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wrote:
That's what you get when you use the surge impedance and compute surge rho. Try steady state impedance for steady state rho. rho = (0-50)/(0+50) = -1 as expected Heh, heh, the steady state impedances are V/I ratios which are results incapable of causing anything. Your logic is a closed loop. The V/I ratios cause the rho to be -1. Therefore, rho=-1 causes the appropriate V/I ratios. Can you name any other result in reality that causes itself? Hint: Only physical impedance discontinuities can cause reflections. Image impedances are incapable of causing anything since they are an end result. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
wrote:
On the other hand... if Vrev = rho * Vfwd then rho^2 = (Vrev^2/Z0) / (Vfwd^2/Z0) So (Vrev^2/Z0) / (Vfwd^2/Z0) can be greater than one. If Sqrt(Pref/Pfwd) can not be greater than 1 then either Pfwd is not equal to (Vfwd^2/Z0) or Prev is not equal to (Vrev^2/Z0) Intriguing result, is it not? Only to the uninitiated. Contradictions don't exist in reality. They only exist in human minds. That should be a clue. |rho1| is not always equal to |rho2|. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
David Robbins wrote:
is that paper on the web somewhere?? Perhaps someone will offer it to you as a .pdf file. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Dr. Slick wrote:
The definition of Rho has been set for "God-knows-how-long!" Actually, 'rho' has contradictory definitions. (Z2-Z1)/(Z2+Z1) is not always the same value as Sqrt(Pref/Pfwd) because of interference energy. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Richard Clark wrote in message . ..
On 3 Sep 2003 23:16:39 -0700, (Dr. Slick) wrote: I think Reg put it best: Hi OM, If you are going to quote him as an authority supporting you, you should at least accept his offer of a bridge to settle this hash shouldn't you? You can't run far on one legged stilts. 73's Richard Clark, KB7QHC Who ever said he was an authority? All he did was open up a can-o-worms by showing how the "normal" Gamma equation is not always less than 1 for passive networks, which is good because i wouldn't have found out about the correct conjugate equation (when Zo is complex). But it is wise to pick and choose your fights, eh? And i'd rather argue with someone who is making sense at any particular point in time. Show us a passive circuit that reflects more power than you feed it (incident), on my Daiwa meter, and i will be VERY impressed. I'm be waiting a long time for that schematic.... Slick |
Dr. Slick wrote:
I can't wait to hook it up to see more reflected power than incident on my DAIWA meter, that would be very interesting. If that's really what you want to observe, connect the meter backwards. :-) -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Cecil Moore wrote:
wrote: That's what you get when you use the surge impedance and compute surge rho. Try steady state impedance for steady state rho. rho = (0-50)/(0+50) = -1 as expected Heh, heh, the steady state impedances are V/I ratios which are results incapable of causing anything. Your logic is a closed loop. The V/I ratios cause the rho to be -1. Therefore, rho=-1 causes the appropriate V/I ratios. Can you name any other result in reality that causes itself? Hint: Only physical impedance discontinuities can cause reflections. Image impedances are incapable of causing anything since they are an end result. Was it not only a few days ago that there was agreement that If it is acceptable to claim that sometimes there is no reflection at an impedance discontinuity, then it must also be acceptable to claim that sometimes there is a reflection where there is no discontinuity. ? ....Keith |
"Dr. Slick" wrote in message m... Cecil Moore wrote in message ... Dr. Slick wrote: If you agree that the Pref/Pfwd ratio cannot be greater than 1 for a passive network, then neither can the [Vref/Vfwd]= rho be greater than 1 either. Sqrt(Pref/Pfwd) cannot be greater than one. (Z2-Z1)/(Z2+Z1) can be greater than one. Both are defined as 'rho' but they are not always equal. (Z2-Z1)/(Z2+Z1) is a physical reflection coefficient. Sqrt(Pref/Pfwd) is an image reflection coefficient. I agree that Sqrt(Pref/Pfwd) cannot be greater than one for a passive network. (Z2-Z1)/(Z2+Z1) can be greater than one, for passive networks and certain combinations of complex Z1 and Z2. I feel this is incorrect usage of this formula, which should be limited to purely real Zo. A [rho] that is greater than one gives meaningless negative SWR data, and is limited to active devices. it only gives negative swr values if you incorrectly use the lossless line approximation to calculate vswr from rho. that is the incorrectly applied formula in this case. that formula is not valid for a lossy line, you must go back to the original definition of VSWR=|Vmax|/|Vmin|. which as we have also noted is not meaningful on a lossy line as Vmax and Vmin are different at each max and min point because of the losses in the line affecting both the forward and reverse waves. |
is that paper on the web somewhere?? i figured it had to be something
with computing powers that was getting mixed in here some how, i think that is the only place you can end up with conjugates in transmission lines. so i assume its not a simple 1 page derivation from basic root principles, it must take a whole new language to express it. ===================================== Yeah ! Trouble is nobody has yet dug up the ancient stone on which the language is carved and translated. There are too many unjustified * 's to make any sense out of thse recently discovered hieroglyphics. I am reminded of my old dear maths master, Mr Stevens. God had blessed him. He was a rare survivor of the machine gun bullets, shrapnel, flame-throwers, and chlorine-gas breathed in without a gas mask while hanging on the barbed wire in no-man's land between the trenches, Shell-fire-Corner, Ypres, Belgium, 1917. He spoke in a hoarse whisper and I always sat in the front row of desks in his classes so I could better hear him. He kept people awake at the back of the class by throwing missiles - sticks of chalk of which he was amply stocked. He referred to Factorial(x) = x! = x Exclamation mark, as "x By Jove" and so endowed on me a lifelong love of the beauty of mathematics. He also taught History in similar vein. --- Reg. |
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