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Parallel coax
On 9/29/2015 3:47 AM, Ian Jackson wrote:
In message , Jerry Stuckle writes On 9/28/2015 7:12 PM, John S wrote: On 9/28/2015 1:51 PM, Jerry Stuckle wrote: On 9/28/2015 1:42 PM, rickman wrote: You said return loss increases with lower SWR. It does not. It does. Sorry, a lower SWR does not increase the amount of loss. Of course it doesn't. No one said it did. It does the opposite, ie a lower SWR gives less loss on the feeder. Please cite a reliable reference that says it does. Even the table Rick cited shows a negative value for return SWR. What is this 'Return SWR'? I'm not familiar with it. Sorry, writing too quickly. I meant return loss. Do you mean Return Loss Ratio (RLR)? This is a simple, easily measurable, and meaningful statement of how strong the returning reflected signal is compared with the outgoing forward signal. The reflected signal is a weaker version of the forward signal. It's expressed as a loss, an attenuation, or relatively how much down the level of the reflection is. You can express this as a numerical ratio - the reflection coefficient (rho) - or (as often more convenient) rho in dB. As others have suggested, what is apparently a negative sign in the chart is presumably more artistic licence than scientific accuracy. If you lose $10, you don't say that you lost 'minus $10'. Similarly, when you lose 10dB of signal, you don't say you lost 'minus 10dB'. Which is greater - 10db or -30db? -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
Parallel coax
On 9/29/2015 12:55 AM, rickman wrote:
On 9/28/2015 8:56 PM, Jerry Stuckle wrote: On 9/28/2015 8:09 PM, rickman wrote: On 9/28/2015 7:55 PM, Jerry Stuckle wrote: On 9/28/2015 5:18 PM, rickman wrote: On 9/28/2015 4:34 PM, Jerry Stuckle wrote: On 9/28/2015 4:14 PM, rickman wrote: On 9/28/2015 4:07 PM, Jerry Stuckle wrote: On 9/28/2015 3:22 PM, Wayne wrote: "Jerry Stuckle" wrote in message ... On 9/28/2015 2:27 PM, Wayne wrote: "Jerry Stuckle" wrote in message ... On 9/28/2015 12:47 PM, rickman wrote: On 9/28/2015 10:38 AM, Jerry Stuckle wrote: On 9/28/2015 12:03 AM, rickman wrote: On 9/27/2015 10:39 PM, Jerry Stuckle wrote: On 9/27/2015 9:46 PM, Wayne wrote: From LUNA web site regarding optical measurements which should be no different from RF... It "shouldn't be" - but optical measurements are handled differently than electrical measurements. Fiber Optics have their own way of measuring loss, reflection and refraction (which doesn't exist in feedlines). That's like applying electrician's color codes to electronics. They both have color codes - but don't hook the electrician's black wire to ground - or the transformer's green wires to safety ground. I thought you would claim optical was different. That's why I included the VSWR vs return loss table link. You didn't comment on that. :) # I didn't because I thought it was obvious. But I guess not to you. # Return loss is calculated with logs. Logs of values 1 are negative. # And -10db is smaller than -5 db. # As the SWR approaches 1:1, the reflected power approaches 0, and the # returned loss approaches NEGATIVE infinity. Note that I said NEGATIVE # infinity. At the same point, the returned power measured in watts is 0. Return loss is a positive number for passive networks. The equation has (P out/P reflected). P out will never be less that P reflected, and thus return loss will never be negative. (for passive networks) As the SWR approaches 1:1, the return loss increases in a positive direction, finally reaching infinity. # No, return loss is calculated as P reflected / P out. P out is the # constant with varying load; P reflected is the variable. The ratio is # always less than one, hence the calculation is always negative DB. # Please point to a reliable source which agrees with you. I have never heard return loss expressed as a negative for passive RF networks. In fields other than RF I suppose anything is possible. Here are some references, searching only for RF definitions: http://www.ab4oj.com/atu/vswr.html http://www.mogami.com/e/cad/vswr.html http://www.microwaves101.com/encyclo...swr-calculator http://www.amphenolrf.com/vswr-conversion-chart/ From wikipedia: https://en.wikipedia.org/wiki/Return_loss Return loss is the negative of the magnitude of the reflection coefficient in dB. Since power is proportional to the square of the voltage, return loss is given by, (couldn't cut/paste the equation) Thus, a large positive return loss indicates the reflected power is small relative to the incident power, which indicates good impedance match from source to load. http://www.spectrum-soft.com/news/fall2009/vswr.shtm The return loss measurement describes the ratio of the power in the reflected wave to the power in the incident wave in units of decibels. The standard output for the return loss is a positive value, so a large return loss value actually means that the power in the reflected wave is small compared to the power in the incident wave and indicates a better impedance match. The return loss can be calculated from the reflection coefficient with the equation: I said RELIABLE SOURCE. Wikipedia and someone's blog are not what I would call reliable. How about IEEE, for instance? I provided the IEEE paper cited by wikipedia. Anyone care to pay for it? IEEE is seldom free. Who else will you accept? I'm not interested. I know what it says. Guess I should have kept up my IEEE membership, but it just wasn't worth it. So share with the rest of us. What does it say? Exactly what your table showed. But you mentioned the resource, not me. You pay for it or you've just once again you're full of it. You said you *know* what the IEEE article says. Why not share with us? You want it - you pay for it. Or once again you prove you're full of it. No, I have not read the article. But I understand the physics and math behind it - unlike you. Someone who thinks magnitude without vector (direction) is valid! ROFLMAO! Ok, so you mispoke when you said, "I know what it says." No, I didn't. I didn't say I read it. I said I know what it says. And I do from other IEEE peer-reviewed articles. I don't need to read it to find out it agrees with other documentation. And if it didn't, it wouldn't have gotten published. You have said repeatedly that the return loss should be calculated by using the power in as the reference and the reflected power as the thing being measured which results in a negative log. I am pretty sure the paper says this is not the correct way to calculate it and many people are making a mistake doing it this way. So you've read other IEEE documentation which supports what you say? I'll see if I can get my hands on the paper. I'm not going to pay for it. If I thought it would get you to admit you were mistaken, I'd pay the $100. But I'm sure you will find a way to berate the authors or twist their logic and I'm not will to pay $100 for that. So stand by. Someone may be getting it for me. I really don't give a damn. You would argue the sun rises in the west. -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
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On 9/29/2015 4:36 AM, Jeff wrote:
On 29/09/2015 00:57, Jerry Stuckle wrote: On 9/28/2015 6:18 PM, Ian Jackson wrote: In message , Roger Hayter writes It would probably make more sense to call return loss "return gain", but, since it is always less than one, that would merely cause a different set of ambiguities. The reflected signal is an attenuated version of the forward (source) signal. It is a 'loss' - in exactly the same way as the signal at the output end of an attenuator is an attenuated version of the source signal at the input end. The RLR (in dB) is the ratio of the what you put in to what you get out. It cannot be less than 1, so the RLR is in positive dB. There is absolutely no ambiguity. No one in RF engineering quotes or uses negative values for RLR (or for attenuators). The greater the RLR, the less signal is reflected. The ratio of output to input can never be greater than one - so the log of that can never be positive. I.E. 100W in and 50W out is -3db, not +3db. Yes, that is a GAIN of -3db. When considering RL you are looking for a reduction (LOSS) in the reflected power so the sign changes. It is just common sense. Jeff OK, so I have a system with gains and losses of +3, +1, +5, +2. What it the total gain of the system? In college we learned to consider everything a "black box". You don't know whether it is active or passive, an amplifier or an attenuator. All you know are the characteristics. If the output has +db over the input, it is an amplifier. -db indicates a loss. The same is true when modeling antennas. The antenna connection is a black box. Returned power is always less than power going into it. -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
Parallel coax
On 9/29/2015 3:58 AM, Ian Jackson wrote:
In message , Jerry Stuckle writes On 9/28/2015 6:21 PM, Ian Jackson wrote: In message , Jerry Stuckle writes On 9/28/2015 3:10 PM, rickman wrote: On 9/28/2015 2:55 PM, Jerry Stuckle wrote: On 9/28/2015 2:19 PM, rickman wrote: On 9/28/2015 2:01 PM, Wayne wrote: "rickman" wrote in message ... On 9/28/2015 11:59 AM, Wayne wrote: "Ian Jackson" wrote in message ... In message , rickman writes Definition of Return Loss In technical terms, RL is the ratio of the light reflected back from a device under test, Pout, to the light launched into that device, Pin, usually expressed as a negative number in dB. RL = 10 log10(Pout/Pin) Here is a link for a table of return loss and VSWR.... http://www.jampro.com/uploads/tech_d.../VSWRChart.pdf It shows a higher return loss (assuming you mean magnitude since the values are all negative) for lower VSWR. I'm surprised to see negative quantities. For 50 years, I've always understood the Return Loss Ratio (RLR) to be exactly what it says on the tin, ie the ratio (in dB) of the LOSS (the attenuation) of the reflected signal wrt the incident signal. This is a +ve quantity. Things are already sufficiently confusing without having to start thinking in unnecessary -ve figures! I think the table headings are using a dash, not a negative sign. Return loss- dB # Look at the equation and you will understand. When the ratio is less # than one, the log is negative. But the ratio is never less than one for passive devices. If all the power forward is reflected, then the power ratio is 1 to 1. That's 0 dB return loss from the equation. Return loss is a positive number. I'm not so sure. It depends on how you define it. What if half the power is reflected? The equation above and *many* other sources say that is 10 log(0.5/1) = -3 dB. A few sources take exception to this and say it is 10 log(1/0.5) = 3 dB. At this point I dunno. Rickman, you are correct. The return loss is calculated as return value (the variable) divided by the output value (the constant). I haven't seen any reliable sources which say otherwise. I haven't seen any reliable sources that say either. Have you? Not for 40 years or so. But then I haven't looked for one since college. I've just dealt with RF engineers, who have used the same terminology. They use negative numbers for loss to figure the gain of the entire system. It doesn't matter which direction the loss is in; loss is a negative number (and gain is a positive number). A loss is only negative if it is being thought of as gain. If it's a loss, its value is positive. Physicists and engineers do not mix gain and loss. Gain is always shown as a positive number and loss as a negative number. Physicists and engineers don't get themselves into situations where 'gain' and 'loss' are used ambiguously. Unfortunately, the same cannot be said of certain radio amateurs. Yes, there are some here who are trying to use the terms ambiguously - saying +10db can be both gain and loss. For instance - a system shows a gain and loss of +3, +5, +2, +1. What is the total gain or loss of the system? A physicists or engineer would never ask such a meaningless question. No, because the physicists and engineers use +db for gain and -db for loss. And they understand -100db is less than -10db. However, they are also math literate, unlike many non-scientists nowadays. -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
Parallel coax
On 9/29/2015 4:40 AM, Jeff wrote:
Physicists and engineers do not mix gain and loss. Gain is always shown as a positive number and loss as a negative number. For instance - a system shows a gain and loss of +3, +5, +2, +1. What is the total gain or loss of the system? Of course they do, particularly when dealing with a quantity that is defined as a LOSS. I have never heard any engineer when asked the question 'what is that attenuator' reply minus 3 dB. It is always 3dB. It is always called a 3dB attenuator, not a minus 3dB attenuator. Jeff Yes, and the power out is never +3db relative to the input. It is always -3db. -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
Parallel coax
In message , Jerry Stuckle
writes On 9/29/2015 3:47 AM, Ian Jackson wrote: In message , Jerry Stuckle writes On 9/28/2015 7:12 PM, John S wrote: On 9/28/2015 1:51 PM, Jerry Stuckle wrote: On 9/28/2015 1:42 PM, rickman wrote: You said return loss increases with lower SWR. It does not. It does. Sorry, a lower SWR does not increase the amount of loss. Of course it doesn't. No one said it did. It does the opposite, ie a lower SWR gives less loss on the feeder. Please cite a reliable reference that says it does. Even the table Rick cited shows a negative value for return SWR. What is this 'Return SWR'? I'm not familiar with it. Sorry, writing too quickly. I meant return loss. Do you mean Return Loss Ratio (RLR)? This is a simple, easily measurable, and meaningful statement of how strong the returning reflected signal is compared with the outgoing forward signal. The reflected signal is a weaker version of the forward signal. It's expressed as a loss, an attenuation, or relatively how much down the level of the reflection is. You can express this as a numerical ratio - the reflection coefficient (rho) - or (as often more convenient) rho in dB. As others have suggested, what is apparently a negative sign in the chart is presumably more artistic licence than scientific accuracy. If you lose $10, you don't say that you lost 'minus $10'. Similarly, when you lose 10dB of signal, you don't say you lost 'minus 10dB'. Which is greater - 10db or -30db? In voltage ratios, they are 1/3 and 1/30 respectively, and in power ratios, 1/10 and 1/000 respectively. But if you lost 30dB down your coax, you'd be losing 20dB MORE than if you were only losing 10dB. But surely even you wouldn't say "My coax has a loss of minus 30dB"? [Or would you?!] -- Ian |
Parallel coax
In message , Jerry Stuckle
writes On 9/29/2015 4:40 AM, Jeff wrote: Physicists and engineers do not mix gain and loss. Gain is always shown as a positive number and loss as a negative number. For instance - a system shows a gain and loss of +3, +5, +2, +1. What is the total gain or loss of the system? Of course they do, particularly when dealing with a quantity that is defined as a LOSS. I have never heard any engineer when asked the question 'what is that attenuator' reply minus 3 dB. It is always 3dB. It is always called a 3dB attenuator, not a minus 3dB attenuator. Jeff Yes, and the power out is never +3db relative to the input. It is always -3db. And that's because it has suffered a positive LOSS of 3dB. As I have already asked you, if you lose $10, do you say "I've lost MINUS $10? -- Ian |
Parallel coax
On 9/29/2015 9:22 AM, Jerry Stuckle wrote:
On 9/29/2015 12:55 AM, rickman wrote: On 9/28/2015 8:56 PM, Jerry Stuckle wrote: On 9/28/2015 8:09 PM, rickman wrote: On 9/28/2015 7:55 PM, Jerry Stuckle wrote: On 9/28/2015 5:18 PM, rickman wrote: On 9/28/2015 4:34 PM, Jerry Stuckle wrote: I'm not interested. I know what it says. Guess I should have kept up my IEEE membership, but it just wasn't worth it. So share with the rest of us. What does it say? Exactly what your table showed. But you mentioned the resource, not me. You pay for it or you've just once again you're full of it. You said you *know* what the IEEE article says. Why not share with us? You want it - you pay for it. Or once again you prove you're full of it. No, I have not read the article. But I understand the physics and math behind it - unlike you. Someone who thinks magnitude without vector (direction) is valid! ROFLMAO! Ok, so you mispoke when you said, "I know what it says." No, I didn't. I didn't say I read it. I said I know what it says. And I do from other IEEE peer-reviewed articles. I don't need to read it to find out it agrees with other documentation. And if it didn't, it wouldn't have gotten published. You have said repeatedly that the return loss should be calculated by using the power in as the reference and the reflected power as the thing being measured which results in a negative log. I am pretty sure the paper says this is not the correct way to calculate it and many people are making a mistake doing it this way. So you've read other IEEE documentation which supports what you say? I'll see if I can get my hands on the paper. I'm not going to pay for it. If I thought it would get you to admit you were mistaken, I'd pay the $100. But I'm sure you will find a way to berate the authors or twist their logic and I'm not will to pay $100 for that. So stand by. Someone may be getting it for me. I really don't give a damn. You would argue the sun rises in the west. I have a copy of the paper. Trevor Bird Editor-in-ehief,Engineering IEEE Transactions on Antennas and Propagation CSIRO leT Centre, PO Box 76 Epping, NSW 1710, Australia Tel: +61 2 9372 4289 Fax: +61 2 9372 4446 E-mail: Definition and Misuse of Return Loss Trevor S. Bird Here is the equation from the article Pin RL = 10 log,10 ( ---- ) dB, (1) Pref The author explicitly states the resulting value will be positive when Pin is greater than Pref. He goes on to say, "That is, return loss is the negative of the reflection coefficient expressed in decibels." He goes on to quote from the "IEEE Standard Dictionary of Electrical and Electronic Terms, Fourth Edition". (1 ) (data transmission) (A) At a discontinuity in a transmission system the difference between the power incident upon the discontinuity. (B) The ratio in deci- bels of the power incident upon the discontinuity to the power reflected from the discontinuity. Note: This ratio is also the square of the reciprocal to the magnitude of the reflection coefficient. (C) More broadly, the return loss is a measure of the dissimilarity between two impedances, being equal to the number of decibels that corresponds to the scalar value of the reciprocal of the reflection coefficient, and hence being expressed by the following formula: |Z1 + Z2| 20 log,10 |-------| decibel |Z1 - Z2| where Z1 and Z2 = the two impedances. (2) (or gain) (waveguide). The ratio of incident to reflected power at a reference plane of a network. So is this what you "knew" the paper said? Seems to be the opposite of what you have been promoting. Anyone feel this paper is incorrect? -- Rick |
Parallel coax
On 9/29/2015 8:14 AM, Jerry Stuckle wrote:
On 9/29/2015 3:27 AM, John S wrote: On 9/28/2015 6:54 PM, Jerry Stuckle wrote: You said return loss increases with lower SWR. It does not. It does. Sorry, a lower SWR does not increase the amount of loss. It increases the amount of loss that is reflected. Hence, return loss. Please cite a reliable reference that says it does. Even the table Rick cited shows a negative value for return SWR. No, the table is correct and does not show negative values for return loss. What is return SWR? No, a 1:1 SWR has no reflection, therefore no reflective loss. And yes, it does show negative values. Don't you see the '-' sign? No. I see a hyphen, not a minus sign. It indicates that the data column is in units of dB. |
Parallel coax
"Jerry Stuckle" wrote in message ... On 9/29/2015 3:58 AM, Ian Jackson wrote: In message , Jerry Stuckle writes On 9/28/2015 6:21 PM, Ian Jackson wrote: In message , Jerry Stuckle writes On 9/28/2015 3:10 PM, rickman wrote: On 9/28/2015 2:55 PM, Jerry Stuckle wrote: On 9/28/2015 2:19 PM, rickman wrote: On 9/28/2015 2:01 PM, Wayne wrote: "rickman" wrote in message ... On 9/28/2015 11:59 AM, Wayne wrote: "Ian Jackson" wrote in message ... In message , rickman writes Definition of Return Loss In technical terms, RL is the ratio of the light reflected back from a device under test, Pout, to the light launched into that device, Pin, usually expressed as a negative number in dB. RL = 10 log10(Pout/Pin) Here is a link for a table of return loss and VSWR.... http://www.jampro.com/uploads/tech_d.../VSWRChart.pdf It shows a higher return loss (assuming you mean magnitude since the values are all negative) for lower VSWR. I'm surprised to see negative quantities. For 50 years, I've always understood the Return Loss Ratio (RLR) to be exactly what it says on the tin, ie the ratio (in dB) of the LOSS (the attenuation) of the reflected signal wrt the incident signal. This is a +ve quantity. Things are already sufficiently confusing without having to start thinking in unnecessary -ve figures! I think the table headings are using a dash, not a negative sign. Return loss- dB # Look at the equation and you will understand. When the ratio is less # than one, the log is negative. But the ratio is never less than one for passive devices. If all the power forward is reflected, then the power ratio is 1 to 1. That's 0 dB return loss from the equation. Return loss is a positive number. I'm not so sure. It depends on how you define it. What if half the power is reflected? The equation above and *many* other sources say that is 10 log(0.5/1) = -3 dB. A few sources take exception to this and say it is 10 log(1/0.5) = 3 dB. At this point I dunno. Rickman, you are correct. The return loss is calculated as return value (the variable) divided by the output value (the constant). I haven't seen any reliable sources which say otherwise. I haven't seen any reliable sources that say either. Have you? Not for 40 years or so. But then I haven't looked for one since college. I've just dealt with RF engineers, who have used the same terminology. They use negative numbers for loss to figure the gain of the entire system. It doesn't matter which direction the loss is in; loss is a negative number (and gain is a positive number). A loss is only negative if it is being thought of as gain. If it's a loss, its value is positive. Physicists and engineers do not mix gain and loss. Gain is always shown as a positive number and loss as a negative number. Physicists and engineers don't get themselves into situations where 'gain' and 'loss' are used ambiguously. Unfortunately, the same cannot be said of certain radio amateurs. # Yes, there are some here who are trying to use the terms ambiguously - # saying +10db can be both gain and loss. I don't have many engineering books left, and my 1950s version of Terman doesn't cover return loss. My perspective on return loss happened late in my career using a network analyzer to evaluate reentry vehicle antennas. The network analyzer we used had RL expressed as a positive number. There is another equation to consider: RL=(Power incident in dBm) minus (Power reflected in dBm) Thus the bigger the RL number, the bigger the difference between forward and reflected power. |
Parallel coax
On 9/29/2015 10:32 AM, rickman wrote:
On 9/29/2015 9:22 AM, Jerry Stuckle wrote: On 9/29/2015 12:55 AM, rickman wrote: On 9/28/2015 8:56 PM, Jerry Stuckle wrote: On 9/28/2015 8:09 PM, rickman wrote: On 9/28/2015 7:55 PM, Jerry Stuckle wrote: On 9/28/2015 5:18 PM, rickman wrote: On 9/28/2015 4:34 PM, Jerry Stuckle wrote: I'm not interested. I know what it says. Guess I should have kept up my IEEE membership, but it just wasn't worth it. So share with the rest of us. What does it say? Exactly what your table showed. But you mentioned the resource, not me. You pay for it or you've just once again you're full of it. You said you *know* what the IEEE article says. Why not share with us? You want it - you pay for it. Or once again you prove you're full of it. No, I have not read the article. But I understand the physics and math behind it - unlike you. Someone who thinks magnitude without vector (direction) is valid! ROFLMAO! Ok, so you mispoke when you said, "I know what it says." No, I didn't. I didn't say I read it. I said I know what it says. And I do from other IEEE peer-reviewed articles. I don't need to read it to find out it agrees with other documentation. And if it didn't, it wouldn't have gotten published. You have said repeatedly that the return loss should be calculated by using the power in as the reference and the reflected power as the thing being measured which results in a negative log. I am pretty sure the paper says this is not the correct way to calculate it and many people are making a mistake doing it this way. So you've read other IEEE documentation which supports what you say? I'll see if I can get my hands on the paper. I'm not going to pay for it. If I thought it would get you to admit you were mistaken, I'd pay the $100. But I'm sure you will find a way to berate the authors or twist their logic and I'm not will to pay $100 for that. So stand by. Someone may be getting it for me. I really don't give a damn. You would argue the sun rises in the west. I have a copy of the paper. Trevor Bird Editor-in-ehief,Engineering IEEE Transactions on Antennas and Propagation CSIRO leT Centre, PO Box 76 Epping, NSW 1710, Australia Tel: +61 2 9372 4289 Fax: +61 2 9372 4446 E-mail: Definition and Misuse of Return Loss Trevor S. Bird Here is the equation from the article Pin RL = 10 log,10 ( ---- ) dB, (1) Pref The author explicitly states the resulting value will be positive when Pin is greater than Pref. He goes on to say, "That is, return loss is the negative of the reflection coefficient expressed in decibels." Since the reflection coefficient is never greater than 1, its value is negative. It follows that the return loss is positive. He goes on to quote from the "IEEE Standard Dictionary of Electrical and Electronic Terms, Fourth Edition". (1 ) (data transmission) (A) At a discontinuity in a transmission system the difference between the power incident upon the discontinuity. (B) The ratio in deci- bels of the power incident upon the discontinuity to the power reflected from the discontinuity. Note: This ratio is also the square of the reciprocal to the magnitude of the reflection coefficient. (C) More broadly, the return loss is a measure of the dissimilarity between two impedances, being equal to the number of decibels that corresponds to the scalar value of the reciprocal of the reflection coefficient, and hence being expressed by the following formula: |Z1 + Z2| 20 log,10 |-------| decibel |Z1 - Z2| where Z1 and Z2 = the two impedances. (2) (or gain) (waveguide). The ratio of incident to reflected power at a reference plane of a network. So is this what you "knew" the paper said? Seems to be the opposite of what you have been promoting. Anyone feel this paper is incorrect? Good find Rick. That should put it to bed. |
Parallel coax
"Ian Jackson" wrote in message ... Which is greater - 10db or -30db? In voltage ratios, they are 1/3 and 1/30 respectively, and in power ratios, 1/10 and 1/000 respectively. But if you lost 30dB down your coax, you'd be losing 20dB MORE than if you were only losing 10dB. But surely even you wouldn't say "My coax has a loss of minus 30dB"? [Or would you?!] dB without a refferance to something like voltage or power is not much if any use. |
Parallel coax
On 9/29/2015 12:45 PM, Ralph Mowery wrote:
"Ian Jackson" wrote in message ... Which is greater - 10db or -30db? In voltage ratios, they are 1/3 and 1/30 respectively, and in power ratios, 1/10 and 1/000 respectively. But if you lost 30dB down your coax, you'd be losing 20dB MORE than if you were only losing 10dB. But surely even you wouldn't say "My coax has a loss of minus 30dB"? [Or would you?!] dB without a refferance to something like voltage or power is not much if any use. Really? Not all measurements are absolute. The return loss we have been discussing is a perfect example. One power level vs. another. No need to consider the units as they factor out. -- Rick |
Parallel coax
"John S" wrote in message ... On 9/29/2015 10:32 AM, rickman wrote: On 9/29/2015 9:22 AM, Jerry Stuckle wrote: On 9/29/2015 12:55 AM, rickman wrote: On 9/28/2015 8:56 PM, Jerry Stuckle wrote: On 9/28/2015 8:09 PM, rickman wrote: On 9/28/2015 7:55 PM, Jerry Stuckle wrote: On 9/28/2015 5:18 PM, rickman wrote: On 9/28/2015 4:34 PM, Jerry Stuckle wrote: I'm not interested. I know what it says. Guess I should have kept up my IEEE membership, but it just wasn't worth it. So share with the rest of us. What does it say? Exactly what your table showed. But you mentioned the resource, not me. You pay for it or you've just once again you're full of it. You said you *know* what the IEEE article says. Why not share with us? You want it - you pay for it. Or once again you prove you're full of it. No, I have not read the article. But I understand the physics and math behind it - unlike you. Someone who thinks magnitude without vector (direction) is valid! ROFLMAO! Ok, so you mispoke when you said, "I know what it says." No, I didn't. I didn't say I read it. I said I know what it says. And I do from other IEEE peer-reviewed articles. I don't need to read it to find out it agrees with other documentation. And if it didn't, it wouldn't have gotten published. You have said repeatedly that the return loss should be calculated by using the power in as the reference and the reflected power as the thing being measured which results in a negative log. I am pretty sure the paper says this is not the correct way to calculate it and many people are making a mistake doing it this way. So you've read other IEEE documentation which supports what you say? I'll see if I can get my hands on the paper. I'm not going to pay for it. If I thought it would get you to admit you were mistaken, I'd pay the $100. But I'm sure you will find a way to berate the authors or twist their logic and I'm not will to pay $100 for that. So stand by. Someone may be getting it for me. I really don't give a damn. You would argue the sun rises in the west. I have a copy of the paper. Trevor Bird Editor-in-ehief,Engineering IEEE Transactions on Antennas and Propagation CSIRO leT Centre, PO Box 76 Epping, NSW 1710, Australia Tel: +61 2 9372 4289 Fax: +61 2 9372 4446 E-mail: Definition and Misuse of Return Loss Trevor S. Bird Here is the equation from the article Pin RL = 10 log,10 ( ---- ) dB, (1) Pref The author explicitly states the resulting value will be positive when Pin is greater than Pref. He goes on to say, "That is, return loss is the negative of the reflection coefficient expressed in decibels." Since the reflection coefficient is never greater than 1, its value is negative. It follows that the return loss is positive. He goes on to quote from the "IEEE Standard Dictionary of Electrical and Electronic Terms, Fourth Edition". (1 ) (data transmission) (A) At a discontinuity in a transmission system the difference between the power incident upon the discontinuity. (B) The ratio in deci- bels of the power incident upon the discontinuity to the power reflected from the discontinuity. Note: This ratio is also the square of the reciprocal to the magnitude of the reflection coefficient. (C) More broadly, the return loss is a measure of the dissimilarity between two impedances, being equal to the number of decibels that corresponds to the scalar value of the reciprocal of the reflection coefficient, and hence being expressed by the following formula: |Z1 + Z2| 20 log,10 |-------| decibel |Z1 - Z2| where Z1 and Z2 = the two impedances. (2) (or gain) (waveguide). The ratio of incident to reflected power at a reference plane of a network. So is this what you "knew" the paper said? Seems to be the opposite of what you have been promoting. Anyone feel this paper is incorrect? # Good find Rick. That should put it to bed. .....LOL :) |
Parallel coax
On 9/29/2015 2:10 PM, Wayne wrote:
"rickman" wrote in message ... On 9/29/2015 9:22 AM, Jerry Stuckle wrote: On 9/29/2015 12:55 AM, rickman wrote: On 9/28/2015 8:56 PM, Jerry Stuckle wrote: On 9/28/2015 8:09 PM, rickman wrote: On 9/28/2015 7:55 PM, Jerry Stuckle wrote: On 9/28/2015 5:18 PM, rickman wrote: On 9/28/2015 4:34 PM, Jerry Stuckle wrote: snip So stand by. Someone may be getting it for me. I really don't give a damn. You would argue the sun rises in the west. I have a copy of the paper. Trevor Bird Editor-in-ehief,Engineering IEEE Transactions on Antennas and Propagation CSIRO leT Centre, PO Box 76 Epping, NSW 1710, Australia Tel: +61 2 9372 4289 Fax: +61 2 9372 4446 E-mail: Definition and Misuse of Return Loss Trevor S. Bird Here is the equation from the article Pin RL = 10 log,10 ( ---- ) dB, (1) Pref The author explicitly states the resulting value will be positive when Pin is greater than Pref. He goes on to say, "That is, return loss is the negative of the reflection coefficient expressed in decibels." He goes on to quote from the "IEEE Standard Dictionary of Electrical and Electronic Terms, Fourth Edition". (1 ) (data transmission) (A) At a discontinuity in a transmission system the difference between the power incident upon the discontinuity. (B) The ratio in deci- bels of the power incident upon the discontinuity to the power reflected from the discontinuity. Note: This ratio is also the square of the reciprocal to the magnitude of the reflection coefficient. (C) More broadly, the return loss is a measure of the dissimilarity between two impedances, being equal to the number of decibels that corresponds to the scalar value of the reciprocal of the reflection coefficient, and hence being expressed by the following formula: |Z1 + Z2| 20 log,10 |-------| decibel |Z1 - Z2| where Z1 and Z2 = the two impedances. (2) (or gain) (waveguide). The ratio of incident to reflected power at a reference plane of a network. So is this what you "knew" the paper said? Seems to be the opposite of what you have been promoting. Anyone feel this paper is incorrect? Well, it is exactly what I thought I learned on the subject. :) To be honest, it seems logical that the power in should be the reference and the reflected power should be the property being measured which is what Jerry is saying. But clearly for this particular term "return loss" this is not the case. Does it seem intuitively correct that "return loss" should be a higher number when the reflection is smaller? -- Rick |
Parallel coax
"rickman" wrote in message ... On 9/29/2015 2:10 PM, Wayne wrote: "rickman" wrote in message ... On 9/29/2015 9:22 AM, Jerry Stuckle wrote: On 9/29/2015 12:55 AM, rickman wrote: On 9/28/2015 8:56 PM, Jerry Stuckle wrote: On 9/28/2015 8:09 PM, rickman wrote: On 9/28/2015 7:55 PM, Jerry Stuckle wrote: On 9/28/2015 5:18 PM, rickman wrote: On 9/28/2015 4:34 PM, Jerry Stuckle wrote: snip So stand by. Someone may be getting it for me. I really don't give a damn. You would argue the sun rises in the west. I have a copy of the paper. Trevor Bird Editor-in-ehief,Engineering IEEE Transactions on Antennas and Propagation CSIRO leT Centre, PO Box 76 Epping, NSW 1710, Australia Tel: +61 2 9372 4289 Fax: +61 2 9372 4446 E-mail: Definition and Misuse of Return Loss Trevor S. Bird Here is the equation from the article Pin RL = 10 log,10 ( ---- ) dB, (1) Pref The author explicitly states the resulting value will be positive when Pin is greater than Pref. He goes on to say, "That is, return loss is the negative of the reflection coefficient expressed in decibels." He goes on to quote from the "IEEE Standard Dictionary of Electrical and Electronic Terms, Fourth Edition". (1 ) (data transmission) (A) At a discontinuity in a transmission system the difference between the power incident upon the discontinuity. (B) The ratio in deci- bels of the power incident upon the discontinuity to the power reflected from the discontinuity. Note: This ratio is also the square of the reciprocal to the magnitude of the reflection coefficient. (C) More broadly, the return loss is a measure of the dissimilarity between two impedances, being equal to the number of decibels that corresponds to the scalar value of the reciprocal of the reflection coefficient, and hence being expressed by the following formula: |Z1 + Z2| 20 log,10 |-------| decibel |Z1 - Z2| where Z1 and Z2 = the two impedances. (2) (or gain) (waveguide). The ratio of incident to reflected power at a reference plane of a network. So is this what you "knew" the paper said? Seems to be the opposite of what you have been promoting. Anyone feel this paper is incorrect? Well, it is exactly what I thought I learned on the subject. :) # To be honest, it seems logical that the power in should be the reference # and the reflected power should be the property being measured which is # what Jerry is saying. But clearly for this particular term "return # loss" this is not the case. # Does it seem intuitively correct that "return loss" should be a higher # number when the reflection is smaller? There seems to be a case for saying it either way. My exposure to RL was "on the job" when playing with antennas and a network analyzer. We always used positive numbers and viewed RL as the difference in dB between the forward and reflected power. Actually, RL was something in the network analyzer data file and printout. We pretty much ignored it and looked at S parameters and SWR. |
Parallel coax
On 9/29/2015 1:45 PM, Wayne wrote:
"rickman" wrote in message ... On 9/29/2015 2:10 PM, Wayne wrote: "rickman" wrote in message ... On 9/29/2015 9:22 AM, Jerry Stuckle wrote: On 9/29/2015 12:55 AM, rickman wrote: On 9/28/2015 8:56 PM, Jerry Stuckle wrote: On 9/28/2015 8:09 PM, rickman wrote: On 9/28/2015 7:55 PM, Jerry Stuckle wrote: On 9/28/2015 5:18 PM, rickman wrote: On 9/28/2015 4:34 PM, Jerry Stuckle wrote: snip So stand by. Someone may be getting it for me. I really don't give a damn. You would argue the sun rises in the west. I have a copy of the paper. Trevor Bird Editor-in-ehief,Engineering IEEE Transactions on Antennas and Propagation CSIRO leT Centre, PO Box 76 Epping, NSW 1710, Australia Tel: +61 2 9372 4289 Fax: +61 2 9372 4446 E-mail: Definition and Misuse of Return Loss Trevor S. Bird Here is the equation from the article Pin RL = 10 log,10 ( ---- ) dB, (1) Pref The author explicitly states the resulting value will be positive when Pin is greater than Pref. He goes on to say, "That is, return loss is the negative of the reflection coefficient expressed in decibels." He goes on to quote from the "IEEE Standard Dictionary of Electrical and Electronic Terms, Fourth Edition". (1 ) (data transmission) (A) At a discontinuity in a transmission system the difference between the power incident upon the discontinuity. (B) The ratio in deci- bels of the power incident upon the discontinuity to the power reflected from the discontinuity. Note: This ratio is also the square of the reciprocal to the magnitude of the reflection coefficient. (C) More broadly, the return loss is a measure of the dissimilarity between two impedances, being equal to the number of decibels that corresponds to the scalar value of the reciprocal of the reflection coefficient, and hence being expressed by the following formula: |Z1 + Z2| 20 log,10 |-------| decibel |Z1 - Z2| where Z1 and Z2 = the two impedances. (2) (or gain) (waveguide). The ratio of incident to reflected power at a reference plane of a network. So is this what you "knew" the paper said? Seems to be the opposite of what you have been promoting. Anyone feel this paper is incorrect? Well, it is exactly what I thought I learned on the subject. :) # To be honest, it seems logical that the power in should be the reference # and the reflected power should be the property being measured which is # what Jerry is saying. But clearly for this particular term "return # loss" this is not the case. # Does it seem intuitively correct that "return loss" should be a higher # number when the reflection is smaller? Yes. It is not necessarily a loss. Consider a loss-less line and a perfect source. You still have a return loss, but there are no losses in the system. There seems to be a case for saying it either way. My exposure to RL was "on the job" when playing with antennas and a network analyzer. We always used positive numbers and viewed RL as the difference in dB between the forward and reflected power. That makes perfect sense. You could never have a negative dB. Actually, RL was something in the network analyzer data file and printout. We pretty much ignored it and looked at S parameters and SWR. |
Parallel coax
In message , rickman
writes Does it seem intuitively correct that "return loss" should be a higher number when the reflection is smaller? After a few milliseconds of serious consideration, "Yes"! -- Ian |
Parallel coax
On 9/27/2015 1:42 PM, John S wrote:
On 9/27/2015 1:20 PM, Wayne wrote: "rickman" wrote in message ... On 9/27/2015 10:41 AM, kg7fu wrote: Matching the antenna won't make the Return Loss go away but it will make the transmitter happy. Can you explain this? I thought matching the antenna would *exactly* make the return loss go away because it would eliminate the mismatch. Not wanting to put words in his mouth.... I read that to mean that the high SWR between the ATU and the antenna would remain, but the transmitter would be happy with the SWR on the transmitter/ATU coax. Rick is correct. If the antenna (load) is matched to the line, there is no return loss, hence no SWR. The ATU will be adjusted (hopefully) to make the transmitter operate properly with the impedance as seen at the transmitter end of the line. I apologize. My statement that "there is no return loss" above is technically incorrect. The return loss with matched conditions is maximum. It will peg your meter. |
Parallel coax
In message , John S
writes On 9/27/2015 1:42 PM, John S wrote: On 9/27/2015 1:20 PM, Wayne wrote: "rickman" wrote in message ... On 9/27/2015 10:41 AM, kg7fu wrote: Matching the antenna won't make the Return Loss go away but it will make the transmitter happy. Can you explain this? I thought matching the antenna would *exactly* make the return loss go away because it would eliminate the mismatch. Not wanting to put words in his mouth.... I read that to mean that the high SWR between the ATU and the antenna would remain, but the transmitter would be happy with the SWR on the transmitter/ATU coax. Rick is correct. If the antenna (load) is matched to the line, there is no return loss, hence no SWR. The ATU will be adjusted (hopefully) to make the transmitter operate properly with the impedance as seen at the transmitter end of the line. I apologize. My statement that "there is no return loss" above is technically incorrect. The return loss with matched conditions is maximum. It will peg your meter. Are you both getting this right? With a perfect match, the return loss is infinite (ie there is absolutely no reflection). Your SWR meter will read 1:1. With a total mismatch, the return loss is zero (ie the reflection is as strong as the forward signal). Your SWR meter will be pegged hard over at FSD. -- Ian |
Parallel coax
In message , John S
writes On 9/29/2015 1:45 PM, Wayne wrote: We always used positive numbers and viewed RL as the difference in dB between the forward and reflected power. That makes perfect sense. You could never have a negative dB. It might be instructive for us all to have a quick look at this information (especially the last sentence!): http://www.microwaves101.com/encyclopedias/vswr "Thus in its correct form, return loss will usually be a positive number. If it's not, you can usually blame measurement error. The exception to the rule is something with negative resistance, which implies that it is an active device (external DC power is converted to RF) and it is potentially unstable (it could oscillate). Not something you have to worry about if you are just looking at coax cables! However, many engineers often omit the minus sign and talk about "-9.5 dB return loss" for example. People that find it necessary to correct engineers who do this have underwear that is too tight." . -- Ian |
Parallel coax
On 9/29/2015 5:05 PM, Ian Jackson wrote:
In message , John S writes On 9/27/2015 1:42 PM, John S wrote: On 9/27/2015 1:20 PM, Wayne wrote: "rickman" wrote in message ... On 9/27/2015 10:41 AM, kg7fu wrote: Matching the antenna won't make the Return Loss go away but it will make the transmitter happy. Can you explain this? I thought matching the antenna would *exactly* make the return loss go away because it would eliminate the mismatch. Not wanting to put words in his mouth.... I read that to mean that the high SWR between the ATU and the antenna would remain, but the transmitter would be happy with the SWR on the transmitter/ATU coax. Rick is correct. If the antenna (load) is matched to the line, there is no return loss, hence no SWR. The ATU will be adjusted (hopefully) to make the transmitter operate properly with the impedance as seen at the transmitter end of the line. I apologize. My statement that "there is no return loss" above is technically incorrect. The return loss with matched conditions is maximum. It will peg your meter. Are you both getting this right? With a perfect match, the return loss is infinite (ie there is absolutely no reflection). Your SWR meter will read 1:1. With a total mismatch, the return loss is zero (ie the reflection is as strong as the forward signal). Your SWR meter will be pegged hard over at FSD. You are correct, of course. I was thinking of a dB meter that reads from 0 to some large value (say, 100) full scale. It is not an SWR meter. I have probably just confused the whole thing. Sorry. |
Parallel coax
On 9/29/2015 8:19 AM, Jerry Stuckle wrote:
On 9/29/2015 3:47 AM, Ian Jackson wrote: In message , Jerry Stuckle writes On 9/28/2015 7:12 PM, John S wrote: On 9/28/2015 1:51 PM, Jerry Stuckle wrote: On 9/28/2015 1:42 PM, rickman wrote: You said return loss increases with lower SWR. It does not. It does. Sorry, a lower SWR does not increase the amount of loss. Of course it doesn't. No one said it did. It does the opposite, ie a lower SWR gives less loss on the feeder. Please cite a reliable reference that says it does. Even the table Rick cited shows a negative value for return SWR. What is this 'Return SWR'? I'm not familiar with it. Sorry, writing too quickly. I meant return loss. Do you mean Return Loss Ratio (RLR)? This is a simple, easily measurable, and meaningful statement of how strong the returning reflected signal is compared with the outgoing forward signal. The reflected signal is a weaker version of the forward signal. It's expressed as a loss, an attenuation, or relatively how much down the level of the reflection is. You can express this as a numerical ratio - the reflection coefficient (rho) - or (as often more convenient) rho in dB. As others have suggested, what is apparently a negative sign in the chart is presumably more artistic licence than scientific accuracy. If you lose $10, you don't say that you lost 'minus $10'. Similarly, when you lose 10dB of signal, you don't say you lost 'minus 10dB'. Which is greater - 10db or -30db? Which is greater: -20dB or +20dB? |
Parallel coax
In message , Jeff writes
On 29/09/2015 14:31, Jerry Stuckle wrote: On 9/29/2015 4:40 AM, Jeff wrote: Physicists and engineers do not mix gain and loss. Gain is always shown as a positive number and loss as a negative number. For instance - a system shows a gain and loss of +3, +5, +2, +1. What is the total gain or loss of the system? Of course they do, particularly when dealing with a quantity that is defined as a LOSS. I have never heard any engineer when asked the question 'what is that attenuator' reply minus 3 dB. It is always 3dB. It is always called a 3dB attenuator, not a minus 3dB attenuator. Jeff Yes, and the power out is never +3db relative to the input. It is always -3db. Which is often referred to as 3dB loss (or 3dB down), ie a positive quantity. And Jerry still hasn't answered my question about him losing -$10. -- Ian |
Parallel coax
On 9/30/2015 10:12 AM, Ian Jackson wrote:
In message , Jeff writes On 29/09/2015 14:31, Jerry Stuckle wrote: On 9/29/2015 4:40 AM, Jeff wrote: Physicists and engineers do not mix gain and loss. Gain is always shown as a positive number and loss as a negative number. For instance - a system shows a gain and loss of +3, +5, +2, +1. What is the total gain or loss of the system? Of course they do, particularly when dealing with a quantity that is defined as a LOSS. I have never heard any engineer when asked the question 'what is that attenuator' reply minus 3 dB. It is always 3dB. It is always called a 3dB attenuator, not a minus 3dB attenuator. Jeff Yes, and the power out is never +3db relative to the input. It is always -3db. Which is often referred to as 3dB loss (or 3dB down), ie a positive quantity. And Jerry still hasn't answered my question about him losing -$10. Ever since he was presented with "reliable sources" he has been silent. I was beginning to think that he would accept no source at all if it didn't agree with him. However, his silence now is a plus for him. |
Parallel coax
"Ian Jackson" wrote in message ... In message , John S writes On 9/29/2015 1:45 PM, Wayne wrote: We always used positive numbers and viewed RL as the difference in dB between the forward and reflected power. That makes perfect sense. You could never have a negative dB. It might be instructive for us all to have a quick look at this information (especially the last sentence!): http://www.microwaves101.com/encyclopedias/vswr "Thus in its correct form, return loss will usually be a positive number. If it's not, you can usually blame measurement error. The exception to the rule is something with negative resistance, which implies that it is an active device (external DC power is converted to RF) and it is potentially unstable (it could oscillate). Not something you have to worry about if you are just looking at coax cables! However, many engineers often omit the minus sign and talk about "-9.5 dB return loss" for example. People that find it necessary to correct engineers who do this have underwear that is too tight." LOL. Yes I have run into people like that. Years ago, I worked with a young engineer who would ask me the following question when I said "SWR". He would ask "Do you mean VSWR?" After a few of his questions, I constructed a nomograph that would convert SWR to VSWR for him. And I even wrote an equation for him. SWR=10(VSWR)/10 hope this doesn't start a new thread :) |
Parallel coax
On 9/30/2015 12:57 PM, John S wrote:
On 9/30/2015 10:12 AM, Ian Jackson wrote: In message , Jeff writes On 29/09/2015 14:31, Jerry Stuckle wrote: On 9/29/2015 4:40 AM, Jeff wrote: Physicists and engineers do not mix gain and loss. Gain is always shown as a positive number and loss as a negative number. For instance - a system shows a gain and loss of +3, +5, +2, +1. What is the total gain or loss of the system? Of course they do, particularly when dealing with a quantity that is defined as a LOSS. I have never heard any engineer when asked the question 'what is that attenuator' reply minus 3 dB. It is always 3dB. It is always called a 3dB attenuator, not a minus 3dB attenuator. Jeff Yes, and the power out is never +3db relative to the input. It is always -3db. Which is often referred to as 3dB loss (or 3dB down), ie a positive quantity. And Jerry still hasn't answered my question about him losing -$10. Ever since he was presented with "reliable sources" he has been silent. I was beginning to think that he would accept no source at all if it didn't agree with him. However, his silence now is a plus for him. And no need to go on about the more dramatic aspects of this conversation. If everyone is happy with the information indicating that return loss is conventionally a positive dB value, let's move on. While I may enjoy showing the "truth" to someone who is being obstinate, I don't wish to make anyone feel like I'm rubbing their nose in it. -- Rick |
Parallel coax
In message , rickman
writes On 9/30/2015 12:57 PM, John S wrote: On 9/30/2015 10:12 AM, Ian Jackson wrote: In message , Jeff writes On 29/09/2015 14:31, Jerry Stuckle wrote: On 9/29/2015 4:40 AM, Jeff wrote: Physicists and engineers do not mix gain and loss. Gain is always shown as a positive number and loss as a negative number. For instance - a system shows a gain and loss of +3, +5, +2, +1. What is the total gain or loss of the system? Of course they do, particularly when dealing with a quantity that is defined as a LOSS. I have never heard any engineer when asked the question 'what is that attenuator' reply minus 3 dB. It is always 3dB. It is always called a 3dB attenuator, not a minus 3dB attenuator. Jeff Yes, and the power out is never +3db relative to the input. It is always -3db. Which is often referred to as 3dB loss (or 3dB down), ie a positive quantity. And Jerry still hasn't answered my question about him losing -$10. Ever since he was presented with "reliable sources" he has been silent. I was beginning to think that he would accept no source at all if it didn't agree with him. However, his silence now is a plus for him. And no need to go on about the more dramatic aspects of this conversation. If everyone is happy with the information indicating that return loss is conventionally a positive dB value, let's move on. While I may enjoy showing the "truth" to someone who is being obstinate, I don't wish to make anyone feel like I'm rubbing their nose in it. So let's get back to the original question. Was it ever really answered? I think it was made slightly more complicated by the fact that the antenna feedpoint impedance was not purely resistive, but was actually around 20-j130 (at 14 MHz), Was there any advantage in having the coax paralleled (both for 20 ohms resistive, and for 20-j130)? -- Ian |
Parallel coax
"Ian Jackson" wrote in message ... In message , rickman writes On 9/30/2015 12:57 PM, John S wrote: On 9/30/2015 10:12 AM, Ian Jackson wrote: In message , Jeff writes On 29/09/2015 14:31, Jerry Stuckle wrote: On 9/29/2015 4:40 AM, Jeff wrote: So let's get back to the original question. Was it ever really answered? I think it was made slightly more complicated by the fact that the antenna feedpoint impedance was not purely resistive, but was actually around 20-j130 (at 14 MHz), Was there any advantage in having the coax paralleled (both for 20 ohms resistive, and for 20-j130)? Expanding on the original question.... Antenna feedpoint approximately 20-j130 The ATU drives the antenna through about 15 feet of coax. Assuming that the ATU provides a +j130 conjugate match, does that leave the coax with a SWR of 50/20= 2.5:1? If so, then I will not bother with considering 2 parallel coax. |
Parallel coax
Wayne wrote:
"Ian Jackson" wrote in message ... In message , rickman writes On 9/30/2015 12:57 PM, John S wrote: On 9/30/2015 10:12 AM, Ian Jackson wrote: In message , Jeff writes On 29/09/2015 14:31, Jerry Stuckle wrote: On 9/29/2015 4:40 AM, Jeff wrote: So let's get back to the original question. Was it ever really answered? I think it was made slightly more complicated by the fact that the antenna feedpoint impedance was not purely resistive, but was actually around 20-j130 (at 14 MHz), Was there any advantage in having the coax paralleled (both for 20 ohms resistive, and for 20-j130)? Expanding on the original question.... Antenna feedpoint approximately 20-j130 The ATU drives the antenna through about 15 feet of coax. Assuming that the ATU provides a +j130 conjugate match, does that leave the coax with a SWR of 50/20= 2.5:1? If so, then I will not bother with considering 2 parallel coax. If I'm not mistaken, the aerial impedance will look a bit different at the other end of about a quarter wave of coax. But that is probably not going to alter your conclusion, just the matching network. -- Roger Hayter |
Parallel coax
"Roger Hayter" wrote in message ... Wayne wrote: "Ian Jackson" wrote in message ... In message , rickman writes On 9/30/2015 12:57 PM, John S wrote: On 9/30/2015 10:12 AM, Ian Jackson wrote: In message , Jeff writes On 29/09/2015 14:31, Jerry Stuckle wrote: On 9/29/2015 4:40 AM, Jeff wrote: So let's get back to the original question. Was it ever really answered? I think it was made slightly more complicated by the fact that the antenna feedpoint impedance was not purely resistive, but was actually around 20-j130 (at 14 MHz), Was there any advantage in having the coax paralleled (both for 20 ohms resistive, and for 20-j130)? Expanding on the original question.... Antenna feedpoint approximately 20-j130 The ATU drives the antenna through about 15 feet of coax. Assuming that the ATU provides a +j130 conjugate match, does that leave the coax with a SWR of 50/20= 2.5:1? If so, then I will not bother with considering 2 parallel coax. # If I'm not mistaken, the aerial impedance will look a bit different at # the other end of about a quarter wave of coax. But that is probably # not going to alter your conclusion, just the matching network. Good point. The impedance will move around the Smith constant swr circle to something else that will need the ATU conjugate match. I'll take a better look at the dimensions. (but going "off the grid" for a while, so it will be a few days) |
Parallel coax
On 9/30/2015 3:31 PM, Wayne wrote:
"Ian Jackson" wrote in message ... In message , rickman writes On 9/30/2015 12:57 PM, John S wrote: On 9/30/2015 10:12 AM, Ian Jackson wrote: In message , Jeff writes On 29/09/2015 14:31, Jerry Stuckle wrote: On 9/29/2015 4:40 AM, Jeff wrote: So let's get back to the original question. Was it ever really answered? I think it was made slightly more complicated by the fact that the antenna feedpoint impedance was not purely resistive, but was actually around 20-j130 (at 14 MHz), Was there any advantage in having the coax paralleled (both for 20 ohms resistive, and for 20-j130)? Expanding on the original question.... Antenna feedpoint approximately 20-j130 The ATU drives the antenna through about 15 feet of coax. Assuming that the ATU provides a +j130 conjugate match, does that leave the coax with a SWR of 50/20= 2.5:1? If so, then I will not bother with considering 2 parallel coax. The ATU will have no effect on the SWR in the 15ft of coax. The SWR is determined entirely by the mismatch between the coax and your antenna. The ATU simply transforms the impedance it sees to give the transmitter the load it needs (50 + j0). Using a Smith chart, I estimate that your load (antenna) will be transformed to 6.67 + j52.5 through the 15ft of coax. So, that's what your ATU will see for a load. The chart also shows about 17:1 SWR in the coax and there is nothing I can do at the transmitter to change it. If I now change the coax to Zo of 25 ohms, the impedance at the transmitter end changes to 1.55 +j18.5 ohms and the coax SWR becomes about 40:1. It seems to me that you would only be hurting yourself. Does this help? |
Parallel coax
On 9/30/2015 9:10 PM, John S wrote:
On 9/30/2015 3:31 PM, Wayne wrote: "Ian Jackson" wrote in message ... In message , rickman writes On 9/30/2015 12:57 PM, John S wrote: On 9/30/2015 10:12 AM, Ian Jackson wrote: In message , Jeff writes On 29/09/2015 14:31, Jerry Stuckle wrote: On 9/29/2015 4:40 AM, Jeff wrote: So let's get back to the original question. Was it ever really answered? I think it was made slightly more complicated by the fact that the antenna feedpoint impedance was not purely resistive, but was actually around 20-j130 (at 14 MHz), Was there any advantage in having the coax paralleled (both for 20 ohms resistive, and for 20-j130)? Expanding on the original question.... Antenna feedpoint approximately 20-j130 The ATU drives the antenna through about 15 feet of coax. Assuming that the ATU provides a +j130 conjugate match, does that leave the coax with a SWR of 50/20= 2.5:1? If so, then I will not bother with considering 2 parallel coax. The ATU will have no effect on the SWR in the 15ft of coax. The SWR is determined entirely by the mismatch between the coax and your antenna. The ATU simply transforms the impedance it sees to give the transmitter the load it needs (50 + j0). I will plead ignorance of ATUs, but wouldn't they also be designed to match the impedance seen at the cable? Using a Smith chart, I estimate that your load (antenna) will be transformed to 6.67 + j52.5 through the 15ft of coax. So, that's what your ATU will see for a load. The chart also shows about 17:1 SWR in the coax and there is nothing I can do at the transmitter to change it. If I now change the coax to Zo of 25 ohms, the impedance at the transmitter end changes to 1.55 +j18.5 ohms and the coax SWR becomes about 40:1. It seems to me that you would only be hurting yourself. Are you calculating the SWR based on the cable to ATU match or the cable to antenna match? Using an online calculator I get 20:1 VSWR with a 50 ohm cable and 36:1 with a 25 ohm cable. Were you just rounding your numbers? I used Z = 20 -j130 ohms for the antenna. -- Rick |
Parallel coax
On 9/30/2015 8:24 PM, rickman wrote:
On 9/30/2015 9:10 PM, John S wrote: On 9/30/2015 3:31 PM, Wayne wrote: "Ian Jackson" wrote in message ... In message , rickman writes On 9/30/2015 12:57 PM, John S wrote: On 9/30/2015 10:12 AM, Ian Jackson wrote: In message , Jeff writes On 29/09/2015 14:31, Jerry Stuckle wrote: On 9/29/2015 4:40 AM, Jeff wrote: So let's get back to the original question. Was it ever really answered? I think it was made slightly more complicated by the fact that the antenna feedpoint impedance was not purely resistive, but was actually around 20-j130 (at 14 MHz), Was there any advantage in having the coax paralleled (both for 20 ohms resistive, and for 20-j130)? Expanding on the original question.... Antenna feedpoint approximately 20-j130 The ATU drives the antenna through about 15 feet of coax. Assuming that the ATU provides a +j130 conjugate match, does that leave the coax with a SWR of 50/20= 2.5:1? If so, then I will not bother with considering 2 parallel coax. The ATU will have no effect on the SWR in the 15ft of coax. The SWR is determined entirely by the mismatch between the coax and your antenna. The ATU simply transforms the impedance it sees to give the transmitter the load it needs (50 + j0). I will plead ignorance of ATUs, but wouldn't they also be designed to match the impedance seen at the cable? Using a Smith chart, I estimate that your load (antenna) will be transformed to 6.67 + j52.5 through the 15ft of coax. So, that's what your ATU will see for a load. The chart also shows about 17:1 SWR in the coax and there is nothing I can do at the transmitter to change it. If I now change the coax to Zo of 25 ohms, the impedance at the transmitter end changes to 1.55 +j18.5 ohms and the coax SWR becomes about 40:1. It seems to me that you would only be hurting yourself. Are you calculating the SWR based on the cable to ATU match or the cable to antenna match? Using an online calculator I get 20:1 VSWR with a 50 ohm cable and 36:1 with a 25 ohm cable. Were you just rounding your numbers? I used Z = 20 -j130 ohms for the antenna. I also used Z = 20 -j130 ohms for the antenna. The cable to ATU match has no effect on SWR nor does it change the impedance seen at the transmitter end of the cable. Does your online calculator include cable loss and velocity factor? I used .0233dB/m and .66 velocity factor with a length of 4.57m. Sorry for the metric numbers, but that's what my chart uses. |
Parallel coax
On 9/30/2015 8:24 PM, rickman wrote:
On 9/30/2015 9:10 PM, John S wrote: On 9/30/2015 3:31 PM, Wayne wrote: "Ian Jackson" wrote in message ... In message , rickman writes On 9/30/2015 12:57 PM, John S wrote: On 9/30/2015 10:12 AM, Ian Jackson wrote: In message , Jeff writes On 29/09/2015 14:31, Jerry Stuckle wrote: On 9/29/2015 4:40 AM, Jeff wrote: So let's get back to the original question. Was it ever really answered? I think it was made slightly more complicated by the fact that the antenna feedpoint impedance was not purely resistive, but was actually around 20-j130 (at 14 MHz), Was there any advantage in having the coax paralleled (both for 20 ohms resistive, and for 20-j130)? Expanding on the original question.... Antenna feedpoint approximately 20-j130 The ATU drives the antenna through about 15 feet of coax. Assuming that the ATU provides a +j130 conjugate match, does that leave the coax with a SWR of 50/20= 2.5:1? If so, then I will not bother with considering 2 parallel coax. The ATU will have no effect on the SWR in the 15ft of coax. The SWR is determined entirely by the mismatch between the coax and your antenna. The ATU simply transforms the impedance it sees to give the transmitter the load it needs (50 + j0). I will plead ignorance of ATUs, but wouldn't they also be designed to match the impedance seen at the cable? Using a Smith chart, I estimate that your load (antenna) will be transformed to 6.67 + j52.5 through the 15ft of coax. So, that's what your ATU will see for a load. The chart also shows about 17:1 SWR in the coax and there is nothing I can do at the transmitter to change it. If I now change the coax to Zo of 25 ohms, the impedance at the transmitter end changes to 1.55 +j18.5 ohms and the coax SWR becomes about 40:1. It seems to me that you would only be hurting yourself. Are you calculating the SWR based on the cable to ATU match or the cable to antenna match? Using an online calculator I get 20:1 VSWR with a 50 ohm cable and 36:1 with a 25 ohm cable. Were you just rounding your numbers? I used Z = 20 -j130 ohms for the antenna. Wow! Is this an interesting discussion, or what? |
Parallel coax
On 9/30/2015 9:37 PM, John S wrote:
On 9/30/2015 8:24 PM, rickman wrote: On 9/30/2015 9:10 PM, John S wrote: On 9/30/2015 3:31 PM, Wayne wrote: "Ian Jackson" wrote in message ... In message , rickman writes On 9/30/2015 12:57 PM, John S wrote: On 9/30/2015 10:12 AM, Ian Jackson wrote: In message , Jeff writes On 29/09/2015 14:31, Jerry Stuckle wrote: On 9/29/2015 4:40 AM, Jeff wrote: So let's get back to the original question. Was it ever really answered? I think it was made slightly more complicated by the fact that the antenna feedpoint impedance was not purely resistive, but was actually around 20-j130 (at 14 MHz), Was there any advantage in having the coax paralleled (both for 20 ohms resistive, and for 20-j130)? Expanding on the original question.... Antenna feedpoint approximately 20-j130 The ATU drives the antenna through about 15 feet of coax. Assuming that the ATU provides a +j130 conjugate match, does that leave the coax with a SWR of 50/20= 2.5:1? If so, then I will not bother with considering 2 parallel coax. The ATU will have no effect on the SWR in the 15ft of coax. The SWR is determined entirely by the mismatch between the coax and your antenna. The ATU simply transforms the impedance it sees to give the transmitter the load it needs (50 + j0). I will plead ignorance of ATUs, but wouldn't they also be designed to match the impedance seen at the cable? Using a Smith chart, I estimate that your load (antenna) will be transformed to 6.67 + j52.5 through the 15ft of coax. So, that's what your ATU will see for a load. The chart also shows about 17:1 SWR in the coax and there is nothing I can do at the transmitter to change it. If I now change the coax to Zo of 25 ohms, the impedance at the transmitter end changes to 1.55 +j18.5 ohms and the coax SWR becomes about 40:1. It seems to me that you would only be hurting yourself. Are you calculating the SWR based on the cable to ATU match or the cable to antenna match? Using an online calculator I get 20:1 VSWR with a 50 ohm cable and 36:1 with a 25 ohm cable. Were you just rounding your numbers? I used Z = 20 -j130 ohms for the antenna. I also used Z = 20 -j130 ohms for the antenna. The cable to ATU match has no effect on SWR nor does it change the impedance seen at the transmitter end of the cable. Does your online calculator include cable loss and velocity factor? I used .0233dB/m and .66 velocity factor with a length of 4.57m. Sorry for the metric numbers, but that's what my chart uses. How do any of those things enter into the VSWR calculation? VSWR is defined solely by the match of the cable and antenna impedances. I don't know much about the calculator I used, some random online thing. http://chemandy.com/calculators/retu...calculator.htm -- Rick |
Parallel coax
"John S" wrote in message ... On 9/30/2015 8:24 PM, rickman wrote: On 9/30/2015 9:10 PM, John S wrote: On 9/30/2015 3:31 PM, Wayne wrote: "Ian Jackson" wrote in message ... In message , rickman writes On 9/30/2015 12:57 PM, John S wrote: On 9/30/2015 10:12 AM, Ian Jackson wrote: In message , Jeff writes On 29/09/2015 14:31, Jerry Stuckle wrote: On 9/29/2015 4:40 AM, Jeff wrote: So let's get back to the original question. Was it ever really answered? I think it was made slightly more complicated by the fact that the antenna feedpoint impedance was not purely resistive, but was actually around 20-j130 (at 14 MHz), Was there any advantage in having the coax paralleled (both for 20 ohms resistive, and for 20-j130)? Expanding on the original question.... Antenna feedpoint approximately 20-j130 The ATU drives the antenna through about 15 feet of coax. Assuming that the ATU provides a +j130 conjugate match, does that leave the coax with a SWR of 50/20= 2.5:1? If so, then I will not bother with considering 2 parallel coax. The ATU will have no effect on the SWR in the 15ft of coax. The SWR is determined entirely by the mismatch between the coax and your antenna. The ATU simply transforms the impedance it sees to give the transmitter the load it needs (50 + j0). I will plead ignorance of ATUs, but wouldn't they also be designed to match the impedance seen at the cable? Using a Smith chart, I estimate that your load (antenna) will be transformed to 6.67 + j52.5 through the 15ft of coax. So, that's what your ATU will see for a load. The chart also shows about 17:1 SWR in the coax and there is nothing I can do at the transmitter to change it. If I now change the coax to Zo of 25 ohms, the impedance at the transmitter end changes to 1.55 +j18.5 ohms and the coax SWR becomes about 40:1. It seems to me that you would only be hurting yourself. Are you calculating the SWR based on the cable to ATU match or the cable to antenna match? Using an online calculator I get 20:1 VSWR with a 50 ohm cable and 36:1 with a 25 ohm cable. Were you just rounding your numbers? I used Z = 20 -j130 ohms for the antenna. # Wow! Is this an interesting discussion, or what? Well, I think it is :) The antenna system has a lossy coax between the antenna feed and the ATU. My understanding is that the SWR on the line is what it is, regardless of the ATU settings. In fact, I use an RF ammeter between the ATU and the antenna and tune for maximum smoke. When I look at the SWR reading on the transmitter panel, it seems to more or less coincide with max RF current. I'll have to drop out of the discussion for a week or so....going off the grid on vacation. |
Parallel coax
On 9/30/2015 8:49 PM, rickman wrote:
On 9/30/2015 9:37 PM, John S wrote: On 9/30/2015 8:24 PM, rickman wrote: On 9/30/2015 9:10 PM, John S wrote: On 9/30/2015 3:31 PM, Wayne wrote: "Ian Jackson" wrote in message ... In message , rickman writes On 9/30/2015 12:57 PM, John S wrote: On 9/30/2015 10:12 AM, Ian Jackson wrote: In message , Jeff writes On 29/09/2015 14:31, Jerry Stuckle wrote: On 9/29/2015 4:40 AM, Jeff wrote: So let's get back to the original question. Was it ever really answered? I think it was made slightly more complicated by the fact that the antenna feedpoint impedance was not purely resistive, but was actually around 20-j130 (at 14 MHz), Was there any advantage in having the coax paralleled (both for 20 ohms resistive, and for 20-j130)? Expanding on the original question.... Antenna feedpoint approximately 20-j130 The ATU drives the antenna through about 15 feet of coax. Assuming that the ATU provides a +j130 conjugate match, does that leave the coax with a SWR of 50/20= 2.5:1? If so, then I will not bother with considering 2 parallel coax. The ATU will have no effect on the SWR in the 15ft of coax. The SWR is determined entirely by the mismatch between the coax and your antenna. The ATU simply transforms the impedance it sees to give the transmitter the load it needs (50 + j0). I will plead ignorance of ATUs, but wouldn't they also be designed to match the impedance seen at the cable? Using a Smith chart, I estimate that your load (antenna) will be transformed to 6.67 + j52.5 through the 15ft of coax. So, that's what your ATU will see for a load. The chart also shows about 17:1 SWR in the coax and there is nothing I can do at the transmitter to change it. If I now change the coax to Zo of 25 ohms, the impedance at the transmitter end changes to 1.55 +j18.5 ohms and the coax SWR becomes about 40:1. It seems to me that you would only be hurting yourself. Are you calculating the SWR based on the cable to ATU match or the cable to antenna match? Using an online calculator I get 20:1 VSWR with a 50 ohm cable and 36:1 with a 25 ohm cable. Were you just rounding your numbers? I used Z = 20 -j130 ohms for the antenna. I also used Z = 20 -j130 ohms for the antenna. The cable to ATU match has no effect on SWR nor does it change the impedance seen at the transmitter end of the cable. Does your online calculator include cable loss and velocity factor? I used .0233dB/m and .66 velocity factor with a length of 4.57m. Sorry for the metric numbers, but that's what my chart uses. How do any of those things enter into the VSWR calculation? VSWR is defined solely by the match of the cable and antenna impedances. You are correct except that the SWR is higher at the antenna end than at the opposite end due to cable attenuation although it might be insignificant for short cables and low frequencies. The velocity factor has no direct effect except it makes the physical length longer than the electrical length which will increase the loss over that which would be estimated if it is not included. I don't know much about the calculator I used, some random online thing. http://chemandy.com/calculators/retu...calculator.htm I encourage you to get a Smith charting program and take some time to get familiar with it. Very enlightening. I have one I recommend, but I think I posted that one before when you were present. http://www.fritz.dellsperger.net/ |
Parallel coax
On 9/30/2015 8:58 PM, Wayne wrote:
"John S" wrote in message ... On 9/30/2015 8:24 PM, rickman wrote: On 9/30/2015 9:10 PM, John S wrote: On 9/30/2015 3:31 PM, Wayne wrote: "Ian Jackson" wrote in message ... In message , rickman writes On 9/30/2015 12:57 PM, John S wrote: On 9/30/2015 10:12 AM, Ian Jackson wrote: In message , Jeff writes On 29/09/2015 14:31, Jerry Stuckle wrote: On 9/29/2015 4:40 AM, Jeff wrote: So let's get back to the original question. Was it ever really answered? I think it was made slightly more complicated by the fact that the antenna feedpoint impedance was not purely resistive, but was actually around 20-j130 (at 14 MHz), Was there any advantage in having the coax paralleled (both for 20 ohms resistive, and for 20-j130)? Expanding on the original question.... Antenna feedpoint approximately 20-j130 The ATU drives the antenna through about 15 feet of coax. Assuming that the ATU provides a +j130 conjugate match, does that leave the coax with a SWR of 50/20= 2.5:1? If so, then I will not bother with considering 2 parallel coax. The ATU will have no effect on the SWR in the 15ft of coax. The SWR is determined entirely by the mismatch between the coax and your antenna. The ATU simply transforms the impedance it sees to give the transmitter the load it needs (50 + j0). I will plead ignorance of ATUs, but wouldn't they also be designed to match the impedance seen at the cable? Using a Smith chart, I estimate that your load (antenna) will be transformed to 6.67 + j52.5 through the 15ft of coax. So, that's what your ATU will see for a load. The chart also shows about 17:1 SWR in the coax and there is nothing I can do at the transmitter to change it. If I now change the coax to Zo of 25 ohms, the impedance at the transmitter end changes to 1.55 +j18.5 ohms and the coax SWR becomes about 40:1. It seems to me that you would only be hurting yourself. Are you calculating the SWR based on the cable to ATU match or the cable to antenna match? Using an online calculator I get 20:1 VSWR with a 50 ohm cable and 36:1 with a 25 ohm cable. Were you just rounding your numbers? I used Z = 20 -j130 ohms for the antenna. # Wow! Is this an interesting discussion, or what? Well, I think it is :) The antenna system has a lossy coax between the antenna feed and the ATU. In Wayne's case it is not a great loss until the SWR gets great. My understanding is that the SWR on the line is what it is, regardless of the ATU settings. Yes, that is so. In fact, I use an RF ammeter between the ATU and the antenna and tune for maximum smoke. When I look at the SWR reading on the transmitter panel, it seems to more or less coincide with max RF current. That's good. I'll have to drop out of the discussion for a week or so....going off the grid on vacation. I wish you a happy and rewarding vacation. Cheers. |
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