWR. A Smith chart is normalized to 1. That is true but is doesn't address the point. There should still be somewhere to represent the source impedance, albeit normalised. The purpose of a Smith chart it to match a SOURCE to a LOAD. EZNEC allows you to set the impedance to anything you want and assumes the transmission line matches the transmitter. Likewise, that is a sweeping statement which evades the point. The main purpose of EZNEC is to design an antenna for amateur radio use and all commercial amateur radio transmitters have an output impedance of 50 Ohms. Neither of those responses address the points. It is clear you cannot support your assertions. EZNEC can set the source impedance to any value one desires, but the default is 50 Ohms as most people are interested in 50 Ohm systems as the majority of transmitters are designed for 50 Ohm loads. EZNEC calculates the SWR for an antenna presented to the SOURCE in the model. For most simulations, the SOURCE is placed at the antenna terminals, which represents what an attached transmission line will see. The transmission line most commonly used is 50 Ohm coax. EZNEC will also allow you to put a SOURCE at one end of a transmission line with the other end of the line at the antenna terminals. In this simulation, EZNEC simulates what the transmitter would see if it were connected to the transmission line/antenna system. SWR is defined in terms of SOURCE impedance and LOAD impedance. SWR = (1 + |r|)/(1 - |r|) Where r = reflection coefficient and r = (Zl - Zo)/(Zl + Zo) Where Zl = complex load impedance, Zo = complex source impedance. The matter at hand isn't SWR it is the output impedance of PAs. That seems to be you sticking point. You are assuming the PA is a transmission line, rather than an active source. You confusion isn't helped by the habit of some manufacturers including SWR in the PA spec. Modern PAs are designed to drive a load of 50 ohms, they don't have a source impedance of 50 ohms. If they are driven into the wrong load, they can operated outside there safe area of operation. If the power isn't reduced, they can be damaged. |
An antenna question--43 ft vertical
On 7/8/2015 1:39 PM, wrote:
John S wrote: On 7/7/2015 1:58 PM, wrote: John S wrote: On 7/5/2015 7:21 PM, wrote: John S wrote: On 7/5/2015 5:24 PM, wrote: Roger Hayter wrote: wrote: The output impedance of an amateur transmitter IS approximately 50 Ohms as is trivially shown by reading the specifications for the transmitter which was designed and manufactured to match a 50 Ohm load. Do you think all those manuals are lies? You are starting with a false premise which makes everything after that false. A quick google demonstrates dozens of specification sheets that say the transmitter is designed for a 50 ohm load, and none that mention its output impedance. If the source impedance were other than 50 Ohms, the SWR with 50 Ohm coax and a 50 Ohm antenna would be high. It is not. Where is the source impedance found on a Smith chart? Also, if you have EZNEC, you will not find a place to specify source impedance but it will show the SWR. A Smith chart is normalized to 1. EZNEC allows you to set the impedance to anything you want and assumes the transmission line matches the transmitter. The EZNEC help file is very comprehensive. Please find any reference to your assertion that there is an assumption of source impedance there and provide information for us to verify your assertion. Why don't you email the author and get his take on your assumptions? Why don't YOU? You are the one in need of knowledge. If I do it and report back here you will just doubt it or find something else to argue about. Better you should do it first-hand. EZNEC calculates the SWR presented to the SOURCE which is usually placed at the antenna terminals. EZNEC also calculates the SWR presented to the SOURCE which can be modeled as a SOURCE at one end of a transmission line and the antenna at the other end. SWR is defined in terms of SOURCE impedance and LOAD impedance. No. It is defined as Vmax/Vmin on the line. Show an equation that defines SWR as the matching of the source to the line. I am tired of typing in the same equations over and over again. Then you are excused from participating. |
An antenna question--43 ft vertical
On 7/8/2015 1:18 PM, wrote:
John S wrote: On 7/7/2015 1:52 PM, wrote: Brian Reay wrote: Do the experiment. Did it decades ago in electromagnetics lab with calibrated test equipmemnt, not with amateur radio equipment. Post the original lab notes, please. That way we cannot challenge the accuracy of your memory. Sorry, that was decades ago. If you are so convinced, do the experiments yourself and post the results. Or you could read an electromagnetics text on transmission lines and show me the errors of my statements. I did, decades ago. The results are that you are wrong. You surely trust my memory as well as I trust yours, yes? |
An antenna question--43 ft vertical
Brian Reay wrote:
On 08/07/15 19:36, wrote: Brian Reay wrote: wrote: Brian Reay wrote: On 06/07/15 01:21, wrote: John S wrote: On 7/5/2015 5:24 PM, wrote: Roger Hayter wrote: wrote: The output impedance of an amateur transmitter IS approximately 50 Ohms as is trivially shown by reading the specifications for the transmitter which was designed and manufactured to match a 50 Ohm load. Do you think all those manuals are lies? You are starting with a false premise which makes everything after that false. A quick google demonstrates dozens of specification sheets that say the transmitter is designed for a 50 ohm load, and none that mention its output impedance. If the source impedance were other than 50 Ohms, the SWR with 50 Ohm coax and a 50 Ohm antenna would be high. It is not. Where is the source impedance found on a Smith chart? Also, if you have EZNEC, you will not find a place to specify source impedance but it will show the SWR. A Smith chart is normalized to 1. That is true but is doesn't address the point. There should still be somewhere to represent the source impedance, albeit normalised. The purpose of a Smith chart it to match a SOURCE to a LOAD. EZNEC allows you to set the impedance to anything you want and assumes the transmission line matches the transmitter. Likewise, that is a sweeping statement which evades the point. The main purpose of EZNEC is to design an antenna for amateur radio use and all commercial amateur radio transmitters have an output impedance of 50 Ohms. Neither of those responses address the points. It is clear you cannot support your assertions. EZNEC can set the source impedance to any value one desires, but the default is 50 Ohms as most people are interested in 50 Ohm systems as the majority of transmitters are designed for 50 Ohm loads. EZNEC calculates the SWR for an antenna presented to the SOURCE in the model. For most simulations, the SOURCE is placed at the antenna terminals, which represents what an attached transmission line will see. The transmission line most commonly used is 50 Ohm coax. EZNEC will also allow you to put a SOURCE at one end of a transmission line with the other end of the line at the antenna terminals. In this simulation, EZNEC simulates what the transmitter would see if it were connected to the transmission line/antenna system. SWR is defined in terms of SOURCE impedance and LOAD impedance. SWR = (1 + |r|)/(1 - |r|) Where r = reflection coefficient and r = (Zl - Zo)/(Zl + Zo) Where Zl = complex load impedance, Zo = complex source impedance. The matter at hand isn't SWR it is the output impedance of PAs. Nope, the matter at hand is the definition of SWR. That seems to be you sticking point. You are assuming the PA is a transmission line, rather than an active source. Nope, the matter at hand is the definition of SWR, which is defined in terms of SOURCE impedance and LOAD impedance. It does not matter in the slightest if the SOURCE impedance is the output of a transmitter or the end of a transmission line. You confusion isn't helped by the habit of some manufacturers including SWR in the PA spec. Manufacturers specify the LOAD impedance for the transmitter, and sometimes the SWR range that the transmitter will handle. Modern PAs are designed to drive a load of 50 ohms, they don't have a source impedance of 50 ohms. If they are driven into the wrong load, they can operated outside there safe area of operation. If the power isn't reduced, they can be damaged. Once again, SWR is defined in terms of SOURCE impedance and LOAD impedance. The normal LOAD for a transmitter is one end of a piece of coax with an antenna on the other end. The SWR at the near end of a piece of coax may or may not be the same as the SWR at the far end of the coax. -- Jim Pennino |
An antenna question--43 ft vertical
On 7/8/2015 10:48 AM, rickman wrote:
On 7/8/2015 10:09 AM, John S wrote: On 7/2/2015 1:38 PM, rickman wrote: On 7/2/2015 1:56 PM, Ralph Mowery wrote: "Jerry Stuckle" wrote in message ... Try this - connect the output of an HF transmitter to an SWR bridge. Now connect a piece of 75 ohm coax such as RG-59 to the output of the SWR meter, and connect that to a 75 ohm resistive load. Do you think the SWR bridge will show a 1:1 SWR? Not a chance. It will be 1.5:1. What you have described is a case of using the wrong swr bridge. You are trying to use a 50 ohm bridge on a 75 ohm system. If a 75 ohm bridge is used it will show a 1:1 SWR. The real SWR is 1:1. With a 75 ohm line and 75 ohm load there is no reflected power. My knowledge of antenna systems is limited, but I do know that this is correct, there will be no reflection from the antenna. If there is no reflections from the antenna, how can there be a loss in the source end? There is NO power returned according to your own statement. I don't see any contradiction. The power comes from the source through the source impedance. The source impedance will create a loss, no? If the transmitter output is 50 ohms there will be a loss in this matching that will result in less power being delivered to the feed line, but that will not result in reflections in the feed line. Why? What causes the loss? The transmitter output resistance? So that would mean that one can never achieve more that 50% efficiency at the transmitter's OUTPUT! And that would mean that a 1000W transmitter is dissipating 500 watts under the BEST circumstances. Good luck on getting that to work to your satisfaction. Maybe "loss" isn't the right term then. The output of a 50 ohm source driving a 75 ohm load will deliver 4% less power into the load than when driving a 50 ohm load. That comes to -0.177 dB. Is there any part of that you disagree with? All of it. Let's say you have a 1A source and it has a 50 ohm impedance in series with its output. With a 50 ohm load it will provide 50W to the load. With a 75 ohm load it will provide 75W to the load. The only difference is that the 50 ohm load will cause the source voltage (before the series impedance) to be 100V while the 75 ohm load will require 112V (before the series impedance). If the series impedance is 0 +/- j75 ohms, it will have no power loss. If the series impedance is 50 + j0 it will have a 50W loss. |
An antenna question--43 ft vertical
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An antenna question--43 ft vertical
On 7/8/2015 1:14 PM, wrote:
John S wrote: On 7/7/2015 1:44 PM, wrote: Ian Jackson wrote: In message , Jerry Stuckle writes Sure, there is ALWAYS VSWR. It may be 1:1, but it's always there. If there's no reflection, there can be no standing wave. So, being pedantic, there's no such thing as an SWR of 1:1! Despite the name, VSWR is defined in terms of complex impedances and wavelengths, not "waves" of any kind. Actually, VSWR is defined as the ratio of Vmax/Vmin. Actually, VSWR can be defined several ways, one of which is: (1 + |r|)/(1 - |r|) Where r is the reflection coefficient which can be defined a: (Zl - Zo)/(Zl + Zo) Where Zl is the complex load impedance and Zo is the complex source impedance. Note that a complex impedance has a frequency dependant part. So, since Vmax/Vmin (the base definition) has no frequency dependent part, does that invalidate it? |
An antenna question--43 ft vertical
On 7/8/2015 2:38 PM, John S wrote:
On 7/8/2015 10:48 AM, rickman wrote: On 7/8/2015 10:09 AM, John S wrote: On 7/2/2015 1:38 PM, rickman wrote: On 7/2/2015 1:56 PM, Ralph Mowery wrote: "Jerry Stuckle" wrote in message ... Try this - connect the output of an HF transmitter to an SWR bridge. Now connect a piece of 75 ohm coax such as RG-59 to the output of the SWR meter, and connect that to a 75 ohm resistive load. Do you think the SWR bridge will show a 1:1 SWR? Not a chance. It will be 1.5:1. What you have described is a case of using the wrong swr bridge. You are trying to use a 50 ohm bridge on a 75 ohm system. If a 75 ohm bridge is used it will show a 1:1 SWR. The real SWR is 1:1. With a 75 ohm line and 75 ohm load there is no reflected power. My knowledge of antenna systems is limited, but I do know that this is correct, there will be no reflection from the antenna. If there is no reflections from the antenna, how can there be a loss in the source end? There is NO power returned according to your own statement. I don't see any contradiction. The power comes from the source through the source impedance. The source impedance will create a loss, no? If the transmitter output is 50 ohms there will be a loss in this matching that will result in less power being delivered to the feed line, but that will not result in reflections in the feed line. Why? What causes the loss? The transmitter output resistance? So that would mean that one can never achieve more that 50% efficiency at the transmitter's OUTPUT! And that would mean that a 1000W transmitter is dissipating 500 watts under the BEST circumstances. Good luck on getting that to work to your satisfaction. Maybe "loss" isn't the right term then. The output of a 50 ohm source driving a 75 ohm load will deliver 4% less power into the load than when driving a 50 ohm load. That comes to -0.177 dB. Is there any part of that you disagree with? All of it. Let's say you have a 1A source and it has a 50 ohm impedance in series with its output. With a 50 ohm load it will provide 50W to the load. With a 75 ohm load it will provide 75W to the load. The only difference is that the 50 ohm load will cause the source voltage (before the series impedance) to be 100V while the 75 ohm load will require 112V (before the series impedance). If the series impedance is 0 +/- j75 ohms, it will have no power loss. If the series impedance is 50 + j0 it will have a 50W loss. Oops! Source voltage will be 70.7V for 50 ohms and 90V for 75 ohms and dissipation-less output impedance. |
An antenna question--43 ft vertical
wrote in message ... Once again, SWR is defined in terms of SOURCE impedance and LOAD impedance. The normal LOAD for a transmitter is one end of a piece of coax with an antenna on the other end. The SWR at the near end of a piece of coax may or may not be the same as the SWR at the far end of the coax. -- Jim Pennino Can you show any place where the SWR definition mentions the Source impedance ? I have never seen anything that mentions the Source impedance. Just the ratio of the voltage or current going forward and reflected. The SWR has to be the same at any point on the coax or transmission line minus the loss in the line. A simple swr meter may show some differance because of the way that kind of meter works. By changing the length of the line , the apparent SWR may be differant at that point. |
An antenna question--43 ft vertical
wrote:
Roger Hayter wrote: Jeff wrote: So, at 1Hz the law has changed, eh? What new law do I need to use? The laws are exactly the same for dc as ac, just at dc the frequency dependant parts tend to the point that they are of no consequence. Jeff And, of course, they all do matter in the short time after switch-on when things are settling. To be pendatic again, there are 3 types of analysis: DC, AC, and transient. In the real world, there are 3 sets of "laws" or equations for each case, with DC being time invariant, AC in the frequency domain, and transient in the time domain. There is only one set of laws, but the maths is simpler for the simpler cases. But the equations for the transient case will still give the right answer for the DC case. -- Roger Hayter |
An antenna question--43 ft vertical
wrote:
John S wrote: On 7/7/2015 1:58 PM, wrote: John S wrote: On 7/5/2015 7:21 PM, wrote: John S wrote: On 7/5/2015 5:24 PM, wrote: Roger Hayter wrote: wrote: The output impedance of an amateur transmitter IS approximately 50 Ohms as is trivially shown by reading the specifications for the transmitter which was designed and manufactured to match a 50 Ohm load. Do you think all those manuals are lies? You are starting with a false premise which makes everything after that false. A quick google demonstrates dozens of specification sheets that say the transmitter is designed for a 50 ohm load, and none that mention its output impedance. If the source impedance were other than 50 Ohms, the SWR with 50 Ohm coax and a 50 Ohm antenna would be high. It is not. Where is the source impedance found on a Smith chart? Also, if you have EZNEC, you will not find a place to specify source impedance but it will show the SWR. A Smith chart is normalized to 1. EZNEC allows you to set the impedance to anything you want and assumes the transmission line matches the transmitter. The EZNEC help file is very comprehensive. Please find any reference to your assertion that there is an assumption of source impedance there and provide information for us to verify your assertion. Why don't you email the author and get his take on your assumptions? Why don't YOU? You are the one in need of knowledge. If I do it and report back here you will just doubt it or find something else to argue about. Better you should do it first-hand. EZNEC calculates the SWR presented to the SOURCE which is usually placed at the antenna terminals. EZNEC also calculates the SWR presented to the SOURCE which can be modeled as a SOURCE at one end of a transmission line and the antenna at the other end. SWR is defined in terms of SOURCE impedance and LOAD impedance. I am tired of typing in the same equations over and over again. Zo is the characteristiic impedance of the transmission line and nothing to do with the source impedance of whatever generator is supplying power to the system. -- Roger Hayter |
An antenna question--43 ft vertical
Roger Hayter wrote:
wrote: Roger Hayter wrote: Jeff wrote: So, at 1Hz the law has changed, eh? What new law do I need to use? The laws are exactly the same for dc as ac, just at dc the frequency dependant parts tend to the point that they are of no consequence. Jeff And, of course, they all do matter in the short time after switch-on when things are settling. To be pendatic again, there are 3 types of analysis: DC, AC, and transient. In the real world, there are 3 sets of "laws" or equations for each case, with DC being time invariant, AC in the frequency domain, and transient in the time domain. There is only one set of laws, but the maths is simpler for the simpler cases. But the equations for the transient case will still give the right answer for the DC case. Only for t = infinity. -- Jim Pennino |
An antenna question--43 ft vertical
John S wrote:
On 7/8/2015 12:47 PM, wrote: John S wrote: So, at 1Hz the law has changed, eh? What new law do I need to use? To be pendatic, there is only one set of physical laws that govern electromagnetics. However for DC all the complex parts of those laws have no effect and all the equations can be simplified to remove the complex parts. In the real, practical world people look upon this as two sets of laws, one for AC and one for DC. A good example of this is the transmission line which does not exist at DC; at DC a transmission line is nothing more than two wires with some resistance that is totally and only due to the ohmic resistance of the material that makes up the wires. So, is .01Hz AC or DC, Jim? How about 1Hz? 10Hz? Where does AC begin and DC end? It is called a limit. If there is NO time varying component, it is DC, otherwise it is AC. Are you playing devil's advocate or are you really that ignorant? -- Jim Pennino |
An antenna question--43 ft vertical
John S wrote:
On 7/8/2015 1:14 PM, wrote: John S wrote: On 7/7/2015 1:44 PM, wrote: Ian Jackson wrote: In message , Jerry Stuckle writes Sure, there is ALWAYS VSWR. It may be 1:1, but it's always there. If there's no reflection, there can be no standing wave. So, being pedantic, there's no such thing as an SWR of 1:1! Despite the name, VSWR is defined in terms of complex impedances and wavelengths, not "waves" of any kind. Actually, VSWR is defined as the ratio of Vmax/Vmin. Actually, VSWR can be defined several ways, one of which is: (1 + |r|)/(1 - |r|) Where r is the reflection coefficient which can be defined a: (Zl - Zo)/(Zl + Zo) Where Zl is the complex load impedance and Zo is the complex source impedance. Note that a complex impedance has a frequency dependant part. So, since Vmax/Vmin (the base definition) has no frequency dependent part, does that invalidate it? The "base definition" can be whatever set of equations you pick that are true. BTW, the Vmax/Vmin DOES have a frequency dependant component that determines WHERE Vmax and Vmin occur. -- Jim Pennino |
An antenna question--43 ft vertical
John S wrote:
On 7/8/2015 1:18 PM, wrote: John S wrote: On 7/7/2015 1:52 PM, wrote: Brian Reay wrote: Do the experiment. Did it decades ago in electromagnetics lab with calibrated test equipmemnt, not with amateur radio equipment. Post the original lab notes, please. That way we cannot challenge the accuracy of your memory. Sorry, that was decades ago. If you are so convinced, do the experiments yourself and post the results. Or you could read an electromagnetics text on transmission lines and show me the errors of my statements. I did, decades ago. The results are that you are wrong. You surely trust my memory as well as I trust yours, yes? What I trust is what I can read in an electromagnetics text. -- Jim Pennino |
An antenna question--43 ft vertical
Ralph Mowery wrote:
wrote in message ... Once again, SWR is defined in terms of SOURCE impedance and LOAD impedance. The normal LOAD for a transmitter is one end of a piece of coax with an antenna on the other end. The SWR at the near end of a piece of coax may or may not be the same as the SWR at the far end of the coax. -- Jim Pennino Can you show any place where the SWR definition mentions the Source impedance ? I have several times now, but once again: SWR = (1 + |r|)/(1 - |r|) Where r = reflection coefficient. r = (Zl - Zo)/(Zl + Zo) Where Zl = complex load impedance and Zo = complex source impedance. https://en.wikipedia.org/wiki/Reflection_coefficient http://www.antenna-theory.com/tutori...nsmission3.php I have never seen anything that mentions the Source impedance. Just the ratio of the voltage or current going forward and reflected. It is generally not mentioned in Amateur publications. The SWR has to be the same at any point on the coax or transmission line minus the loss in the line. A simple swr meter may show some differance because of the way that kind of meter works. By changing the length of the line , the apparent SWR may be differant at that point. There is no such thing as apparent SWR. It is what it is in a given place. Transmission line transformers. http://highfrequencyelectronics.com/...TraskPart2.pdf Impedance matching. https://en.wikipedia.org/wiki/Impedance_matching -- Jim Pennino |
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