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#41
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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 ================== |
#42
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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 ================== |
#43
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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 |
#44
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Parallel coax
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 |
#45
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Parallel coax
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 |
#46
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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 |
#47
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Parallel coax
"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: |
#48
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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 ================== |
#49
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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 ================== |
#50
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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 ================== |
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