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Parallel coax
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. -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
Parallel coax
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. -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
Parallel coax
On 9/28/2015 2:51 PM, Jerry Stuckle wrote:
On 9/28/2015 1:42 PM, rickman wrote: On 9/28/2015 12:54 PM, Jerry Stuckle wrote: 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: "John S" wrote in message ... 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. # Yes, the SWR due to mismatch of the antenna (load) and line will remain. # Even if the real part of your load impedance is matched to the line, you # will still have a high SWR if the reactance remains. # Does this make sense? Yes. That's what I was trying to say using SWR instead of return loss. Return loss numbers get bigger with lower SWR. For example: SWR 1:1 = infinite return loss. Incorrect. Return loss increases with an increased SWR. An SWR of 1:1 has no return loss because there is no returned signal to lose. 100% of the signal is radiated. 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. I believe that is exactly what I said in the portions of my post which you trimmed. These values for RF return loss match exactly the equation which you are saying is not used in RF. So which is it, the return loss table is correct with negative values of return loss or the equation I posted is incorrect even though it gives the values in the table? You said return loss increases with lower SWR. It does not. Are you being pedantic that -1 dB is not lower than -10 dB? It is not numerically lower in value, but is lower in magnitude and it is a lower loss. I even referred to the magnitude in my post. But then that was the same part you snipped which I referred to earlier. "It shows a higher return loss (assuming you mean magnitude since the values are all negative) for lower VSWR." -- Rick |
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In message , Jerry Stuckle
writes You said return loss increases with lower SWR. It does not. It does if you are used to +ve values of RLR! -- Ian |
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On 9/28/2015 2:53 PM, Jerry Stuckle wrote:
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'd like to see a reliable source for either position. From the references in the wikipedia article... http://ieeexplore.ieee.org/xpl/freea...number=5162049 https://en.wikipedia.org/wiki/Return_loss I didn't pay for the article, but it seems pretty clear that the authors are saying the term is often incorrectly used. I have found other loss equations in the form of Ls = 10 log (pt/pa) where pt is the input power and pa is the output power. Even this is not quite the same since it refers to the measured power at each end rather than the "lost" power. The point is the input power is in the numerator yielding a positive dB result for all cases. I'd be happy to find anything authoritative. -- Rick |
Parallel coax
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? -- Rick |
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"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: |
Parallel coax
On 9/28/2015 3:02 PM, rickman wrote:
On 9/28/2015 2:51 PM, Jerry Stuckle wrote: On 9/28/2015 1:42 PM, rickman wrote: On 9/28/2015 12:54 PM, Jerry Stuckle wrote: 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: "John S" wrote in message ... 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. # Yes, the SWR due to mismatch of the antenna (load) and line will remain. # Even if the real part of your load impedance is matched to the line, you # will still have a high SWR if the reactance remains. # Does this make sense? Yes. That's what I was trying to say using SWR instead of return loss. Return loss numbers get bigger with lower SWR. For example: SWR 1:1 = infinite return loss. Incorrect. Return loss increases with an increased SWR. An SWR of 1:1 has no return loss because there is no returned signal to lose. 100% of the signal is radiated. 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. I believe that is exactly what I said in the portions of my post which you trimmed. These values for RF return loss match exactly the equation which you are saying is not used in RF. So which is it, the return loss table is correct with negative values of return loss or the equation I posted is incorrect even though it gives the values in the table? You said return loss increases with lower SWR. It does not. Are you being pedantic that -1 dB is not lower than -10 dB? It is not numerically lower in value, but is lower in magnitude and it is a lower loss. I even referred to the magnitude in my post. But then that was the same part you snipped which I referred to earlier. "It shows a higher return loss (assuming you mean magnitude since the values are all negative) for lower VSWR." No, I am not being pedantic. -1 is greater than -10. The fact it is a negative number makes all the difference - as any engineer who knows what he's talking about will tell you. What you claim is like saying 1/100 is greater than 1/10. -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
Parallel coax
On 9/28/2015 3:09 PM, rickman wrote:
On 9/28/2015 2:53 PM, Jerry Stuckle wrote: 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'd like to see a reliable source for either position. From the references in the wikipedia article... http://ieeexplore.ieee.org/xpl/freea...number=5162049 https://en.wikipedia.org/wiki/Return_loss I didn't pay for the article, but it seems pretty clear that the authors are saying the term is often incorrectly used. I have found other loss equations in the form of Ls = 10 log (pt/pa) where pt is the input power and pa is the output power. Even this is not quite the same since it refers to the measured power at each end rather than the "lost" power. The point is the input power is in the numerator yielding a positive dB result for all cases. I'd be happy to find anything authoritative. You'll probably have to pay for the article to get something authoritative. Or you could swing by a university bookstore and check out some of the EE texts. You might find one in the local library, but I doubt it. -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
Parallel coax
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? -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
Parallel coax
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). -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
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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 This page is clear, but it is just one guy citing no references, not exactly authoritative. http://www.mogami.com/e/cad/vswr.html This page is not at all clear and only expresses return loss in terms of the impedances. Following through on the calculations give negative values for RL. http://www.microwaves101.com/encyclo...swr-calculator Offers a definition of RL in terms of gamma. Far from authoritative. Definition of gamma is in terms of VSWR and VSWR in terms of gamma. http://www.amphenolrf.com/vswr-conversion-chart/ Convention is to give positive values in tables. 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. You didn't cite the relevant section... Sign Properly, loss quantities, when expressed in decibels, should be positive numbers.[note 1] However, return loss has historically been expressed as a negative number, and this convention is still widely found in the literature.[1] 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: This one is a bit confusing. The words in the first sentence say one thing and the rest say something different. I don't see an equation in terms of the power that I can understand without interpreting their programming language. At this point I guess the issue is very muddy with usage in both camps. I think the wiki page indicates that with the section on Sign. What really bugs me is why return loss is in terms of the loss while other loss figures are in terms of the remaining power. Or did I find an anomalous reference for that one? -- Rick |
Parallel coax
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? -- Rick |
Parallel coax
On 9/28/2015 4:02 PM, Jerry Stuckle wrote:
On 9/28/2015 3:02 PM, rickman wrote: On 9/28/2015 2:51 PM, Jerry Stuckle wrote: On 9/28/2015 1:42 PM, rickman wrote: On 9/28/2015 12:54 PM, Jerry Stuckle wrote: 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: "John S" wrote in message ... 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. # Yes, the SWR due to mismatch of the antenna (load) and line will remain. # Even if the real part of your load impedance is matched to the line, you # will still have a high SWR if the reactance remains. # Does this make sense? Yes. That's what I was trying to say using SWR instead of return loss. Return loss numbers get bigger with lower SWR. For example: SWR 1:1 = infinite return loss. Incorrect. Return loss increases with an increased SWR. An SWR of 1:1 has no return loss because there is no returned signal to lose. 100% of the signal is radiated. 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. I believe that is exactly what I said in the portions of my post which you trimmed. These values for RF return loss match exactly the equation which you are saying is not used in RF. So which is it, the return loss table is correct with negative values of return loss or the equation I posted is incorrect even though it gives the values in the table? You said return loss increases with lower SWR. It does not. Are you being pedantic that -1 dB is not lower than -10 dB? It is not numerically lower in value, but is lower in magnitude and it is a lower loss. I even referred to the magnitude in my post. But then that was the same part you snipped which I referred to earlier. "It shows a higher return loss (assuming you mean magnitude since the values are all negative) for lower VSWR." No, I am not being pedantic. -1 is greater than -10. The fact it is a negative number makes all the difference - as any engineer who knows what he's talking about will tell you. Except that I clearly stated I was referring to the magnitude. -- Rick |
Parallel coax
On 9/28/2015 4:15 PM, rickman wrote:
On 9/28/2015 4:02 PM, Jerry Stuckle wrote: On 9/28/2015 3:02 PM, rickman wrote: On 9/28/2015 2:51 PM, Jerry Stuckle wrote: On 9/28/2015 1:42 PM, rickman wrote: On 9/28/2015 12:54 PM, Jerry Stuckle wrote: 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: "John S" wrote in message ... 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. # Yes, the SWR due to mismatch of the antenna (load) and line will remain. # Even if the real part of your load impedance is matched to the line, you # will still have a high SWR if the reactance remains. # Does this make sense? Yes. That's what I was trying to say using SWR instead of return loss. Return loss numbers get bigger with lower SWR. For example: SWR 1:1 = infinite return loss. Incorrect. Return loss increases with an increased SWR. An SWR of 1:1 has no return loss because there is no returned signal to lose. 100% of the signal is radiated. 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. I believe that is exactly what I said in the portions of my post which you trimmed. These values for RF return loss match exactly the equation which you are saying is not used in RF. So which is it, the return loss table is correct with negative values of return loss or the equation I posted is incorrect even though it gives the values in the table? You said return loss increases with lower SWR. It does not. Are you being pedantic that -1 dB is not lower than -10 dB? It is not numerically lower in value, but is lower in magnitude and it is a lower loss. I even referred to the magnitude in my post. But then that was the same part you snipped which I referred to earlier. "It shows a higher return loss (assuming you mean magnitude since the values are all negative) for lower VSWR." No, I am not being pedantic. -1 is greater than -10. The fact it is a negative number makes all the difference - as any engineer who knows what he's talking about will tell you. Except that I clearly stated I was referring to the magnitude. Magnitude without direction is meaningless. -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
Parallel coax
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. -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
Parallel coax
rickman 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 This page is clear, but it is just one guy citing no references, not exactly authoritative. http://www.mogami.com/e/cad/vswr.html This page is not at all clear and only expresses return loss in terms of the impedances. Following through on the calculations give negative values for RL. http://www.microwaves101.com/encyclo...swr-calculator Offers a definition of RL in terms of gamma. Far from authoritative. Definition of gamma is in terms of VSWR and VSWR in terms of gamma. http://www.amphenolrf.com/vswr-conversion-chart/ Convention is to give positive values in tables. 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. You didn't cite the relevant section... Sign Properly, loss quantities, when expressed in decibels, should be positive numbers.[note 1] However, return loss has historically been expressed as a negative number, and this convention is still widely found in the literature.[1] 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: This one is a bit confusing. The words in the first sentence say one thing and the rest say something different. I don't see an equation in terms of the power that I can understand without interpreting their programming language. At this point I guess the issue is very muddy with usage in both camps. I think the wiki page indicates that with the section on Sign. What really bugs me is why return loss is in terms of the loss while other loss figures are in terms of the remaining power. Or did I find an anomalous reference for that one? 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. -- Roger Hayter |
Parallel coax
On 9/28/2015 4:33 PM, Jerry Stuckle wrote:
On 9/28/2015 4:15 PM, rickman wrote: On 9/28/2015 4:02 PM, Jerry Stuckle wrote: On 9/28/2015 3:02 PM, rickman wrote: On 9/28/2015 2:51 PM, Jerry Stuckle wrote: On 9/28/2015 1:42 PM, rickman wrote: On 9/28/2015 12:54 PM, Jerry Stuckle wrote: 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: "John S" wrote in message ... 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. # Yes, the SWR due to mismatch of the antenna (load) and line will remain. # Even if the real part of your load impedance is matched to the line, you # will still have a high SWR if the reactance remains. # Does this make sense? Yes. That's what I was trying to say using SWR instead of return loss. Return loss numbers get bigger with lower SWR. For example: SWR 1:1 = infinite return loss. Incorrect. Return loss increases with an increased SWR. An SWR of 1:1 has no return loss because there is no returned signal to lose. 100% of the signal is radiated. 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. I believe that is exactly what I said in the portions of my post which you trimmed. These values for RF return loss match exactly the equation which you are saying is not used in RF. So which is it, the return loss table is correct with negative values of return loss or the equation I posted is incorrect even though it gives the values in the table? You said return loss increases with lower SWR. It does not. Are you being pedantic that -1 dB is not lower than -10 dB? It is not numerically lower in value, but is lower in magnitude and it is a lower loss. I even referred to the magnitude in my post. But then that was the same part you snipped which I referred to earlier. "It shows a higher return loss (assuming you mean magnitude since the values are all negative) for lower VSWR." No, I am not being pedantic. -1 is greater than -10. The fact it is a negative number makes all the difference - as any engineer who knows what he's talking about will tell you. Except that I clearly stated I was referring to the magnitude. Magnitude without direction is meaningless. Lol. There is only one direction for return loss. -- Rick |
Parallel coax
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? -- Rick |
Parallel coax
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. -- Ian |
Parallel coax
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. -- Ian |
Parallel coax
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. I have seen a number of references that give the formula as Pout - Pin. Clearly both forms are in use. I can offer a book that shows a graph of negative RL values. https://books.google.com/books?id=nH...0ratio&f=false -- Rick |
Parallel coax
On 9/28/2015 1:51 PM, Jerry Stuckle wrote:
On 9/28/2015 1:42 PM, rickman wrote: On 9/28/2015 12:54 PM, Jerry Stuckle wrote: 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: "John S" wrote in message ... 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. # Yes, the SWR due to mismatch of the antenna (load) and line will remain. # Even if the real part of your load impedance is matched to the line, you # will still have a high SWR if the reactance remains. # Does this make sense? Yes. That's what I was trying to say using SWR instead of return loss. Return loss numbers get bigger with lower SWR. For example: SWR 1:1 = infinite return loss. Incorrect. Return loss increases with an increased SWR. An SWR of 1:1 has no return loss because there is no returned signal to lose. 100% of the signal is radiated. 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. I believe that is exactly what I said in the portions of my post which you trimmed. These values for RF return loss match exactly the equation which you are saying is not used in RF. So which is it, the return loss table is correct with negative values of return loss or the equation I posted is incorrect even though it gives the values in the table? You said return loss increases with lower SWR. It does not. It does. |
Parallel coax
On 9/28/2015 5:17 PM, rickman wrote:
On 9/28/2015 4:33 PM, Jerry Stuckle wrote: On 9/28/2015 4:15 PM, rickman wrote: On 9/28/2015 4:02 PM, Jerry Stuckle wrote: On 9/28/2015 3:02 PM, rickman wrote: On 9/28/2015 2:51 PM, Jerry Stuckle wrote: On 9/28/2015 1:42 PM, rickman wrote: On 9/28/2015 12:54 PM, Jerry Stuckle wrote: 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: "John S" wrote in message ... 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. # Yes, the SWR due to mismatch of the antenna (load) and line will remain. # Even if the real part of your load impedance is matched to the line, you # will still have a high SWR if the reactance remains. # Does this make sense? Yes. That's what I was trying to say using SWR instead of return loss. Return loss numbers get bigger with lower SWR. For example: SWR 1:1 = infinite return loss. Incorrect. Return loss increases with an increased SWR. An SWR of 1:1 has no return loss because there is no returned signal to lose. 100% of the signal is radiated. 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. I believe that is exactly what I said in the portions of my post which you trimmed. These values for RF return loss match exactly the equation which you are saying is not used in RF. So which is it, the return loss table is correct with negative values of return loss or the equation I posted is incorrect even though it gives the values in the table? You said return loss increases with lower SWR. It does not. Are you being pedantic that -1 dB is not lower than -10 dB? It is not numerically lower in value, but is lower in magnitude and it is a lower loss. I even referred to the magnitude in my post. But then that was the same part you snipped which I referred to earlier. "It shows a higher return loss (assuming you mean magnitude since the values are all negative) for lower VSWR." No, I am not being pedantic. -1 is greater than -10. The fact it is a negative number makes all the difference - as any engineer who knows what he's talking about will tell you. Except that I clearly stated I was referring to the magnitude. Magnitude without direction is meaningless. Lol. There is only one direction for return loss. Yes - the loss is a negative gain - hence the use of negative db values. You can't just give a magnitude. You need to specify the direction, also. You seem to have a very hard time understanding such a simple concept - one that was taught in freshman physics. -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
Parallel coax
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: On 9/28/2015 12:54 PM, Jerry Stuckle wrote: 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: "John S" wrote in message ... 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. # Yes, the SWR due to mismatch of the antenna (load) and line will remain. # Even if the real part of your load impedance is matched to the line, you # will still have a high SWR if the reactance remains. # Does this make sense? Yes. That's what I was trying to say using SWR instead of return loss. Return loss numbers get bigger with lower SWR. For example: SWR 1:1 = infinite return loss. Incorrect. Return loss increases with an increased SWR. An SWR of 1:1 has no return loss because there is no returned signal to lose. 100% of the signal is radiated. 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. I believe that is exactly what I said in the portions of my post which you trimmed. These values for RF return loss match exactly the equation which you are saying is not used in RF. So which is it, the return loss table is correct with negative values of return loss or the equation I posted is incorrect even though it gives the values in the table? You said return loss increases with lower SWR. It does not. It does. Sorry, a lower SWR does not increase the amount of loss. Please cite a reliable reference that says it does. Even the table Rick cited shows a negative value for return SWR. -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
Parallel coax
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. -- ================== Remove the "x" from my email address Jerry Stuckle ================== |
Parallel coax
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. -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
Parallel coax
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. For instance - a system shows a gain and loss of +3, +5, +2, +1. What is the total gain or loss of the system? -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
Parallel coax
On 9/28/2015 7:52 PM, Jerry Stuckle wrote:
On 9/28/2015 5:17 PM, rickman wrote: On 9/28/2015 4:33 PM, Jerry Stuckle wrote: On 9/28/2015 4:15 PM, rickman wrote: On 9/28/2015 4:02 PM, Jerry Stuckle wrote: On 9/28/2015 3:02 PM, rickman wrote: On 9/28/2015 2:51 PM, Jerry Stuckle wrote: On 9/28/2015 1:42 PM, rickman wrote: On 9/28/2015 12:54 PM, Jerry Stuckle wrote: 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: "John S" wrote in message ... 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. # Yes, the SWR due to mismatch of the antenna (load) and line will remain. # Even if the real part of your load impedance is matched to the line, you # will still have a high SWR if the reactance remains. # Does this make sense? Yes. That's what I was trying to say using SWR instead of return loss. Return loss numbers get bigger with lower SWR. For example: SWR 1:1 = infinite return loss. Incorrect. Return loss increases with an increased SWR. An SWR of 1:1 has no return loss because there is no returned signal to lose. 100% of the signal is radiated. 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. I believe that is exactly what I said in the portions of my post which you trimmed. These values for RF return loss match exactly the equation which you are saying is not used in RF. So which is it, the return loss table is correct with negative values of return loss or the equation I posted is incorrect even though it gives the values in the table? You said return loss increases with lower SWR. It does not. Are you being pedantic that -1 dB is not lower than -10 dB? It is not numerically lower in value, but is lower in magnitude and it is a lower loss. I even referred to the magnitude in my post. But then that was the same part you snipped which I referred to earlier. "It shows a higher return loss (assuming you mean magnitude since the values are all negative) for lower VSWR." No, I am not being pedantic. -1 is greater than -10. The fact it is a negative number makes all the difference - as any engineer who knows what he's talking about will tell you. Except that I clearly stated I was referring to the magnitude. Magnitude without direction is meaningless. Lol. There is only one direction for return loss. Yes - the loss is a negative gain - hence the use of negative db values. You can't just give a magnitude. You need to specify the direction, also. You seem to have a very hard time understanding such a simple concept - one that was taught in freshman physics. Ok Jerry. You just said the value can only be negative, so if the magnitude is specified it is a complete specification. I'll toss in the towel as you will never stop arguing such a silly point. -- Rick |
Parallel coax
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? -- Rick |
Parallel coax
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! -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
Parallel coax
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." 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. 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. -- Rick |
Parallel coax
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? |
Parallel coax
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. Please cite a reliable reference that says it does. From the ARRL: http://www.arrl.org/news/amateur-radio-quiz-a-log-of-dbs "9) D -- Higher positive values of Return Loss (RL) in dB indicate less power returning from a load, indicating a lower SWR." |
Parallel coax
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. Please cite a reliable reference that says it does. Even the table Rick cited shows a negative value for return SWR. OH! I see your problem. You think the column heading dashes are minus signs. They are not. Loss - dB indicates that the column data have the units of dB, not that they are negative. I usually use Loss (dB) for my column headings. |
Parallel coax
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. 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'. -- Ian |
Parallel coax
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. 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. -- Ian |
Parallel coax
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. Please cite a reliable reference that says it does. Even the table Rick cited shows a negative value for return SWR. Create your own chart that shows you are incorrect: http://www.jampro.com/uploads/tech_calc/vswr.htm |
Parallel coax
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? -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
Parallel coax
On 9/29/2015 3:36 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. Please cite a reliable reference that says it does. From the ARRL: http://www.arrl.org/news/amateur-radio-quiz-a-log-of-dbs "9) D -- Higher positive values of Return Loss (RL) in dB indicate less power returning from a load, indicating a lower SWR." OK, I'll have to take that up with N0AX. My university professors and the IEEE disagree with him from an engineering view. However, hams often try to make things easier - and not always correctly. -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
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