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An antenna question--43 ft vertical
John S wrote:
On 7/9/2015 12:58 PM, wrote: rickman wrote: 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? It has been my observation that when the subject matter is long established science, such as transmission line theory, wiki is normally correct. Wiki is subject to the same errors you make because the information is usually supplied by people like you. You mean people that know what they are doing as opposed to people like you that just pull crap out of their ass based on old wive's tales? -- Jim Pennino |
An antenna question--43 ft vertical
On 7/10/2015 12:29 AM, wrote:
John S wrote: On 7/9/2015 12:32 AM, wrote: 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. Where can one be purchased? At the ideal battery store. C'mon, jimp, what concession from me will it take to get us back on track so we can discuss this topic in an adult and gentlemanly manner? |
An antenna question--43 ft vertical
wrote:
John S wrote: On 7/9/2015 1:01 PM, wrote: 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 Most people take the source impdedance to be the system impedance, i.e. the impedance for which everything else is designed for. Most *engineers* take the source impedance to be the impedance of the *generator*. Which, as in most cases is purchased and has a fixed impedance of 50 Ohms, and thus defines the system impedance. Please find one specification for a transmitter that says it *has* an output impedance of 50 ohms. You will find plenty that say they are designed to drive a load impedance of 50 ohms, but few that state their output impedance. Of those I have seen that do, none are amateur transmitters and the impedance they mention is much lower than 50 ohms. In fact, perhaps the rest of us should call it the generator rather than the source so that we can communicate with you on your level. Babbling horse****. -- Roger Hayter |
An antenna question--43 ft vertical
John S wrote:
On 7/10/2015 12:29 AM, wrote: John S wrote: On 7/9/2015 12:32 AM, wrote: 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. Where can one be purchased? At the ideal battery store. C'mon, jimp, what concession from me will it take to get us back on track so we can discuss this topic in an adult and gentlemanly manner? When one analyzes circuits, it is done with ideal components. If the real world properties are important, they are in turn modeled with additional ideal components. For example, an ideal voltage source has constant voltage and zero source resistance forever. If the source resistance is important to the circuit, then it is modeled by putting an ideal resistor in series with the voltage source. Your statement: "Then there is no such thing as DC because even a battery looses voltage over a period of time. DC voltage sources have noise." is either just a childish strawman or you have no real clue how circuits, including electromagnetic circuits, are modeled. The only time where the fact that a real battery discharges would be of any significance is if you were analyzing a circuit for it's performance over a voltage range, in which casee one would step the voltage source. -- Jim Pennino |
An antenna question--43 ft vertical
Jeff wrote:
On 09/07/2015 18:35, wrote: 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). Unless you have a very long feeder at 160m you cannot have a complete voltage maxima and minima from the standing wave on the line so VSWR is meaningless. That is not to say that you cannot calculate an 'effective' VSWR from other quantities such as return loss, S11, by measuring the forward and reflected signals as you would with a Network Analyser or SWR bridge. Jeff Nope, VSWR is always meaningful and you have the cart before the horse. VSWR is a consequence of an impedance match and standing waves are a consequence of a VSWR greater than 1:1 on a transmission line. Attach a SWR meter directly to the output of YOUR transmitter and a 1 Ohm resistor directly to the other end of the SWR meter. The meter reading will be the same as the calculated value, there will be no standing waves as there is no transmission line, but the results WILL be meaninful to your transmitter. -- Jim Pennino |
An antenna question--43 ft vertical
Jeff wrote:
So who exactly declared which set of definitions is the one and true definition of VSWR? Is P=EI or P=E^2R? It was defined when the quantity was invented and is obvious from the name, ie the Ratio of the Voltage of the Standing Wave. Vmax/Vmin of the standing wave. There was nothing invented; there was something observed. A name may or may not be meaningfull. While the name does discribe what happens on a transmission line, standing waves are a consequence of a SWR greater than 1:1 on a transmission line. SWR exists no matter what the physical impedances are and do NOT have to be transmission lines. If the impedances are not transmission lines, there are no standing waves as there is no place for them to exist. -- Jim Pennino |
An antenna question--43 ft vertical
Jeff wrote:
Nowhere is it written in stone that the Vmax/Vmin is the one, true, only and holy definition of SWR. Wave transmission by Connor: "The standing wave ratio s is defined as the ratio of |Vmax| to |Vmin| ....The standing wave s is thus directly related to |p|...." |p| being the reflection coefficient. Antenna Engineering Handbook by Jasik: "Standing Wave ratio(SWR) is expressed as decibels or as a voltage ratio (VSWR). This is expressed as follows: SWR(db)= 20 log VSWR = 20 log Vmax/Vmin." Hence it follows that VSWR = Vmax/Vmin. plus many other references to VSWR=Vmax/Vmin and no mention of any other definition. Two standard works, and I am sure that I could dig out many more if I could be bothered. Jeff And I can find just as many that define SWR in terms of reflection coefficient and impedance. All the definitions are equally valid. All the definitions are equally valid. All the definitions are equally valid. All the definitions are equally valid. All the definitions are equally valid. All the definitions are equally valid. All the definitions are equally valid. All the definitions are equally valid. All the definitions are equally valid. All the definitions are equally valid. If you believe that SWR is only valid and relevant to transmission lines, stick a 0.5 Ohm resistor into the output connector of your transmitter. -- Jim Pennino |
An antenna question--43 ft vertical
Jeff wrote:
All the definitions are equally valid. You seem totally incapable of differentiating between a definition of a quantity and a formula that links that definition to another quantity. Further discussion seems pointless. You seem totally incapable of understanding that ALL the equations are equally valid, that NONE of them is the formal definition, and that it is irrelevant WHAT the formal definition is as long as the equations are valid. -- Jim Pennino |
An antenna question--43 ft vertical
Roger Hayter wrote:
wrote: John S wrote: On 7/9/2015 1:01 PM, wrote: 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 Most people take the source impdedance to be the system impedance, i.e. the impedance for which everything else is designed for. Most *engineers* take the source impedance to be the impedance of the *generator*. Which, as in most cases is purchased and has a fixed impedance of 50 Ohms, and thus defines the system impedance. Please find one specification for a transmitter that says it *has* an output impedance of 50 ohms. You will find plenty that say they are designed to drive a load impedance of 50 ohms, but few that state their output impedance. Of those I have seen that do, none are amateur transmitters and the impedance they mention is much lower than 50 ohms. Notice I did not use the word "transmitter" in my post. I was speaking from an engineering point of view, not from an Amateur radio operator point of view. Notice that the post I was responding to used the words "engineer", not "Amateur" and "generator" not "transmitter". -- Jim Pennino |
An antenna question--43 ft vertical
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An antenna question--43 ft vertical
Jeff wrote:
That is correct, but not the situation that we are discussing, we are talking about matching a load to a 50 ohm transmission line. In that case changing the length of line will NEVER result in a match. Using a *different impedance* length of coax as a transmission line transformer is a totally different case, and as you say will result is a standing wave on the line and associated losses. Jeff So you are only interested in special cases? No. I am commenting on the original proposal that changing the length of a line of the same impedance as the system impedance can result in a good match when one did not exist to start with. Jeff OK, but that has been settled long ago. BTW, for a real world transmission line transformer, the transformer section where the impedance mismatch occurs will be of a very short length compared to the total transmission line and thus will have low loss. -- Jim Pennino |
An antenna question--43 ft vertical
Jeff wrote:
How about a section of transmission line with one impedance of some length attached to a section of transmission line with a different impedance of random length? That is not the same situation as changing the length of a transmission line of the same impedance which is what we are discussing. Jeff OK; I thought that was totally settled. -- Jim Pennino |
An antenna question--43 ft vertical
On 7/10/2015 12:29 PM, wrote:
John S wrote: On 7/10/2015 12:29 AM, wrote: John S wrote: On 7/9/2015 12:32 AM, wrote: 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. Where can one be purchased? At the ideal battery store. C'mon, jimp, what concession from me will it take to get us back on track so we can discuss this topic in an adult and gentlemanly manner? When one analyzes circuits, it is done with ideal components. Yes, until you throw in the requirement that one must go get an off-the-shelf ham transmitter for measurements. If the real world properties are important, they are in turn modeled with additional ideal components. Yes. For example, an ideal voltage source has constant voltage and zero source resistance forever. Yes. If the source resistance is important to the circuit, then it is modeled by putting an ideal resistor in series with the voltage source. Yes. Your statement: "Then there is no such thing as DC because even a battery looses voltage over a period of time. DC voltage sources have noise." is either just a childish strawman or you have no real clue how circuits, including electromagnetic circuits, are modeled. This was in response to your asinine statement "At the ideal battery store." And, the whole thing started when I asked you to define the difference between AC and DC. There is no breakpoint, jimp, it is dependent on which tool we use to analyze the problem at hand. The only time where the fact that a real battery discharges would be of any significance is if you were analyzing a circuit for it's performance over a voltage range, in which casee one would step the voltage source. Okay, jimp. You did not answer my direct question, so I must assume you have no further interest in continuing this childish argument. I am tired of your nit-picky responses that we can all see through. They are designed exclusively to satisfy your ego and we don't need that in spite of the fact that you can be a valuable source of information. Yeah, I needled you right back because you refused to continue in a reasonable manner. From now on, I will not respond to you. I'm not saying that I will killfile you. I will just not respond to any of your childish pranks. I really don't have time to argue with your ego. That's too bad. |
An antenna question--43 ft vertical
John S wrote:
On 7/10/2015 12:29 PM, wrote: John S wrote: On 7/10/2015 12:29 AM, wrote: John S wrote: On 7/9/2015 12:32 AM, wrote: 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. Where can one be purchased? At the ideal battery store. C'mon, jimp, what concession from me will it take to get us back on track so we can discuss this topic in an adult and gentlemanly manner? When one analyzes circuits, it is done with ideal components. Yes, until you throw in the requirement that one must go get an off-the-shelf ham transmitter for measurements. Babbling nonsense. Amateur radio equipment is designed by the same engineering tools that design everything else. If the real world properties are important, they are in turn modeled with additional ideal components. Yes. For example, an ideal voltage source has constant voltage and zero source resistance forever. Yes. If the source resistance is important to the circuit, then it is modeled by putting an ideal resistor in series with the voltage source. Yes. Your statement: "Then there is no such thing as DC because even a battery looses voltage over a period of time. DC voltage sources have noise." is either just a childish strawman or you have no real clue how circuits, including electromagnetic circuits, are modeled. This was in response to your asinine statement "At the ideal battery store." And, the whole thing started when I asked you to define the difference between AC and DC. There is no breakpoint, jimp, it is dependent on which tool we use to analyze the problem at hand. Nope; In mathematics it is called a lower limit. My "asinine statement" was appropriate for your asinine question. The tools used at DC may have been derived from that same base principals as used at AC, but they are much simplified and many things that exist at AC simply do not exist at DC, e.g. capacitors, inductors, transmission lines, propagating fields, etc. The only time where the fact that a real battery discharges would be of any significance is if you were analyzing a circuit for it's performance over a voltage range, in which casee one would step the voltage source. Okay, jimp. You did not answer my direct question, so I must assume you have no further interest in continuing this childish argument. I am tired of your nit-picky responses that we can all see through. They are designed exclusively to satisfy your ego and we don't need that in spite of the fact that you can be a valuable source of information. Yeah, I needled you right back because you refused to continue in a reasonable manner. I am responding as an engineer would respond. An engineer doesn't arm wave crap about battery discharge disproving DC theory. -- Jim Pennino |
An antenna question--43 ft vertical
rickman wrote:
On 7/10/2015 1:39 PM, wrote: standing waves are a consequence of a VSWR greater than 1:1 on a transmission line. Did you really write that? The standing waves are a consequence of a standing wave ratio of greater than 1:1? As VSWR is a consequence of an impedance mismatch, yes. You keep ignoring or failing to understand that part. No matter what you call it, a value of greater than 1:1 indicates an IMPEDANCE mismatch. If the equations that describe VSWR were discovered by Dr. Snagpuss and the effect were named after him, then the statement would be that a Snagpuss greater than 1:1 causes standing waves on a transmission line. An IMPEDANCE mismatch on a TRANSMISSION LINE results in standing waves. It is irrelevant what you call the value, only the value itself is of any importance. -- Jim Pennino |
An antenna question--43 ft vertical
In message ,
writes Jeff wrote: On 09/07/2015 18:35, wrote: 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). Unless you have a very long feeder at 160m you cannot have a complete voltage maxima and minima from the standing wave on the line so VSWR is meaningless. That is not to say that you cannot calculate an 'effective' VSWR from other quantities such as return loss, S11, by measuring the forward and reflected signals as you would with a Network Analyser or SWR bridge. Jeff Nope, VSWR is always meaningful and you have the cart before the horse. VSWR is a consequence of an impedance match and standing waves are a consequence of a VSWR greater than 1:1 on a transmission line. Attach a SWR meter directly to the output of YOUR transmitter and a 1 Ohm resistor directly to the other end of the SWR meter. The meter reading will be the same as the calculated value, there will be no standing waves as there is no transmission line, but the results WILL be meaninful to your transmitter. Even when the only transmission line consists the output connector of the SWR meter, and maybe an inch of internal coax, there will still BE a standing wave - but it will only be a tiny portion of longer one. However, surely an SWR meter is really measuring the ratio of the go and return signals - ie the RLR? If so, would it not end all this essentially esoteric argument if we called it an RLR meter? -- Ian |
An antenna question--43 ft vertical
Ian Jackson wrote:
In message , writes Jeff wrote: On 09/07/2015 18:35, wrote: 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). Unless you have a very long feeder at 160m you cannot have a complete voltage maxima and minima from the standing wave on the line so VSWR is meaningless. That is not to say that you cannot calculate an 'effective' VSWR from other quantities such as return loss, S11, by measuring the forward and reflected signals as you would with a Network Analyser or SWR bridge. Jeff Nope, VSWR is always meaningful and you have the cart before the horse. VSWR is a consequence of an impedance match and standing waves are a consequence of a VSWR greater than 1:1 on a transmission line. Attach a SWR meter directly to the output of YOUR transmitter and a 1 Ohm resistor directly to the other end of the SWR meter. The meter reading will be the same as the calculated value, there will be no standing waves as there is no transmission line, but the results WILL be meaninful to your transmitter. Even when the only transmission line consists the output connector of the SWR meter, and maybe an inch of internal coax, there will still BE a standing wave - but it will only be a tiny portion of longer one. There will NOT be standing waves and there will not be a voltage maximum and a voltage minimum unless there is a transmission line. However, that is irrelevant to the point. From the transmitter point of view, there is no difference between a directly connected 1 Ohm resistor and a 1 Ohm resistor at the end of a lossless 1 Ohm transmission line 1,000,000 meters long. However, surely an SWR meter is really measuring the ratio of the go and return signals - ie the RLR? If so, would it not end all this essentially esoteric argument if we called it an RLR meter? The SWR meter will read the IMPEDANCE MISMATCH. Here's a hot flash; people in the engineering world have been using the consept and equations of VSWR for many decades for many things that do NOT include transmission lines with no problems whatsoever. It seems only some Amateur radio operators are hung up on terminology like "standing waves". -- Jim Pennino |
An antenna question--43 ft vertical
On 7/10/2015 2:34 PM, wrote:
rickman wrote: On 7/10/2015 1:39 PM, wrote: standing waves are a consequence of a VSWR greater than 1:1 on a transmission line. Did you really write that? The standing waves are a consequence of a standing wave ratio of greater than 1:1? An IMPEDANCE mismatch on a TRANSMISSION LINE results in standing waves. First you say the standing waves are a result of the SWR being greater than 1:1, now you say it is a result of the impedance mismatch. I'm *so* confused.... :( I'm getting the impression you are being water boarded and will say anything you think will make it end! Give us a location and we will save you! -- Rick |
An antenna question--43 ft vertical
|
An antenna question--43 ft vertical
On 7/10/2015 4:10 PM, Ian Jackson wrote:
In message , writes Ian Jackson wrote: Even when the only transmission line consists the output connector of the SWR meter, and maybe an inch of internal coax, there will still BE a standing wave - but it will only be a tiny portion of longer one. There will NOT be standing waves and there will not be a voltage maximum and a voltage minimum unless there is a transmission line. Are you saying that for a standing wave to qualify as a standing wave, the transmission line needs to be long enough for there to be a voltage maximum a voltage minimum? It seems to be very much like watching a train wreck. We are fascinated by the event even though it is terrible to watch. I'm starting to get a bit tired of it though. I can only watch the train run off the track so many times. -- Rick |
An antenna question--43 ft vertical
rickman wrote:
On 7/10/2015 2:34 PM, wrote: rickman wrote: On 7/10/2015 1:39 PM, wrote: standing waves are a consequence of a VSWR greater than 1:1 on a transmission line. Did you really write that? The standing waves are a consequence of a standing wave ratio of greater than 1:1? An IMPEDANCE mismatch on a TRANSMISSION LINE results in standing waves. First you say the standing waves are a result of the SWR being greater than 1:1, now you say it is a result of the impedance mismatch. I'm *so* confused.... :( Are you serious? SWR is a measure of impedance match. A SWR greater than 1:1 indicates an impedance mismatch. An impedance mismatch results in standing waves on a transmission line. What is so difficult to understand about that or are you fixated on the "SW" standing for "standing wave" in the name of the measurement? I'm getting the impression you are being water boarded and will say anything you think will make it end! Give us a location and we will save you! I'm getting the impression you are playing stupid just for the sake of arguing. -- Jim Pennino |
An antenna question--43 ft vertical
Ian Jackson wrote:
In message , writes Ian Jackson wrote: Even when the only transmission line consists the output connector of the SWR meter, and maybe an inch of internal coax, there will still BE a standing wave - but it will only be a tiny portion of longer one. There will NOT be standing waves and there will not be a voltage maximum and a voltage minimum unless there is a transmission line. Are you saying that for a standing wave to qualify as a standing wave, the transmission line needs to be long enough for there to be a voltage maximum a voltage minimum? I am saying that if the connection between the two things of interest is short enough in terms of wavelengths at the frequency of interest, the connection no longer functions as a transmission line, there are no standing waves, but the measurment we call SWR still exists. -- Jim Pennino |
An antenna question--43 ft vertical
On 7/10/2015 4:28 PM, wrote:
rickman wrote: On 7/10/2015 2:34 PM, wrote: rickman wrote: On 7/10/2015 1:39 PM, wrote: standing waves are a consequence of a VSWR greater than 1:1 on a transmission line. Did you really write that? The standing waves are a consequence of a standing wave ratio of greater than 1:1? An IMPEDANCE mismatch on a TRANSMISSION LINE results in standing waves. First you say the standing waves are a result of the SWR being greater than 1:1, now you say it is a result of the impedance mismatch. I'm *so* confused.... :( Are you serious? SWR is a measure of impedance match. A SWR greater than 1:1 indicates an impedance mismatch. An impedance mismatch results in standing waves on a transmission line. What is so difficult to understand about that or are you fixated on the "SW" standing for "standing wave" in the name of the measurement? I'm getting the impression you are being water boarded and will say anything you think will make it end! Give us a location and we will save you! I'm getting the impression you are playing stupid just for the sake of arguing. Guilty as charged. How about you? -- Rick |
An antenna question--43 ft vertical
rickman wrote:
On 7/10/2015 4:10 PM, Ian Jackson wrote: In message , writes Ian Jackson wrote: Even when the only transmission line consists the output connector of the SWR meter, and maybe an inch of internal coax, there will still BE a standing wave - but it will only be a tiny portion of longer one. There will NOT be standing waves and there will not be a voltage maximum and a voltage minimum unless there is a transmission line. Are you saying that for a standing wave to qualify as a standing wave, the transmission line needs to be long enough for there to be a voltage maximum a voltage minimum? It seems to be very much like watching a train wreck. We are fascinated by the event even though it is terrible to watch. I'm starting to get a bit tired of it though. I can only watch the train run off the track so many times. Yes, it looks like a train wreck to me also. A very big problem is people fixated on the term "standing wave". Another is people who do not understand what a transmission line is. To function as a transmission line, the conductors have to be a significant fraction of a wavelength long. The general rule of thumb is that the connection must be greater than 1/10 of a wavelength at the frequency of interest to be regarded as a transmission line. A 10 mm wire carrying a 1 MHz signal is NOT a transmission line even if the wire is RG-8 coaxial cable. A transmission line carries the electromagnetic energy in the electromagnetic field between the conductors that make up the line, not in the conductors. A very short connection can not generate an internal electomagnetic field. This is true for ALL transmission lines, whether they be parallel lines. coaxial lines, or wave guides. Standing waves only occur on transmission lines. SWR is a measurement of impedance and only depends on the impedances of the connection, be it a wire or a transmission line. -- Jim Pennino |
An antenna question--43 ft vertical
rickman wrote:
On 7/10/2015 4:28 PM, wrote: rickman wrote: On 7/10/2015 2:34 PM, wrote: rickman wrote: On 7/10/2015 1:39 PM, wrote: standing waves are a consequence of a VSWR greater than 1:1 on a transmission line. Did you really write that? The standing waves are a consequence of a standing wave ratio of greater than 1:1? An IMPEDANCE mismatch on a TRANSMISSION LINE results in standing waves. First you say the standing waves are a result of the SWR being greater than 1:1, now you say it is a result of the impedance mismatch. I'm *so* confused.... :( Are you serious? SWR is a measure of impedance match. A SWR greater than 1:1 indicates an impedance mismatch. An impedance mismatch results in standing waves on a transmission line. What is so difficult to understand about that or are you fixated on the "SW" standing for "standing wave" in the name of the measurement? I'm getting the impression you are being water boarded and will say anything you think will make it end! Give us a location and we will save you! I'm getting the impression you are playing stupid just for the sake of arguing. Guilty as charged. How about you? At least I know what a transmission line actually is while you obviously do not. -- Jim Pennino |
An antenna question--43 ft vertical
In message ,
writes Ian Jackson wrote: In message , writes Ian Jackson wrote: Even when the only transmission line consists the output connector of the SWR meter, and maybe an inch of internal coax, there will still BE a standing wave - but it will only be a tiny portion of longer one. There will NOT be standing waves and there will not be a voltage maximum and a voltage minimum unless there is a transmission line. Are you saying that for a standing wave to qualify as a standing wave, the transmission line needs to be long enough for there to be a voltage maximum a voltage minimum? I am saying that if the connection between the two things of interest is short enough in terms of wavelengths at the frequency of interest, the connection no longer functions as a transmission line, there are no standing waves, but the measurment we call SWR still exists. Pray tell me exactly (in wavelengths) when something which is too short to be a transmission line suddenly changes into something which IS long enough to be a transmission line. -- Ian |
An antenna question--43 ft vertical
Ian Jackson wrote:
In message , writes Ian Jackson wrote: In message , writes Ian Jackson wrote: Even when the only transmission line consists the output connector of the SWR meter, and maybe an inch of internal coax, there will still BE a standing wave - but it will only be a tiny portion of longer one. There will NOT be standing waves and there will not be a voltage maximum and a voltage minimum unless there is a transmission line. Are you saying that for a standing wave to qualify as a standing wave, the transmission line needs to be long enough for there to be a voltage maximum a voltage minimum? I am saying that if the connection between the two things of interest is short enough in terms of wavelengths at the frequency of interest, the connection no longer functions as a transmission line, there are no standing waves, but the measurment we call SWR still exists. Pray tell me exactly (in wavelengths) when something which is too short to be a transmission line suddenly changes into something which IS long enough to be a transmission line. Sure. A transmission line is distinguished from a wire by the fact that a transmission line carries the energy in the form of an electromagnetic field contained by the structure of the transmission line while a wire carries the energy in the form of conduction in the wire. This is true for all transmission lines, be they parallel, coaxial, wave guide, microstrip, stripline, or any other type of transmission line. A conducting structure becomes a transmission line when it's length in wavelengths becomes long enough to allow the establishment of an electromagnetic field within it's structure. The general rule of thumb is that the dividing point is about 1/10 of a wavelength. For the pendatic, this does NOT mean that at exactly 1/10 of a wave length things suddenly change, it means that in general transmission line effects become negligable below 1/10 of a wave length. A piece of coax will not function as a transmission line at 1/100 of a wavelength even though it is constructed to be a transmission line because it is too small to establish an electromagnetic field between the center conductor and the shield. Note: This is a slightly simplified explaination, for details and mathematical derivations: https://en.wikipedia.org/wiki/Transmission_line http://www.antenna-theory.com/tutori...ine.php#txline http://www.ece.uci.edu/docs/hspice/h...001_2-269.html http://www.allaboutcircuits.com/text...mission-lines/ Standing waves only occur on a transmission line and are due to reflections on the line. If the line length is too short to act as a transmission line, there can be no reflections and no standing waves. -- Jim Pennino |
An antenna question--43 ft vertical
On 7/10/2015 4:10 PM, Ian Jackson wrote:
In message , writes Ian Jackson wrote: Even when the only transmission line consists the output connector of the SWR meter, and maybe an inch of internal coax, there will still BE a standing wave - but it will only be a tiny portion of longer one. There will NOT be standing waves and there will not be a voltage maximum and a voltage minimum unless there is a transmission line. Are you saying that for a standing wave to qualify as a standing wave, the transmission line needs to be long enough for there to be a voltage maximum a voltage minimum? Ian, It's been very interesting to follow the stupid posting of some who claim to know the laws of physics. But then nothing I've have seen in this thread has surprised me in the least. -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
An antenna question--43 ft vertical
In message ,
writes Ian Jackson wrote: In message , writes Ian Jackson wrote: In message , writes Ian Jackson wrote: Even when the only transmission line consists the output connector of the SWR meter, and maybe an inch of internal coax, there will still BE a standing wave - but it will only be a tiny portion of longer one. There will NOT be standing waves and there will not be a voltage maximum and a voltage minimum unless there is a transmission line. Are you saying that for a standing wave to qualify as a standing wave, the transmission line needs to be long enough for there to be a voltage maximum a voltage minimum? I am saying that if the connection between the two things of interest is short enough in terms of wavelengths at the frequency of interest, the connection no longer functions as a transmission line, there are no standing waves, but the measurment we call SWR still exists. Pray tell me exactly (in wavelengths) when something which is too short to be a transmission line suddenly changes into something which IS long enough to be a transmission line. Sure. A transmission line is distinguished from a wire by the fact that a transmission line carries the energy in the form of an electromagnetic field contained by the structure of the transmission line while a wire carries the energy in the form of conduction in the wire. This is true for all transmission lines, be they parallel, coaxial, wave guide, microstrip, stripline, or any other type of transmission line. A conducting structure becomes a transmission line when it's length in wavelengths becomes long enough to allow the establishment of an electromagnetic field within it's structure. The general rule of thumb is that the dividing point is about 1/10 of a wavelength. For the pendatic, this does NOT mean that at exactly 1/10 of a wave length things suddenly change, it means that in general transmission line effects become negligable below 1/10 of a wave length. A piece of coax will not function as a transmission line at 1/100 of a wavelength even though it is constructed to be a transmission line because it is too small to establish an electromagnetic field between the center conductor and the shield. Note: This is a slightly simplified explaination, for details and mathematical derivations: https://en.wikipedia.org/wiki/Transmission_line http://www.antenna-theory.com/tutori...ine.php#txline http://www.ece.uci.edu/docs/hspice/h...001_2-269.html http://www.allaboutcircuits.com/text...nt/chpt-14/lon g-and-short-transmission-lines/ Standing waves only occur on a transmission line and are due to reflections on the line. If the line length is too short to act as a transmission line, there can be no reflections and no standing waves. I haven't checked those references yet, but regardless of what they may say, if that 10' of coax between my TX and my 160m ATU is NOT a transmission line - just what IS it? Do different laws of physics apply? -- Ian |
An antenna question--43 ft vertical
In message , Jeff writes
A load in isolation without any transmission line connected cannot have a standing wave, but it is still common to quote the mismatch as a VSWR which is plain wrong, but still very common. But as I've said (nitpickingly), any length of connection (no matter how short) where the load is not a perfect match for its characteristic impedance, will have a very tiny portion of a standing wave on it. And as I've also said, the normal SWR meter DOESN'T measure (respond) to SWR. It is a reflectometer, and it responds independently to the forward-going signal and the reverse-going signal. It's really telling you what the return loss ratio (RLR) is - but it's still perfectly legitimate for it to be scaled in terms of SWR. It's a darned sight easier way of finding out what the equivalent SWR would be than to try and measure the Vmax and Vmin 'for real' along a long line. -- Ian |
An antenna question--43 ft vertical
On 7/11/2015 5:38 AM, Ian Jackson wrote:
In message , Jeff writes A load in isolation without any transmission line connected cannot have a standing wave, but it is still common to quote the mismatch as a VSWR which is plain wrong, but still very common. But as I've said (nitpickingly), any length of connection (no matter how short) where the load is not a perfect match for its characteristic impedance, will have a very tiny portion of a standing wave on it. And as I've also said, the normal SWR meter DOESN'T measure (respond) to SWR. It is a reflectometer, and it responds independently to the forward-going signal and the reverse-going signal. It's really telling you what the return loss ratio (RLR) is - but it's still perfectly legitimate for it to be scaled in terms of SWR. It's a darned sight easier way of finding out what the equivalent SWR would be than to try and measure the Vmax and Vmin 'for real' along a long line. Why don't we use the RLR in all these measurements instead of SWR? Isn't that what we are really after? -- Rick |
An antenna question--43 ft vertical
On 7/11/2015 10:49 AM, Jeff wrote:
Why don't we use the RLR in all these measurements instead of SWR? Isn't that what we are really after? A very good question. One possible answer is that RL is normally quoted in dB, and VSWR linear scales are perhaps easier to envisage. eg 3:1 ~6dB RL 2:1 ~9.5dB RL 1.5:1 ~14dB RL 1.1:1 ~26dB RL Personally I find log scales more intuitive for most things as they more closely relates to factors of significance, no? But I see right away that RL scales the non-intuitive way, a larger number is a less significant value. While SWR scales the right way with 1 being no effect. SWR can also be given in dB which would make the numbers very intuitive. -- Rick |
An antenna question--43 ft vertical
In message , Jeff writes
Why don't we use the RLR in all these measurements instead of SWR? Tradition! Isn't that what we are really after? As long as you know what you're after, and get close to it, it doesn't really matter. A very good question. One possible answer is that RL is normally quoted in dB, and VSWR linear scales are perhaps easier to envisage. eg 3:1 ~6dB RL 2:1 ~9.5dB RL 1.5:1 ~14dB RL 1.1:1 ~26dB RL Isn't there somehow something sort-of unnatural about trying to aim for an infinite value? -- Ian |
An antenna question--43 ft vertical
Ian Jackson wrote:
In message , writes Ian Jackson wrote: In message , writes Ian Jackson wrote: In message , writes Ian Jackson wrote: Even when the only transmission line consists the output connector of the SWR meter, and maybe an inch of internal coax, there will still BE a standing wave - but it will only be a tiny portion of longer one. There will NOT be standing waves and there will not be a voltage maximum and a voltage minimum unless there is a transmission line. Are you saying that for a standing wave to qualify as a standing wave, the transmission line needs to be long enough for there to be a voltage maximum a voltage minimum? I am saying that if the connection between the two things of interest is short enough in terms of wavelengths at the frequency of interest, the connection no longer functions as a transmission line, there are no standing waves, but the measurment we call SWR still exists. Pray tell me exactly (in wavelengths) when something which is too short to be a transmission line suddenly changes into something which IS long enough to be a transmission line. Sure. A transmission line is distinguished from a wire by the fact that a transmission line carries the energy in the form of an electromagnetic field contained by the structure of the transmission line while a wire carries the energy in the form of conduction in the wire. This is true for all transmission lines, be they parallel, coaxial, wave guide, microstrip, stripline, or any other type of transmission line. A conducting structure becomes a transmission line when it's length in wavelengths becomes long enough to allow the establishment of an electromagnetic field within it's structure. The general rule of thumb is that the dividing point is about 1/10 of a wavelength. For the pendatic, this does NOT mean that at exactly 1/10 of a wave length things suddenly change, it means that in general transmission line effects become negligable below 1/10 of a wave length. A piece of coax will not function as a transmission line at 1/100 of a wavelength even though it is constructed to be a transmission line because it is too small to establish an electromagnetic field between the center conductor and the shield. Note: This is a slightly simplified explaination, for details and mathematical derivations: https://en.wikipedia.org/wiki/Transmission_line http://www.antenna-theory.com/tutori...ine.php#txline http://www.ece.uci.edu/docs/hspice/h...001_2-269.html http://www.allaboutcircuits.com/text...nt/chpt-14/lon g-and-short-transmission-lines/ Standing waves only occur on a transmission line and are due to reflections on the line. If the line length is too short to act as a transmission line, there can be no reflections and no standing waves. I haven't checked those references yet, but regardless of what they may say, if that 10' of coax between my TX and my 160m ATU is NOT a transmission line - just what IS it? Do different laws of physics apply? No, it means that you can view the coax as just a wire and that the transmission line effects are negligable. About the only practical consequence of this that I can think of off the top of my head at the moment, is that a very short, in wavelengths, piece of coax does not have a characteristic impedance so it would not matter what kind of coax you use. To put it another way, if your system is the common 50 Ohms and you had the very best of lab grade test equipment, for very short lengths you would see no difference between using 50 Ohm coax and 100 Ohm coax. As you increase the length, you begin to see differences, and at some length around 1/10 of a wavelength the differences become big enough to be significant. -- Jim Pennino |
An antenna question--43 ft vertical
Jeff wrote:
There will NOT be standing waves and there will not be a voltage maximum and a voltage minimum unless there is a transmission line. Are you saying that for a standing wave to qualify as a standing wave, the transmission line needs to be long enough for there to be a voltage maximum a voltage minimum? No, I think the point is that VSWR is the wrong quantity to be using under those circumstances. It is possible to calculate what the swr *would have been* IF the line had been long enough to observe a max and min, but by its very name it is clear that it is not possible to measure it directly and see a ratio of the standing wave due to the shortness of the line. No, the point is that VSWR, according to the laws of physics, can be shown to be a voltage ratio under the conditions where such voltages exist, AND and impedance ratio that has no dependance on line length. In those circumstances a better solution would be to use return loss, reflection coefficient or S11 etc. The fact that lots of people use VSWR as a measure of a mismatch does not make it correct when it is not possible to measure the VSWR directly by observing the ratio of the standing wave. It can be shown by the laws of physics the return loss, reflection coefficient, or S11 etc. can be converted to VSWR. Which convention you use for the measurement is relevant only to what equipment you have on hand to do the measurement. If you tell the grocer you want a pound of banannas and he gives you 2.2 kilograms of banannas because his scale is calibrated in kilograms, are you going to get into a ****ing contest with him? A load in isolation without any transmission line connected cannot have a standing wave, but it is still common to quote the mismatch as a VSWR which is plain wrong, but still very common. A load in isolation has no source and is thus irrelevant to anything in this discussion. -- Jim Pennino |
An antenna question--43 ft vertical
Ian Jackson wrote:
In message , Jeff writes A load in isolation without any transmission line connected cannot have a standing wave, but it is still common to quote the mismatch as a VSWR which is plain wrong, but still very common. But as I've said (nitpickingly), any length of connection (no matter how short) where the load is not a perfect match for its characteristic impedance, will have a very tiny portion of a standing wave on it. I guess you could look at it that way, but the point is that such effects are so small they are not measurable and irrelevant. And as I've also said, the normal SWR meter DOESN'T measure (respond) to SWR. It is a reflectometer, and it responds independently to the forward-going signal and the reverse-going signal. It's really telling you what the return loss ratio (RLR) is - but it's still perfectly legitimate for it to be scaled in terms of SWR. It's a darned sight easier way of finding out what the equivalent SWR would be than to try and measure the Vmax and Vmin 'for real' along a long line. Most scales do not measure weight, they measure the length of spring extension, but they are calibrated to show pounds or kilograms. Does that mean the measurement is not "real"? The laws of physics allow one to both convert the forward and reverse power measurements to VSWR and to convert spring deflection to weight. As you say, it does not matter what a measurement device is actually measuring, all that matters is that it is correctely calibrated to display the information in the form you desire. -- Jim Pennino |
An antenna question--43 ft vertical
rickman wrote:
On 7/11/2015 5:38 AM, Ian Jackson wrote: In message , Jeff writes A load in isolation without any transmission line connected cannot have a standing wave, but it is still common to quote the mismatch as a VSWR which is plain wrong, but still very common. But as I've said (nitpickingly), any length of connection (no matter how short) where the load is not a perfect match for its characteristic impedance, will have a very tiny portion of a standing wave on it. And as I've also said, the normal SWR meter DOESN'T measure (respond) to SWR. It is a reflectometer, and it responds independently to the forward-going signal and the reverse-going signal. It's really telling you what the return loss ratio (RLR) is - but it's still perfectly legitimate for it to be scaled in terms of SWR. It's a darned sight easier way of finding out what the equivalent SWR would be than to try and measure the Vmax and Vmin 'for real' along a long line. Why don't we use the RLR in all these measurements instead of SWR? Isn't that what we are really after? What we are really after is a convenient way to determine the quality of an impedance match. VSWR is about as convenient as there is. -- Jim Pennino |
An antenna question--43 ft vertical
Jeff wrote:
Why don't we use the RLR in all these measurements instead of SWR? Isn't that what we are really after? A very good question. One possible answer is that RL is normally quoted in dB, and VSWR linear scales are perhaps easier to envisage. eg 3:1 ~6dB RL 2:1 ~9.5dB RL 1.5:1 ~14dB RL 1.1:1 ~26dB RL Jeff A lot easier for most people, which is why it is so commonly used. -- Jim Pennino |
An antenna question--43 ft vertical
rickman wrote:
On 7/11/2015 10:49 AM, Jeff wrote: Why don't we use the RLR in all these measurements instead of SWR? Isn't that what we are really after? A very good question. One possible answer is that RL is normally quoted in dB, and VSWR linear scales are perhaps easier to envisage. eg 3:1 ~6dB RL 2:1 ~9.5dB RL 1.5:1 ~14dB RL 1.1:1 ~26dB RL Personally I find log scales more intuitive for most things as they more closely relates to factors of significance, no? But I see right away that RL scales the non-intuitive way, a larger number is a less significant value. While SWR scales the right way with 1 being no effect. SWR can also be given in dB which would make the numbers very intuitive. Perhaps the world is ready for the Rickman, where Rickman = 10 * log (VSWR). 0 Rickman = 1:1 VSWR. 1.76 Rickman = 1.5:1 VSWR. 3.01 Rickman = 2:1 VSWR. At the very least, it would eliminate any arm waving about standing waves. -- Jim Pennino |
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