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Parallel coax
In message , Wayne
writes "Ian Jackson" wrote in message ... In message , rickman writes On 9/30/2015 12:57 PM, John S wrote: On 9/30/2015 10:12 AM, Ian Jackson wrote: In message , Jeff writes On 29/09/2015 14:31, Jerry Stuckle wrote: On 9/29/2015 4:40 AM, Jeff wrote: So let's get back to the original question. Was it ever really answered? I think it was made slightly more complicated by the fact that the antenna feedpoint impedance was not purely resistive, but was actually around 20-j130 (at 14 MHz), Was there any advantage in having the coax paralleled (both for 20 ohms resistive, and for 20-j130)? A quick model. A vertical antenna about 4.2m long with a wire radius of 0.5mm approximates to what you have . The devil is in the j130 If you use a 1.5uH series L the SWR for the parallel 50ohm line is 1.19: 1 and not a bad match over the band Expanding on the original question.... Antenna feedpoint approximately 20-j130 The ATU drives the antenna through about 15 feet of coax. Assuming that the ATU provides a +j130 conjugate match, does that leave the coax with a SWR of 50/20= 2.5:1? If so, then I will not bother with considering 2 parallel coax. No the line SWR is still about 36:1. If you run a bit of poke, you might melt the coax. If you want a single 50 ohm feed, extend the antenna to 5.7m ( ~50+ j87) and use a series C ~130pf to take out the the +j87. You have to do this at the antenna end. It goes without saying you need a good ground. Brian -- Brian Howie |
Parallel coax
On 10/1/2015 4:24 AM, Jeff wrote:
Does your online calculator include cable loss and velocity factor? I used .0233dB/m and .66 velocity factor with a length of 4.57m. Sorry for the metric numbers, but that's what my chart uses. How do any of those things enter into the VSWR calculation? VSWR is defined solely by the match of the cable and antenna impedances. I don't know much about the calculator I used, some random online thing. http://chemandy.com/calculators/retu...calculator.htm Yes they do have an effect. Cable loss will improve the VSWR as the cable gets longer and the loss increases so the VSWR reduces. You are referring to the VSWR at the ATU rather than at the antenna? John already mentioned that. The velocity factor determines how long the cable appears electrically compared to the physical length. The calculator that you linked to takes no account of the cable length and what the mismatch, or vswr, at the load looks like at the other end of a length of cable. If the cable impedance matched the system impedance, 50 ohms in this case, and it were totally lossless then the VSWR at the tx end of the cable would be the same as at the antenna end, BUT the phase of the mismatch would change, ie your -j120 could end up as being +j120 or anywhere in between depending on cable length. Now if the feeder is an impedance other than 50 ohms it act as a transformer and both the real and imaginary parts of the impedance seen at the Tx end will change. Depending on length and what impedance the feeder is it may or may not improve things. Jeff -- Rick |
Parallel coax
On 10/1/2015 3:29 AM, Brian Howie wrote:
In message , Wayne writes "Ian Jackson" wrote in message ... In message , rickman writes On 9/30/2015 12:57 PM, John S wrote: On 9/30/2015 10:12 AM, Ian Jackson wrote: In message , Jeff writes On 29/09/2015 14:31, Jerry Stuckle wrote: On 9/29/2015 4:40 AM, Jeff wrote: So let's get back to the original question. Was it ever really answered? I think it was made slightly more complicated by the fact that the antenna feedpoint impedance was not purely resistive, but was actually around 20-j130 (at 14 MHz), Was there any advantage in having the coax paralleled (both for 20 ohms resistive, and for 20-j130)? A quick model. A vertical antenna about 4.2m long with a wire radius of 0.5mm approximates to what you have . The devil is in the j130 If you use a 1.5uH series L the SWR for the parallel 50ohm line is 1.19: 1 and not a bad match over the band Expanding on the original question.... Antenna feedpoint approximately 20-j130 The ATU drives the antenna through about 15 feet of coax. Assuming that the ATU provides a +j130 conjugate match, does that leave the coax with a SWR of 50/20= 2.5:1? If so, then I will not bother with considering 2 parallel coax. No the line SWR is still about 36:1. If you run a bit of poke, you might melt the coax. If you want a single 50 ohm feed, extend the antenna to 5.7m ( ~50+ j87) and use a series C ~130pf to take out the the +j87. You have to do this at the antenna end. It goes without saying you need a good ground. Brian What do you think of this while leaving his antenna unchanged? 2.3uH ___ '-----o-----UUU--------------- .-. | ^ | | | | 20 | | C| ' '-' C| 4uH | C| 50 ohms | | --- | . -130--- | | | | v '-----o------------------------ (created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de) |
Parallel coax
In message , John S
writes On 10/1/2015 3:29 AM, Brian Howie wrote: In message , Wayne writes "Ian Jackson" wrote in message ... In message , rickman writes On 9/30/2015 12:57 PM, John S wrote: On 9/30/2015 10:12 AM, Ian Jackson wrote: In message , Jeff writes On 29/09/2015 14:31, Jerry Stuckle wrote: On 9/29/2015 4:40 AM, Jeff wrote: So let's get back to the original question. Was it ever really answered? I think it was made slightly more complicated by the fact that the antenna feedpoint impedance was not purely resistive, but was actually around 20-j130 (at 14 MHz), Was there any advantage in having the coax paralleled (both for 20 ohms resistive, and for 20-j130)? A quick model. A vertical antenna about 4.2m long with a wire radius of 0.5mm approximates to what you have . The devil is in the j130 If you use a 1.5uH series L the SWR for the parallel 50ohm line is 1.19: 1 and not a bad match over the band Expanding on the original question.... Antenna feedpoint approximately 20-j130 The ATU drives the antenna through about 15 feet of coax. Assuming that the ATU provides a +j130 conjugate match, does that leave the coax with a SWR of 50/20= 2.5:1? If so, then I will not bother with considering 2 parallel coax. No the line SWR is still about 36:1. If you run a bit of poke, you might melt the coax. If you want a single 50 ohm feed, extend the antenna to 5.7m ( ~50+ j87) and use a series C ~130pf to take out the the +j87. You have to do this at the antenna end. It goes without saying you need a good ground. Brian What do you think of this while leaving his antenna unchanged? 2.3uH ___ '-----o-----UUU--------------- .-. | ^ | | | | 20 | | C| ' '-' C| 4uH | C| 50 ohms | | --- | . -130--- | | | | v '-----o------------------------ (created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de) Yes that works too. Lots of ways to do it. Brian -- Brian Howie |
Parallel coax
On 10/1/2015 11:56 AM, Brian Howie wrote:
In message , John S writes On 10/1/2015 3:29 AM, Brian Howie wrote: In message , Wayne writes "Ian Jackson" wrote in message ... In message , rickman writes On 9/30/2015 12:57 PM, John S wrote: On 9/30/2015 10:12 AM, Ian Jackson wrote: In message , Jeff writes On 29/09/2015 14:31, Jerry Stuckle wrote: On 9/29/2015 4:40 AM, Jeff wrote: So let's get back to the original question. Was it ever really answered? I think it was made slightly more complicated by the fact that the antenna feedpoint impedance was not purely resistive, but was actually around 20-j130 (at 14 MHz), Was there any advantage in having the coax paralleled (both for 20 ohms resistive, and for 20-j130)? A quick model. A vertical antenna about 4.2m long with a wire radius of 0.5mm approximates to what you have . The devil is in the j130 If you use a 1.5uH series L the SWR for the parallel 50ohm line is 1.19: 1 and not a bad match over the band Expanding on the original question.... Antenna feedpoint approximately 20-j130 The ATU drives the antenna through about 15 feet of coax. Assuming that the ATU provides a +j130 conjugate match, does that leave the coax with a SWR of 50/20= 2.5:1? If so, then I will not bother with considering 2 parallel coax. No the line SWR is still about 36:1. If you run a bit of poke, you might melt the coax. If you want a single 50 ohm feed, extend the antenna to 5.7m ( ~50+ j87) and use a series C ~130pf to take out the the +j87. You have to do this at the antenna end. It goes without saying you need a good ground. Brian What do you think of this while leaving his antenna unchanged? 2.3uH ___ '-----o-----UUU--------------- .-. | ^ | | | | 20 | | C| ' '-' C| 4uH | C| 50 ohms | | --- | . -130--- | | | | v '-----o------------------------ (created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de) Yes that works too. Lots of ways to do it. Brian Oh, yes of course. I try to avoid capacitors as much as I can because one end can float with static voltage while an inductor does not. I've had issues with nearby static lightning discharges. Just my paranoia. To each his own. |
Parallel coax
On 10/1/2015 1:09 PM, John S wrote:
On 10/1/2015 11:56 AM, Brian Howie wrote: In message , John S writes On 10/1/2015 3:29 AM, Brian Howie wrote: In message , Wayne writes "Ian Jackson" wrote in message ... In message , rickman writes On 9/30/2015 12:57 PM, John S wrote: On 9/30/2015 10:12 AM, Ian Jackson wrote: In message , Jeff writes On 29/09/2015 14:31, Jerry Stuckle wrote: On 9/29/2015 4:40 AM, Jeff wrote: So let's get back to the original question. Was it ever really answered? I think it was made slightly more complicated by the fact that the antenna feedpoint impedance was not purely resistive, but was actually around 20-j130 (at 14 MHz), Was there any advantage in having the coax paralleled (both for 20 ohms resistive, and for 20-j130)? A quick model. A vertical antenna about 4.2m long with a wire radius of 0.5mm approximates to what you have . The devil is in the j130 If you use a 1.5uH series L the SWR for the parallel 50ohm line is 1.19: 1 and not a bad match over the band Expanding on the original question.... Antenna feedpoint approximately 20-j130 The ATU drives the antenna through about 15 feet of coax. Assuming that the ATU provides a +j130 conjugate match, does that leave the coax with a SWR of 50/20= 2.5:1? If so, then I will not bother with considering 2 parallel coax. No the line SWR is still about 36:1. If you run a bit of poke, you might melt the coax. If you want a single 50 ohm feed, extend the antenna to 5.7m ( ~50+ j87) and use a series C ~130pf to take out the the +j87. You have to do this at the antenna end. It goes without saying you need a good ground. Brian What do you think of this while leaving his antenna unchanged? 2.3uH ___ '-----o-----UUU--------------- .-. | ^ | | | | 20 | | C| ' '-' C| 4uH | C| 50 ohms | | --- | . -130--- | | | | v '-----o------------------------ (created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de) Yes that works too. Lots of ways to do it. Brian Oh, yes of course. I try to avoid capacitors as much as I can because one end can float with static voltage while an inductor does not. I've had issues with nearby static lightning discharges. Just my paranoia. To each his own. But it is connected by a 20 ohm resistor. How bad can that be? -- Rick |
Parallel coax
On 10/1/2015 12:18 PM, rickman wrote:
On 10/1/2015 1:09 PM, John S wrote: On 10/1/2015 11:56 AM, Brian Howie wrote: In message , John S writes On 10/1/2015 3:29 AM, Brian Howie wrote: In message , Wayne writes "Ian Jackson" wrote in message ... In message , rickman writes On 9/30/2015 12:57 PM, John S wrote: On 9/30/2015 10:12 AM, Ian Jackson wrote: In message , Jeff writes On 29/09/2015 14:31, Jerry Stuckle wrote: On 9/29/2015 4:40 AM, Jeff wrote: So let's get back to the original question. Was it ever really answered? I think it was made slightly more complicated by the fact that the antenna feedpoint impedance was not purely resistive, but was actually around 20-j130 (at 14 MHz), Was there any advantage in having the coax paralleled (both for 20 ohms resistive, and for 20-j130)? A quick model. A vertical antenna about 4.2m long with a wire radius of 0.5mm approximates to what you have . The devil is in the j130 If you use a 1.5uH series L the SWR for the parallel 50ohm line is 1.19: 1 and not a bad match over the band Expanding on the original question.... Antenna feedpoint approximately 20-j130 The ATU drives the antenna through about 15 feet of coax. Assuming that the ATU provides a +j130 conjugate match, does that leave the coax with a SWR of 50/20= 2.5:1? If so, then I will not bother with considering 2 parallel coax. No the line SWR is still about 36:1. If you run a bit of poke, you might melt the coax. If you want a single 50 ohm feed, extend the antenna to 5.7m ( ~50+ j87) and use a series C ~130pf to take out the the +j87. You have to do this at the antenna end. It goes without saying you need a good ground. Brian What do you think of this while leaving his antenna unchanged? 2.3uH ___ '-----o-----UUU--------------- .-. | ^ | | | | 20 | | C| ' '-' C| 4uH | C| 50 ohms | | --- | . -130--- | | | | v '-----o------------------------ (created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de) Yes that works too. Lots of ways to do it. Brian Oh, yes of course. I try to avoid capacitors as much as I can because one end can float with static voltage while an inductor does not. I've had issues with nearby static lightning discharges. Just my paranoia. To each his own. But it is connected by a 20 ohm resistor. How bad can that be? I don't understand. Do you mean the antenna's feed point resistance of 20 ohms? My understanding of the installation is that the antenna is not directly connected to ground. Am I off track here? |
Parallel coax
On 10/1/2015 1:21 PM, John S wrote:
On 10/1/2015 12:18 PM, rickman wrote: On 10/1/2015 1:09 PM, John S wrote: On 10/1/2015 11:56 AM, Brian Howie wrote: In message , John S writes On 10/1/2015 3:29 AM, Brian Howie wrote: In message , Wayne writes "Ian Jackson" wrote in message ... In message , rickman writes On 9/30/2015 12:57 PM, John S wrote: On 9/30/2015 10:12 AM, Ian Jackson wrote: In message , Jeff writes On 29/09/2015 14:31, Jerry Stuckle wrote: On 9/29/2015 4:40 AM, Jeff wrote: So let's get back to the original question. Was it ever really answered? I think it was made slightly more complicated by the fact that the antenna feedpoint impedance was not purely resistive, but was actually around 20-j130 (at 14 MHz), Was there any advantage in having the coax paralleled (both for 20 ohms resistive, and for 20-j130)? A quick model. A vertical antenna about 4.2m long with a wire radius of 0.5mm approximates to what you have . The devil is in the j130 If you use a 1.5uH series L the SWR for the parallel 50ohm line is 1.19: 1 and not a bad match over the band Expanding on the original question.... Antenna feedpoint approximately 20-j130 The ATU drives the antenna through about 15 feet of coax. Assuming that the ATU provides a +j130 conjugate match, does that leave the coax with a SWR of 50/20= 2.5:1? If so, then I will not bother with considering 2 parallel coax. No the line SWR is still about 36:1. If you run a bit of poke, you might melt the coax. If you want a single 50 ohm feed, extend the antenna to 5.7m ( ~50+ j87) and use a series C ~130pf to take out the the +j87. You have to do this at the antenna end. It goes without saying you need a good ground. Brian What do you think of this while leaving his antenna unchanged? 2.3uH ___ '-----o-----UUU--------------- .-. | ^ | | | | 20 | | C| ' '-' C| 4uH | C| 50 ohms | | --- | . -130--- | | | | v '-----o------------------------ (created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de) Yes that works too. Lots of ways to do it. Brian Oh, yes of course. I try to avoid capacitors as much as I can because one end can float with static voltage while an inductor does not. I've had issues with nearby static lightning discharges. Just my paranoia. To each his own. But it is connected by a 20 ohm resistor. How bad can that be? I don't understand. Do you mean the antenna's feed point resistance of 20 ohms? My understanding of the installation is that the antenna is not directly connected to ground. Am I off track here? My bad. I didn't realize that was the antenna. But the capacitor could be bypassed with a large value resistor if static charge is your concern. A kohm should do the job without impacting the circuit significantly. But wait! Isn't the -130 cap also the antenna then? -- Rick |
Parallel coax
In message , John S
writes On 10/1/2015 11:56 AM, Brian Howie wrote: In message , John S writes On 10/1/2015 3:29 AM, Brian Howie wrote: In message , Wayne writes "Ian Jackson" wrote in message ... In message , rickman writes On 9/30/2015 12:57 PM, John S wrote: On 9/30/2015 10:12 AM, Ian Jackson wrote: In message , Jeff writes On 29/09/2015 14:31, Jerry Stuckle wrote: On 9/29/2015 4:40 AM, Jeff wrote: So let's get back to the original question. Was it ever really answered? I think it was made slightly more complicated by the fact that the antenna feedpoint impedance was not purely resistive, but was actually around 20-j130 (at 14 MHz), Was there any advantage in having the coax paralleled (both for 20 ohms resistive, and for 20-j130)? A quick model. A vertical antenna about 4.2m long with a wire radius of 0.5mm approximates to what you have . The devil is in the j130 If you use a 1.5uH series L the SWR for the parallel 50ohm line is 1.19: 1 and not a bad match over the band Expanding on the original question.... Antenna feedpoint approximately 20-j130 The ATU drives the antenna through about 15 feet of coax. Assuming that the ATU provides a +j130 conjugate match, does that leave the coax with a SWR of 50/20= 2.5:1? If so, then I will not bother with considering 2 parallel coax. No the line SWR is still about 36:1. If you run a bit of poke, you might melt the coax. If you want a single 50 ohm feed, extend the antenna to 5.7m ( ~50+ j87) and use a series C ~130pf to take out the the +j87. You have to do this at the antenna end. It goes without saying you need a good ground. Brian What do you think of this while leaving his antenna unchanged? 2.3uH ___ '-----o-----UUU--------------- .-. | ^ | | | | 20 | | C| ' '-' C| 4uH | C| 50 ohms | | --- | . -130--- | | | | v '-----o------------------------ (created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de) Yes that works too. Lots of ways to do it. Brian Oh, yes of course. I try to avoid capacitors as much as I can because one end can float with static voltage while an inductor does not. I've had issues with nearby static lightning discharges. Just my paranoia. To each his own. Good point. Brian -- Brian Howie |
Parallel coax
On 10/1/2015 12:27 PM, rickman wrote:
On 10/1/2015 1:21 PM, John S wrote: On 10/1/2015 12:18 PM, rickman wrote: On 10/1/2015 1:09 PM, John S wrote: On 10/1/2015 11:56 AM, Brian Howie wrote: In message , John S writes On 10/1/2015 3:29 AM, Brian Howie wrote: In message , Wayne writes "Ian Jackson" wrote in message ... In message , rickman writes On 9/30/2015 12:57 PM, John S wrote: On 9/30/2015 10:12 AM, Ian Jackson wrote: In message , Jeff writes On 29/09/2015 14:31, Jerry Stuckle wrote: On 9/29/2015 4:40 AM, Jeff wrote: So let's get back to the original question. Was it ever really answered? I think it was made slightly more complicated by the fact that the antenna feedpoint impedance was not purely resistive, but was actually around 20-j130 (at 14 MHz), Was there any advantage in having the coax paralleled (both for 20 ohms resistive, and for 20-j130)? A quick model. A vertical antenna about 4.2m long with a wire radius of 0.5mm approximates to what you have . The devil is in the j130 If you use a 1.5uH series L the SWR for the parallel 50ohm line is 1.19: 1 and not a bad match over the band Expanding on the original question.... Antenna feedpoint approximately 20-j130 The ATU drives the antenna through about 15 feet of coax. Assuming that the ATU provides a +j130 conjugate match, does that leave the coax with a SWR of 50/20= 2.5:1? If so, then I will not bother with considering 2 parallel coax. No the line SWR is still about 36:1. If you run a bit of poke, you might melt the coax. If you want a single 50 ohm feed, extend the antenna to 5.7m ( ~50+ j87) and use a series C ~130pf to take out the the +j87. You have to do this at the antenna end. It goes without saying you need a good ground. Brian What do you think of this while leaving his antenna unchanged? 2.3uH ___ '-----o-----UUU--------------- .-. | ^ | | | | 20 | | C| ' '-' C| 4uH | C| 50 ohms | | --- | . -130--- | | | | v '-----o------------------------ (created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de) Yes that works too. Lots of ways to do it. Brian Oh, yes of course. I try to avoid capacitors as much as I can because one end can float with static voltage while an inductor does not. I've had issues with nearby static lightning discharges. Just my paranoia. To each his own. But it is connected by a 20 ohm resistor. How bad can that be? I don't understand. Do you mean the antenna's feed point resistance of 20 ohms? My understanding of the installation is that the antenna is not directly connected to ground. Am I off track here? My bad. I didn't realize that was the antenna. But the capacitor could be bypassed with a large value resistor if static charge is your concern. A kohm should do the job without impacting the circuit significantly. But wait! Isn't the -130 cap also the antenna then? Yes. I guess I should have enclosed the combination in a box to represent the antenna. Sorry. |
Parallel coax
In message , Brian Howie
writes In message , John S writes On 10/1/2015 3:29 AM, Brian Howie wrote: In message , Wayne writes "Ian Jackson" wrote in message ... In message , rickman writes On 9/30/2015 12:57 PM, John S wrote: On 9/30/2015 10:12 AM, Ian Jackson wrote: In message , Jeff writes On 29/09/2015 14:31, Jerry Stuckle wrote: On 9/29/2015 4:40 AM, Jeff wrote: So let's get back to the original question. Was it ever really answered? I think it was made slightly more complicated by the fact that the antenna feedpoint impedance was not purely resistive, but was actually around 20-j130 (at 14 MHz), Was there any advantage in having the coax paralleled (both for 20 ohms resistive, and for 20-j130)? A quick model. A vertical antenna about 4.2m long with a wire radius of 0.5mm approximates to what you have . The devil is in the j130 If you use a 1.5uH series L the SWR for the parallel 50ohm line is 1.19: 1 and not a bad match over the band Expanding on the original question.... Antenna feedpoint approximately 20-j130 The ATU drives the antenna through about 15 feet of coax. Assuming that the ATU provides a +j130 conjugate match, does that leave the coax with a SWR of 50/20= 2.5:1? If so, then I will not bother with considering 2 parallel coax. No the line SWR is still about 36:1. If you run a bit of poke, you might melt the coax. If you want a single 50 ohm feed, extend the antenna to 5.7m ( ~50+ j87) and use a series C ~130pf to take out the the +j87. You have to do this at the antenna end. It goes without saying you need a good ground. Brian What do you think of this while leaving his antenna unchanged? 2.3uH ___ '-----o-----UUU--------------- .-. | ^ | | | | 20 | | C| ' '-' C| 4uH | C| 50 ohms | | --- | . -130--- | | | | v '-----o------------------------ (created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de) Yes that works too. Lots of ways to do it. I'm taking the opportunity to refresh my admittedly rudimentary skills with the Smith Chart - and in particular, single- and double-stub matching (little used for over 50 years!!). However, in Wayne's situation, the length of the coax is only 15'. On 14MHz, that's just over a quarterwave (taking the velocity factor into account). But even with a horrendous SWR, how much loss does this length of 213 coax have? It might be a lot more convenient to do all the matching in the warmth and comfort of the shack. I have done exactly this with a 130' inverted-L Marconi-type antenna, fed at the far end directly with around 100' of old (early 1960s) semi-airspaced TV trunk cable (with a good ground there). It worked fine on 160-80-40m (the bands I was interested in working), but it loaded up fine up to 10m - and as it seemed lively enough on receive, I'm sure it would have put out a reasonable signal. Although I eventually treated myself to a remote automatic ATU, I'm not convinced the system works any better than it did with the direct coax connection. -- Ian |
Parallel coax
On 10/1/2015 2:37 PM, John S wrote:
On 10/1/2015 12:27 PM, rickman wrote: On 10/1/2015 1:21 PM, John S wrote: On 10/1/2015 12:18 PM, rickman wrote: On 10/1/2015 1:09 PM, John S wrote: On 10/1/2015 11:56 AM, Brian Howie wrote: In message , John S writes On 10/1/2015 3:29 AM, Brian Howie wrote: In message , Wayne writes "Ian Jackson" wrote in message ... In message , rickman writes On 9/30/2015 12:57 PM, John S wrote: On 9/30/2015 10:12 AM, Ian Jackson wrote: In message , Jeff writes On 29/09/2015 14:31, Jerry Stuckle wrote: On 9/29/2015 4:40 AM, Jeff wrote: So let's get back to the original question. Was it ever really answered? I think it was made slightly more complicated by the fact that the antenna feedpoint impedance was not purely resistive, but was actually around 20-j130 (at 14 MHz), Was there any advantage in having the coax paralleled (both for 20 ohms resistive, and for 20-j130)? A quick model. A vertical antenna about 4.2m long with a wire radius of 0.5mm approximates to what you have . The devil is in the j130 If you use a 1.5uH series L the SWR for the parallel 50ohm line is 1.19: 1 and not a bad match over the band Expanding on the original question.... Antenna feedpoint approximately 20-j130 The ATU drives the antenna through about 15 feet of coax. Assuming that the ATU provides a +j130 conjugate match, does that leave the coax with a SWR of 50/20= 2.5:1? If so, then I will not bother with considering 2 parallel coax. No the line SWR is still about 36:1. If you run a bit of poke, you might melt the coax. If you want a single 50 ohm feed, extend the antenna to 5.7m ( ~50+ j87) and use a series C ~130pf to take out the the +j87. You have to do this at the antenna end. It goes without saying you need a good ground. Brian What do you think of this while leaving his antenna unchanged? 2.3uH ___ '-----o-----UUU--------------- .-. | ^ | | | | 20 | | C| ' '-' C| 4uH | C| 50 ohms | | --- | . -130--- | | | | v '-----o------------------------ (created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de) Yes that works too. Lots of ways to do it. Brian Oh, yes of course. I try to avoid capacitors as much as I can because one end can float with static voltage while an inductor does not. I've had issues with nearby static lightning discharges. Just my paranoia. To each his own. But it is connected by a 20 ohm resistor. How bad can that be? I don't understand. Do you mean the antenna's feed point resistance of 20 ohms? My understanding of the installation is that the antenna is not directly connected to ground. Am I off track here? My bad. I didn't realize that was the antenna. But the capacitor could be bypassed with a large value resistor if static charge is your concern. A kohm should do the job without impacting the circuit significantly. But wait! Isn't the -130 cap also the antenna then? Yes. I guess I should have enclosed the combination in a box to represent the antenna. Sorry. So there is no cap, right? No cap, no worry. Oh, wait again. I see there are two circuits being discussed. So *that* cap can be bypassed with a 1 kohm resistor and not impact the circuit, right? -- Rick |
Parallel coax
On 10/1/2015 2:41 PM, rickman wrote:
On 10/1/2015 2:37 PM, John S wrote: On 10/1/2015 12:27 PM, rickman wrote: On 10/1/2015 1:21 PM, John S wrote: On 10/1/2015 12:18 PM, rickman wrote: On 10/1/2015 1:09 PM, John S wrote: On 10/1/2015 11:56 AM, Brian Howie wrote: In message , John S writes On 10/1/2015 3:29 AM, Brian Howie wrote: In message , Wayne writes "Ian Jackson" wrote in message ... In message , rickman writes On 9/30/2015 12:57 PM, John S wrote: On 9/30/2015 10:12 AM, Ian Jackson wrote: In message , Jeff writes On 29/09/2015 14:31, Jerry Stuckle wrote: On 9/29/2015 4:40 AM, Jeff wrote: So let's get back to the original question. Was it ever really answered? I think it was made slightly more complicated by the fact that the antenna feedpoint impedance was not purely resistive, but was actually around 20-j130 (at 14 MHz), Was there any advantage in having the coax paralleled (both for 20 ohms resistive, and for 20-j130)? A quick model. A vertical antenna about 4.2m long with a wire radius of 0.5mm approximates to what you have . The devil is in the j130 If you use a 1.5uH series L the SWR for the parallel 50ohm line is 1.19: 1 and not a bad match over the band Expanding on the original question.... Antenna feedpoint approximately 20-j130 The ATU drives the antenna through about 15 feet of coax. Assuming that the ATU provides a +j130 conjugate match, does that leave the coax with a SWR of 50/20= 2.5:1? If so, then I will not bother with considering 2 parallel coax. No the line SWR is still about 36:1. If you run a bit of poke, you might melt the coax. If you want a single 50 ohm feed, extend the antenna to 5.7m ( ~50+ j87) and use a series C ~130pf to take out the the +j87. You have to do this at the antenna end. It goes without saying you need a good ground. Brian What do you think of this while leaving his antenna unchanged? 2.3uH ___ '-----o-----UUU--------------- .-. | ^ | | | | 20 | | C| ' '-' C| 4uH | C| 50 ohms | | --- | . -130--- | | | | v '-----o------------------------ (created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de) Yes that works too. Lots of ways to do it. Brian Oh, yes of course. I try to avoid capacitors as much as I can because one end can float with static voltage while an inductor does not. I've had issues with nearby static lightning discharges. Just my paranoia. To each his own. But it is connected by a 20 ohm resistor. How bad can that be? I don't understand. Do you mean the antenna's feed point resistance of 20 ohms? My understanding of the installation is that the antenna is not directly connected to ground. Am I off track here? My bad. I didn't realize that was the antenna. But the capacitor could be bypassed with a large value resistor if static charge is your concern. A kohm should do the job without impacting the circuit significantly. But wait! Isn't the -130 cap also the antenna then? Yes. I guess I should have enclosed the combination in a box to represent the antenna. Sorry. So there is no cap, right? No cap, no worry. Oh, wait again. I see there are two circuits being discussed. So *that* cap can be bypassed with a 1 kohm resistor and not impact the circuit, right? No, there is only one cap. The 20 ohm resistor in series with the -130 ohm cap represents the antenna impedance of 20 - j130 ohms. |
Parallel coax
On 10/1/2015 1:56 PM, Ian Jackson wrote:
In message , Brian Howie writes In message , John S writes On 10/1/2015 3:29 AM, Brian Howie wrote: In message , Wayne writes "Ian Jackson" wrote in message ... In message , rickman writes On 9/30/2015 12:57 PM, John S wrote: On 9/30/2015 10:12 AM, Ian Jackson wrote: In message , Jeff writes On 29/09/2015 14:31, Jerry Stuckle wrote: On 9/29/2015 4:40 AM, Jeff wrote: So let's get back to the original question. Was it ever really answered? I think it was made slightly more complicated by the fact that the antenna feedpoint impedance was not purely resistive, but was actually around 20-j130 (at 14 MHz), Was there any advantage in having the coax paralleled (both for 20 ohms resistive, and for 20-j130)? A quick model. A vertical antenna about 4.2m long with a wire radius of 0.5mm approximates to what you have . The devil is in the j130 If you use a 1.5uH series L the SWR for the parallel 50ohm line is 1.19: 1 and not a bad match over the band Expanding on the original question.... Antenna feedpoint approximately 20-j130 The ATU drives the antenna through about 15 feet of coax. Assuming that the ATU provides a +j130 conjugate match, does that leave the coax with a SWR of 50/20= 2.5:1? If so, then I will not bother with considering 2 parallel coax. No the line SWR is still about 36:1. If you run a bit of poke, you might melt the coax. If you want a single 50 ohm feed, extend the antenna to 5.7m ( ~50+ j87) and use a series C ~130pf to take out the the +j87. You have to do this at the antenna end. It goes without saying you need a good ground. Brian What do you think of this while leaving his antenna unchanged? 2.3uH ___ '-----o-----UUU--------------- .-. | ^ | | | | 20 | | C| ' '-' C| 4uH | C| 50 ohms | | --- | . -130--- | | | | v '-----o------------------------ (created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de) Yes that works too. Lots of ways to do it. I'm taking the opportunity to refresh my admittedly rudimentary skills with the Smith Chart - and in particular, single- and double-stub matching (little used for over 50 years!!). However, in Wayne's situation, the length of the coax is only 15'. On 14MHz, that's just over a quarterwave (taking the velocity factor into account). But even with a horrendous SWR, how much loss does this length of 213 coax have? It might be a lot more convenient to do all the matching in the warmth and comfort of the shack. He is already doing that with his ATU and is pretty happy with it. And he has indicated that he doesn't want to add matching components up at the feed point. Brian and I were just bouncing ideas back and forth. I have done exactly this with a 130' inverted-L Marconi-type antenna, fed at the far end directly with around 100' of old (early 1960s) semi-airspaced TV trunk cable (with a good ground there). It worked fine on 160-80-40m (the bands I was interested in working), but it loaded up fine up to 10m - and as it seemed lively enough on receive, I'm sure it would have put out a reasonable signal. Although I eventually treated myself to a remote automatic ATU, I'm not convinced the system works any better than it did with the direct coax connection. |
Parallel coax
On 10/1/2015 9:16 PM, John S wrote:
On 10/1/2015 2:41 PM, rickman wrote: On 10/1/2015 2:37 PM, John S wrote: On 10/1/2015 12:27 PM, rickman wrote: On 10/1/2015 1:21 PM, John S wrote: On 10/1/2015 12:18 PM, rickman wrote: On 10/1/2015 1:09 PM, John S wrote: On 10/1/2015 11:56 AM, Brian Howie wrote: In message , John S writes On 10/1/2015 3:29 AM, Brian Howie wrote: In message , Wayne writes "Ian Jackson" wrote in message ... In message , rickman writes On 9/30/2015 12:57 PM, John S wrote: On 9/30/2015 10:12 AM, Ian Jackson wrote: In message , Jeff writes On 29/09/2015 14:31, Jerry Stuckle wrote: On 9/29/2015 4:40 AM, Jeff wrote: So let's get back to the original question. Was it ever really answered? I think it was made slightly more complicated by the fact that the antenna feedpoint impedance was not purely resistive, but was actually around 20-j130 (at 14 MHz), Was there any advantage in having the coax paralleled (both for 20 ohms resistive, and for 20-j130)? A quick model. A vertical antenna about 4.2m long with a wire radius of 0.5mm approximates to what you have . The devil is in the j130 If you use a 1.5uH series L the SWR for the parallel 50ohm line is 1.19: 1 and not a bad match over the band Expanding on the original question.... Antenna feedpoint approximately 20-j130 The ATU drives the antenna through about 15 feet of coax. Assuming that the ATU provides a +j130 conjugate match, does that leave the coax with a SWR of 50/20= 2.5:1? If so, then I will not bother with considering 2 parallel coax. No the line SWR is still about 36:1. If you run a bit of poke, you might melt the coax. If you want a single 50 ohm feed, extend the antenna to 5.7m ( ~50+ j87) and use a series C ~130pf to take out the the +j87. You have to do this at the antenna end. It goes without saying you need a good ground. Brian What do you think of this while leaving his antenna unchanged? 2.3uH ___ '-----o-----UUU--------------- .-. | ^ | | | | 20 | | C| ' '-' C| 4uH | C| 50 ohms | | --- | . -130--- | | | | v '-----o------------------------ (created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de) Yes that works too. Lots of ways to do it. Brian Oh, yes of course. I try to avoid capacitors as much as I can because one end can float with static voltage while an inductor does not. I've had issues with nearby static lightning discharges. Just my paranoia. To each his own. But it is connected by a 20 ohm resistor. How bad can that be? I don't understand. Do you mean the antenna's feed point resistance of 20 ohms? My understanding of the installation is that the antenna is not directly connected to ground. Am I off track here? My bad. I didn't realize that was the antenna. But the capacitor could be bypassed with a large value resistor if static charge is your concern. A kohm should do the job without impacting the circuit significantly. But wait! Isn't the -130 cap also the antenna then? Yes. I guess I should have enclosed the combination in a box to represent the antenna. Sorry. So there is no cap, right? No cap, no worry. Oh, wait again. I see there are two circuits being discussed. So *that* cap can be bypassed with a 1 kohm resistor and not impact the circuit, right? No, there is only one cap. The 20 ohm resistor in series with the -130 ohm cap represents the antenna impedance of 20 - j130 ohms. Way back in the thread... "If you want a single 50 ohm feed, extend the antenna to 5.7m (~50+ j87) and use a series C ~130pf to take out the the +j87. You have to do this at the antenna end." Otherwise there is no cap to worry about blowing up. -- Rick |
Parallel coax
On 10/2/2015 2:34 AM, Jeff wrote:
___ '-----o-----UUU--------------- .-. | ^ | | | | 20 | | C| ' '-' C| 4uH | C| 50 ohms | | --- | . -130--- | | | | v '-----o------------------------ (created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de) Yes that works too. Lots of ways to do it. Brian Oh, yes of course. I try to avoid capacitors as much as I can because one end can float with static voltage while an inductor does not. I've had issues with nearby static lightning discharges. Just my paranoia. To each his own. But it is connected by a 20 ohm resistor. How bad can that be? I don't understand. Do you mean the antenna's feed point resistance of 20 ohms? My understanding of the installation is that the antenna is not directly connected to ground. Am I off track here? My bad. I didn't realize that was the antenna. But the capacitor could be bypassed with a large value resistor if static charge is your concern. A kohm should do the job without impacting the circuit significantly. But wait! Isn't the -130 cap also the antenna then? What capacitor??????????? The 20-j130 is the impedance presented by the antenna at the feed point to gnd, there is no physical capacitor, the 4uH to gnd will provide a dc path for any static. Yes, that is what I said. "Isn't the -130 cap also the antenna then?" I was confused because of poor trimming when someone posted about a different matching network using a capacitor and a reply saying they don't like capacitors because they fail. With the context not being clear I thought they were talking about the capacitor in the diagram with the inductive matching network, but as you say, this capacitance is just part of the antenna. -- Rick |
Parallel coax
On 10/2/2015 8:14 AM, Jeff wrote:
On 02/10/2015 07:47, Jeff wrote: However, in Wayne's situation, the length of the coax is only 15'. On 14MHz, that's just over a quarterwave (taking the velocity factor into account). But even with a horrendous SWR, how much loss does this length of 213 coax have? It might be a lot more convenient to do all the matching in the warmth and comfort of the shack. You still have to bear in mind that the 20-j130 load gives something like a 20:1 mismatch at the interface between the coax and the load. This means that only about 17% of your froward power will reach the antenna on its 1st trip up and down the coax, the other 83% will be reflected back down the coax towards the ATU and suffer the loos in the coax again. This will be repeated again and again, with 83% of what is re-reflected being re-reflected again, and so on. So you can see that even with a small loss in the coax if is amplified by the multiple trips up and down the coax. This higher the VSWR the more power is re-reflected to undergo multiple losses. Jeff Having done a few quick calculations; with a vswr of 20:1 at the antenna, and a feeder having a loss of 0.5dB, and an ATU matching at the Tx end, the total power wasted in the cable with a 100W transmitter is over 55W, with only about 45W actually reaching the antenna. Even reducing the cable loss to 0.1dB results in about 20W being lost in the cable. Reducing the VSWR reduces the loss significantly. What is the reflection coefficient at the ATU/feedline interface? -- Rick |
Parallel coax
On 10/3/2015 1:29 AM, Jeff wrote:
Having done a few quick calculations; with a vswr of 20:1 at the antenna, and a feeder having a loss of 0.5dB, and an ATU matching at the Tx end, the total power wasted in the cable with a 100W transmitter is over 55W, with only about 45W actually reaching the antenna. Even reducing the cable loss to 0.1dB results in about 20W being lost in the cable. Reducing the VSWR reduces the loss significantly. What is the reflection coefficient at the ATU/feedline interface? Assuming a 'perfect' ATU and a prefect conjugate match, giving 1:1 at the TX, then it is 1. ie all of the reflected power that reaches the ATU is re-reflected back up towards the antenna. Is that good? If the cable length is 1/4 wavelength (as it is in the info provided for this case) the reflected power is nearly 180 degrees out of phase with the initial power at the antenna. I think reducing this through cable losses would not be so bad, or better to dump it in the ATU? -- Rick |
Parallel coax
On 10/4/2015 4:48 AM, Jeff wrote:
What is the reflection coefficient at the ATU/feedline interface? Assuming a 'perfect' ATU and a prefect conjugate match, giving 1:1 at the TX, then it is 1. ie all of the reflected power that reaches the ATU is re-reflected back up towards the antenna. Is that good? If the cable length is 1/4 wavelength (as it is in the info provided for this case) the reflected power is nearly 180 degrees out of phase with the initial power at the antenna. I think reducing this through cable losses would not be so bad, or better to dump it in the ATU? It is good in as much as some of the re-reflected power is radiated (and some re-re-reflected) since the ATU causes the phase of the re-reflected wave to be 'in-phase' at the antenna. It is bad in as much as the reflected power suffers 2 times the cable loss, and dissipates that in heat, on each return trip, up and down the coax. With a high VSWR at the antenna there will be many return trips before the re-reflected power drops to a negligible level. No power is 'dumped' in the ATU; although there will be losses, but that is another story. If not the ATU, then the transmitter. I'm sure not all of the power is reflected back from the ATU. Exactly what is the phase of the reflected power from the ATU? I haven't seen an actual circuit for the ATU in question. For that matter, what is the phase of the power reflected from the antenna? I'm pretty confident we are not looking at the return of the reflected wave in phase with the incident wave. -- Rick |
Parallel coax
rickman wrote:
On 10/4/2015 4:48 AM, Jeff wrote: What is the reflection coefficient at the ATU/feedline interface? Assuming a 'perfect' ATU and a prefect conjugate match, giving 1:1 at the TX, then it is 1. ie all of the reflected power that reaches the ATU is re-reflected back up towards the antenna. Is that good? If the cable length is 1/4 wavelength (as it is in the info provided for this case) the reflected power is nearly 180 degrees out of phase with the initial power at the antenna. I think reducing this through cable losses would not be so bad, or better to dump it in the ATU? It is good in as much as some of the re-reflected power is radiated (and some re-re-reflected) since the ATU causes the phase of the re-reflected wave to be 'in-phase' at the antenna. It is bad in as much as the reflected power suffers 2 times the cable loss, and dissipates that in heat, on each return trip, up and down the coax. With a high VSWR at the antenna there will be many return trips before the re-reflected power drops to a negligible level. No power is 'dumped' in the ATU; although there will be losses, but that is another story. If not the ATU, then the transmitter. I'm sure not all of the power is reflected back from the ATU. Exactly what is the phase of the reflected power from the ATU? I haven't seen an actual circuit for the ATU in question. For that matter, what is the phase of the power reflected from the antenna? I'm pretty confident we are not looking at the return of the reflected wave in phase with the incident wave. Having specified the transmitter power output, by definiton, no power can be "lost" in the amplifier, because transmitter power output is *defined* as the net power it actually manages to get out of its output socket net of any reflections. The mismatch may make it harder for the transmitter to achieve that; but, by definition, whatever it does achieve is its power output. You can't draw a valid defiinition of a difference between power it never produced at all and power sent back to it. -- Roger Hayter |
Parallel coax
Roger Hayter wrote:
rickman wrote: On 10/4/2015 4:48 AM, Jeff wrote: What is the reflection coefficient at the ATU/feedline interface? Assuming a 'perfect' ATU and a prefect conjugate match, giving 1:1 at the TX, then it is 1. ie all of the reflected power that reaches the ATU is re-reflected back up towards the antenna. Is that good? If the cable length is 1/4 wavelength (as it is in the info provided for this case) the reflected power is nearly 180 degrees out of phase with the initial power at the antenna. I think reducing this through cable losses would not be so bad, or better to dump it in the ATU? It is good in as much as some of the re-reflected power is radiated (and some re-re-reflected) since the ATU causes the phase of the re-reflected wave to be 'in-phase' at the antenna. It is bad in as much as the reflected power suffers 2 times the cable loss, and dissipates that in heat, on each return trip, up and down the coax. With a high VSWR at the antenna there will be many return trips before the re-reflected power drops to a negligible level. No power is 'dumped' in the ATU; although there will be losses, but that is another story. If not the ATU, then the transmitter. I'm sure not all of the power is reflected back from the ATU. Exactly what is the phase of the reflected power from the ATU? I haven't seen an actual circuit for the ATU in question. For that matter, what is the phase of the power reflected from the antenna? I'm pretty confident we are not looking at the return of the reflected wave in phase with the incident wave. Having specified the transmitter power output, by definiton, no power can be "lost" in the amplifier, because transmitter power output is *defined* as the net power it actually manages to get out of its output socket net of any reflections. The mismatch may make it harder for the transmitter to achieve that; but, by definition, whatever it does achieve is its power output. You can't draw a valid defiinition of a difference between power it never produced at all and power sent back to it. Sorry, typo. I mean you can't draw a valid distinction between power never produced and power reflected. Whatever power the transmitter will produce in a given load *is* its power output. -- Roger Hayter |
Parallel coax
In message , Jeff writes
Is that good? If the cable length is 1/4 wavelength (as it is in the info provided for this case) the reflected power is nearly 180 degrees out of phase with the initial power at the antenna. I think reducing this through cable losses would not be so bad, or better to dump it in the ATU? It is good in as much as some of the re-reflected power is radiated (and some re-re-reflected) since the ATU causes the phase of the re-reflected wave to be 'in-phase' at the antenna. It is bad in as much as the reflected power suffers 2 times the cable loss, and dissipates that in heat, on each return trip, up and down the coax. With a high VSWR at the antenna there will be many return trips before the re-reflected power drops to a negligible level. No power is 'dumped' in the ATU; although there will be losses, but that is another story. If not the ATU, then the transmitter. I'm sure not all of the power is reflected back from the ATU. Exactly what is the phase of the reflected power from the ATU? I haven't seen an actual circuit for the ATU in question. For that matter, what is the phase of the power reflected from the antenna? I'm pretty confident we are not looking at the return of the reflected wave in phase with the incident wave. If the ATU is adjusted so that the Tx sees a 1:1 match then no power is reflected to the transmitter. A 1:1 match means that there is no power reflected to the Tx by definition (and can be proved by measurement). By conservation of energy then all of the power must be radiated or lost as heat, (mostly in the coax). When the ATU gives a conjugate match, (ie the Tx sees a 1:1 vswr) the phase of the re-reflected wave at the antenna is 0, ie it is in-phase with the original forward wave and so any power that is not re-re-reflected again adds to the power supplied to the antenna. If the coax were lossless, and there were no losses in-the ATU, then ALL of the power that was supplied by the tx would be radiated regardless of the mismatch at coax to antenna interface. However, in reality even small coax losses add up to a significant loss when the mismatch at the coax to antenna interface is high due to the number of times that the power bounces up & down the coax suffering loss on each trip.. A concise explanation. What is the easiest way of calculating the power loss (say, assuming it's all in coax, and none in the ATU)? Is it simply a case of adding up a large number of diminishing losses as the signal repeatedly rattles up and down the coax (until it becomes so small that the losses can be ignored), or are there some more-elegant (and accurate) methods? -- Ian |
Parallel coax
In message , Jeff writes
If the ATU is adjusted so that the Tx sees a 1:1 match then no power is reflected to the transmitter. A 1:1 match means that there is no power reflected to the Tx by definition (and can be proved by measurement). By conservation of energy then all of the power must be radiated or lost as heat, (mostly in the coax). When the ATU gives a conjugate match, (ie the Tx sees a 1:1 vswr) the phase of the re-reflected wave at the antenna is 0, ie it is in-phase with the original forward wave and so any power that is not re-re-reflected again adds to the power supplied to the antenna. If the coax were lossless, and there were no losses in-the ATU, then ALL of the power that was supplied by the tx would be radiated regardless of the mismatch at coax to antenna interface. However, in reality even small coax losses add up to a significant loss when the mismatch at the coax to antenna interface is high due to the number of times that the power bounces up & down the coax suffering loss on each trip.. A concise explanation. What is the easiest way of calculating the power loss (say, assuming it's all in coax, and none in the ATU)? Is it simply a case of adding up a large number of diminishing losses as the signal repeatedly rattles up and down the coax (until it becomes so small that the losses can be ignored), or are there some more-elegant (and accurate) methods? The easiest way is to make an Excel spreadsheet. Convert the VSWR to return loss and convert to a ratio. Take the Tx power subtract the line loss and multiply the RL ratio to get the reflected power. Subtract that power from the incident power to find the power transmitted. Take the reflected power and subtract twice the line loss, then multiply that incident power by the RL ratio and continue as above, adding the proportion that goes to that antenna to the transmitted power. Also add the 2x line loss the the original line loss to get the power dissipated in the coax. Repeat above until the reflected power becomes insignificant; probably at least six ox seven time if the vswr is very high. Ah yes! Thanks. It's a while since I honed my skills in driving an Excel spreadsheet! I'll certainly give it a go. -- Ian |
Parallel coax
On 10/4/2015 5:51 AM, Jeff wrote:
Is that good? If the cable length is 1/4 wavelength (as it is in the info provided for this case) the reflected power is nearly 180 degrees out of phase with the initial power at the antenna. I think reducing this through cable losses would not be so bad, or better to dump it in the ATU? It is good in as much as some of the re-reflected power is radiated (and some re-re-reflected) since the ATU causes the phase of the re-reflected wave to be 'in-phase' at the antenna. It is bad in as much as the reflected power suffers 2 times the cable loss, and dissipates that in heat, on each return trip, up and down the coax. With a high VSWR at the antenna there will be many return trips before the re-reflected power drops to a negligible level. No power is 'dumped' in the ATU; although there will be losses, but that is another story. If not the ATU, then the transmitter. I'm sure not all of the power is reflected back from the ATU. Exactly what is the phase of the reflected power from the ATU? I haven't seen an actual circuit for the ATU in question. For that matter, what is the phase of the power reflected from the antenna? I'm pretty confident we are not looking at the return of the reflected wave in phase with the incident wave. If the ATU is adjusted so that the Tx sees a 1:1 match then no power is reflected to the transmitter. A 1:1 match means that there is no power reflected to the Tx by definition (and can be proved by measurement). Fine, but not related to my question if that is what you were responding to. By conservation of energy then all of the power must be radiated or lost as heat, (mostly in the coax). Uh, I'm not sure of that since there are other paths and I'm pretty sure power can be sent back into the driver of the transmitter. When the ATU gives a conjugate match, (ie the Tx sees a 1:1 vswr) the phase of the re-reflected wave at the antenna is 0, ie it is in-phase with the original forward wave and so any power that is not re-re-reflected again adds to the power supplied to the antenna. You seem to be mixing two things here. We have been talking about an ATU at the transmitter. The feed line cable is between the ATU and the antenna. You then talk about reflections both at the TX to ATU interface and the cable to antenna interface. You also make assumptions about the reflected power phase when it returns to the antenna. If you want to talk about the phase at that point you have to take into account the electrical length of the cable and *three* reflections. If the coax were lossless, and there were no losses in-the ATU, then ALL of the power that was supplied by the tx would be radiated regardless of the mismatch at coax to antenna interface. However, in reality even small coax losses add up to a significant loss when the mismatch at the coax to antenna interface is high due to the number of times that the power bounces up & down the coax suffering loss on each trip.. One poster has tried to say the reflections are all such that the phase of the reflected power ends up in phase with the original signal delivered to the antenna. But no one has done a power phase analysis. You seem to be saying the same thing. Power is reflected from the complex impedance of the antenna. Is that power not out of phase from the incident signal by some amount determined by the complex impedance? On traveling down the cable there is a phase shift by an amount related to the ratio of the electrical length and the wavelength. The ATU reflects some of the signal back up the cable and some is passed directly to the TX (since there is no or a very short cable between them) and is partially reflected by the TX, again with some phase shift determined by the two impedances (TX output and the ATU input). Another phase shift by the cable and the reflected wave arrives at the antenna. Is anyone going to tell me that all this has to add up to a reflected signal arriving *in phase* with the incident signal? -- Rick |
Parallel coax
In message , rickman
writes On 10/4/2015 5:51 AM, Jeff wrote: Is that good? If the cable length is 1/4 wavelength (as it is in the info provided for this case) the reflected power is nearly 180 degrees out of phase with the initial power at the antenna. I think reducing this through cable losses would not be so bad, or better to dump it in the ATU? It is good in as much as some of the re-reflected power is radiated (and some re-re-reflected) since the ATU causes the phase of the re-reflected wave to be 'in-phase' at the antenna. It is bad in as much as the reflected power suffers 2 times the cable loss, and dissipates that in heat, on each return trip, up and down the coax. With a high VSWR at the antenna there will be many return trips before the re-reflected power drops to a negligible level. No power is 'dumped' in the ATU; although there will be losses, but that is another story. If not the ATU, then the transmitter. I'm sure not all of the power is reflected back from the ATU. Exactly what is the phase of the reflected power from the ATU? I haven't seen an actual circuit for the ATU in question. For that matter, what is the phase of the power reflected from the antenna? I'm pretty confident we are not looking at the return of the reflected wave in phase with the incident wave. If the ATU is adjusted so that the Tx sees a 1:1 match then no power is reflected to the transmitter. A 1:1 match means that there is no power reflected to the Tx by definition (and can be proved by measurement). Fine, but not related to my question if that is what you were responding to. By conservation of energy then all of the power must be radiated or lost as heat, (mostly in the coax). Uh, I'm not sure of that since there are other paths and I'm pretty sure power can be sent back into the driver of the transmitter. When the ATU gives a conjugate match, (ie the Tx sees a 1:1 vswr) the phase of the re-reflected wave at the antenna is 0, ie it is in-phase with the original forward wave and so any power that is not re-re-reflected again adds to the power supplied to the antenna. You seem to be mixing two things here. We have been talking about an ATU at the transmitter. The feed line cable is between the ATU and the antenna. You then talk about reflections both at the TX to ATU interface and the cable to antenna interface. You also make assumptions about the reflected power phase when it returns to the antenna. If you want to talk about the phase at that point you have to take into account the electrical length of the cable and *three* reflections. If the coax were lossless, and there were no losses in-the ATU, then ALL of the power that was supplied by the tx would be radiated regardless of the mismatch at coax to antenna interface. However, in reality even small coax losses add up to a significant loss when the mismatch at the coax to antenna interface is high due to the number of times that the power bounces up & down the coax suffering loss on each trip.. One poster has tried to say the reflections are all such that the phase of the reflected power ends up in phase with the original signal delivered to the antenna. But no one has done a power phase analysis. You seem to be saying the same thing. Power is reflected from the complex impedance of the antenna. Is that power not out of phase from the incident signal by some amount determined by the complex impedance? On traveling down the cable there is a phase shift by an amount related to the ratio of the electrical length and the wavelength. The ATU reflects some of the signal back up the cable and some is passed directly to the TX (since there is no or a very short cable between them) and is partially reflected by the TX, again with some phase shift determined by the two impedances (TX output and the ATU input). Another phase shift by the cable and the reflected wave arrives at the antenna. Is anyone going to tell me that all this has to add up to a reflected signal arriving *in phase* with the incident signal? I don't think you need to think about phase. The forward-going signal has no need to know about the reflected signal, so just think of them as being independent. One simply passes through the other. I understand Jeff correctly, if the ATU input presents a perfect 1-to-1 SWR to the transmitter output, it follows that none of power in the multiple reflections in the coax is getting back into the TX. Therefore, if we ignore the losses in the ATU, all the TX output power must eventually, after countless ever-diminishing reflections between the antenna and the ATU output, be either reaching the antenna, or being lost in heating up the coax. -- Ian |
Parallel coax
On 10/4/2015 3:00 PM, Ian Jackson wrote:
In message , rickman writes On 10/4/2015 5:51 AM, Jeff wrote: Is that good? If the cable length is 1/4 wavelength (as it is in the info provided for this case) the reflected power is nearly 180 degrees out of phase with the initial power at the antenna. I think reducing this through cable losses would not be so bad, or better to dump it in the ATU? It is good in as much as some of the re-reflected power is radiated (and some re-re-reflected) since the ATU causes the phase of the re-reflected wave to be 'in-phase' at the antenna. It is bad in as much as the reflected power suffers 2 times the cable loss, and dissipates that in heat, on each return trip, up and down the coax. With a high VSWR at the antenna there will be many return trips before the re-reflected power drops to a negligible level. No power is 'dumped' in the ATU; although there will be losses, but that is another story. If not the ATU, then the transmitter. I'm sure not all of the power is reflected back from the ATU. Exactly what is the phase of the reflected power from the ATU? I haven't seen an actual circuit for the ATU in question. For that matter, what is the phase of the power reflected from the antenna? I'm pretty confident we are not looking at the return of the reflected wave in phase with the incident wave. If the ATU is adjusted so that the Tx sees a 1:1 match then no power is reflected to the transmitter. A 1:1 match means that there is no power reflected to the Tx by definition (and can be proved by measurement). Fine, but not related to my question if that is what you were responding to. By conservation of energy then all of the power must be radiated or lost as heat, (mostly in the coax). Uh, I'm not sure of that since there are other paths and I'm pretty sure power can be sent back into the driver of the transmitter. When the ATU gives a conjugate match, (ie the Tx sees a 1:1 vswr) the phase of the re-reflected wave at the antenna is 0, ie it is in-phase with the original forward wave and so any power that is not re-re-reflected again adds to the power supplied to the antenna. You seem to be mixing two things here. We have been talking about an ATU at the transmitter. The feed line cable is between the ATU and the antenna. You then talk about reflections both at the TX to ATU interface and the cable to antenna interface. You also make assumptions about the reflected power phase when it returns to the antenna. If you want to talk about the phase at that point you have to take into account the electrical length of the cable and *three* reflections. If the coax were lossless, and there were no losses in-the ATU, then ALL of the power that was supplied by the tx would be radiated regardless of the mismatch at coax to antenna interface. However, in reality even small coax losses add up to a significant loss when the mismatch at the coax to antenna interface is high due to the number of times that the power bounces up & down the coax suffering loss on each trip.. One poster has tried to say the reflections are all such that the phase of the reflected power ends up in phase with the original signal delivered to the antenna. But no one has done a power phase analysis. You seem to be saying the same thing. Power is reflected from the complex impedance of the antenna. Is that power not out of phase from the incident signal by some amount determined by the complex impedance? On traveling down the cable there is a phase shift by an amount related to the ratio of the electrical length and the wavelength. The ATU reflects some of the signal back up the cable and some is passed directly to the TX (since there is no or a very short cable between them) and is partially reflected by the TX, again with some phase shift determined by the two impedances (TX output and the ATU input). Another phase shift by the cable and the reflected wave arrives at the antenna. Is anyone going to tell me that all this has to add up to a reflected signal arriving *in phase* with the incident signal? I don't think you need to think about phase. The forward-going signal has no need to know about the reflected signal, so just think of them as being independent. One simply passes through the other. When they are reaching the antenna you damn well *do* need to know about phase. If the incident wave and the reflected wave returning to the antenna are 180 degrees out of phase the power subtracts! I understand Jeff correctly, if the ATU input presents a perfect 1-to-1 SWR to the transmitter output, it follows that none of power in the multiple reflections in the coax is getting back into the TX. I think you do *not* understand Jeff. The TX to ATU match has nothing to do with the energy reaching the TX, just the opposite, it means all power from the antenna passing through the ATU is absorbed by the TX. But I have not seen anyone say the ATU matches the TX impedance, only that it is a conjugate to the antenna. There are three interfaces, TX/ATU, ATU/cable, cable/antenna. Each one can have reflections and each of the four parts of the system can absorb (or radiate in the case of the antenna) power. This is not so easy to analyze fully. At least until some of the fundamentals are revealed that we have not talked about. Therefore, if we ignore the losses in the ATU, all the TX output power must eventually, after countless ever-diminishing reflections between the antenna and the ATU output, be either reaching the antenna, or being lost in heating up the coax. A given reflection may "reach" the antenna, but will it contribute positively to the radiated field or counter it? -- Rick |
Parallel coax
In message , rickman
writes On 10/4/2015 3:00 PM, Ian Jackson wrote: In message , rickman writes On 10/4/2015 5:51 AM, Jeff wrote: Is that good? If the cable length is 1/4 wavelength (as it is in the info provided for this case) the reflected power is nearly 180 degrees out of phase with the initial power at the antenna. I think reducing this through cable losses would not be so bad, or better to dump it in the ATU? It is good in as much as some of the re-reflected power is radiated (and some re-re-reflected) since the ATU causes the phase of the re-reflected wave to be 'in-phase' at the antenna. It is bad in as much as the reflected power suffers 2 times the cable loss, and dissipates that in heat, on each return trip, up and down the coax. With a high VSWR at the antenna there will be many return trips before the re-reflected power drops to a negligible level. No power is 'dumped' in the ATU; although there will be losses, but that is another story. If not the ATU, then the transmitter. I'm sure not all of the power is reflected back from the ATU. Exactly what is the phase of the reflected power from the ATU? I haven't seen an actual circuit for the ATU in question. For that matter, what is the phase of the power reflected from the antenna? I'm pretty confident we are not looking at the return of the reflected wave in phase with the incident wave. If the ATU is adjusted so that the Tx sees a 1:1 match then no power is reflected to the transmitter. A 1:1 match means that there is no power reflected to the Tx by definition (and can be proved by measurement). Fine, but not related to my question if that is what you were responding to. By conservation of energy then all of the power must be radiated or lost as heat, (mostly in the coax). Uh, I'm not sure of that since there are other paths and I'm pretty sure power can be sent back into the driver of the transmitter. When the ATU gives a conjugate match, (ie the Tx sees a 1:1 vswr) the phase of the re-reflected wave at the antenna is 0, ie it is in-phase with the original forward wave and so any power that is not re-re-reflected again adds to the power supplied to the antenna. You seem to be mixing two things here. We have been talking about an ATU at the transmitter. The feed line cable is between the ATU and the antenna. You then talk about reflections both at the TX to ATU interface and the cable to antenna interface. You also make assumptions about the reflected power phase when it returns to the antenna. If you want to talk about the phase at that point you have to take into account the electrical length of the cable and *three* reflections. If the coax were lossless, and there were no losses in-the ATU, then ALL of the power that was supplied by the tx would be radiated regardless of the mismatch at coax to antenna interface. However, in reality even small coax losses add up to a significant loss when the mismatch at the coax to antenna interface is high due to the number of times that the power bounces up & down the coax suffering loss on each trip.. One poster has tried to say the reflections are all such that the phase of the reflected power ends up in phase with the original signal delivered to the antenna. But no one has done a power phase analysis. You seem to be saying the same thing. Power is reflected from the complex impedance of the antenna. Is that power not out of phase from the incident signal by some amount determined by the complex impedance? On traveling down the cable there is a phase shift by an amount related to the ratio of the electrical length and the wavelength. The ATU reflects some of the signal back up the cable and some is passed directly to the TX (since there is no or a very short cable between them) and is partially reflected by the TX, again with some phase shift determined by the two impedances (TX output and the ATU input). Another phase shift by the cable and the reflected wave arrives at the antenna. Is anyone going to tell me that all this has to add up to a reflected signal arriving *in phase* with the incident signal? I don't think you need to think about phase. The forward-going signal has no need to know about the reflected signal, so just think of them as being independent. One simply passes through the other. When they are reaching the antenna you damn well *do* need to know about phase. If the incident wave and the reflected wave returning to the antenna are 180 degrees out of phase the power subtracts! I understand Jeff correctly, if the ATU input presents a perfect 1-to-1 SWR to the transmitter output, it follows that none of power in the multiple reflections in the coax is getting back into the TX. I think you do *not* understand Jeff. The TX to ATU match has nothing to do with the energy reaching the TX, just the opposite, it means all power from the antenna passing through the ATU is absorbed by the TX. But I have not seen anyone say the ATU matches the TX impedance, only that it is a conjugate to the antenna. Should be not be a conjugate match to the impedance as seen looking into the coax at the TX end? There are three interfaces, TX/ATU, ATU/cable, cable/antenna. Each one can have reflections and each of the four parts of the system can absorb (or radiate in the case of the antenna) power. This is not so easy to analyze fully. At least until some of the fundamentals are revealed that we have not talked about. Therefore, if we ignore the losses in the ATU, all the TX output power must eventually, after countless ever-diminishing reflections between the antenna and the ATU output, be either reaching the antenna, or being lost in heating up the coax. A given reflection may "reach" the antenna, but will it contribute positively to the radiated field or counter it? I think we are looking at the scenario in two different ways. You are trying to see a 'snapshot' of the vectorial additions of all the forward and reverse signals, and I am simply looking at them as being independent entities, rattling up and down the coax, and progressively either escaping via the antenna feedpoint and being radiated, or heating up the coax. -- Ian |
Parallel coax
On 10/4/2015 6:01 PM, Ian Jackson wrote:
In message , rickman writes On 10/4/2015 3:00 PM, Ian Jackson wrote: In message , rickman writes On 10/4/2015 5:51 AM, Jeff wrote: Is that good? If the cable length is 1/4 wavelength (as it is in the info provided for this case) the reflected power is nearly 180 degrees out of phase with the initial power at the antenna. I think reducing this through cable losses would not be so bad, or better to dump it in the ATU? It is good in as much as some of the re-reflected power is radiated (and some re-re-reflected) since the ATU causes the phase of the re-reflected wave to be 'in-phase' at the antenna. It is bad in as much as the reflected power suffers 2 times the cable loss, and dissipates that in heat, on each return trip, up and down the coax. With a high VSWR at the antenna there will be many return trips before the re-reflected power drops to a negligible level. No power is 'dumped' in the ATU; although there will be losses, but that is another story. If not the ATU, then the transmitter. I'm sure not all of the power is reflected back from the ATU. Exactly what is the phase of the reflected power from the ATU? I haven't seen an actual circuit for the ATU in question. For that matter, what is the phase of the power reflected from the antenna? I'm pretty confident we are not looking at the return of the reflected wave in phase with the incident wave. If the ATU is adjusted so that the Tx sees a 1:1 match then no power is reflected to the transmitter. A 1:1 match means that there is no power reflected to the Tx by definition (and can be proved by measurement). Fine, but not related to my question if that is what you were responding to. By conservation of energy then all of the power must be radiated or lost as heat, (mostly in the coax). Uh, I'm not sure of that since there are other paths and I'm pretty sure power can be sent back into the driver of the transmitter. When the ATU gives a conjugate match, (ie the Tx sees a 1:1 vswr) the phase of the re-reflected wave at the antenna is 0, ie it is in-phase with the original forward wave and so any power that is not re-re-reflected again adds to the power supplied to the antenna. You seem to be mixing two things here. We have been talking about an ATU at the transmitter. The feed line cable is between the ATU and the antenna. You then talk about reflections both at the TX to ATU interface and the cable to antenna interface. You also make assumptions about the reflected power phase when it returns to the antenna. If you want to talk about the phase at that point you have to take into account the electrical length of the cable and *three* reflections. If the coax were lossless, and there were no losses in-the ATU, then ALL of the power that was supplied by the tx would be radiated regardless of the mismatch at coax to antenna interface. However, in reality even small coax losses add up to a significant loss when the mismatch at the coax to antenna interface is high due to the number of times that the power bounces up & down the coax suffering loss on each trip.. One poster has tried to say the reflections are all such that the phase of the reflected power ends up in phase with the original signal delivered to the antenna. But no one has done a power phase analysis. You seem to be saying the same thing. Power is reflected from the complex impedance of the antenna. Is that power not out of phase from the incident signal by some amount determined by the complex impedance? On traveling down the cable there is a phase shift by an amount related to the ratio of the electrical length and the wavelength. The ATU reflects some of the signal back up the cable and some is passed directly to the TX (since there is no or a very short cable between them) and is partially reflected by the TX, again with some phase shift determined by the two impedances (TX output and the ATU input). Another phase shift by the cable and the reflected wave arrives at the antenna. Is anyone going to tell me that all this has to add up to a reflected signal arriving *in phase* with the incident signal? I don't think you need to think about phase. The forward-going signal has no need to know about the reflected signal, so just think of them as being independent. One simply passes through the other. When they are reaching the antenna you damn well *do* need to know about phase. If the incident wave and the reflected wave returning to the antenna are 180 degrees out of phase the power subtracts! I understand Jeff correctly, if the ATU input presents a perfect 1-to-1 SWR to the transmitter output, it follows that none of power in the multiple reflections in the coax is getting back into the TX. I think you do *not* understand Jeff. The TX to ATU match has nothing to do with the energy reaching the TX, just the opposite, it means all power from the antenna passing through the ATU is absorbed by the TX. But I have not seen anyone say the ATU matches the TX impedance, only that it is a conjugate to the antenna. Should be not be a conjugate match to the impedance as seen looking into the coax at the TX end? There are three interfaces, TX/ATU, ATU/cable, cable/antenna. Each one can have reflections and each of the four parts of the system can absorb (or radiate in the case of the antenna) power. This is not so easy to analyze fully. At least until some of the fundamentals are revealed that we have not talked about. Therefore, if we ignore the losses in the ATU, all the TX output power must eventually, after countless ever-diminishing reflections between the antenna and the ATU output, be either reaching the antenna, or being lost in heating up the coax. A given reflection may "reach" the antenna, but will it contribute positively to the radiated field or counter it? I think we are looking at the scenario in two different ways. You are trying to see a 'snapshot' of the vectorial additions of all the forward and reverse signals, and I am simply looking at them as being independent entities, rattling up and down the coax, and progressively either escaping via the antenna feedpoint and being radiated, or heating up the coax. You seem to want to ignore the possibility that the reflected waves can interfere with your incident wave. This is entirely possible. Even if reflected energy is radiated it can cancel the incident wave and reduce your signal strength. -- Rick |
Parallel coax
On 10/5/2015 2:48 AM, Jeff wrote:
If the ATU is adjusted so that the Tx sees a 1:1 match then no power is reflected to the transmitter. A 1:1 match means that there is no power reflected to the Tx by definition (and can be proved by measurement). Fine, but not related to my question if that is what you were responding to. Of course it is. If the ATU is adjusted for a 1:1 match then NO power is sent back to the Tx. By conservation of energy then all of the power must be radiated or lost as heat, (mostly in the coax). Uh, I'm not sure of that since there are other paths and I'm pretty sure power can be sent back into the driver of the transmitter. NO, as above, if the ATU is adjusted for a 1:1 match then there is no power reflected back to the TX. You are saying the ATU doesn't reflect power back from the TX. If the match is 1:1 that is true. But you aren't considering the power reflected from the antenna. The antenna reflects power back to the ATU and there is nothing in the ATU to prevent that power from being handed to the transmitter. When the ATU gives a conjugate match, (ie the Tx sees a 1:1 vswr) the phase of the re-reflected wave at the antenna is 0, ie it is in-phase with the original forward wave and so any power that is not re-re-reflected again adds to the power supplied to the antenna. You seem to be mixing two things here. We have been talking about an ATU at the transmitter. The feed line cable is between the ATU and the antenna. You then talk about reflections both at the TX to ATU interface and the cable to antenna interface. You also make assumptions about the reflected power phase when it returns to the antenna. If you want to talk about the phase at that point you have to take into account the electrical length of the cable and *three* reflections. No, I and talking about a Tx then an ATU then a feeder to a mismatched antenna. I believe that is what we are all talking about. When adjusted for a 1:1 match there are no reflections between the Tx and ATU, but there are multiple reflections between the ATU and teh antenna. That is where a lot of power is dissipated in the antenna is a poor match to the feeder impedance. No one is talking about the power from the TX being reflected back to the TX by the ATU. One poster has tried to say the reflections are all such that the phase of the reflected power ends up in phase with the original signal delivered to the antenna. But no one has done a power phase analysis. You seem to be saying the same thing. The reflected wave will start off at the antenna out of phase with the forward wave, (the actual phase depends on the complex impedance of the mismatch), what the conjugate match that the ATU provides, when adjusted so that there is a 1:1 match and no power reflected power sent to the Tx, This doesn't cover the reflected power from the antenna, just the power from the TX. is a phase shift such that the re-reflected wave from the ATU towards the antenna is in phase with the original forward wave, so when it reaches the antenna the portion of that re-reflected wave that is not bounced back again down the coax by the mismatch is delivered to the antenna. How does that work? There is phase shift in the reflections which may be compensated for by the ATU, but there is also phase shift in the cable. Power is reflected from the complex impedance of the antenna. Is that power not out of phase from the incident signal by some amount determined by the complex impedance? On traveling down the cable there is a phase shift by an amount related to the ratio of the electrical length and the wavelength. The ATU reflects some of the signal back up the cable and some is passed directly to the TX (since there is no or a very short cable between them) and is partially reflected by the TX, again with some phase shift determined by the two impedances (TX output and the ATU input). Another phase shift by the cable and the reflected wave arrives at the antenna. NO, no power gets back to the tx if the ATU is adjusted for a 1:1 match. after all that is the definition of a 1:1 match; no reflected power. You keep saying that the 1:1 match between the TX and the ATU prevents any power from being sent to the TX which is not true. You are confusing the power from the TX which is not reflected and the power reflected from the antenna which passes through the ATU to the TX. Is anyone going to tell me that all this has to add up to a reflected signal arriving *in phase* with the incident signal? YES. I may have to look at the math for this. How does the ATU reflect all the power from the antenna back to the antenna? I thought only an open or a short can reflect all the power. -- Rick |
Parallel coax
In message , rickman
writes You keep saying that the 1:1 match between the TX and the ATU prevents any power from being sent to the TX which is not true. You are confusing the power from the TX which is not reflected and the power reflected from the antenna which passes through the ATU to the TX. If the SWR meter between the TX output indicates a 1:1 SWR, then there can be NO power travelling between the ATU input and the TX output - ie there IS no reflected power. QED, surely? If you ignore the losses in the ATU, all the power that the mismatched antenna reflects, and that makes it back to the ATU output, MUST be re-reflected by the ATU output impedance, and head off back towards the antenna. This is because the reflected signal cannot heat up a lossless ATU, and the SWR meter says it isn't coming back through the ATU. It simply has nowhere to go except back down the coax. -- Ian |
Parallel coax
On 10/5/2015 4:17 AM, Ian Jackson wrote:
In message , rickman writes You keep saying that the 1:1 match between the TX and the ATU prevents any power from being sent to the TX which is not true. You are confusing the power from the TX which is not reflected and the power reflected from the antenna which passes through the ATU to the TX. If the SWR meter between the TX output indicates a 1:1 SWR, then there can be NO power travelling between the ATU input and the TX output - ie there IS no reflected power. QED, surely? If you ignore the losses in the ATU, all the power that the mismatched antenna reflects, and that makes it back to the ATU output, MUST be re-reflected by the ATU output impedance, and head off back towards the antenna. This is because the reflected signal cannot heat up a lossless ATU, and the SWR meter says it isn't coming back through the ATU. It simply has nowhere to go except back down the coax. You saying something is true or imagining a SWR reading is not the same as understanding what is going on. What SWR reading are you imagining? Can you explain this in terms of the circuit analysis? The ATU consists of what circuit? The TX has some source impedance, what would that be? I don't think you can design an ATU circuit that will isolate the real source impedance of the TX from the reflected wave from the antenna. -- Rick |
Parallel coax
rickman wrote:
On 10/4/2015 6:01 PM, Ian Jackson wrote: In message , rickman writes On 10/4/2015 3:00 PM, Ian Jackson wrote: In message , rickman writes On 10/4/2015 5:51 AM, Jeff wrote: Is that good? If the cable length is 1/4 wavelength (as it is in the info provided for this case) the reflected power is nearly 180 degrees out of phase with the initial power at the antenna. I think reducing this through cable losses would not be so bad, or better to dump it in the ATU? It is good in as much as some of the re-reflected power is radiated (and some re-re-reflected) since the ATU causes the phase of the re-reflected wave to be 'in-phase' at the antenna. It is bad in as much as the reflected power suffers 2 times the cable loss, and dissipates that in heat, on each return trip, up and down the coax. With a high VSWR at the antenna there will be many return trips before the re-reflected power drops to a negligible level. No power is 'dumped' in the ATU; although there will be losses, but that is another story. If not the ATU, then the transmitter. I'm sure not all of the power is reflected back from the ATU. Exactly what is the phase of the reflected power from the ATU? I haven't seen an actual circuit for the ATU in question. For that matter, what is the phase of the power reflected from the antenna? I'm pretty confident we are not looking at the return of the reflected wave in phase with the incident wave. If the ATU is adjusted so that the Tx sees a 1:1 match then no power is reflected to the transmitter. A 1:1 match means that there is no power reflected to the Tx by definition (and can be proved by measurement). Fine, but not related to my question if that is what you were responding to. By conservation of energy then all of the power must be radiated or lost as heat, (mostly in the coax). Uh, I'm not sure of that since there are other paths and I'm pretty sure power can be sent back into the driver of the transmitter. When the ATU gives a conjugate match, (ie the Tx sees a 1:1 vswr) the phase of the re-reflected wave at the antenna is 0, ie it is in-phase with the original forward wave and so any power that is not re-re-reflected again adds to the power supplied to the antenna. You seem to be mixing two things here. We have been talking about an ATU at the transmitter. The feed line cable is between the ATU and the antenna. You then talk about reflections both at the TX to ATU interface and the cable to antenna interface. You also make assumptions about the reflected power phase when it returns to the antenna. If you want to talk about the phase at that point you have to take into account the electrical length of the cable and *three* reflections. If the coax were lossless, and there were no losses in-the ATU, then ALL of the power that was supplied by the tx would be radiated regardless of the mismatch at coax to antenna interface. However, in reality even small coax losses add up to a significant loss when the mismatch at the coax to antenna interface is high due to the number of times that the power bounces up & down the coax suffering loss on each trip.. One poster has tried to say the reflections are all such that the phase of the reflected power ends up in phase with the original signal delivered to the antenna. But no one has done a power phase analysis. You seem to be saying the same thing. Power is reflected from the complex impedance of the antenna. Is that power not out of phase from the incident signal by some amount determined by the complex impedance? On traveling down the cable there is a phase shift by an amount related to the ratio of the electrical length and the wavelength. The ATU reflects some of the signal back up the cable and some is passed directly to the TX (since there is no or a very short cable between them) and is partially reflected by the TX, again with some phase shift determined by the two impedances (TX output and the ATU input). Another phase shift by the cable and the reflected wave arrives at the antenna. Is anyone going to tell me that all this has to add up to a reflected signal arriving *in phase* with the incident signal? I don't think you need to think about phase. The forward-going signal has no need to know about the reflected signal, so just think of them as being independent. One simply passes through the other. When they are reaching the antenna you damn well *do* need to know about phase. If the incident wave and the reflected wave returning to the antenna are 180 degrees out of phase the power subtracts! I understand Jeff correctly, if the ATU input presents a perfect 1-to-1 SWR to the transmitter output, it follows that none of power in the multiple reflections in the coax is getting back into the TX. I think you do *not* understand Jeff. The TX to ATU match has nothing to do with the energy reaching the TX, just the opposite, it means all power from the antenna passing through the ATU is absorbed by the TX. But I have not seen anyone say the ATU matches the TX impedance, only that it is a conjugate to the antenna. Should be not be a conjugate match to the impedance as seen looking into the coax at the TX end? There are three interfaces, TX/ATU, ATU/cable, cable/antenna. Each one can have reflections and each of the four parts of the system can absorb (or radiate in the case of the antenna) power. This is not so easy to analyze fully. At least until some of the fundamentals are revealed that we have not talked about. Therefore, if we ignore the losses in the ATU, all the TX output power must eventually, after countless ever-diminishing reflections between the antenna and the ATU output, be either reaching the antenna, or being lost in heating up the coax. A given reflection may "reach" the antenna, but will it contribute positively to the radiated field or counter it? I think we are looking at the scenario in two different ways. You are trying to see a 'snapshot' of the vectorial additions of all the forward and reverse signals, and I am simply looking at them as being independent entities, rattling up and down the coax, and progressively either escaping via the antenna feedpoint and being radiated, or heating up the coax. You seem to want to ignore the possibility that the reflected waves can interfere with your incident wave. This is entirely possible. Even if reflected energy is radiated it can cancel the incident wave and reduce your signal strength. Conservation of energy suggests that it can only reduce it in one direction by increasing it in another. If feeder reflections can have this effect it should be possible to demonstrate it in small model aerials at high frequencies. -- Roger Hayter |
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rickman wrote:
On 10/5/2015 4:17 AM, Ian Jackson wrote: In message , rickman writes You keep saying that the 1:1 match between the TX and the ATU prevents any power from being sent to the TX which is not true. You are confusing the power from the TX which is not reflected and the power reflected from the antenna which passes through the ATU to the TX. If the SWR meter between the TX output indicates a 1:1 SWR, then there can be NO power travelling between the ATU input and the TX output - ie there IS no reflected power. QED, surely? If you ignore the losses in the ATU, all the power that the mismatched antenna reflects, and that makes it back to the ATU output, MUST be re-reflected by the ATU output impedance, and head off back towards the antenna. This is because the reflected signal cannot heat up a lossless ATU, and the SWR meter says it isn't coming back through the ATU. It simply has nowhere to go except back down the coax. You saying something is true or imagining a SWR reading is not the same as understanding what is going on. What SWR reading are you imagining? Can you explain this in terms of the circuit analysis? The ATU consists of what circuit? The TX has some source impedance, what would that be? I don't think you can design an ATU circuit that will isolate the real source impedance of the TX from the reflected wave from the antenna. So, when you tune your transmitter-end ATU for minimum SWR *between the Tx and the ATU*, what exactly are you doing? You are making the feeder/aerilal/ATU combination look like a resistive load from the transmitter side, what are you making the ATU/transmitter combination look like from the aerial feeder side? It must be something that reflects back incident waves, otherwise your SWR meter wouldn't be reading 1.0. That seems to be the argument, and it sounds moderately convincing to me. -- Roger Hayter |
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In message , rickman
writes On 10/5/2015 4:17 AM, Ian Jackson wrote: In message , rickman writes You keep saying that the 1:1 match between the TX and the ATU prevents any power from being sent to the TX which is not true. You are confusing the power from the TX which is not reflected and the power reflected from the antenna which passes through the ATU to the TX. If the SWR meter between the TX output indicates a 1:1 SWR, then there can be NO power travelling between the ATU input and the TX output - ie there IS no reflected power. QED, surely? If you ignore the losses in the ATU, all the power that the mismatched antenna reflects, and that makes it back to the ATU output, MUST be re-reflected by the ATU output impedance, and head off back towards the antenna. This is because the reflected signal cannot heat up a lossless ATU, and the SWR meter says it isn't coming back through the ATU. It simply has nowhere to go except back down the coax. You saying something is true or imagining a SWR reading is not the same as understanding what is going on. What SWR reading are you imagining? A typical HF station setup is TX - SWR meter - ATU - feeder - antenna. [Please no one start saying they don't use this configuration.] Can you explain this in terms of the circuit analysis? The only analysis required is to ask yourself why, after much careful twiddling with the knobs, you eventually get the SWR meter to show 1:1. Does this not indicate that no power is coming back to the TX from the ATU? [If not, what does it indicate?] The ATU consists of what circuit? I haven't a clue. It's a large shiny black box with three silver knobs and two RF connectors. All the spec says is "Insertion loss 0dB". [Mind you, it was very, very expensive.] The TX has some source impedance, what would that be? Somewhere between not-a-lot and probably not-too-high (because, in operation, the TX doesn't get unduly hot). It really doesn't matter what it is. The SWR meter shows that zero power is coming back out the ATU input, so there is nothing to get re-reflected from the TX output. I don't think you can design an ATU circuit that will isolate the real source impedance of the TX from the reflected wave from the antenna. The purpose of what many of us (often slightly incorrectly) call an ATU is to present the TX output with its design load impedance. In most cases, this is 50 ohms resistive. The TX output impedance is really of no concern to the designers of the ATU. As long as they can produce an ATU that will present the TX with a load of 50 ohms resistive, then it will do its job perfectly - and this will be indicated by the 50 ohm SWR meter between the TX and the ATU showing 1:1. And when an SWR meter shows 1:1, it means that there is no power coming back towards the TX output. -- Ian |
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In message , Jeff writes
NO, as above, if the ATU is adjusted for a 1:1 match then there is no power reflected back to the TX. You are saying the ATU doesn't reflect power back from the TX. If the match is 1:1 that is true. But you aren't considering the power reflected from the antenna. The antenna reflects power back to the ATU and there is nothing in the ATU to prevent that power from being handed to the transmitter. The whole function of the ATU is to provide a 1:1 match for the Tx, when that is the case NO POWER GOES BACK TO THE TX. Yes the antenna reflects power back towards the ATU, BUT the ATU then reflects ALL of that power back again towards the antenna and none into the TX (assuming that the ATU achieved a conjugate match.) When adjusted for a 1:1 match there are no reflections between the Tx and ATU, but there are multiple reflections between the ATU and teh antenna. That is where a lot of power is dissipated in the antenna is a poor match to the feeder impedance. No one is talking about the power from the TX being reflected back to the TX by the ATU. The reflected wave will start off at the antenna out of phase with the forward wave, (the actual phase depends on the complex impedance of the mismatch), what the conjugate match that the ATU provides, when adjusted so that there is a 1:1 match and no power reflected power sent to the Tx, This doesn't cover the reflected power from the antenna, just the power from the TX. NO. It covers the reflected power from the antenna. There is no reflected power between the ATU and Tx if the match is 1:1. is a phase shift such that the re-reflected wave from the ATU towards the antenna is in phase with the original forward wave, so when it reaches the antenna the portion of that re-reflected wave that is not bounced back again down the coax by the mismatch is delivered to the antenna. How does that work? There is phase shift in the reflections which may be compensated for by the ATU, but there is also phase shift in the cable. The ATU applies a conjugate match at the end of the cable, the impedance that it sees, and applies the conjugate of, is the impedance of the antenna modified by the length of cable. It just so happens that the conditions for a 1:1 vswr, and that of a conjugate match, are that match causes the phase at the antenna end of the cable to be the same phase as the original forward wave. That is the physics of a conjugate match. NO, no power gets back to the tx if the ATU is adjusted for a 1:1 match. after all that is the definition of a 1:1 match; no reflected power. You keep saying that the 1:1 match between the TX and the ATU prevents any power from being sent to the TX which is not true. You are confusing the power from the TX which is not reflected and the power reflected from the antenna which passes through the ATU to the TX. YES IT IS TRUE. I am not confusing anything. Is anyone going to tell me that all this has to add up to a reflected signal arriving *in phase* with the incident signal? YES. I may have to look at the math for this. How does the ATU reflect all the power from the antenna back to the antenna? I thought only an open or a short can reflect all the power. Please look it up, do some experiments yourself. Try some simulations in Spice or similar. You will find that I am correct. While I've being saying "Let's ignore the losses in the ATU", presumably you can assign a loss to it, and the reflected signal will suffer this loss each time it bounces off the ATU output. If so, the loss can be treated in the same way as the loss in the coax. In fact as 15' of decent coax will only have (say) 0.5dB matched loss on 14MHz, the summation of the 'return and go' ATU losses (say 2dB each time?) could be more significant than the 1dB each time the signal traverses the coax. -- Ian |
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On 09/27/2015 11:22 AM, John S wrote: 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. Connector losses are static up to UHF at around .5db per connector for a typical PL-259/SO-239 pair. Insertion losses for professional grade connectors on high quality rigid and semi-rigid coax such as Andrew Heliax are around .05db. Matching the antenna won't make the Return Loss go away but it will make the transmitter happy. I should have said matching the line. 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 |
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On 10/5/2015 5:29 AM, Roger Hayter wrote:
rickman wrote: On 10/4/2015 6:01 PM, Ian Jackson wrote: In message , rickman writes On 10/4/2015 3:00 PM, Ian Jackson wrote: In message , rickman writes On 10/4/2015 5:51 AM, Jeff wrote: Is that good? If the cable length is 1/4 wavelength (as it is in the info provided for this case) the reflected power is nearly 180 degrees out of phase with the initial power at the antenna. I think reducing this through cable losses would not be so bad, or better to dump it in the ATU? It is good in as much as some of the re-reflected power is radiated (and some re-re-reflected) since the ATU causes the phase of the re-reflected wave to be 'in-phase' at the antenna. It is bad in as much as the reflected power suffers 2 times the cable loss, and dissipates that in heat, on each return trip, up and down the coax. With a high VSWR at the antenna there will be many return trips before the re-reflected power drops to a negligible level. No power is 'dumped' in the ATU; although there will be losses, but that is another story. If not the ATU, then the transmitter. I'm sure not all of the power is reflected back from the ATU. Exactly what is the phase of the reflected power from the ATU? I haven't seen an actual circuit for the ATU in question. For that matter, what is the phase of the power reflected from the antenna? I'm pretty confident we are not looking at the return of the reflected wave in phase with the incident wave. If the ATU is adjusted so that the Tx sees a 1:1 match then no power is reflected to the transmitter. A 1:1 match means that there is no power reflected to the Tx by definition (and can be proved by measurement). Fine, but not related to my question if that is what you were responding to. By conservation of energy then all of the power must be radiated or lost as heat, (mostly in the coax). Uh, I'm not sure of that since there are other paths and I'm pretty sure power can be sent back into the driver of the transmitter. When the ATU gives a conjugate match, (ie the Tx sees a 1:1 vswr) the phase of the re-reflected wave at the antenna is 0, ie it is in-phase with the original forward wave and so any power that is not re-re-reflected again adds to the power supplied to the antenna. You seem to be mixing two things here. We have been talking about an ATU at the transmitter. The feed line cable is between the ATU and the antenna. You then talk about reflections both at the TX to ATU interface and the cable to antenna interface. You also make assumptions about the reflected power phase when it returns to the antenna. If you want to talk about the phase at that point you have to take into account the electrical length of the cable and *three* reflections. If the coax were lossless, and there were no losses in-the ATU, then ALL of the power that was supplied by the tx would be radiated regardless of the mismatch at coax to antenna interface. However, in reality even small coax losses add up to a significant loss when the mismatch at the coax to antenna interface is high due to the number of times that the power bounces up & down the coax suffering loss on each trip.. One poster has tried to say the reflections are all such that the phase of the reflected power ends up in phase with the original signal delivered to the antenna. But no one has done a power phase analysis. You seem to be saying the same thing. Power is reflected from the complex impedance of the antenna. Is that power not out of phase from the incident signal by some amount determined by the complex impedance? On traveling down the cable there is a phase shift by an amount related to the ratio of the electrical length and the wavelength. The ATU reflects some of the signal back up the cable and some is passed directly to the TX (since there is no or a very short cable between them) and is partially reflected by the TX, again with some phase shift determined by the two impedances (TX output and the ATU input). Another phase shift by the cable and the reflected wave arrives at the antenna. Is anyone going to tell me that all this has to add up to a reflected signal arriving *in phase* with the incident signal? I don't think you need to think about phase. The forward-going signal has no need to know about the reflected signal, so just think of them as being independent. One simply passes through the other. When they are reaching the antenna you damn well *do* need to know about phase. If the incident wave and the reflected wave returning to the antenna are 180 degrees out of phase the power subtracts! I understand Jeff correctly, if the ATU input presents a perfect 1-to-1 SWR to the transmitter output, it follows that none of power in the multiple reflections in the coax is getting back into the TX. I think you do *not* understand Jeff. The TX to ATU match has nothing to do with the energy reaching the TX, just the opposite, it means all power from the antenna passing through the ATU is absorbed by the TX. But I have not seen anyone say the ATU matches the TX impedance, only that it is a conjugate to the antenna. Should be not be a conjugate match to the impedance as seen looking into the coax at the TX end? There are three interfaces, TX/ATU, ATU/cable, cable/antenna. Each one can have reflections and each of the four parts of the system can absorb (or radiate in the case of the antenna) power. This is not so easy to analyze fully. At least until some of the fundamentals are revealed that we have not talked about. Therefore, if we ignore the losses in the ATU, all the TX output power must eventually, after countless ever-diminishing reflections between the antenna and the ATU output, be either reaching the antenna, or being lost in heating up the coax. A given reflection may "reach" the antenna, but will it contribute positively to the radiated field or counter it? I think we are looking at the scenario in two different ways. You are trying to see a 'snapshot' of the vectorial additions of all the forward and reverse signals, and I am simply looking at them as being independent entities, rattling up and down the coax, and progressively either escaping via the antenna feedpoint and being radiated, or heating up the coax. You seem to want to ignore the possibility that the reflected waves can interfere with your incident wave. This is entirely possible. Even if reflected energy is radiated it can cancel the incident wave and reduce your signal strength. Conservation of energy suggests that it can only reduce it in one direction by increasing it in another. If feeder reflections can have this effect it should be possible to demonstrate it in small model aerials at high frequencies. Conservation of energy? Just show the mechanics. The energy will take care of itself. -- Rick |
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On 10/5/2015 5:43 AM, Roger Hayter wrote:
rickman wrote: On 10/5/2015 4:17 AM, Ian Jackson wrote: In message , rickman writes You keep saying that the 1:1 match between the TX and the ATU prevents any power from being sent to the TX which is not true. You are confusing the power from the TX which is not reflected and the power reflected from the antenna which passes through the ATU to the TX. If the SWR meter between the TX output indicates a 1:1 SWR, then there can be NO power travelling between the ATU input and the TX output - ie there IS no reflected power. QED, surely? If you ignore the losses in the ATU, all the power that the mismatched antenna reflects, and that makes it back to the ATU output, MUST be re-reflected by the ATU output impedance, and head off back towards the antenna. This is because the reflected signal cannot heat up a lossless ATU, and the SWR meter says it isn't coming back through the ATU. It simply has nowhere to go except back down the coax. You saying something is true or imagining a SWR reading is not the same as understanding what is going on. What SWR reading are you imagining? Can you explain this in terms of the circuit analysis? The ATU consists of what circuit? The TX has some source impedance, what would that be? I don't think you can design an ATU circuit that will isolate the real source impedance of the TX from the reflected wave from the antenna. So, when you tune your transmitter-end ATU for minimum SWR *between the Tx and the ATU*, what exactly are you doing? You are making the feeder/aerilal/ATU combination look like a resistive load from the transmitter side, what are you making the ATU/transmitter combination look like from the aerial feeder side? It must be something that reflects back incident waves, otherwise your SWR meter wouldn't be reading 1.0. That seems to be the argument, and it sounds moderately convincing to me. I don't know what you are doing. I am talking about analysis of the circuits involved. If analyzing the circuits is not appropriate we can just drop this discussion. Any chance of coming up with a specific circuit? Forget tuning, let's set parameters for the TX, cable, antenna, calculate the circuit for the ATU and then we can see what happens at the various interfaces. -- Rick |
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On 10/5/2015 5:58 AM, Ian Jackson wrote:
In message , rickman writes On 10/5/2015 4:17 AM, Ian Jackson wrote: In message , rickman writes You keep saying that the 1:1 match between the TX and the ATU prevents any power from being sent to the TX which is not true. You are confusing the power from the TX which is not reflected and the power reflected from the antenna which passes through the ATU to the TX. If the SWR meter between the TX output indicates a 1:1 SWR, then there can be NO power travelling between the ATU input and the TX output - ie there IS no reflected power. QED, surely? If you ignore the losses in the ATU, all the power that the mismatched antenna reflects, and that makes it back to the ATU output, MUST be re-reflected by the ATU output impedance, and head off back towards the antenna. This is because the reflected signal cannot heat up a lossless ATU, and the SWR meter says it isn't coming back through the ATU. It simply has nowhere to go except back down the coax. You saying something is true or imagining a SWR reading is not the same as understanding what is going on. What SWR reading are you imagining? A typical HF station setup is TX - SWR meter - ATU - feeder - antenna. [Please no one start saying they don't use this configuration.] Can you explain this in terms of the circuit analysis? The only analysis required is to ask yourself why, after much careful twiddling with the knobs, you eventually get the SWR meter to show 1:1. Does this not indicate that no power is coming back to the TX from the ATU? [If not, what does it indicate?] Not sending power to the TX is not the same as reflecting all power back to the antenna. The ATU consists of what circuit? I haven't a clue. It's a large shiny black box with three silver knobs and two RF connectors. All the spec says is "Insertion loss 0dB". [Mind you, it was very, very expensive.] The TX has some source impedance, what would that be? Somewhere between not-a-lot and probably not-too-high (because, in operation, the TX doesn't get unduly hot). It really doesn't matter what it is. The SWR meter shows that zero power is coming back out the ATU input, so there is nothing to get re-reflected from the TX output. I don't think you can design an ATU circuit that will isolate the real source impedance of the TX from the reflected wave from the antenna. The purpose of what many of us (often slightly incorrectly) call an ATU is to present the TX output with its design load impedance. In most cases, this is 50 ohms resistive. The TX output impedance is really of no concern to the designers of the ATU. As long as they can produce an ATU that will present the TX with a load of 50 ohms resistive, then it will do its job perfectly - and this will be indicated by the 50 ohm SWR meter between the TX and the ATU showing 1:1. And when an SWR meter shows 1:1, it means that there is no power coming back towards the TX output. -- Rick |
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On 10/5/2015 7:29 AM, Jeff wrote:
NO, as above, if the ATU is adjusted for a 1:1 match then there is no power reflected back to the TX. You are saying the ATU doesn't reflect power back from the TX. If the match is 1:1 that is true. But you aren't considering the power reflected from the antenna. The antenna reflects power back to the ATU and there is nothing in the ATU to prevent that power from being handed to the transmitter. The whole function of the ATU is to provide a 1:1 match for the Tx, when that is the case NO POWER GOES BACK TO THE TX. That is not technically correct. A 1:1 match prevents TX power from being reflected, it does not prevent other power from being sent to the TX. Yes the antenna reflects power back towards the ATU, BUT the ATU then reflects ALL of that power back again towards the antenna and none into the TX (assuming that the ATU achieved a conjugate match.) That is the part I am trying to understand. No one seems to be able to explain how the ATU does all this, present a 50 ohm resistive load to the TX, reflect all power back toward the antenna *and* align the phase of that reflected signal so it is phase aligned to the incident radiation. I am told it is a "black box" with a knob. Ok, I'll accept that this is possible, but I'd like to see the math that shows it. So far no one seems to understand the math. If that is true, then I'll stop talking about this. I'm not trying to bug people. When adjusted for a 1:1 match there are no reflections between the Tx and ATU, but there are multiple reflections between the ATU and teh antenna. That is where a lot of power is dissipated in the antenna is a poor match to the feeder impedance. No one is talking about the power from the TX being reflected back to the TX by the ATU. The reflected wave will start off at the antenna out of phase with the forward wave, (the actual phase depends on the complex impedance of the mismatch), what the conjugate match that the ATU provides, when adjusted so that there is a 1:1 match and no power reflected power sent to the Tx, This doesn't cover the reflected power from the antenna, just the power from the TX. NO. It covers the reflected power from the antenna. There is no reflected power between the ATU and Tx if the match is 1:1. You keep saying that, but it isn't true. Power can come through the ATU circuit into the TX even when there is a 1:1 impedance match. In a silly case lightning can strike the antenna. I don't think the 1:1 impedance match will prevent that power from reaching the TX. is a phase shift such that the re-reflected wave from the ATU towards the antenna is in phase with the original forward wave, so when it reaches the antenna the portion of that re-reflected wave that is not bounced back again down the coax by the mismatch is delivered to the antenna. How does that work? There is phase shift in the reflections which may be compensated for by the ATU, but there is also phase shift in the cable. The ATU applies a conjugate match at the end of the cable, the impedance that it sees, and applies the conjugate of, is the impedance of the antenna modified by the length of cable. It just so happens that the conditions for a 1:1 vswr, and that of a conjugate match, are that match causes the phase at the antenna end of the cable to be the same phase as the original forward wave. That is the physics of a conjugate match. This is what I'd like to see. Any math that you can share? NO, no power gets back to the tx if the ATU is adjusted for a 1:1 match. after all that is the definition of a 1:1 match; no reflected power. You keep saying that the 1:1 match between the TX and the ATU prevents any power from being sent to the TX which is not true. You are confusing the power from the TX which is not reflected and the power reflected from the antenna which passes through the ATU to the TX. YES IT IS TRUE. I am not confusing anything. Yes, what you say it true, no reflected power... "read my lips, no new taxes". But the 1:1 match does not imply anything other than the reflected power from the TX. Is anyone going to tell me that all this has to add up to a reflected signal arriving *in phase* with the incident signal? YES. I may have to look at the math for this. How does the ATU reflect all the power from the antenna back to the antenna? I thought only an open or a short can reflect all the power. Please look it up, do some experiments yourself. Try some simulations in Spice or similar. You will find that I am correct. Jeff -- Rick |
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