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Jim Kelley wrote:
Try to keep your mind focussed on the fact that we're discussing the steady state. If one wants to know where the energy in the transmission line came from, one must consider the initial transient state. Let's take a simple example like the one W7EL uses in his food for thought #1, a simple stub, except let's make it one second long and lossless. 100W------one second long lossless feedline------open During the first two seconds, the source pours 200 joules into the stub. After two seconds, steady-state is reached and there is no net transfer of energy in either direction at the source. Where are the 200 joules that were sourced and not dissipated? -- 73, Cecil http://www.qsl.net/w5dxp |
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Cecil Moore wrote: Jim Kelley wrote: How about we just get back to the question. Energy in your 1 sec. transmission line, only with no load on the circulator (and no autotuner). How much? Explain. Try it! I honestly don't know what happens when the third port of the circulator is unterminated You should think about it because amateur radios don't usually have circulators on their output, and the result illuminates a corner of the discussion that you've been pretty closed minded about. Without a load on the circulator, you must know that no energy can be flowing to it or through it. No power is being dissipated by it. What effect should that have on the total energy in the transmission line, and why? I hope it's not true about Reg. That would be a huge loss. Perhaps somebody will archive his website. 73, ac6xg |
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On Wed, 30 Aug 2006 10:17:44 -0700, Jim Kelley
wrote: I hope it's not true about Reg. That would be a huge loss. Perhaps somebody will archive his website. Hi Jim, The announcement came from his account, and his computer. I've archived his site, but for others who wish to do the same, the best tool for that purpose can be found at: http://www.httrack.com This is a website harvesting robot that will replicate an entire website into the directory of your choice (changing links so that you can browse it on your system). 73's Richard Clark, KB7QHC |
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Jim Kelley wrote:
Cecil Moore wrote: I honestly don't know what happens when the third port of the circulator is unterminated You should think about it because amateur radios don't usually have circulators on their output, and the result illuminates a corner of the discussion that you've been pretty closed minded about. I know what happens if the circulator is not present. I just don't don't know what happens with an unterminated circulator. Maybe you can help me out here. If the load resistors are removed from the circulators in the previous experiment, what is the forward power and reflected power readings on the one second long lossless transmission line during steady-state? Without a load on the circulator, you must know that no energy can be flowing to it or through it. No power is being dissipated by it. What effect should that have on the total energy in the transmission line, and why? Let's remove the circulators and try to figure out what happens. Let's assume the signal generator is a Thevenin equivalent with an internal resistance of 50 ohms and not equipped with a circulator. 200W SG---one second long 50 ohm lossless feedline---100W SG Would you agree with me that during the first second, the feedline is loaded with 300 joules because there have been no reflections? What happens after that depends upon the reflection coefficients at the signal generators. But the $64k question is what happens to the original 300 joules? I hope it's not true about Reg. That would be a huge loss. Perhaps somebody will archive his website. One of the posters to the news of Reg's death was whom I suspect is Reg's son (or nephew). Humans are mortal and can die at any time. My son died when he was 8 months old. I wish he had lived as long as Reg who certainly had a full, useful, and colorful life. -- 73, Cecil http://www.qsl.net/w5dxp |
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On Wed, 30 Aug 2006 12:12:20 -0700, Richard Clark wrote:
On Wed, 30 Aug 2006 10:17:44 -0700, Jim Kelley wrote: I hope it's not true about Reg. That would be a huge loss. Perhaps somebody will archive his website. Hi Jim, The announcement came from his account, and his computer. I've archived his site, but for others who wish to do the same, the best tool for that purpose can be found at: http://www.httrack.com This is a website harvesting robot that will replicate an entire website into the directory of your choice (changing links so that you can browse it on your system). 73's Richard Clark, KB7QHC Richard, I've just reviewed the url above, and found that I don't know how to use it to download Reg's web page. I see that you've already downloaded it, so could I download it from your copy? Walt, W2DU |
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On Wed, 30 Aug 2006 15:52:52 -0400, Walter Maxwell
wrote: On Wed, 30 Aug 2006 12:12:20 -0700, Richard Clark wrote: On Wed, 30 Aug 2006 10:17:44 -0700, Jim Kelley wrote: I hope it's not true about Reg. That would be a huge loss. Perhaps somebody will archive his website. Hi Jim, The announcement came from his account, and his computer. I've archived his site, but for others who wish to do the same, the best tool for that purpose can be found at: http://www.httrack.com This is a website harvesting robot that will replicate an entire website into the directory of your choice (changing links so that you can browse it on your system). 73's Richard Clark, KB7QHC Richard, I've just reviewed the url above, and found that I don't know how to use it to download Reg's web page. I see that you've already downloaded it, so could I download it from your copy? Walt, W2DU Hi Walt, The paths are tied intimately to my file system hierarchy (which is pretty deep). Rather, I will give you a walk-thru. In spite of the apparent complexity (it is a technician's tool), it is quite simple to use with only two or three particulars to satisfy: 1. As directed on the front page, press the NEXT button; 2. On the next page for Project Name, enter Reg Edwards G4FGQ; 3. Below that (skip the category), click the ellipses button to open a storage path and select an existing folder the website will be stored here in a folder named Reg Edwards G4FGQ; 4. Press the NEXT button at the bottom; 5. leave the ACTION selection at "Download web site(s)"; 6. past Reggie's top level page, http://www.btinternet.com/~g4fgq.regp, into the Web Addresses text box; 7. Press the NEXT button at the bottom; 8. At the next page, Press the FINISH button at the bottom. This will start the robots harvesting with a view of them on a new page that shows each robot in its own thread - about half a dozen of them running simultaneously. Depending on the load at the server, the entire process should take 5 minutes or so at T1 speeds. The total download is 4.33MB. The robot activity screen will disappear at the end of the harvesting. I can zip up a copy (sorry Reg) and mail it for those who want a copy that is located at the drive root (I did this down load again to confirm the steps described above). 73's Richard Clark, KB7QHC |
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Jim Kelley wrote:
You should think about it because amateur radios don't usually have circulators on their output, and the result illuminates a corner of the discussion that you've been pretty closed minded about. Here's a brainteaser for you, Jim. Assume the following example under steady-state conditions: XMTR--X--tuner--Y--one second long lossless feedline---291.42 ohms Ps=100W Pfor=200W-- --100W=Pref PL=100W The lossless tuner is tuned for a Z0-match so there is zero reflected power at point 'X'. The Z0 of the feedline is 50 ohms. The voltage reflection coefficient at the load is 0.707 making the power reflection coefficient 0.5 We suddenly disconnect the source at point 'Y'. After the source is disconnected, how many joules of energy are delivered to the load? _____ How many seconds does it take to dissipate all the energy in the feedline? _____ On a second by second basis, how much power is delivered to the load after the disconnect? _____ _____ _____ _____ _____ -- 73, Cecil, http://www.qsl.net/w5dxp |
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Cecil Moore wrote: Jim Kelley wrote: You should think about it because amateur radios don't usually have circulators on their output, and the result illuminates a corner of the discussion that you've been pretty closed minded about. Here's a brainteaser for you, Jim. Assume the following example under steady-state conditions: XMTR--X--tuner--Y--one second long lossless feedline---291.42 ohms Ps=100W Pfor=200W-- --100W=Pref PL=100W If you aren't going to solve the problem I posed, then I don't see why I should feel obliged to bother with any more of yours. Only seems fair. There is a significant point to my question that you still have not addressed. Would you have us believe that some new revelation is being presented by this latest problem of yours? Or perhaps it's simply intended to divert attention away, once again, from my question. Or maybe it was just too hard for you. How about a simpler problem then: XMTR--X--one second long lossless feedline---infinite ohms PS=100W How many Joules are stored in the transmission line? Why? 73, ac6xg |
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On Fri, 01 Sep 2006 11:03:39 -0700, Jim Kelley
wrote: How many Joules are stored in the transmission line? 100 Why? Spillage. The cup is two halves full (or no halves empty for the pessimists in the crowd) |
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Richard Clark wrote:
On Fri, 01 Sep 2006 11:03:39 -0700, Jim Kelley wrote: How many Joules are stored in the transmission line? 100 That may be the correct solution to a somewhat different problem. Why? Spillage. The cup is two halves full (or no halves empty for the pessimists in the crowd) Pessimists would report that "there are a disasterous and insurmountable number of full halves", when in fact there are just two. 73, ac6xg |
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