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Brainteaser
Last week I posted a brainteaser over on qrz.com. Nobody figured
it out. I wonder how many on r.r.a.a can figure it out. Given: A one second long lossless transmission line with a steady-state forward power of 200 watts and a reflected power of 100 watts. How many joules are contained in that feedline? -- 73, Cecil, W5DXP |
Brainteaser
I had Beaker set up an experiment to test this, and after he put out
the fire he started in his hair when the first load he tried melted a hole in the lab bench, he finally got it set up right. He determined that the mass of the line increased by about 3.338 femtograms when powered as compared with it unpowered. We're now looking into what to do with the line, as it's cluttering up the lab, and we're open to suggestions. Regards, Bunsen Cecil Moore wrote: Last week I posted a brainteaser over on qrz.com. Nobody figured it out. I wonder how many on r.r.a.a can figure it out. Given: A one second long lossless transmission line with a steady-state forward power of 200 watts and a reflected power of 100 watts. How many joules are contained in that feedline? -- 73, Cecil, W5DXP |
Brainteaser
Cecil, W5DXP wrote:
"How many joules are contained in that feedline?" I`ll speculate that after one second, 200 joules are contained in the forward wave on that line. Then, after two seconds, another 100 joules has been reflected back toward the line feedpoint where it opposes growth of power input to the line. Total joules on the line is 300. Forward power minus the reflected power equals 100 watts being supplied by the generator to the load with 200 watts forward power and 100 watts reflrcted power in the line. Best regards, Richard Harrison, KB5WZI |
Brainteaser
Richard Harrison wrote:
Cecil, W5DXP wrote: "How many joules are contained in that feedline?" Total joules on the line is 300. That can be shown to be true by noting that during the transient buildup to steady-state, 300 joules sourced by the generator have not yet reached the load. That remains true until the generator is powered down, i.e. all during steady-state. Forward power minus the reflected power equals 100 watts being supplied by the generator to the load with 200 watts forward power and 100 watts reflrcted power in the line. But what about the people who say there's no energy in the reflected wave? Reckon how they sweep all those joules, whose energy must be conserved, under the rug? :-) -- 73, Cecil http://www.qsl.net/w5dxp |
Brainteaser
One should also carefully consider the
more interesting variation of the problem: an open transmission line. In the steady state we have 100 watts forward, 100 watts reflected, 200 Joules in the line, and 0 watts being sourced by the generator. :-) ac6xg Cecil Moore wrote: Richard Harrison wrote: Cecil, W5DXP wrote: "How many joules are contained in that feedline?" Total joules on the line is 300. That can be shown to be true by noting that during the transient buildup to steady-state, 300 joules sourced by the generator have not yet reached the load. That remains true until the generator is powered down, i.e. all during steady-state. Forward power minus the reflected power equals 100 watts being supplied by the generator to the load with 200 watts forward power and 100 watts reflrcted power in the line. But what about the people who say there's no energy in the reflected wave? Reckon how they sweep all those joules, whose energy must be conserved, under the rug? :-) |
Brainteaser
Jim Kelley wrote:
One should also carefully consider the more interesting variation of the problem: an open transmission line. In the steady state we have 100 watts forward, 100 watts reflected, 200 Joules in the line, and 0 watts being sourced by the generator. :-) Yes, but the 200 joules in the line was previously sourced by the generator during the transient state. It's hard to sweep 200 joules under the reflected power rug. -- 73, Cecil http://www.qsl.net/w5dxp |
Brainteaser
Jim Kelley wrote:
One should also carefully consider the more interesting variation of the problem: an open transmission line. In the steady state we have 100 watts forward, 100 watts reflected, 200 Joules in the line, and 0 watts being sourced by the generator. :-) Expanding on my earlier response - For the first two seconds, the source doesn't know it is looking into an open transmission line so a 100 watt source would faithfully output 200 joules into a one second long open circuit transmission line. That 200 joules cannot be destroyed. Is it mere coincidence that the forward and reflected waves are 100 joules/sec*(one second), exactly equal to the 200 joules supplied by the source? -- 73, Cecil http://www.qsl.net/w5dxp |
Brainteaser
Dr. Honeydew wrote:
I had Beaker set up an experiment to test this, and after he put out the fire he started in his hair when the first load he tried melted a hole in the lab bench, he finally got it set up right. He determined that the mass of the line increased by about 3.338 femtograms when powered as compared with it unpowered. We're now looking into what to do with the line, as it's cluttering up the lab, and we're open to suggestions. Regards, Bunsen That is the best response we will get. Even if there are correct ones! tom K0TAR |
Brainteaser
On Thu, 25 May 2006 22:04:12 GMT, Cecil Moore
wrote: But what about the people who say there's no energy in the reflected wave? Reckon how they sweep all those joules, whose energy must be conserved, under the rug? :-) with a jouler's brush? -- 73 for now Buck N4PGW |
Brainteaser
Buck wrote:
Cecil Moore wrote: But what about the people who say there's no energy in the reflected wave? Reckon how they sweep all those joules, whose energy must be conserved, under the rug? :-) with a jouler's brush? All kidding aside, whatever number of joules of energy are required to support the forward power and reflected power is provided by the source during the transient condition following power up. If the source power is 100 watts, the forward power is 200 watts, the reflected power is 100 watts, and the load power is 100 watts, all the joules per second needed to support that forward and reflected power was supplied by the source before steady-state was reached. Anything else would violate the conservation of energy principle. This is easily proven using a one second long lossless transmission line as a conceptual training aid for the uninitiated. -- 73, Cecil http://www.qsl.net/w5dxp |
Brainteaser
On Fri, 26 May 2006 03:16:38 GMT, Cecil Moore
wrote: If the source power is 100 watts, the forward power is 200 watts, the reflected power is 100 watts, and the load power is 100 watts, all the joules per second needed to support that forward and reflected power was supplied by the source before steady-state was reached. Anything else would violate the conservation of energy principle. This is easily proven using a one second long lossless transmission line as a conceptual training aid for the uninitiated. I gathered that from an earlier post, but I think the confusion comes in the second second, doesn't it? 100 watts of power is generated from the generator for two seconds. the first second, 100 watts travels to the antenna and only 100 watts is in the transmission line. Then for the second second, 100 forward watts is moving towards the antenna and 50 reflected watts are returning to the transmitter for a total of 150 watts in the transmission line? Of course, since the transmitter isn't matched to the antenna, the reflected power is reflected again for 25 watts being reflected back to the antenna bringing the power from 100 forward watts to 125 watts (the reason power meters go up in wattage when SWR rises) thus the antenna reflects 62.5 watts which adds to the 125 for a total of 187.5 watts in the transmission line. Sooner or later one will have a meltdown that will make Chernoble look like a firecracker compared to a fireworks display. lol -- 73 for now Buck N4PGW |
Brainteaser
is that line 1 second long, or 1/2 second long?? a 1 second long line would
take 2 seconds worth of energy from the generator before the reflection returned to let it know the line is terminated. this is of course very important if you are measuring lines with tdr's where you can really see those returned waves, of course the time it takes them to get back to the tdr is double the one-way travel time... mess it up and you are looking for faults twice as far down the line as you calculate. "Cecil Moore" wrote in message . com... Jim Kelley wrote: One should also carefully consider the more interesting variation of the problem: an open transmission line. In the steady state we have 100 watts forward, 100 watts reflected, 200 Joules in the line, and 0 watts being sourced by the generator. :-) Expanding on my earlier response - For the first two seconds, the source doesn't know it is looking into an open transmission line so a 100 watt source would faithfully output 200 joules into a one second long open circuit transmission line. That 200 joules cannot be destroyed. Is it mere coincidence that the forward and reflected waves are 100 joules/sec*(one second), exactly equal to the 200 joules supplied by the source? -- 73, Cecil http://www.qsl.net/w5dxp |
Brainteaser
Buck wrote:
I gathered that from an earlier post, but I think the confusion comes in the second second, doesn't it? 100 watts of power is generated from the generator for two seconds. the first second, 100 watts travels to the antenna and only 100 watts is in the transmission line. Then for the second second, 100 forward watts is moving towards the antenna and 50 reflected watts are returning to the transmitter for a total of 150 watts in the transmission line? Of course, since the transmitter isn't matched to the antenna, the reflected power is reflected again for 25 watts being reflected back to the antenna bringing the power from 100 forward watts to 125 watts (the reason power meters go up in wattage when SWR rises) thus the antenna reflects 62.5 watts which adds to the 125 for a total of 187.5 watts in the transmission line. Sooner or later one will have a meltdown that will make Chernoble look like a firecracker compared to a fireworks display. lol Actually, the forward power levels off at 200 watts during steady- state. The reflected power levels off at 100 watts. 300 joules of energy exist that have been sourced but not dissipated in the load. 300 joules are exactly what the forward wave and reflected wave need to support their existence. Here's the brainteaser configuration: 100W-SGAT---one second long lossless 50 ohm feedline---291.42 ohm load The 'SGAT' is a signal generator equipped with a super fast auto-tuner that re-reflects all the reflected power back toward the load so there are no losses at the source end. The only losses in the entire system are in the 291.42 ohm load which, to make the math easier, was chosen for a power reflection coefficient of 0.5, i.e. half the incident power is reflected at the load. I will generate an EXCEL spreadsheet today that will give a second by second summary of the powers and total energy. In the meantime, here is a more simplified version of the brainteaser. 100W-SGCL---one second long lossless 50 ohm feedline---291.42 ohm load The 'SGCL' is a signal generator equipped with a circulator and 50 ohm resistor. During steady-state, half the generated power is dissipated in the load and half in the circulator resistor. There are no re- reflections at the source. During the 1st second, the source supplies 100 joules into the feedline. There is no dissipation either in the load or circulator resistor. During the 2nd second, the source supplies another 100 joules into the feedline, 50 joules of which are converted to heat in the load and 50 joules of which are reflected from the load as 50 watts of reflected power. At the end of the 2nd second, the feedline contains 150 joules of energy. During the 3rd second, steady-state is reached. Of the 100 watts sourced, 50 watts are dissipated in the load and 50 watts are dissipated in the circulator resistor. At the end of the 3rd second, the source has generated 300 joules, 100 of which have made it to the load, and 50 of which have made it to the circulator resistor. 300-100-50 = 150 joules stored in the feedline. 100 joules are stored in the 100W forward wave. 100 joules/sec = 100W 50 joules are stored in the 50W reflected wave. 50 joules/sec = 50W This conservation of energy works for any length feedline. If the feedline were one microsecond long instead of one second long, there would be 150 microjoules stored in the feedline, 100 microjoules in the forward wave and 50 microjoules in the reflected wave. -- 73, Cecil http://www.qsl.net/w5dxp |
Brainteaser
Dave wrote:
is that line 1 second long, or 1/2 second long?? a 1 second long line would take 2 seconds worth of energy from the generator before the reflection returned to let it know the line is terminated. this is of course very important if you are measuring lines with tdr's where you can really see those returned waves, of course the time it takes them to get back to the tdr is double the one-way travel time... mess it up and you are looking for faults twice as far down the line as you calculate. The line is 1 second long. The source is generating 100 watts. So the number of joules generated starting at t=0 is 100N where N is the total number of seconds. At the end of 30 seconds, the generator will have sourced 3000 joules which must be conserved. I will generate an EXCEL spreadsheet at work today and post it to my web page when I get home. It will cover the first 30 seconds in 1 second increments. It will show that of the 3000 joules sourced by the generator during that first 30 seconds, only 2700 joules have reached the load. The other 300 joules are contained in the 200W forward wave and 100W reflected wave. (200 watts)(one second) = 200 joules in the forward wave (100 watts)(one second) = 100 joules in the reflected wave 200 joules + 100 joules = 300 joules not delivered to the load -- 73, Cecil http://www.qsl.net/w5dxp |
Brainteaser
On Fri, 26 May 2006 12:41:45 GMT, Cecil Moore
wrote: Buck wrote: All this technical detail is interesting, but in the last thirty years of hamming, I have taken note to know the following: My radio is most happy when matched with an SWR of less than 1.8:1 If my antenna exceeds that, either adjust it or use a tuner The higher I raise my antenna, the better. A yagi works better than a dipole given the same height and direction. A dipole works better to the west than a beam pointed north. i would rather have a high gain antenna than a high power amp. Most importantly: If band conditions permit, I can talk across the ocean on an antenna with a 30 db loss, but with certain band conditions, even a 30 db gain antenna won't make the contact! I can't tell you how many joules are in my coax, but, if you ask, I can tell you how many jewels are in my logbook. :) -- 73 for now Buck N4PGW |
Brainteaser
"Buck" wrote:
I can't tell you how many joules are in my coax, but, if you ask, I can tell you how many jewels are in my logbook. :) This is not aimed at people who don't care about such things. This is aimed at people who do care about such things especially those who are disseminating false technical information about such things. Some say that since there's 100 watts in and 100 watts out during steady-state, there's nothing left over for the reflected waves. They apparently forgot about the time immediately following power up when there was 100 watts in and less than 100 watts out. -- 73, Cecil, W5DXP |
Brainteaser
I've lost the original question.
But there is a lossless transmission line 1 second long. With a velocity factor of 1.0 it is 30,000 kilometres long. Let Zo = 100 ohms. Let applied volts = 100 VDC. Let terminating resistance be 100 ohms. Line Inductance = 10,000 Henrys. Line Capacitance = 1 Farad. Under steady state conditions - Line current = 1 amp. Energy stored in inductance = Sqr(I)*L/2 = 5000 Joules. Volts across capacitance = 100 volts. Energy stored in capacitance = SqrV)*C/2 = 5000 Joules. Total energy stored = 10, 000 Joules = 10,000 watt.seconds. Which has nothing whatsoever to do with all this nonsense about reflections. Note that energy in inductance equals energy in capacitance. (I trust I have not made a mistake with the arithmetic.) ---- Reg. "Cecil Moore" wrote in message . net... "Buck" wrote: I can't tell you how many joules are in my coax, but, if you ask, I can tell you how many jewels are in my logbook. :) This is not aimed at people who don't care about such things. This is aimed at people who do care about such things especially those who are disseminating false technical information about such things. Some say that since there's 100 watts in and 100 watts out during steady-state, there's nothing left over for the reflected waves. They apparently forgot about the time immediately following power up when there was 100 watts in and less than 100 watts out. -- 73, Cecil, W5DXP |
Brainteaser
Reg Edwards wrote:
Let Zo = 100 ohms. For what range of frequencies is the above true? -- 73, Cecil http://www.qsl.net/w5dxp |
Brainteaser
For what range of frequencies is the above true? -- 73, Cecil http://www.qsl.net/w5dxp ==================================== If you ask silly questions you can expect silly answers. Are you looking for a reason to introduce your irrelevant reflections again. --- Reg. |
Brainteaser
Reg Edwards wrote:
I've lost the original question. But there is a lossless transmission line 1 second long. With a velocity factor of 1.0 it is 30,000 kilometres long. Let Zo = 100 ohms. Let applied volts = 100 VDC. Let terminating resistance be 100 ohms. Line Inductance = 10,000 Henrys. Line Capacitance = 1 Farad. Under steady state conditions - Line current = 1 amp. Energy stored in inductance = Sqr(I)*L/2 = 5000 Joules. Volts across capacitance = 100 volts. Energy stored in capacitance = SqrV)*C/2 = 5000 Joules. Total energy stored = 10, 000 Joules = 10,000 watt.seconds. Which has nothing whatsoever to do with all this nonsense about reflections. Note that energy in inductance equals energy in capacitance. Yep, you don't have to mess with those pesky reflections when the line is terminated in a load equal to its characteristic impedance. Then the problem reduces to the trivial one you've posed. Simple problem, simple solution. Roy Lewallen, W7EL |
Brainteaser
"Roy Lewallen" wrote Yep, you don't have to mess with those pesky reflections when the line is terminated in a load equal to its characteristic impedance. Then the problem reduces to the trivial one you've posed. Simple problem, simple solution. ======================================= As I say, I've lost the original question. But if a steady-state answer is needed for the original question then the answer can be found by the same simple means without the unnecessary complication of reflections. It's the Devil which makes Cecil pose such loaded questions. ---- Reg. |
Brainteaser
Reg Edwards wrote:
If you ask silly questions you can expect silly answers. Are you looking for a reason to introduce your irrelevant reflections again. No, just trying to correlate your example to mine. In your example, how long does it take from power up for the load to accept 100 joules/second? -- 73, Cecil http://www.qsl.net/w5dxp |
Brainteaser
Roy Lewallen wrote:
Yep, you don't have to mess with those pesky reflections when the line is terminated in a load equal to its characteristic impedance. Which begs the question of what is to be done when "those pesky reflections" exist in a mismatched system. -- 73, Cecil http://www.qsl.net/w5dxp |
Brainteaser
Sigh. Yes, I believe you HAVE made a mistake in your calculations,
right at the beginning. C = tau/Z0 = 1/100 farad; L = tau*Z0 = 100 henries. But of course, the concept is perfect. 100V, 1A, 100W, E^2*C/2 = 50 joules stored in the capacitance. I^2*L/2 = 50 joules stored in the inductance. The two are equal only for a termination equal to the line Z0 of course. Too bad Cec doesn't understand that lossless line has unvarying Z0 as a function of frequency from DC up, as long as TEM mode is supported. Cheers, Tom Reg Edwards wrote: I've lost the original question. But there is a lossless transmission line 1 second long. With a velocity factor of 1.0 it is 30,000 kilometres long. Let Zo = 100 ohms. Let applied volts = 100 VDC. Let terminating resistance be 100 ohms. Line Inductance = 10,000 Henrys. Line Capacitance = 1 Farad. Under steady state conditions - Line current = 1 amp. Energy stored in inductance = Sqr(I)*L/2 = 5000 Joules. Volts across capacitance = 100 volts. Energy stored in capacitance = SqrV)*C/2 = 5000 Joules. Total energy stored = 10, 000 Joules = 10,000 watt.seconds. Which has nothing whatsoever to do with all this nonsense about reflections. Note that energy in inductance equals energy in capacitance. (I trust I have not made a mistake with the arithmetic.) ---- Reg. "Cecil Moore" wrote in message . net... "Buck" wrote: I can't tell you how many joules are in my coax, but, if you ask, I can tell you how many jewels are in my logbook. :) This is not aimed at people who don't care about such things. This is aimed at people who do care about such things especially those who are disseminating false technical information about such things. Some say that since there's 100 watts in and 100 watts out during steady-state, there's nothing left over for the reflected waves. They apparently forgot about the time immediately following power up when there was 100 watts in and less than 100 watts out. -- 73, Cecil, W5DXP |
Brainteaser
Reg Edwards wrote:
It's the Devil which makes Cecil pose such loaded questions. It's only loaded with the conservation of energy principle, Reg. Here's a simple example. The SG-CR is a 50 ohm signal generator equipped with a circulator resistor to dissipate all reflected power incident upon the source. 200W SG-CR---one second long lossless 50 ohm line---291.42 ohm SWR is 5.83:1 rho^2 = 0.5 Steady-state forward power is 200 watts. Steady-state reflected power is 100 watts. The question is: After steady-state has been reached, how many joules of energy have been generated but have not been dissipated in the load resistor or circulator resistor? -- 73, Cecil http://www.qsl.net/w5dxp |
Brainteaser
K7ITM wrote:
Sigh. Yes, I believe you HAVE made a mistake in your calculations, right at the beginning. C = tau/Z0 = 1/100 farad; L = tau*Z0 = 100 henries. But of course, the concept is perfect. 100V, 1A, 100W, E^2*C/2 = 50 joules stored in the capacitance. I^2*L/2 = 50 joules stored in the inductance. The two are equal only for a termination equal to the line Z0 of course. Too bad Cec doesn't understand that lossless line has unvarying Z0 as a function of frequency from DC up, as long as TEM mode is supported. I do understand that. That's why Reg's 10,000 joules don't make any sense to me. Now if the Z0 was one ohm at DC, it would make sense for 100 volts to be able to supply 10,000 joules in one second. -- 73, Cecil http://www.qsl.net/w5dxp |
Brainteaser
Reg Edwards wrote: I've lost the original question. .... Actually, it was "200 watts forward, 100 watts reflected" -- but it works the same way: 100 ohm line, 1 second long: source is DC 200*sqrt(2) volts in series with 100 ohms. Load is such that it absorbs 100 watts from that source: 100*(3 +/- sqrt(8)) ohms. About 17.16 ohms works OK. Then the line voltage is 41.42V and the energy stored in the 1/100F line capacitance is 8.579 joules, and the line current is 2.414A and the energy stored in the 100H line inductance is 291.421 joules. Total energy stored is 300 joules. It's somewhat comforting that it's just as you'd expect from power delivered versus time. It's an illustration of something I believe Ian White was trying to get across in another thread: there are commonly multiple ways to analyze a problem, and if they agree, you MAY have the right answer. If they disagree, you have at least as many wrong answers as the number of disagreements (or perhaps the disagreements only seem to be disagreements). Now, I have no idea who it is that might be saying that there's no energy stored in the fields in a line, so I really don't know what the to-do is all about. Maybe it's just another misinterpretation of what people have actually said...perhaps about lines that are very far indeed from being lossless. Cheers, Tom |
Brainteaser
"K7ITM" wrote in message ups.com... Sigh. Yes, I believe you HAVE made a mistake in your calculations, right at the beginning. C = tau/Z0 = 1/100 farad; L = tau*Z0 = 100 henries. But of course, the concept is perfect. 100V, 1A, 100W, E^2*C/2 = 50 joules stored in the capacitance. I^2*L/2 = 50 joules stored in the inductance. The two are equal only for a termination equal to the line Z0 of course. ================================= Tom, Furthermore, the energy dissipated each second in the 100-ohm termination (100 watts) is equal to the energy stored in the line. Damned decimal points. I just KNEW I had made a mistake somewhere along the line. Thank you. By the way - what is tau ? ---- Reg. |
Brainteaser
K7ITM wrote:
Actually, it was "200 watts forward, 100 watts reflected" -- but it works the same way: 100 ohm line, 1 second long: source is DC 200*sqrt(2) volts in series with 100 ohms. Load is such that it absorbs 100 watts from that source: 100*(3 +/- sqrt(8)) ohms. About 17.16 ohms works OK. Then the line voltage is 41.42V and the energy stored in the 1/100F line capacitance is 8.579 joules, and the line current is 2.414A and the energy stored in the 100H line inductance is 291.421 joules. Total energy stored is 300 joules. As can be seen from http://www.qsl.net/w5dxp/1secsgcr.gif Do you think it's just a coincidence that 200 watts of forward power in a one second long line would require 200 joules and 100 watts of reflected power in a one second long line would require 100 joules for a total of 300 joules? Now, I have no idea who it is that might be saying that there's no energy stored in the fields in a line, so I really don't know what the to-do is all about. Please read what W7EL says about forward and reflected power in his "food for thought" writings. In particular: "Some inventive people have supposed they can separate forward and reverse power with a circulator. That really sounds attractive, particularly with an open or short circuited load. In that condition, the forward and reverse powers are each 100 watts, yet the transmitter (or the transmitter voltage source) doesn't have to produce any power at all." What W7EL doesn't seem to realize is that the source produced the power in the forward and reverse power waves while the feedline was charging up during the transient period before the reverse power waves reached the source so there is bona fide energy in those waves. He also says there is no model for handling such yet one exists in detail in the magazine article on my web page. -- 73, Cecil http://www.qsl.net/w5dxp |
Brainteaser
Reg Edwards wrote:
Damned decimal points. I just KNEW I had made a mistake somewhere along the line. Now do you see why my question wasn't simple or loaded? I just wondered how you shoved 10,000 joules down a Z0=100 ohm feedline in one second using a fixed 100 volt source. -- 73, Cecil http://www.qsl.net/w5dxp |
Brainteaser
K7ITM wrote:
It's somewhat comforting that it's just as you'd expect from power delivered versus time. It's an illustration of something I believe Ian White was trying to get across in another thread: there are commonly multiple ways to analyze a problem, and if they agree, you MAY have the right answer. If they disagree, you have at least as many wrong answers as the number of disagreements (or perhaps the disagreements only seem to be disagreements). What I said was that there are commonly multiple methods to analyse the same problem, but that all correct methods MUST agree, because they are only different views of the same physical reality. (I think that amounts to the same as you said above, Tom.) -- 73 from Ian GM3SEK 'In Practice' columnist for RadCom (RSGB) http://www.ifwtech.co.uk/g3sek |
Brainteaser
Ian White GM3SEK wrote:
What I said was that there are commonly multiple methods to analyse the same problem, but that all correct methods MUST agree, because they are only different views of the same physical reality. That attitude is certainly an improvement from the earlier labeling of an alternate valid analysis as "Gobbledygook". -- 73, Cecil http://www.qsl.net/w5dxp |
Brainteaser
Cecil Moore wrote:
Ian White GM3SEK wrote: What I said was that there are commonly multiple methods to analyse the same problem, but that all correct methods MUST agree, because they are only different views of the same physical reality. That attitude is certainly an improvement from the earlier labeling of an alternate valid analysis as "Gobbledygook". "Improvement" implies a change; but I have always held the same view as quoted above. It is one of the main reasons why I remain sceptical about your theories. Where they ought to be agreeing with other analyses of the same problem, they don't. Also, "gobbledygook" is not a word I use... it's one of yours, I believe. -- 73 from Ian GM3SEK 'In Practice' columnist for RadCom (RSGB) http://www.ifwtech.co.uk/g3sek |
Brainteaser
On 26 May 2006 15:50:07 -0700, "K7ITM" wrote:
To heck with the red wine...think I'll have another shot or two of the Scapa single-malt this evening. Hi Tom, Is the Scapa flowing? 73's Richard Clark, KB7QHC |
Brainteaser
Ian White GM3SEK wrote:
It is one of the main reasons why I remain sceptical about your theories. Where they ought to be agreeing with other analyses of the same problem, they don't. No chance the "other analyses" could be wrong? Could you please give me an example of what you are talking about? Also, "gobbledygook" is not a word I use... it's one of yours, I believe. I didn't say you used it, Ian, but you didn't object when it was used. -- 73, Cecil http://www.qsl.net/w5dxp |
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Well, after communing this evening with some hard cherry wood from my
back yard that's going to become part of a loom, I decided to go with the Ledaig instead. Cheers, Tom |
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Ian wrote, "(I think that amounts to the same as you said above, Tom.)"
Well, I hope so...I'd certainly not intentionally misstate it. I do think your way of putting it is a bit more positive than mine, though. Cheers, Tom |
Brainteaser
In a previous posting, I noted that for the 100 ohm line driven by
200*sqrt(2) volts DC through 100 ohms, there are two load resistors that dissipate 100 watts: 100*(3 +/- sqrt(8) ) ohms. Continuing the observation about energy stored in electric versus energy stored in magnetic fields, for the low resistance, the steady state load and line voltage is low and the current is high; for the high resistance, the steady state voltage is high and the current is low. In the first case, 8.5786 joules is stored in the electric field and 291.42 in the magnetic; in the second those are reversed and 291.42 are stored in the electric and 8.5786 in the magnetic. The ratio of the two energies is exactly the same as the ratio of the load resistances. In our lossless line, if the excitation is AC and steady-state has been reached, then the total energy stored per unit length varies as a function of time and distance in a somewhat complex manner. The ratio of electric to magnetic varies along the length of the line and as a function of time, with extremes being in a ratio equal to the square of the SWR, occuring of course at voltage and current maxima, spaced 1/4 wave apart, alternating down the line. I'm not sure that those observations will be any use to anyone, but maybe they are of academic interest. Who knows what I might come up with after some Glenmorangie. Cheers, Tom |
Brainteaser
Cecil Moore wrote: Jim Kelley wrote: One should also carefully consider the more interesting variation of the problem: an open transmission line. In the steady state we have 100 watts forward, 100 watts reflected, 200 Joules in the line, and 0 watts being sourced by the generator. :-) Yes, but the 200 joules in the line was previously sourced by the generator during the transient state. It's hard to sweep 200 joules under the reflected power rug. So is this your proof that Joules of energy are likewise reflected from antireflective surfaces? ac6xg |
Brainteaser
Cecil Moore wrote: Jim Kelley wrote: One should also carefully consider the more interesting variation of the problem: an open transmission line. In the steady state we have 100 watts forward, 100 watts reflected, 200 Joules in the line, and 0 watts being sourced by the generator. :-) Expanding on my earlier response - For the first two seconds, the source doesn't know it is looking into an open transmission line so a 100 watt source would faithfully output 200 joules into a one second long open circuit transmission line. That 200 joules cannot be destroyed. Is it mere coincidence that the forward and reflected waves are 100 joules/sec*(one second), exactly equal to the 200 joules supplied by the source? But you're missing, or trying to circumvent, the most interesting aspect of the problem. It's the one which highlights the very core of our disagreement. The energy stored in the line, remains stored in the line as long as steady state is maintained without a single Joule of additional energy moving into or out of the line. To me, this illustrates clearly how the fields at the impedance interfaces of a matching transformer can be maintained without requiring multiple rereflections of energy. I'm hoping some day you'll see it too. 73, ac6xg |
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