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Question on SWR
I make here reference to the well-known chart, shown in almost all the ARRL
Antenna Handbooks, setting the relationship among the real SWR (i.e. that measured at antenna), the SWR measured at transmitter and the "line loss". In the text they explain that the "line loss" to be considered when using the chart is the line loss under matched conditions (i.e. that given by the line manufacturer), that is without taking into account the extra loss caused by SWR. They make the example of a line having a (matched) loss of 1.0dB with an SWR of 4.5 at antenna. The graph shows that the corresponding SWR at transmitter is 3.0. Unless I am wrong, a simple calculation shows that, in the considered example, the SWR at transmitter is about 2.3 rather than 3.0. Here it goes (please note that, for the assumed SWR at antenna of 4.5, the extra loss caused by SWR is just 1.0dB): - actual loss on the forward wave: 1.0dB (matched) + 1.0dB (extra by SWR), for a total of 2.0dB - return loss corresponding to an SWR at antenna of 4.5: 4.0dB - actual loss on the reflected wave: 1.0dB (matched) + 1.0dB (extra by SWR), for a total of 2.0dB - return loss at transmitter: 8.0dB - SWR measured at transmitter (corresponding to a return loss of 8.0dB): about 2.3 It is interesting to note that the chart would give an identical result if by "line loss" they would mean the total line loss (that is also including the extra loss due to SWR) rather than just the matched line loss (as they state in the text). Any comment? Thanks and 73 Tony I0JX |
#2
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Question on SWR
I realized my error!
In my budget I counted the extra line loss caused by SWR twice: - first time when I have added 1 dB to the loss of the forward and the return wave - second time when I took some power out of the antenna (to account for an SWR of 4.5) and delivered it back to the transmitter ARRL is always correct! Sorry for the useless noise Tony I0JX |
#3
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Question on SWR
"Antonio Vernucci" wrote in message ... I realized my error! In my budget I counted the extra line loss caused by SWR twice: - first time when I have added 1 dB to the loss of the forward and the return wave - second time when I took some power out of the antenna (to account for an SWR of 4.5) and delivered it back to the transmitter ARRL is always correct! All Hail the ARRL! |
#4
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Question on SWR
"Antonio Vernucci" wrote in
: I realized my error! .... ARRL is always correct! The ARRL information on "extra loss due to VSWR" is may be incomplete in that it may not the assumptions that underly the formula used for the graphs. The very concept that SWR necessarily increases loss from the matched line loss figure is flawed. Try the line loss calculator at http://www.vk1od.net/tl/tllc.php to calculate the loss in 1m of RG58 at say 2MHz with loads of 5 and 500 ohms (both VSWR=10). Now refer to the ARRL... does it explain the difference? Owen |
#5
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Question on SWR
Owen Duffy wrote:
The ARRL information on "extra loss due to VSWR" is may be incomplete in that it may not the assumptions that underly the formula used for the graphs. It is possible for a feedline with a high SWR to have lower loss than the matched-line loss. For instance, if we have 1/8WL of feedline with a current miminum in the middle of the line, the losses at HF will be lower than matched line loss because I^2*R losses tend to dominate at HF. -- 73, Cecil http://www.w5dxp.com "According to the general theory of relativity, space without ether is unthinkable." Albert Einstein |
#6
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Question on SWR
In message , Cecil Moore
writes Owen Duffy wrote: The ARRL information on "extra loss due to VSWR" is may be incomplete in that it may not the assumptions that underly the formula used for the graphs. It is possible for a feedline with a high SWR to have lower loss than the matched-line loss. For instance, if we have 1/8WL of feedline with a current miminum in the middle of the line, the losses at HF will be lower than matched line loss because I^2*R losses tend to dominate at HF. I'd never thought of that. I suppose it applies to any situation where the feeder is electrically short, and the majority of the current is less than it would be when matched. I presume that the moral is that formulas only really work when the feeder is electrically long enough for you to be concerned about what the losses might be. -- Ian. |
#7
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Question on SWR
I'd never thought of that. I suppose it applies to any situation where the
feeder is electrically short, and the majority of the current is less than it would be when matched. I presume that the moral is that formulas only really work when the feeder is electrically long enough for you to be concerned about what the losses might be. -- Conversely, for a very short line closed on 5 ohm (instead of 500 ohm), the extra loss caused by SWR would be higher than that shown on the ARRL graph (apart from the fact that, when attenuation is so low, the extra attenuation is generally not of much interest, nor it can be read on the ARRL chart). Evidently the ARRL chart shows some average between the two cases. On the other hand they probably had no better way to synthetically illustrate a concept without giving too many details. 73 Tony I0JX |
#8
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Question on SWR
Ian Jackson wrote:
In message , Cecil Moore writes Owen Duffy wrote: The ARRL information on "extra loss due to VSWR" is may be incomplete in that it may not the assumptions that underly the formula used for the graphs. It is possible for a feedline with a high SWR to have lower loss than the matched-line loss. For instance, if we have 1/8WL of feedline with a current miminum in the middle of the line, the losses at HF will be lower than matched line loss because I^2*R losses tend to dominate at HF. I'd never thought of that. I suppose it applies to any situation where the feeder is electrically short, and the majority of the current is less than it would be when matched. I presume that the moral is that formulas only really work when the feeder is electrically long enough for you to be concerned about what the losses might be. That's a good way of putting it, but it only applies to the generalized ARRL chart which takes no account of the actual load impedance or the actual feedline length. Owen's explicit method should get it right in all cases. If you select say 0.125 wavelengths of RG213 in Owen's online calculator, the predicted loss with a 100 ohm load resistance is *less* than the matched loss. If you change the load to 25 ohms, the predicted loss is *greater* than the matched loss. Both of these results make perfect physical sense because the largest part of the loss is proportional to the square of the current, which will be greater with the lower-resistance load. The two different resistances correctly give different results, yet they both have a VSWR of 2 (based on the 50-ohm system impedance). This shows that VSWR does not contain sufficient information to give an explicit single-valued result. Owen's program will accept a VSWR input, but it correctly posts a bold red warning that the result is an estimate. If you let the program select the worst-case load impedance for the supplied value of VSWR, you're back on track and it can calculate an explicit result. Although we're debating fractions of a milliBel here, the debate has shown how often the terms "VSWR" and "load impedance" are used interchangeably - which they aren't. It isn't a big mistake here, but it can be in other applications. For example, a solid-state PA designed for a 50 ohm load will respond very differently to load *impedances* of 100 or 25 ohms, yet the load *VSWR* is the same in both cases. -- 73 from Ian GM3SEK |
#9
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Question on SWR
Ian Jackson wrote:
I'd never thought of that. I suppose it applies to any situation where the feeder is electrically short, and the majority of the current is less than it would be when matched. I presume that the moral is that formulas only really work when the feeder is electrically long enough for you to be concerned about what the losses might be. I suspect the ARRL additional loss due to SWR charts are based on 1/2WL increments of feedlines. -- 73, Cecil http://www.w5dxp.com "According to the general theory of relativity, space without ether is unthinkable." Albert Einstein |
#10
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Question on SWR
"Owen Duffy" wrote in message ... "Antonio Vernucci" wrote in : I realized my error! ... ARRL is always correct! The ARRL information on "extra loss due to VSWR" is may be incomplete in that it may not the assumptions that underly the formula used for the graphs. The very concept that SWR necessarily increases loss from the matched line loss figure is flawed. Try the line loss calculator at http://www.vk1od.net/tl/tllc.php to calculate the loss in 1m of RG58 at say 2MHz with loads of 5 and 500 ohms (both VSWR=10). Now refer to the ARRL... does it explain the difference? Owen yeah, when you use the full complex Z0 and probably the full transmission line equations it gets a bit more complex. but for amateur use that graph is close enough. the difference between 5 and 500 ohm loads of .07db or so for 100m just ain't worth quibbling about for normal amateur hf use. unless you want to argue it out with cecil. |
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