|
An antenna question--43 ft vertical
Roger Hayter wrote:
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. No, Zo is the source impedance and it is irrelevant what the source physically is and has nothing to do with power. https://en.wikipedia.org/wiki/Reflection_coefficient http://www.antenna-theory.com/tutori...nsmission3.php -- Jim Pennino |
An antenna question--43 ft vertical
wrote in message ... Ralph Mowery wrote: 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 YOu have just proven my point. Read carefully from your refernce to Wikipedia : "The reflection coefficient of a load is determined by its impedance and the impedance toward the source." Notice it says TOWARD and not THE SOURCE. From the second referaence notice that it says load impedance and impedance of the transmission line. Nothing mentions the source at all: "The reflection coefficient is usually denoted by the symbol gamma. Note that the magnitude of the reflection coefficient does not depend on the length of the line, only the load impedance and the impedance of the transmission line. Also, note that if ZL=Z0, then the line is "matched". In this case, there is no mismatch loss and all power is transferred to the load." |
An antenna question--43 ft vertical
Ralph Mowery wrote:
wrote in message ... Ralph Mowery wrote: 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 YOu have just proven my point. Read carefully from your refernce to Wikipedia : "The reflection coefficient of a load is determined by its impedance and the impedance toward the source." Notice it says TOWARD and not THE SOURCE. Notice it actually says "the impedance toward the source". From the second referaence notice that it says load impedance and impedance of the transmission line. Nothing mentions the source at all: What the hell do you think the transmission line is in this case if not the source? "The reflection coefficient is usually denoted by the symbol gamma. Note that the magnitude of the reflection coefficient does not depend on the length of the line, only the load impedance and the impedance of the transmission line. Also, note that if ZL=Z0, then the line is "matched". In this case, there is no mismatch loss and all power is transferred to the load." Perhaps you would like the second link better as it has pictures. Of maybe this one that explains it all starting with lumped equivelant circuits. http://www.maximintegrated.com/en/ap...dex.mvp/id/742 Notice that ALL the links talk about the source impedance. -- Jim Pennino |
An antenna question--43 ft vertical
On 7/8/2015 3:38 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. I was referring to a voltage source. -- Rick |
An antenna question--43 ft vertical
|
An antenna question--43 ft vertical
rickman wrote:
On 7/8/2015 3:38 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. I was referring to a voltage source. Instead of arguing about it, one can download QUCS for free which will simulate the whole thing and one can see what really happens. Download QUCS for your operating system: http://qucs.sourceforge.net/ Generate a model consisting of a voltage source with a series resistance of a few Ohms to simulate a solid state source or a much higher resistance to simulate a vacuum tube source. Chose a convienient frequency for the source. Go to: http://home.sandiego.edu/~ekim/e194r.../matcher2.html to calculate an impedance matching network to match the resistance you've chosen to 50 Ohms. Put the matching circuit in the model. Add a transmission line to the model. Terminate the transmission line with a 50 Ohms resistor. Add a fixed frequency AC simulation at the desired frequency. Change various parameters to your heart's content to see what happens. Change the matching network such that the output of your transmitter is no longer 50 Ohms and see what happens. When the QUCS output disagrees with your beliefs, you can argue with the program. -- Jim Pennino |
An antenna question--43 ft vertical
Wayne wrote:
"John S" wrote in message ... 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. That's also my understanding of the definition. That is one definition. In fact since SWR is defined as the maximum to minimum voltage ratio, the "V" in VSWR is redundant. There are other kinds of "SWR", but you will never see them refered to in Amateur literature. But, using the voltage ratio definition, you can work yourself back around to various source and load impedances, reflection coefficients, etc. Or you can start from source and load impedances. The whole Vmax/Vmin definition is a leftover from the early days when RF measurments were done on Lecher lines or slotted lines and you actually looked for the min and max points on the line. I doubt few under about 50 have heard of slotted lines or Lecher lines much less ever used one. -- Jim Pennino |
An antenna question--43 ft vertical
On 7/8/2015 4:48 PM, wrote:
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? Then there is no such thing as DC because even a battery looses voltage over a period of time. DC voltage sources have noise. Are just being argumentative or are you really that ignorant? |
An antenna question--43 ft vertical
On 7/8/2015 4:51 PM, wrote:
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. You are just being argumentative. The WHERE doesn't matter in measuring VSWR. You still measure correct VSWR wherever the locations. |
An antenna question--43 ft vertical
On 7/8/2015 7:27 PM, Wayne wrote:
"John S" wrote in message ... 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. That's also my understanding of the definition. In fact since SWR is defined as the maximum to minimum voltage ratio, the "V" in VSWR is redundant. Sort of. There is also ISWR but it is not used frequently. But, using the voltage ratio definition, you can work yourself back around to various source and load impedances, reflection coefficients, etc. Sure. In fact I once built a slotted line to measure load impedances because I could not afford to buy a vector voltmeter at the time. |
An antenna question--43 ft vertical
On 7/8/2015 4:52 PM, wrote:
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. Fine. Then you should confine your arguments to those sources rather than referring to decades old measurements to you may not remember correctly. |
An antenna question--43 ft vertical
On 7/8/2015 8:34 PM, wrote:
rickman wrote: On 7/8/2015 3:38 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. I was referring to a voltage source. Instead of arguing about it, one can download QUCS for free which will simulate the whole thing and one can see what really happens. Download QUCS for your operating system: http://qucs.sourceforge.net/ Generate a model consisting of a voltage source with a series resistance of a few Ohms to simulate a solid state source or a much higher resistance to simulate a vacuum tube source. Chose a convienient frequency for the source. Go to: http://home.sandiego.edu/~ekim/e194r.../matcher2.html to calculate an impedance matching network to match the resistance you've chosen to 50 Ohms. Put the matching circuit in the model. Add a transmission line to the model. Terminate the transmission line with a 50 Ohms resistor. Add a fixed frequency AC simulation at the desired frequency. Change various parameters to your heart's content to see what happens. Change the matching network such that the output of your transmitter is no longer 50 Ohms and see what happens. When the QUCS output disagrees with your beliefs, you can argue with the program. Uh, I already did a simulation using LTspice. No one even commented on the simulation as I recall. Besides, I don't really need a simulation to measure the power delivered to a load. That is a *very* simple circuit to calculate in a few seconds. But first we have to agree on what we are discussing. I'm not going to retrace this entire conversation, but someone said a matching network which presents a complex impedance with a non-zero real part and a zero imaginary part would reflect 100% of the wave which had been reflected from the antenna back to the transmitter output. I was trying to nail down this example so I could do some calculations on it. If you want to do that I am happy to work on the problem. If not, that's fine too. -- Rick |
An antenna question--43 ft vertical
On 7/8/2015 6:43 PM, wrote:
Ralph Mowery wrote: wrote in message ... Ralph Mowery wrote: 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 YOu have just proven my point. Read carefully from your refernce to Wikipedia : "The reflection coefficient of a load is determined by its impedance and the impedance toward the source." Notice it says TOWARD and not THE SOURCE. Notice it actually says "the impedance toward the source". From the second referaence notice that it says load impedance and impedance of the transmission line. Nothing mentions the source at all: What the hell do you think the transmission line is in this case if not the source? "The reflection coefficient is usually denoted by the symbol gamma. Note that the magnitude of the reflection coefficient does not depend on the length of the line, only the load impedance and the impedance of the transmission line. Also, note that if ZL=Z0, then the line is "matched". In this case, there is no mismatch loss and all power is transferred to the load." Perhaps you would like the second link better as it has pictures. Of maybe this one that explains it all starting with lumped equivelant circuits. http://www.maximintegrated.com/en/ap...dex.mvp/id/742 Notice that ALL the links talk about the source impedance. So, you are saying Zo is the source impedance while every one else thinks it is the characteristic impedance of the line. Go back to your books and look up the definition of Zo. |
An antenna question--43 ft vertical
On 7/8/2015 7:43 PM, wrote:
Ralph Mowery wrote: wrote in message ... Ralph Mowery wrote: 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 You might check that again. I don't see Zo being defined as the complex source impedance, but rather as the transmission line characteristic impedance... not the same thing at all. YOu have just proven my point. Read carefully from your refernce to Wikipedia : "The reflection coefficient of a load is determined by its impedance and the impedance toward the source." Notice it says TOWARD and not THE SOURCE. Notice it actually says "the impedance toward the source". From the second referaence notice that it says load impedance and impedance of the transmission line. Nothing mentions the source at all: What the hell do you think the transmission line is in this case if not the source? "The reflection coefficient is usually denoted by the symbol gamma. Note that the magnitude of the reflection coefficient does not depend on the length of the line, only the load impedance and the impedance of the transmission line. Also, note that if ZL=Z0, then the line is "matched". In this case, there is no mismatch loss and all power is transferred to the load." Perhaps you would like the second link better as it has pictures. Of maybe this one that explains it all starting with lumped equivelant circuits. http://www.maximintegrated.com/en/ap...dex.mvp/id/742 Notice that ALL the links talk about the source impedance. How about this one? https://en.wikipedia.org/wiki/Standi...dance_matching I think this has some very interesting analysis, very specifically referring to "purely resistive load impedance". -- Rick |
An antenna question--43 ft vertical
rickman wrote:
On 7/8/2015 8:34 PM, wrote: rickman wrote: On 7/8/2015 3:38 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. I was referring to a voltage source. Instead of arguing about it, one can download QUCS for free which will simulate the whole thing and one can see what really happens. Download QUCS for your operating system: http://qucs.sourceforge.net/ Generate a model consisting of a voltage source with a series resistance of a few Ohms to simulate a solid state source or a much higher resistance to simulate a vacuum tube source. Chose a convienient frequency for the source. Go to: http://home.sandiego.edu/~ekim/e194r.../matcher2.html to calculate an impedance matching network to match the resistance you've chosen to 50 Ohms. Put the matching circuit in the model. Add a transmission line to the model. Terminate the transmission line with a 50 Ohms resistor. Add a fixed frequency AC simulation at the desired frequency. Change various parameters to your heart's content to see what happens. Change the matching network such that the output of your transmitter is no longer 50 Ohms and see what happens. When the QUCS output disagrees with your beliefs, you can argue with the program. Uh, I already did a simulation using LTspice. No one even commented on the simulation as I recall. Besides, I don't really need a simulation to measure the power delivered to a load. That is a *very* simple circuit to calculate in a few seconds. But first we have to agree on what we are discussing. Does LTspice do transmission lines? I'm not going to retrace this entire conversation, but someone said a matching network which presents a complex impedance with a non-zero real part and a zero imaginary part would reflect 100% of the wave which had been reflected from the antenna back to the transmitter output. I was trying to nail down this example so I could do some calculations on it. If you want to do that I am happy to work on the problem. If not, that's fine too. Nope, I know what happens. BTW, my response was not directed at any particular person other than those that do not understand the conditions for maximum power transfer given a fixed source. -- Jim Pennino |
An antenna question--43 ft vertical
John S wrote:
On 7/8/2015 4:48 PM, wrote: 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? Then there is no such thing as DC because even a battery looses voltage over a period of time. DC voltage sources have noise. An ideal battery doesn't. -- Jim Pennino |
An antenna question--43 ft vertical
John S wrote:
On 7/8/2015 4:51 PM, wrote: 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. You are just being argumentative. The WHERE doesn't matter in measuring VSWR. You still measure correct VSWR wherever the locations. Can you measure VSWR on a 1 meter long Lecher line at 1 MHz? -- Jim Pennino |
An antenna question--43 ft vertical
John S wrote:
On 7/8/2015 7:27 PM, Wayne wrote: "John S" wrote in message ... 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. That's also my understanding of the definition. In fact since SWR is defined as the maximum to minimum voltage ratio, the "V" in VSWR is redundant. Sort of. There is also ISWR but it is not used frequently. Not sort of, but is. There is also PSWR. -- Jim Pennino |
An antenna question--43 ft vertical
John S wrote:
On 7/8/2015 4:52 PM, wrote: 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. Fine. Then you should confine your arguments to those sources rather than referring to decades old measurements to you may not remember correctly. Instead of arguing about it, one can download QUCS for free which will simulate the whole thing and one can see what really happens. Download QUCS for your operating system: http://qucs.sourceforge.net/ Generate a model consisting of a voltage source with a series resistance of a few Ohms to simulate a solid state source or a much higher resistance to simulate a vacuum tube source. Chose a convienient frequency for the simulation. Go to: http://home.sandiego.edu/~ekim/e194r.../matcher2.html to calculate an impedance matching network to match the resistance you've chosen to 50 Ohms at the choosen frequency. Put the matching circuit in the model. Add a 50 Ohm transmission line to the model. Terminate the transmission line with a 50 Ohm resistor. Add a fixed frequency AC simulation at the choosen frequency. Change various parameters to your heart's content to see what happens. Change the matching network such that the output of your transmitter is no longer 50 Ohms and see what happens. When the QUCS output disagrees with your beliefs, you can argue with the program. -- Jim Pennino |
An antenna question--43 ft vertical
John S wrote:
On 7/8/2015 6:43 PM, wrote: Ralph Mowery wrote: wrote in message ... Ralph Mowery wrote: 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 YOu have just proven my point. Read carefully from your refernce to Wikipedia : "The reflection coefficient of a load is determined by its impedance and the impedance toward the source." Notice it says TOWARD and not THE SOURCE. Notice it actually says "the impedance toward the source". From the second referaence notice that it says load impedance and impedance of the transmission line. Nothing mentions the source at all: What the hell do you think the transmission line is in this case if not the source? "The reflection coefficient is usually denoted by the symbol gamma. Note that the magnitude of the reflection coefficient does not depend on the length of the line, only the load impedance and the impedance of the transmission line. Also, note that if ZL=Z0, then the line is "matched". In this case, there is no mismatch loss and all power is transferred to the load." Perhaps you would like the second link better as it has pictures. Of maybe this one that explains it all starting with lumped equivelant circuits. http://www.maximintegrated.com/en/ap...dex.mvp/id/742 Notice that ALL the links talk about the source impedance. So, you are saying Zo is the source impedance while every one else thinks it is the characteristic impedance of the line. Go back to your books and look up the definition of Zo. When a transmission line is connected to a load, the source for the load IS the end of the transmission line. Where else would you think the source is? Instead of arguing about it, one can download QUCS for free which will simulate the whole thing and one can see what really happens. Download QUCS for your operating system: http://qucs.sourceforge.net/ Generate a model consisting of a voltage source with a series resistance of a few Ohms to simulate a solid state source or a much higher resistance to simulate a vacuum tube source. Chose a convienient frequency for the simulation. Go to: http://home.sandiego.edu/~ekim/e194r.../matcher2.html to calculate an impedance matching network to match the resistance you've chosen to 50 Ohms at the choosen frequency. Put the matching circuit in the model. Add a 50 Ohm transmission line to the model. Terminate the transmission line with a 50 Ohm resistor. Add a fixed frequency AC simulation at the choosen frequency. Change various parameters to your heart's content to see what happens. Change the matching network such that the output of your transmitter is no longer 50 Ohms and see what happens. When the QUCS output disagrees with your beliefs, you can argue with the program. -- Jim Pennino |
An antenna question--43 ft vertical
|
An antenna question--43 ft vertical
John S wrote:
On 7/8/2015 4:48 PM, wrote: 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? Then there is no such thing as DC because even a battery looses voltage over a period of time. DC voltage sources have noise. Are just being argumentative or are you really that ignorant? What you say is true in a literal sense. However there is certainly such a thing as DC from the POV of what you want to know about a system being calculated or measured with sufficient and relevant accuracy and precision by treating it as DC. In real life this is remarkably common. -- Roger Hayter |
An antenna question--43 ft vertical
Jeff wrote:
On 08/07/2015 19:14, 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. Note the the definition of VSWR uses the magnitude of the reflection coefficient, |r|, which removes the phase and frequency dependant parts. Jeff The magnitude remains frequency dependent. -- Roger Hayter |
An antenna question--43 ft vertical
On 7/8/2015 9:07 PM, John S wrote:
On 7/8/2015 4:48 PM, wrote: 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? Then there is no such thing as DC because even a battery looses voltage over a period of time. DC voltage sources have noise. Are just being argumentative or are you really that ignorant? Even if you have a theoretical voltage source, there are no circuits (other than imaginary) that have been on since before the big bang and will be on for all time in the future. -- Rick |
An antenna question--43 ft vertical
"Jeff" wrote in message ... 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. By 'apparent SWR' he means as indicated SWR on the meter, and yes it can change at various point on the line due to inadequacies in the meter; the 'real' VSWR will of course remain the same at any point on a lossless line. Jeff That is what I mean Jeff. If there is any SWR, by changing the length of the line, the voltage/current changes in such a maner that at certain points you may get a 50 ohm match at that point. |
An antenna question--43 ft vertical
On 7/9/2015 9:14 AM, Ralph Mowery wrote:
"Jeff" wrote in message ... 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. By 'apparent SWR' he means as indicated SWR on the meter, and yes it can change at various point on the line due to inadequacies in the meter; the 'real' VSWR will of course remain the same at any point on a lossless line. Jeff That is what I mean Jeff. If there is any SWR, by changing the length of the line, the voltage/current changes in such a maner that at certain points you may get a 50 ohm match at that point. https://en.wikipedia.org/wiki/Standi...dance_matching "if there is a perfect match between the load impedance Zload and the source impedance Zsource=Z*load, that perfect match will remain if the source and load are connected through a transmission line with an electrical length of one half wavelength (or a multiple of one half wavelengths) using a transmission line of any characteristic impedance Z0." This wiki article has a lot of good info in it. I have seen a lot of stuff posted here that this article directly contradicts.... I wonder who is right? -- Rick |
An antenna question--43 ft vertical
On 7/9/2015 12:31 AM, wrote:
rickman wrote: On 7/8/2015 8:34 PM, wrote: rickman wrote: On 7/8/2015 3:38 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. I was referring to a voltage source. Instead of arguing about it, one can download QUCS for free which will simulate the whole thing and one can see what really happens. Download QUCS for your operating system: http://qucs.sourceforge.net/ Generate a model consisting of a voltage source with a series resistance of a few Ohms to simulate a solid state source or a much higher resistance to simulate a vacuum tube source. Chose a convienient frequency for the source. Go to: http://home.sandiego.edu/~ekim/e194r.../matcher2.html to calculate an impedance matching network to match the resistance you've chosen to 50 Ohms. Put the matching circuit in the model. Add a transmission line to the model. Terminate the transmission line with a 50 Ohms resistor. Add a fixed frequency AC simulation at the desired frequency. Change various parameters to your heart's content to see what happens. Change the matching network such that the output of your transmitter is no longer 50 Ohms and see what happens. When the QUCS output disagrees with your beliefs, you can argue with the program. Uh, I already did a simulation using LTspice. No one even commented on the simulation as I recall. Besides, I don't really need a simulation to measure the power delivered to a load. That is a *very* simple circuit to calculate in a few seconds. But first we have to agree on what we are discussing. Does LTspice do transmission lines? Yes. I'm not going to retrace this entire conversation, but someone said a matching network which presents a complex impedance with a non-zero real part and a zero imaginary part would reflect 100% of the wave which had been reflected from the antenna back to the transmitter output. I was trying to nail down this example so I could do some calculations on it. If you want to do that I am happy to work on the problem. If not, that's fine too. Nope, I know what happens. BTW, my response was not directed at any particular person other than those that do not understand the conditions for maximum power transfer given a fixed source. |
An antenna question--43 ft vertical
wrote in message ... John S wrote: On 7/8/2015 7:27 PM, Wayne wrote: "John S" wrote in message ... 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. That's also my understanding of the definition. In fact since SWR is defined as the maximum to minimum voltage ratio, the "V" in VSWR is redundant. Sort of. There is also ISWR but it is not used frequently. # Not sort of, but is. # There is also PSWR. And both go back to the Vmax/Vmin definition. The PSWR is a tricky one because you can end up with a power ratio instead of a voltage ratio. |
An antenna question--43 ft vertical
On 7/8/2015 8:25 PM, rickman wrote:
On 7/8/2015 8:34 PM, wrote: rickman wrote: On 7/8/2015 3:38 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. I was referring to a voltage source. Instead of arguing about it, one can download QUCS for free which will simulate the whole thing and one can see what really happens. Download QUCS for your operating system: http://qucs.sourceforge.net/ Generate a model consisting of a voltage source with a series resistance of a few Ohms to simulate a solid state source or a much higher resistance to simulate a vacuum tube source. Chose a convienient frequency for the source. Go to: http://home.sandiego.edu/~ekim/e194r.../matcher2.html to calculate an impedance matching network to match the resistance you've chosen to 50 Ohms. Put the matching circuit in the model. Add a transmission line to the model. Terminate the transmission line with a 50 Ohms resistor. Add a fixed frequency AC simulation at the desired frequency. Change various parameters to your heart's content to see what happens. Change the matching network such that the output of your transmitter is no longer 50 Ohms and see what happens. When the QUCS output disagrees with your beliefs, you can argue with the program. Uh, I already did a simulation using LTspice. No one even commented on the simulation as I recall. Besides, I don't really need a simulation to measure the power delivered to a load. That is a *very* simple circuit to calculate in a few seconds. But first we have to agree on what we are discussing. Did you include a transmission line? If not, see TransmissionLineInverter.asc in the \Program Files (x86)\LTC\LTspiceIV\examples\Educational folder to get you started. Do you have EZNEC? I'm not going to retrace this entire conversation, but someone said a matching network which presents a complex impedance with a non-zero real part and a zero imaginary part would reflect 100% of the wave which had been reflected from the antenna back to the transmitter output. I was trying to nail down this example so I could do some calculations on it. If you want to do that I am happy to work on the problem. If not, that's fine too. |
An antenna question--43 ft vertical
"Jeff" wrote in message ... By 'apparent SWR' he means as indicated SWR on the meter, and yes it can change at various point on the line due to inadequacies in the meter; the 'real' VSWR will of course remain the same at any point on a lossless line. Jeff That is what I mean Jeff. If there is any SWR, by changing the length of the line, the voltage/current changes in such a maner that at certain points you may get a 50 ohm match at that point. Absolutely NOT. By changing the length of a transmission you will NEVER create the situation where you get a 50 ohm match from an initial mismatch. This is clearly demonstrable on a Smith chart. Take any starting point other than a pure 50 ohms and add a length of transmission line. What you will find is that as you increase the length of line your point will merely rotate around the chart at a fixed radius (known as a constant VSWR circle), it will never spiral into the centre which is 50 ohms and where it must be for a perfect match. The only time that it will start to spiral inwards is if the line is lossy, but you will need a very long length, and the spiralling inwards is due to the loss in the coax NOT any matching characteristics due to the length of line. If such an effect as you are talking about is observed it is merely due to the finite, and often poor, directivity of the SWR meter giving you a false reading. Also it is worth noting that achieving 50 ohms as a magnitude |Z| of the complex impedance (Sqrt(R^2+X^2)) is not the same as getting a good match with 50 ohms resistive. Even if |Z| = 50 ohms it will have a VSWR greater than 1 if Z0. Again, plot the point on a Smith chart and you will see that it can never be in the centre of the chart. Jeff That is easy to disprove Jeff. If I have a 50 ohm load and use a 1/2 wave of any impedance line other than 50 ohms, the swr will be greater than 1:1, except at 1/2 wave multiplies of the line. At this point there will be a 50 ohm match. The swr of the line will not actually be 1:1 but some greater value. |
An antenna question--43 ft vertical
On 7/9/2015 11:40 AM, Jeff wrote:
you may get a 50 ohm match at that point. https://en.wikipedia.org/wiki/Standi...dance_matching "if there is a perfect match between the load impedance Zload and the source impedance Zsource=Z*load, that perfect match will remain if the source and load are connected through a transmission line with an electrical length of one half wavelength (or a multiple of one half wavelengths) using a transmission line of any characteristic impedance Z0." This wiki article has a lot of good info in it. I have seen a lot of stuff posted here that this article directly contradicts.... I wonder who is right? That is a very specific case where the source is not at the system impedance and happens to be equal to the load impedance, there will also be standing waves on the transmission line and associated losses as the VSWR on the line will be equal to the magnitude of the mismatch between the transmission line impedance and the load impedance. Jeff Yes, this is a specific case just as you indicated in the section you both wrote and snipped... By 'apparent SWR' he means as indicated SWR on the meter, and yes it can change at various point on the line due to inadequacies in the meter; the 'real' VSWR will of course remain the same at any point on a lossless line. Jeff "It can change at various points on the line". That's all I was attempting to indicate. -- Rick |
An antenna question--43 ft vertical
rickman wrote:
On 7/8/2015 9:07 PM, John S wrote: On 7/8/2015 4:48 PM, wrote: 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? Then there is no such thing as DC because even a battery looses voltage over a period of time. DC voltage sources have noise. Are just being argumentative or are you really that ignorant? Even if you have a theoretical voltage source, there are no circuits (other than imaginary) that have been on since before the big bang and will be on for all time in the future. So what? Is there some point to all this other than to be argumentative? How long before someone brings up the fact that a resistor generates AC signals as some kind of straw man objection to DC theory? -- Jim Pennino |
An antenna question--43 ft vertical
Jeff wrote:
Can you measure VSWR on a 1 meter long Lecher line at 1 MHz? VSWR is not meaningful in such a situation, however, you can measure return loss and Reflection Coefficient etc.. Of course that in not to say that VSWR is not used in situations where it is not appropriate in order to indicate how good a match is, when RL or Reflection Coefficient would be more appropriate. Jeff Jeff Are you trying to say that VSWR is not meaningfull at 160M (to put it in an Amateur context)? For those that don't know, a Lecher wire is just a carefully contructed, rigid parallel transmission line upon which one would slide a high impedance sensor to find voltage minimum, maximum, and where they occured. That and a Smith chart were used to solve transmission line and impedance matching problems and were often home built by Amateurs in the early VHF days. Today you would use a VNA (Vector Network Analyzer). -- Jim Pennino |
An antenna question--43 ft vertical
Jeff wrote:
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? No, of course not, the other equations are NOT a definition of VSWR, they are formulas the link other quantities to VSWR. So who exactly declared which set of definitions is the one and true definition of VSWR? Is P=EI or P=E^2R? Taking Reflection Coefficient for example, it requires the phase information to be removed before conversion to VSWR by using only its magnitude. So what? To emphasise this VSRW is phase independent and on a lossless transmission line; its value does not change anywhere along that line; that is equivalent to rotating around a constant VSWR circle on a Smith Chart. Jeff -- Jim Pennino |
An antenna question--43 ft vertical
Jeff wrote:
On 08/07/2015 19:14, 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. Note the the definition of VSWR uses the magnitude of the reflection coefficient, |r|, which removes the phase and frequency dependant parts. Jeff The magnitude DEPENDS on the frequency dependant parts. -- Jim Pennino |
An antenna question--43 ft vertical
Jeff wrote:
Note the the definition of VSWR uses the magnitude of the reflection coefficient, |r|, which removes the phase and frequency dependant parts. Jeff The magnitude remains frequency dependent. It may or may not be depending on the load and how the phase of the reflection coefficient changes with frequency. Ergo it DOES remain frequency dependent in the general case. -- Jim Pennino |
An antenna question--43 ft vertical
Wayne wrote:
wrote in message ... John S wrote: On 7/8/2015 7:27 PM, Wayne wrote: "John S" wrote in message ... 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. That's also my understanding of the definition. In fact since SWR is defined as the maximum to minimum voltage ratio, the "V" in VSWR is redundant. Sort of. There is also ISWR but it is not used frequently. # Not sort of, but is. # There is also PSWR. And both go back to the Vmax/Vmin definition. The PSWR is a tricky one because you can end up with a power ratio instead of a voltage ratio. Actually, no, PSWR has nothing to do with power ratios as in RF power, rather it has to do with power ratios as in values raised to the second power. -- Jim Pennino |
An antenna question--43 ft vertical
rickman wrote:
On 7/8/2015 7:43 PM, wrote: Ralph Mowery wrote: wrote in message ... Ralph Mowery wrote: 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 You might check that again. I don't see Zo being defined as the complex source impedance, but rather as the transmission line characteristic impedance... not the same thing at all. YOu have just proven my point. Read carefully from your refernce to Wikipedia : "The reflection coefficient of a load is determined by its impedance and the impedance toward the source." Notice it says TOWARD and not THE SOURCE. Notice it actually says "the impedance toward the source". From the second referaence notice that it says load impedance and impedance of the transmission line. Nothing mentions the source at all: What the hell do you think the transmission line is in this case if not the source? "The reflection coefficient is usually denoted by the symbol gamma. Note that the magnitude of the reflection coefficient does not depend on the length of the line, only the load impedance and the impedance of the transmission line. Also, note that if ZL=Z0, then the line is "matched". In this case, there is no mismatch loss and all power is transferred to the load." Perhaps you would like the second link better as it has pictures. Of maybe this one that explains it all starting with lumped equivelant circuits. http://www.maximintegrated.com/en/ap...dex.mvp/id/742 Notice that ALL the links talk about the source impedance. How about this one? https://en.wikipedia.org/wiki/Standi...dance_matching I think this has some very interesting analysis, very specifically referring to "purely resistive load impedance". So what? A purely resistive anything is a special case of the general problem. -- Jim Pennino |
An antenna question--43 ft vertical
On 7/9/2015 1:27 PM, wrote:
rickman wrote: On 7/8/2015 9:07 PM, John S wrote: On 7/8/2015 4:48 PM, wrote: 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? Then there is no such thing as DC because even a battery looses voltage over a period of time. DC voltage sources have noise. Are just being argumentative or are you really that ignorant? Even if you have a theoretical voltage source, there are no circuits (other than imaginary) that have been on since before the big bang and will be on for all time in the future. So what? Is there some point to all this other than to be argumentative? How long before someone brings up the fact that a resistor generates AC signals as some kind of straw man objection to DC theory? The point is that separating DC and AC as being ruled by separate "laws" is pointless. Just discuss the topic of interest rather than digressing onto pointless diversions. -- Rick |
| All times are GMT +1. The time now is 08:56 AM. |
Powered by vBulletin® Copyright ©2000 - 2025, Jelsoft Enterprises Ltd.
RadioBanter.com