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swr question
Assuming I have an antenna that is perfect on 3.8 MHz. Perfect being
defined, as I am feeding it with exactly 1/2 electrical wave length of 50 Ohm feedline and it is 1:1 SWR measured at the source end. What would the SWR be if I substituted the 50 Ohm feedline with a 1/2 wavelength of 72 Ohm feedline? The head of the pin is now open for dancing... |
Fred W4JLE wrote: Assuming I have an antenna that is perfect on 3.8 MHz. Perfect being defined, as I am feeding it with exactly 1/2 electrical wave length of 50 Ohm feedline and it is 1:1 SWR measured at the source end. What would the SWR be if I substituted the 50 Ohm feedline with a 1/2 wavelength of 72 Ohm feedline? The head of the pin is now open for dancing... Hi Fred, It would still be 1:1. Doesnt matter about the characteristic impedance of the transmission line as long as you use 1/2 electrical wavelength. Gary N4AST |
Sounds right to me, the 50 ohm load of the ant is "reflected" back by a
1/2 wave line... .... so, I am probably wrong... grin Warmest regards, John wrote in message ups.com... Fred W4JLE wrote: Assuming I have an antenna that is perfect on 3.8 MHz. Perfect being defined, as I am feeding it with exactly 1/2 electrical wave length of 50 Ohm feedline and it is 1:1 SWR measured at the source end. What would the SWR be if I substituted the 50 Ohm feedline with a 1/2 wavelength of 72 Ohm feedline? The head of the pin is now open for dancing... Hi Fred, It would still be 1:1. Doesnt matter about the characteristic impedance of the transmission line as long as you use 1/2 electrical wavelength. Gary N4AST |
"Fred W4JLE" wrote in message ... Assuming I have an antenna that is perfect on 3.8 MHz. Perfect being defined, as I am feeding it with exactly 1/2 electrical wave length of 50 Ohm feedline and it is 1:1 SWR measured at the source end. What would the SWR be if I substituted the 50 Ohm feedline with a 1/2 wavelength of 72 Ohm feedline? The head of the pin is now open for dancing... Hi Fred, Assuming that when you say the antenna is 'perfect' with a 1:1 SWR on a 50-ohm line it means that the input impedance to the line is 50 + j0 ohms. Since the line is 1/2 wl,and presumed lossless, the terminal impedance of the antenna is also 50 + j0. However, when you replace the 50-ohm line with one of 72 ohms the SWR on the line changes from 1:1 to 1.44:1. But since it's length is 1/2 wl the input impedance is still 50 + j0 ohms. The 1.44 SWR results from the 1.44 mismatch between the 72-ohm line and the 50 + j0 ohm impedance of the termination, the terminal impedance of the antenna. The 1/2 wl consists of two 1/4 wl lines in series. At the point 1/4 wl toward the input the line impedance is transformed up to103.68 + j0, (72 x 1.44) which is then transformed back to 50 + j0 (72/1.44) along the 1/4 wl section returning to the input of the line. Walt, W2DU |
This is a good illustration of one of Reg's hot buttons -- that "SWR
meters" don't actually measure the SWR on a transmission line, but rather are reporting the degree of match or mismatch at the meter's insertion point. In the case of 50 ohm line, it turns out that the "SWR" reported by the meter is actually the SWR on the 50 ohm transmission line, assuming that the meter is designed for use in a 50 ohm system and reads properly in that environment. When you substitute the 72 ohm line, the SWR meter will still read 1:1 because the impedance at its insertion point is still 50 ohms, but the SWR on the 72 ohm transmission line is actually 1.44:1. So the answer depends on what you mean by "what would be the SWR": The SWR on the 72 ohm transmission line will be 1.44:1 The SWR meter will read 1:1 If there is a 50 ohm line between your rig and the 72 ohm line, its SWR will be 1:1. Roy Lewallen, W7EL Fred W4JLE wrote: Assuming I have an antenna that is perfect on 3.8 MHz. Perfect being defined, as I am feeding it with exactly 1/2 electrical wave length of 50 Ohm feedline and it is 1:1 SWR measured at the source end. What would the SWR be if I substituted the 50 Ohm feedline with a 1/2 wavelength of 72 Ohm feedline? The head of the pin is now open for dancing... |
"The head of the pin is now open for dancing..."
.... ahhh, missed this, so you now much choose: 1) I am going to use this for a practical application--so I really don't care how it works (like time)--I simply need something which works and functions on a equiv to a 1:1, frequency specific, voltage transformer. 2) I want to debate the physics of one-half wavelength lines. .... since my basic motto/bmod (basic method of operation) is "whatever works!" ... my choice is clear... grin Warmest regards, John "Fred W4JLE" wrote in message ... Assuming I have an antenna that is perfect on 3.8 MHz. Perfect being defined, as I am feeding it with exactly 1/2 electrical wave length of 50 Ohm feedline and it is 1:1 SWR measured at the source end. What would the SWR be if I substituted the 50 Ohm feedline with a 1/2 wavelength of 72 Ohm feedline? The head of the pin is now open for dancing... |
Fred W4JLE wrote:
Assuming I have an antenna that is perfect on 3.8 MHz. Perfect being defined, as I am feeding it with exactly 1/2 electrical wave length of 50 Ohm feedline and it is 1:1 SWR measured at the source end. What would the SWR be if I substituted the 50 Ohm feedline with a 1/2 wavelength of 72 Ohm feedline? The SWR on the 72 ohm feedline would measure 1.44:1. You do have an SWR meter calibrated for 72 ohms, don't you? :-) Your 50 ohm SWR meter will measure 1:1, but as Reg says, it is merely measuring the degree of match to your transmitter designed for 50 ohm loads. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
Cecil Moore wrote: Fred W4JLE wrote: Assuming I have an antenna that is perfect on 3.8 MHz. Perfect being defined, as I am feeding it with exactly 1/2 electrical wave length of 50 Ohm feedline and it is 1:1 SWR measured at the source end. What would the SWR be if I substituted the 50 Ohm feedline with a 1/2 wavelength of 72 Ohm feedline? The SWR on the 72 ohm feedline would measure 1.44:1. You do have an SWR meter calibrated for 72 ohms, don't you? :-) Your 50 ohm SWR meter will measure 1:1, but as Reg says, it is merely measuring the degree of match to your transmitter designed for 50 ohm loads. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- It was stipulated in the original post that the swr was being measured at the source end for the 50 ohm line. If all one changes is the impedance of a 1/2 electrical wavelength line, and nothing else, the answer is 1:1. Did not say anything about measuring the swr anywhere but at the source. Whatever the antenna Z is in the 50 ohm example that gives a 1:1 match, determines the swr bridge characteristics. If the antenna Z was 72 ohms, then the bridge is 72 ohms for a 1:1 match. Change the transmission line to 72 ohms, still a 1:1. Gary N4AST |
wrote:
It was stipulated in the original post that the swr was being measured at the source end for the 50 ohm line. If all one changes is the impedance of a 1/2 electrical wavelength line, and nothing else, the answer is 1:1. Nope, not if a 72 ohm SWR meter is being used. An SWR meter calibrated for the transmission line Z0 of 72 ohms will read 1.44:1 even if the 50 ohm transmitter is happy with the 50+j0 ohm virtual impedance being presented to it. I have SWR meters calibrated for 50, 75, 300, 450, and 600 ohms - doesn't everybody? -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
Cecil Moore wrote: Cecil Moore wrote: wrote: It was stipulated in the original post that the swr was being measured at the source end for the 50 ohm line. If all one changes is the impedance of a 1/2 electrical wavelength line, and nothing else, the answer is 1:1. Nope, not if a 72 ohm SWR meter is being used. An SWR meter calibrated for the transmission line Z0 of 72 ohms will read 1.44:1 even if the 50 ohm transmitter is happy with the 50+j0 ohm virtual impedance being presented to it. I have SWR meters calibrated for 50, 75, 300, 450, and 600 ohms - doesn't everybody? Please note that it was ***NOT*** stipulated in the original post that the SWR meter was calibrated for 50 ohms. "What is the SWR?" was the question. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- SWR meter reads 1:1 at the source with 50 ohm line, that will tell you that the swr bridge Z and the antenna Z are the same initially (1/2 wave line). Changing to 72 ohm line will still be 1:1. Gary N4AST |
wrote:
Cecil Moore wrote: Fred W4JLE wrote: Assuming I have an antenna that is perfect on 3.8 MHz. Perfect being defined, as I am feeding it with exactly 1/2 electrical wave length of 50 Ohm feedline and it is 1:1 SWR measured at the source end. What would the SWR be if I substituted the 50 Ohm feedline with a 1/2 wavelength of 72 Ohm feedline? The SWR on the 72 ohm feedline would measure 1.44:1. You do have an SWR meter calibrated for 72 ohms, don't you? :-) Your 50 ohm SWR meter will measure 1:1, but as Reg says, it is merely measuring the degree of match to your transmitter designed for 50 ohm loads. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- It was stipulated in the original post that the swr was being measured at the source end for the 50 ohm line. If all one changes is the impedance of a 1/2 electrical wavelength line, and nothing else, the answer is 1:1. Did not say anything about measuring the swr anywhere but at the source. Whatever the antenna Z is in the 50 ohm example that gives a 1:1 match, determines the swr bridge characteristics. If the antenna Z was 72 ohms, then the bridge is 72 ohms for a 1:1 match. Change the transmission line to 72 ohms, still a 1:1. Gary N4AST VSWR stands for Voltage Standing Wave Ratio. It's supposed to be the ratio of the greatest voltage on a transmission line to the least voltage on the same line. On the line with a 72 ohm Z0, and a 50 ohm load, there exists a standing wave, and the ratio of maximum to minimum is 1.44 whether you measure it with a 50 ohm bridge at the beginning or not. If you define SWR as what a 50 ohm SWR bridge measures, you haven't quite grasped the concept. 73, Tom Donaly, KA6RUH |
wrote:
SWR meter reads 1:1 at the source with 50 ohm line, that will tell you that the swr bridge Z and the antenna Z are the same initially (1/2 wave line). Changing to 72 ohm line will still be 1:1. Here is the original question again: "What would the SWR be if I substituted the 50 Ohm feedline with a 1/2 wavelength of 72 Ohm feedline?" Did he ask what would a 50 ohm SWR meter read? NO! He asked "WHAT WOULD THE SWR BE ..." The SWR would be 72/50=1.44:1 on the 72 ohm feedline. That answers the original question. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
It was stipulated in the original post that the swr was being
measured at the source end for the 50 ohm line. He said, "WHAT WOULD THE SWR BE ...?" not what would a 50 ohm SWR meter measure. The SWR would be 1.44:1 on the 72 ohm feedline no matter what the 50 ohm SWR meter erroneously measured. If all one changes is the impedance of a 1/2 electrical wavelength line, and nothing else, the answer is 1:1. Did not say anything about measuring the swr anywhere but at the source. Go back and read the question again, Tom. He asked: "What would the SWR be if I substituted the 50 Ohm feedline with a 1/2 wavelength of 72 Ohm feedline?" Did he say, "What SWR would be measured?" NO!!! Did he say, "What would the SWR be ...?" YES!!! ***THE SWR WOULD BE 1.44:1*** on the 72 ohm feedline. You guys are not answering the question that he asked. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
Reg Edwards wrote:
You will be pleased to hear I'm back on Sierra Valley, Californian Red tonight. Peter Vella Merlot for me. My doctor told me to drink two glasses of red wine a day so I use ~400 ml iced-tea glasses. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
Tom Donaly wrote:
VSWR stands for Voltage Standing Wave Ratio. It's supposed to be the ratio of the greatest voltage on a transmission line to the least voltage on the same line. On the line with a 72 ohm Z0, and a 50 ohm load, there exists a standing wave, and the ratio of maximum to minimum is 1.44 whether you measure it with a 50 ohm bridge at the beginning or not. If you define SWR as what a 50 ohm SWR bridge measures, you haven't quite grasped the concept. And Fred didn't ask what an SWR bridge would measure. He asked: "What would the SWR be ... ?" You're right. The SWR would be 1.44:1. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
Cecil Moore wrote:
Go back and read the question again, Tom. He asked: Sorry, Tom, somehow I screwed up the attributes. I don't think you posted what I responded to. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
OK CECIL!!!
Leave it up to you to pull the rug out from under me... If I have a transmitter that has 50 ohm out, and it is going to hook to a 50 ohm cable (and I can't see how this coax is terminated) why would I ever choose anything other than a 50 ohm calibrated swr meter to measure it with? Is that what I have seen on FS meters before (a rise in apparent radiation from the coax shield--and yet match looks good) and the 1/2 wave coax is really now part of the antenna? And, if the meter didn't give me the right reading, and cooked my "BEEG LEENEAIR" could I sue the manufacturer, buy a yacht and live in the Bahamas, drinking Peter Vella Merlot? grin Warmest regards, John "Cecil Moore" wrote in message ... Tom Donaly wrote: VSWR stands for Voltage Standing Wave Ratio. It's supposed to be the ratio of the greatest voltage on a transmission line to the least voltage on the same line. On the line with a 72 ohm Z0, and a 50 ohm load, there exists a standing wave, and the ratio of maximum to minimum is 1.44 whether you measure it with a 50 ohm bridge at the beginning or not. If you define SWR as what a 50 ohm SWR bridge measures, you haven't quite grasped the concept. And Fred didn't ask what an SWR bridge would measure. He asked: "What would the SWR be ... ?" You're right. The SWR would be 1.44:1. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
Cecil:
I screwed that up... in the above, should have been... "If I have a transmitter that has 50 ohm out, and it is going to hook to a 72 ohm cable..." The 72 ohm cable being t he only difference... and point being--don't I only care I am presenting a 50 ohm load to the transmitter? Warmest regards, John "John Smith" wrote in message ... OK CECIL!!! Leave it up to you to pull the rug out from under me... If I have a transmitter that has 50 ohm out, and it is going to hook to a 50 ohm cable (and I can't see how this coax is terminated) why would I ever choose anything other than a 50 ohm calibrated swr meter to measure it with? Is that what I have seen on FS meters before (a rise in apparent radiation from the coax shield--and yet match looks good) and the 1/2 wave coax is really now part of the antenna? And, if the meter didn't give me the right reading, and cooked my "BEEG LEENEAIR" could I sue the manufacturer, buy a yacht and live in the Bahamas, drinking Peter Vella Merlot? grin Warmest regards, John "Cecil Moore" wrote in message ... Tom Donaly wrote: VSWR stands for Voltage Standing Wave Ratio. It's supposed to be the ratio of the greatest voltage on a transmission line to the least voltage on the same line. On the line with a 72 ohm Z0, and a 50 ohm load, there exists a standing wave, and the ratio of maximum to minimum is 1.44 whether you measure it with a 50 ohm bridge at the beginning or not. If you define SWR as what a 50 ohm SWR bridge measures, you haven't quite grasped the concept. And Fred didn't ask what an SWR bridge would measure. He asked: "What would the SWR be ... ?" You're right. The SWR would be 1.44:1. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
John Smith wrote:
Leave it up to you to pull the rug out from under me... If I have a transmitter that has 50 ohm out, and it is going to hook to a 50 ohm cable (and I can't see how this coax is terminated) why would I ever choose anything other than a 50 ohm calibrated swr meter to measure it with? Some of us want to know what the SWR on the feedline is. That's how we calculate feedline losses. I get a kick out of some ham saying, "I'm running a 66' dipole on 75m and my SWR is 1.1:1." All that means is that the virtual impedance at the tuner input is probably 45 ohms or 55 ohms. But what is the SWR at the output of the tuner where it matters the most? -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
John Smith wrote:
The 72 ohm cable being t he only difference... and point being--don't I only care I am presenting a 50 ohm load to the transmitter? I won't presume to know what you care about. I care about the SWR on my feedline. I have homebrew SWR meters calibrated for 300, 450, and 600 ohms. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
Cecil:
Darn!!! I was hoping ya cared... just kidding I think you took that the wrong way... I didn't mean to insinuate there was nothing to worry about... or that I would not be interested in the long run about losses... but either way, the answer is just as appreciated... grin Warmest regards, John "Cecil Moore" wrote in message ... John Smith wrote: The 72 ohm cable being t he only difference... and point being--don't I only care I am presenting a 50 ohm load to the transmitter? I won't presume to know what you care about. I care about the SWR on my feedline. I have homebrew SWR meters calibrated for 300, 450, and 600 ohms. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
John Smith wrote:
Cecil: Darn!!! I was hoping ya cared... just kidding I didn't say I didn't care. I said I didn't presume. :-) -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
"Cecil Moore" wrote in message ... Reg Edwards wrote: You will be pleased to hear I'm back on Sierra Valley, Californian Red tonight. Peter Vella Merlot for me. My doctor told me to drink two glasses of red wine a day so I use ~400 ml iced-tea glasses. -- 73, Cecil http://www.qsl.net/w5dxp Cecil, if I drank that much wine every day I'd believe the SWR at the input of the 72-ohm line is 1:1 too, but it ain't. I'm concerned with the guys who think it is. So please let me be totally elementary in explaining why the SWR on the 72-ohm line is 1.44 everywhere on the line, including the input, yet still has a 50-ohm input impedance. Step 1. We have a lossless 50-ohm line of random length feeding a 50-ohm source to an antenna with terminal impedance of 50 + j0 ohms. The input voltage is 1 v and the input current is 0.02 a. Z = E/R, and 1/0.02 = 50 verifying that the Zo of the line is 50-ohms. The power absorbed by the antenna is IE = E*/R = 0.02 watts. The SWR is 1:1 everywhere. Step 2. We now insert a lossless 72-ohm, 1/2 wl line between the antenna and the output end of the 50-ohm line. Does the power absorbed in the antenna change? NO. Do the forward voltage and current on the 50-ohm line change? NO. Are there any reflections on the 50-ohm line? NO. Are the forward voltage and current in the 75-ohm line 1 v and 0.02 a.? NO. Are there reflections on the 72-ohm line? YES. Is the input impedance of the 72-ohm line 50 + j0 ohms? YES. Is the SWR at the input of the 72-ohm line1:1? NO. Is the SWR at the input of the 72-ohm line 1.44:1? YES. Does the antenna still absorb 0.02 watts? YES. Step 3. To understand why the input impedance of the 72-ohm line is 50 + j0 with an SWR = 1.44 we need to understand the voltage and current reflection coefficients, rho as well as the voltage and current transmission coefficients, tau. Step 4. In going to a higher line impedance the voltage reflection coefficient rhoE is positive, and the current reflection coefficient rhoI is negative; The reverse is true when going to a lower line impedance, as when going from the 72-ohm line into the 50-ohm load. Step 5. In going to a higher line impedance the voltage transmission coefficient tauE = (rhoE + 1) and the current transmission coefficient tauI = (rhoI - 1). The reverse is true when going to a lower line impedance. Step 6. For an SWR of 1.44, rho = 0.1803. Therefore, when going from a 50-ohm line to a 72-ohm line, tauE = 1.1803 and tauI = 0.8197. Consequently, the forward voltage in the 72-ohm line is 1.1803v and the forward current is 0.02 x 0.8197 = 0.01639 a. Note that 1.180/0.01639 = 71.996, which would be 72 if we used more significant figures. Step 7. When forward voltage 1.1803 v reaches the 50-ohm antenna load, the reflection coefficient 0.1803 x 1.1803 = 0.2128 v, which subtracts from the 1.1803 v at the input of the 72-ohm line, making the total voltage at the input 0.9675 v. Step 8. When forward current 0.01639 a reaches the 50-ohm antenna load, the current reflection coefficient 0.1803 x 0.0.01639 a = 0.002956 a, which adds to the 0.01639 a, making the total current at the input of the 72-ohm line 0.01935 a. Step 9. Dividing the total voltage at the input of the 72-ohm line by the total current, we get 0.9675/0.01935 = 50.01 ohms, the input resistance of the 72-ohm line with a 1.44:1 SWR on the line. Step 10. Consequently, we see that the addition of the reflected voltage and current to the forward voltage and current yield an impedance at the input of the 72-ohm line that is 50 ohms, not 72 ohms. It is the effect of the 1.44:1 SWR on the line that has changed the line impedance (not the characteristic impedance) to 50 ohms. I hope this helps in understanding why the SWR at the input of the 72-ohm line is 1.44:1, and NOT 1:1. Walt, W2DU |
Walter Maxwell wrote:
I hope this helps in understanding why the SWR at the input of the 72-ohm line is 1.44:1, and NOT 1:1. Very good stuff, Walt, as usual. It is possible that Gary misunderstood the question. He apparently thought the question was: What SWR will a 50 ohm SWR meter indicate and of course, a 50 ohm SWR meter will erroneously report an SWR of 1:1 at the 50 ohm (current maximum) point on the 72 ohm feedline. If we move the 50 ohm SWR meter to the 103.7 ohm (current minimum) point on the 72 ohm feedline, it will erroneously report an SWR of 2.1:1. But, on the 72 ohm feedline, the SWR is, of course, 1.44:1 at both the 50 ohm point and the 103.7 point and at all other points up and down the feedline (neglecting losses). -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
My question was totally answered by Walt W2DU. Thank you for the detailed
explaination. |
On Mon, 30 May 2005 16:12:44 -0500, Cecil Moore
wrote: Fred W4JLE wrote: Assuming I have an antenna that is perfect on 3.8 MHz. Perfect being defined, as I am feeding it with exactly 1/2 electrical wave length of 50 Ohm feedline and it is 1:1 SWR measured at the source end. What would the SWR be if I substituted the 50 Ohm feedline with a 1/2 wavelength of 72 Ohm feedline? The SWR on the 72 ohm feedline would measure 1.44:1. You do have an SWR meter calibrated for 72 ohms, don't you? :-) Your 50 ohm SWR meter will measure 1:1, but as Reg says, it is merely measuring the degree of match to your transmitter designed for 50 ohm loads. Gentlemen, I know most of you are much more knowledgeable in some of this, but with all your nit-picking over the impedances, etc., may I interject that something is being overlooked? The antenna feedline has some loss, however little, that may affect the swr at the "source", which I interpret to be at the radio. 100 watts may be transmitted, 90 watts received at the antenna, 9 watts reflected, and 8.1 watts received back at the rig for reflected power. My numbers are hypothetical, but you get the idea. However, your discussions concerning SWR bridge impedance have led me to wonder how accurate the swr meter in my rig is concerning the antennas I use. I feed a dipole with 72 ohm into my rig, an IC-706 MKII which expects a 50 ohm load. My attitude is that it must be ok if the rig sees it as low as it should measure it as it sees it. -- 73 for now Buck N4PGW |
Cecil Moore wrote: Walter Maxwell wrote: I hope this helps in understanding why the SWR at the input of the 72-ohm line is 1.44:1, and NOT 1:1. Very good stuff, Walt, as usual. It is possible that Gary misunderstood the question. He apparently thought the question was: What SWR will a 50 ohm SWR meter indicate and of course, a 50 ohm SWR meter will erroneously report an SWR of 1:1 at the 50 ohm (current maximum) point on the 72 ohm feedline. If we move the 50 ohm SWR meter to the 103.7 ohm (current minimum) point on the 72 ohm feedline, it will erroneously report an SWR of 2.1:1. But, on the 72 ohm feedline, the SWR is, of course, 1.44:1 at both the 50 ohm point and the 103.7 point and at all other points up and down the feedline (neglecting losses). -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- Yep, it is possible I did, won't be the first time. The orginal question stated that the swr was somehow measured at the source, how else could one know it was 1:1. The measurement instrument was not specified. Assume that it is the average ham with a 50 ohm swr bridge, an antenna analyzer, and a 50 ohm RLB. If one changes to 1/2 wave 72 ohm cable, using the same measurement instrument you get the same results. I understand the mechanics of the actual VSWR on mismatched lines, but I intrepreted the the question differently I guess. My handy Smith Chart program reports a 1:1 VSWR, of course it is only at the source, at 90 degrees it reports 2:1. Wonder what that program uses to calculate swr? Gary N4AST Gary N4AST |
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