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Parallel coax
I use a short vertical antenna that has a low feedpoint impedance.
Would there be any advantage to running two parallel 50 ohm coax to reduce losses to the shack tuner? |
Parallel coax
On 9/25/2015 4:31 PM, Wayne wrote:
I use a short vertical antenna that has a low feedpoint impedance. Would there be any advantage to running two parallel 50 ohm coax to reduce losses to the shack tuner? Single band or multi band? How does the feed point change with frequency. 450 ladder line is the usual low loss wire used for multi band systems. But what do know, I've never put up a ham frequency antenna. Mikek |
Parallel coax
On 9/25/2015 4:31 PM, Wayne wrote:
I use a short vertical antenna that has a low feedpoint impedance. Would there be any advantage to running two parallel 50 ohm coax to reduce losses to the shack tuner? If your antenna is 25+j0, paralleling two 50 ohm coax lines would reduce the SWR from 2:1 to 1:1. Is that what you want to know? |
Parallel coax
"John S" wrote in message ... On 9/25/2015 4:31 PM, Wayne wrote: I use a short vertical antenna that has a low feedpoint impedance. Would there be any advantage to running two parallel 50 ohm coax to reduce losses to the shack tuner? If your antenna is 25+j0, paralleling two 50 ohm coax lines would reduce the SWR from 2:1 to 1:1. Is that what you want to know? Yes, more or less. But, I wouldn't even bother trying to reduce a 2:1 unless the rig wouldn't tolerate it. The antenna feedpoint impedance is around 20-j130 at 14 MHz, approximately a 21:1 SWR. Some years back there was a discussion on paralleled coax, and I seem to recall that it wasn't considered to accomplish much even under ideal conditions. |
Parallel coax
On 9/26/2015 2:24 PM, Wayne wrote:
"John S" wrote in message ... On 9/25/2015 4:31 PM, Wayne wrote: I use a short vertical antenna that has a low feedpoint impedance. Would there be any advantage to running two parallel 50 ohm coax to reduce losses to the shack tuner? If your antenna is 25+j0, paralleling two 50 ohm coax lines would reduce the SWR from 2:1 to 1:1. Is that what you want to know? Yes, more or less. But, I wouldn't even bother trying to reduce a 2:1 unless the rig wouldn't tolerate it. The antenna feedpoint impedance is around 20-j130 at 14 MHz, approximately a 21:1 SWR. Some years back there was a discussion on paralleled coax, and I seem to recall that it wasn't considered to accomplish much even under ideal conditions. Oh. Well, then, it won't help. Are you opposed to putting an L and a C up there at the base? |
Parallel coax
Doubling the number of feedlines would double the losses. Not only that
but each connector in the system inserts losses so that number would be 4x. Increasing the coax diameter would effectively reduce the loss. Matching the antenna won't make the Return Loss go away but it will make the transmitter happy. KG7FU On 09/25/2015 02:31 PM, Wayne wrote: I use a short vertical antenna that has a low feedpoint impedance. Would there be any advantage to running two parallel 50 ohm coax to reduce losses to the shack tuner? |
Parallel coax
"John S" wrote in message ... On 9/26/2015 2:24 PM, Wayne wrote: "John S" wrote in message ... On 9/25/2015 4:31 PM, Wayne wrote: I use a short vertical antenna that has a low feedpoint impedance. Would there be any advantage to running two parallel 50 ohm coax to reduce losses to the shack tuner? If your antenna is 25+j0, paralleling two 50 ohm coax lines would reduce the SWR from 2:1 to 1:1. Is that what you want to know? Yes, more or less. But, I wouldn't even bother trying to reduce a 2:1 unless the rig wouldn't tolerate it. The antenna feedpoint impedance is around 20-j130 at 14 MHz, approximately a 21:1 SWR. Some years back there was a discussion on paralleled coax, and I seem to recall that it wasn't considered to accomplish much even under ideal conditions. # Oh. Well, then, it won't help. Are you opposed to putting an L and a C # up there at the base? Yes I'm opposed to that, but not for technical reasons. It's a problem of an old man on a ladder :) |
Parallel coax
On 9/27/2015 10:46 AM, Wayne wrote:
"John S" wrote in message ... On 9/26/2015 2:24 PM, Wayne wrote: "John S" wrote in message ... On 9/25/2015 4:31 PM, Wayne wrote: I use a short vertical antenna that has a low feedpoint impedance. Would there be any advantage to running two parallel 50 ohm coax to reduce losses to the shack tuner? If your antenna is 25+j0, paralleling two 50 ohm coax lines would reduce the SWR from 2:1 to 1:1. Is that what you want to know? Yes, more or less. But, I wouldn't even bother trying to reduce a 2:1 unless the rig wouldn't tolerate it. The antenna feedpoint impedance is around 20-j130 at 14 MHz, approximately a 21:1 SWR. Some years back there was a discussion on paralleled coax, and I seem to recall that it wasn't considered to accomplish much even under ideal conditions. # Oh. Well, then, it won't help. Are you opposed to putting an L and a C # up there at the base? Yes I'm opposed to that, but not for technical reasons. It's a problem of an old man on a ladder :) In that case, how would you manage to add a parallel run of coax? Is that somehow easier? |
Parallel coax
"John S" wrote in message ... On 9/27/2015 10:46 AM, Wayne wrote: "John S" wrote in message ... On 9/26/2015 2:24 PM, Wayne wrote: "John S" wrote in message ... On 9/25/2015 4:31 PM, Wayne wrote: I use a short vertical antenna that has a low feedpoint impedance. Would there be any advantage to running two parallel 50 ohm coax to reduce losses to the shack tuner? If your antenna is 25+j0, paralleling two 50 ohm coax lines would reduce the SWR from 2:1 to 1:1. Is that what you want to know? Yes, more or less. But, I wouldn't even bother trying to reduce a 2:1 unless the rig wouldn't tolerate it. The antenna feedpoint impedance is around 20-j130 at 14 MHz, approximately a 21:1 SWR. Some years back there was a discussion on paralleled coax, and I seem to recall that it wasn't considered to accomplish much even under ideal conditions. # Oh. Well, then, it won't help. Are you opposed to putting an L and a C # up there at the base? Yes I'm opposed to that, but not for technical reasons. It's a problem of an old man on a ladder :) # In that case, how would you manage to add a parallel run of coax? Is # that somehow easier? For the setup, yes that would be easier and can be done by the xyl :) |
Parallel coax
On 9/27/2015 11:50 AM, Wayne wrote:
"John S" wrote in message ... On 9/27/2015 10:46 AM, Wayne wrote: "John S" wrote in message ... On 9/26/2015 2:24 PM, Wayne wrote: "John S" wrote in message ... On 9/25/2015 4:31 PM, Wayne wrote: I use a short vertical antenna that has a low feedpoint impedance. Would there be any advantage to running two parallel 50 ohm coax to reduce losses to the shack tuner? If your antenna is 25+j0, paralleling two 50 ohm coax lines would reduce the SWR from 2:1 to 1:1. Is that what you want to know? Yes, more or less. But, I wouldn't even bother trying to reduce a 2:1 unless the rig wouldn't tolerate it. The antenna feedpoint impedance is around 20-j130 at 14 MHz, approximately a 21:1 SWR. Some years back there was a discussion on paralleled coax, and I seem to recall that it wasn't considered to accomplish much even under ideal conditions. # Oh. Well, then, it won't help. Are you opposed to putting an L and a C # up there at the base? Yes I'm opposed to that, but not for technical reasons. It's a problem of an old man on a ladder :) # In that case, how would you manage to add a parallel run of coax? Is # that somehow easier? For the setup, yes that would be easier and can be done by the xyl :) Well, I can imagine possibilities where a network could be constructed in your shack and then installed by your xyl. But, I won't belabor the subject. I wish you good luck in this quest. |
Parallel coax
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. -- Rick |
Parallel coax
"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. |
Parallel coax
On 9/27/2015 9:41 AM, kg7fu wrote:
Doubling the number of feedlines would double the losses. Not only that but each connector in the system inserts losses so that number would be 4x. How can that be? Each line carries half the power. Connector loss at 14MHz is insignificant. Increasing the coax diameter would effectively reduce the loss. No argument here. Except that it costs much more. Matching the antenna won't make the Return Loss go away but it will make the transmitter happy. Maybe you are working from the assumption that a mismatch between the source and line causes return loss. It does not. Only the mismatch between the load and line causes return loss. KG7FU On 09/25/2015 02:31 PM, Wayne wrote: I use a short vertical antenna that has a low feedpoint impedance. Would there be any advantage to running two parallel 50 ohm coax to reduce losses to the shack tuner? |
Parallel coax
On 9/27/2015 2:20 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. Are you suggesting the ATU is at the transmitter rather than at the antenna? The losses are in the coax and transmitter. If the ATU is at the transmitter you still have coax losses. If the ATU is at the antenna there are then no losses other than due to non-ideality of the ATU. -- Rick |
Parallel coax
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? |
Parallel coax
rickman wrote:
On 9/27/2015 2:20 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. Are you suggesting the ATU is at the transmitter rather than at the antenna? The losses are in the coax and transmitter. If the ATU is at the transmitter you still have coax losses. If the ATU is at the antenna there are then no losses other than due to non-ideality of the ATU. It was stated in the original post the ATU is at the transmitter end. -- Jim Pennino |
Parallel coax
"rickman" wrote in message ... On 9/27/2015 2:20 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. # Are you suggesting the ATU is at the transmitter rather than at the # antenna? Yes. It is a whip with a 15 foot run of 213 to an ATU in the shack. The coax loss is low enough that I can live with a higher SWR. |
Parallel coax
"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. But I assumed that Rick was talking about the reflected power used in the return loss calculation. That part goes to zero for a perfect match, hence the infinite return loss. Since my ATU is closer to the transmitter than the antenna, I tune for lowest SWR from the transmitter to the ATU and don't worry about the ATU to antenna SWR. I believe we are all talking about the same thing. |
Parallel coax
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. But I assumed that Rick was talking about the reflected power used in the return loss calculation. That part goes to zero for a perfect match, hence the infinite return loss. You cannot have a return loss when there is no returned signal. 0 divided by anything is still 0. Since my ATU is closer to the transmitter than the antenna, I tune for lowest SWR from the transmitter to the ATU and don't worry about the ATU to antenna SWR. But that is where the loss occurs. The loss will be dependent on the SWR and the length of the coax. If your coax is short, you won't have a significant loss with a reasonable SWR. I believe we are all talking about the same thing. I don't think so. -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
Parallel coax
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... 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. But I assumed that Rick was talking about the reflected power used in the return loss calculation. That part goes to zero for a perfect match, hence the infinite return loss. You cannot have a return loss when there is no returned signal. 0 divided by anything is still 0. Yes, that is why the value of return loss goes to minus infinity, log of zero is not technically defined, but in the limit, it goes to negative infinity. Since my ATU is closer to the transmitter than the antenna, I tune for lowest SWR from the transmitter to the ATU and don't worry about the ATU to antenna SWR. But that is where the loss occurs. The loss will be dependent on the SWR and the length of the coax. If your coax is short, you won't have a significant loss with a reasonable SWR. Return loss doesn't refer to the loss of signal in the cable. It refers to the loss of signal due to the reflection from the antenna rather than being transferred to the antenna. It is true that some of that signal may be reflected again from the transmitter or other irregularities, but that is not relevant to the return loss measurement. I believe we are all talking about the same thing. I don't think so. I agree that we are not all on the same page. Then some of us much prefer to argue rather than discuss. Is there anything about Waynes post you like? Are the facts more clear now at least? -- Rick |
Parallel coax
On 9/28/2015 2:40 AM, Jeff wrote:
I agree that we are not all on the same page. Then some of us much prefer to argue rather than discuss. Is there anything about Waynes post you like? Are the facts more clear now at least? The situation is that with the ATU at the Tx end there are 2 mismatches to consider. 1, the mismatch at the ATU output ( which hopefully has been adjusted by the ATU to be small to keep the Tx happy.) 2. the mismatch between the antenna and the characteristic impedance of the feeder. Mismatch 2 causes the part of the signal to be reflected back down the feeder towards the ATU which then re-reflects it back to the antenna, which of course then re-reflects part of it back towards the ATU and so on ad infinitum. The ad infinitum can be expressed with a simple ratio. The reflections constitute an infinite series that approaches a limit easily calculable. Regardless, any incident energy reaching the antenna will be reflected in the ratio determined by the impedance mismatch and measured by the VSWR or the return loss ratio. It doesn't matter where the energy comes from so talking bout infinite reflections serves no point when discussing return loss ratio. Of course each time the signal passes along the feeder it suffers the loss due to the feeder. As the amount of signal that is reflected at the antenna depends on the SWR (or return loss) it is obvious that as the mismatch between the feeder and antenna increases more of the power is subjected to multiple trips along the feeder and so encounters more loss. If you had lossless coax there would not be a problem, but even quite low loss in the coax will cause significant loss when the mismatch between the feeder and antenna gets a little high. So getting the feeder impedance closer to that of the antenna can help. Why can't the ATU also match the impedance of the cable eliminating reflections there? -- Rick |
Parallel coax
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. -- ================== Remove the "x" from my email address Jerry Stuckle ================== |
Parallel coax
"Jerry Stuckle" wrote in message ... 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. Return loss is the difference in dB between the forward power and reflected power. Less reflected makes a bigger difference and the return loss goes up. I did a very quick Google and came up with this page that will calculate SWR, return loss, and reflection coefficient. Give it a try with SWR of 3, 2, and 1. http://cgi.www.telestrian.co.uk/cgi-....co.uk/vswr.pl |
Parallel coax
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! -- Ian |
Parallel coax
"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 |
Parallel coax
On 9/28/2015 10:47 AM, Ian Jackson wrote:
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 you are correct. I think the confusion is the word *loss*. If you have a positive *loss* number, the return signal is reduced. To have a negative return loss number, you need to refer to *gain*. For example, a return *loss* of 20dB is the same as a return *gain* of -20dB. |
Parallel coax
On 9/28/2015 7:23 AM, Jeff wrote:
Why can't the ATU also match the impedance of the cable eliminating reflections there? The ATU is only in one location, if it is directly at the antenna feed point the there is only 1 mismatch to tune out. If you put the ATU at the Tx end and have a feeder between the ATU and the antenna then the ATU can do nothing about the mismatch between the antenna and the characteristic impedance of the feeder; all it can do is make whatever impedance there is at the Tx end of the feeder acceptable for the Tx. I wasn't refering to reflections at the antenna. I was referring to the reflections from the ATU to cable interface. The ATU can match both the transmitter and the cable, no? -- Rick |
Parallel coax
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. :) -- Rick |
Parallel coax
On 9/28/2015 11:47 AM, Ian Jackson wrote:
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! The ratio is not larger than one. So the log will be zero or less. In dB the value will be negative for any reflection other than none. -- Rick |
Parallel coax
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. -- Rick |
Parallel coax
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. However, I guess it's just too difficult for you to understand negative numbers and how to relate db to watts. -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
Parallel coax
On 9/28/2015 11:47 AM, Ian Jackson wrote:
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! Ian, Yes, you are correct. But the return signal can never be greater than the incident signal, so the actual ratio of the reflected signal loss to the incident signal must always be 1. And the log of a value 1 is a negative number. And yes, it is specified as a negative quantity because the result of the calculation is a negative quantity. Physicists and engineers specify loss as a negative number rather than confuse issues by changing signs to suit them. Changing signs only leads to problems in later calculations. Non-professionals do change signs because it's often easier to understand that a 10db loss is greater than a 5 db loss. Not so easy to grasp quickly is that a -10db gain is more lossy (less gain) than a -5db gain. -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
Parallel coax
On 9/28/2015 12:05 PM, John S wrote:
On 9/28/2015 10:47 AM, Ian Jackson wrote: 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 you are correct. I think the confusion is the word *loss*. If you have a positive *loss* number, the return signal is reduced. To have a negative return loss number, you need to refer to *gain*. For example, a return *loss* of 20dB is the same as a return *gain* of -20dB. In my other replies on this I mistakenly said the return loss in dB is negative for all but zero reflection. That should have been for all but 100% reflection. :( However, I did a bit more research on the topic and I noticed that Wikipedia says for the general case a loss ratio in decibels should be a positive number. But they have a paragraph explaining that by convention the return loss ratio is a negative number which makes it identical to the reflection coefficient. I could not check the reference wikipedia gives for this as it is a subscription document. I did find in the Electronic Engineers' Handbook a formula for "system" losses starting at the antenna terminal which are expressed as Ls = 10 log (pt/pa) = Pt - Pa, dB where Ls is the loss in decibels, pt is the power delivered to the antenna terminals, pa is the power at the receiving antenna. The capital form of these power levels are in dB. I think the reason we see a different formula for return loss is because in the above equation the ratio is the power sent to the power received. The formula for the return loss is the ratio of the power provided to the power reflected which is not at all the same ratio. To make the loss equations measure the same thing return loss would be Lr = 10 log (pt/(pt-pr)) where pr is the power reflected yielding the same ratio as pt/pa. This would still be a positive value in dB. So I'm not at all sure why the convention of negative dB for return loss got started. Maybe it is one of those issues where there is a lot of misinformation that never gets corrected properly. So I withdraw my statements since the issue is very confused in the literature. Clearly there is no universal convention. -- Rick |
Parallel coax
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? -- Rick |
Parallel coax
"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. |
Parallel coax
In message , Wayne
writes "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 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 Phew - what a relief! -- Ian |
Parallel coax
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. -- Rick |
Parallel coax
"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. |
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. Where do you get the equation? Do you have a reference? -- Rick |
Parallel coax
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. -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
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