|
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 |
| All times are GMT +1. The time now is 01:21 AM. |
Powered by vBulletin® Copyright ©2000 - 2025, Jelsoft Enterprises Ltd.
RadioBanter.com