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W5DXP wrote:
Jim Kelley wrote: I've never disputed the solutions produce the same number. I disputed the validity of the approach you took to arrive at the equation. Here's a question for you, Jim. How much power is transferred from the mismatched load rearward by Pref2? First I have to know where it is being transferred. I don't see two ends of a path. Is there a load resistor hiding in a circulator somewhere? Is the 33.33 watts extra power that I could drive another antenna with, but is otherwise going unused? Or is it instead, simply a field that is excited by the boundary conditions at the load that is being measured. source---50 ohm feedline---+---150 ohm feedline---load Pfwd1=200W -- Pfwd2=133.33W -- -- Pref1=100W -- Pref2=33.33W If, as you have asserted, Pref2 can transfer zero watts when it measures 33.33W, can it also transfer 100W when it measures 33.33W? The meter measures voltage, assumes a transfer of energy, and displays power based on that assumption. 73, ac6xg |
Jim Kelley wrote:
First I have to know where it is being transferred. This is the basic flaw in your argument. The power goes where it goes. That you need to know something indicates that you believe what you know dictates reality. Sorry, Jim, you are NOT that powerful. Mother Nature doesn't care what you know or don't know. Reality is exactly the same either way and will be exactly the same way after you are six feet under. Why do you think you are so important that what you know affects reality? -- 73, Cecil http://www.qsl.net/w5dxp "One thing I have learned in a long life: that all our science, measured against reality, is primitive and childlike ..." Albert Einstein -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 80,000 Newsgroups - 16 Different Servers! =----- |
W5DXP wrote: Jim Kelley wrote: First I have to know where it is being transferred. This is the basic flaw in your argument. The power goes where it goes. So, where does the 33.33 watts in your question go? 73, ac6xg |
Jim Kelley wrote:
But where does that 33.33 watts actually go? If you have 100 watts coming from the source, and 100 watts traversing the first and second boundaries, where does the 33.33 watts go? Through a two step process, it merges in phase with the forward wave at the Z0-match point. A directional wattmeter reads 133.33W forward power. 100W + 33.33W = 133.33W. 100W/|1-rho|^2 = 133.33W Forward power - reflected power = power delivered to the load. The standing waves confirm that there is indeed 133.33W forward and 33.33W reflected. Let's look at the following system with two sources. Each source is a signal generator equipped with a circulator and load. SGCL1 is equipped with a 50 ohm circulator and load and SGCL2 is equipped with a 150 ohm circulator and load. The two signal generators are phase locked but can be turned on and off independently. rho=0.5 100W SGCL1---50 ohm feedline---+---150 ohm feedline---33.33W SGCL2 Pfwd1-- Pfwd2-- --Pref1 --Pref2 Using the principle of superposition: With SGCL1 on and SGCL2 off, Pfwd1=100W, Pref1=25W, Pfwd2=75W, and Pref2=0W. SGCL2 dissipates 75W (Pfwd1)(|rho|^2) and SGCL1 dissipates 25W (Pfwd1)(1-|rho|^2). This is how the s11 and s21 s-parameters are measured. With SGCL2 on and SGCL1 off, Pfwd1=0W, Pref1=25W, Pfwd2=8.33W, Pref2=33.33W. SGCL2 dissipates 8.33W (Pref2)(|rho|^2) and SGCL1 dissipates 25W (Pref2)( 1-|rho|^2). This is how the s12 and s22 s-parameters are measured. Note that SGCL1 dissipates 25W in both of the above cases. With SGCL1 and SGCL2 both on, Pfwd1=100W, Pref1=0W, Pfwd2=133.33W, and Pref2=33.33W. This is similar to the earlier single source example. Note that the two 25W waves obviously engage in wave cancellation and their combined 50W of destructive interference joins the forward wave as constructive interference. Assume there is a set of feedline lengths that will accomplish the given values. Vref2 must be 180 degrees out of phase with Vfwd1 at the impedance discontinuity. I think 1WL of 50 ohm feedline and 1/2WL of 150 ohm feedline will do that if the voltages from the two sources are max+ at the same time. b1 = s11*a1 + s12*a2 = 0V s11*a1 and s12*a2 engage in destructive interference. Pref1 = Pfwd1(|rho|^2) + Pref2(1-|rho|^2) - destructive interference 0W = 25W + 25W - 50W Pfwd2 = Pfwd1(1-|rho|^2) + Pref2(|rho|^2) + constructive interference 133.33W = 75W + 8.33W = 50W -- 73, Cecil http://www.qsl.net/w5dxp "One thing I have learned in a long life: that all our science, measured against reality, is primitive and childlike ..." Albert Einstein -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 80,000 Newsgroups - 16 Different Servers! =----- |
Cecil, W5DXP wrote:
"Let`s look at the following system with two sources," Is this tempest about creation of an oppositely phased wave to cancel a reflection? A way to avoid reflected waves within a waveguide is mentioned by Terman on page 148 of his 1955 edition: "Create a reflected wave near the load that is equal in magnitude but opposite in phase from the wave reflected by the load; in this way the two reflected waves cancel each other." "Some of these (matching arrangements) are analogous to the impedance matching arrangements employed in transmission lines (described in Sec. 4-11) while others are unique to waveguides." Best regards, Richard Harrison, KB5WZI |
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W5DXP wrote: A directional wattmeter reads 133.33W forward power. 100W + 33.33W = 133.33W. 100W/|1-rho|^2 = 133.33W Forward power - reflected power = power delivered to the load. The standing waves confirm that there is indeed 133.33W forward and 33.33W reflected. So where does the 33.33 watts actually go? The 100 watt number seems to account for all of the transfer of energy. Let's look at the following system with two sources. Each source is a signal generator equipped with a circulator and load. SGCL1 is equipped with a 50 ohm circulator and load and SGCL2 is equipped with a 150 ohm circulator and load. The two signal generators are phase locked but can be turned on and off independently. rho=0.5 100W SGCL1---50 ohm feedline---+---150 ohm feedline---33.33W SGCL2 Pfwd1-- Pfwd2-- --Pref1 --Pref2 Using the principle of superposition: With SGCL1 on and SGCL2 off, Pfwd1=100W, Pref1=25W, Pfwd2=75W, and Pref2=0W. SGCL2 dissipates 75W (Pfwd1)(|rho|^2) and SGCL1 dissipates 25W (Pfwd1)(1-|rho|^2). With SGCL2 on and SGCL1 off, Pfwd1=0W, Pref1=25W, Pfwd2=8.33W, Pref2=33.33W. SGCL2 dissipates 8.33W (Pref2)(|rho|^2) and SGCL1 dissipates 25W (Pref2)( 1-|rho|^2). This is how the s12 and s22 s-parameters are measured. Note that SGCL1 dissipates 25W in both of the above cases. With SGCL1 and SGCL2 both on, Pfwd1=100W, Pref1=0W, Pfwd2=133.33W, and Pref2=33.33W. This is similar to the earlier single source example. Except for the amount of power input. Although you didn't mention it explicitly, I assume SGCL2 inputs 33.33 watts. Obviously there's that much more power being input to the network in this example that the others we've worked on. Note that the two 25W waves obviously engage in wave cancellation and their combined 50W of destructive interference joins the forward wave as constructive interference. I note that interference takes place. Assume there is a set of feedline lengths that will accomplish the given values. Vref2 must be 180 degrees out of phase with Vfwd1 at the impedance discontinuity. I think 1WL of 50 ohm feedline and 1/2WL of 150 ohm feedline will do that if the voltages from the two sources are max+ at the same time. Well, you won't get a phase reversal from reflection at the 150/50 ohm boundary as you would with the 150/450 ohm boundary in our other problem, so I think the 150 ohm feedline would need to be an odd number of 1/4WL's in order to put Vref1 out of phase with Vfwd1 at the first boundary. But since we're really interested in the relative phase between Vref2 and Vref1, and Vref1 becomes inverted at the 50/150 ohm boundary, you would want to maintain the phase of Vref2 by using an even multiple of 1/4 wave lengths of the 150 ohm line. b1 = s11*a1 + s12*a2 = 0V s11*a1 and s12*a2 engage in destructive interference. Pref1 = Pfwd1(|rho|^2) + Pref2(1-|rho|^2) - destructive interference 0W = 25W + 25W - 50W Pfwd2 = Pfwd1(1-|rho|^2) + Pref2(|rho|^2) + constructive interference 133.33W = 75W + 8.33W = 50W I'll trade you a plus sign for your equal sign. ;-) We also know that 133.33W = 100W + 33.33W. So 100W = 100W + 33.33W - 33.33W. Which is equally revealing. ;-) 73, Jim AC6XG |
Jim Kelley wrote:
So where does the 33.33 watts actually go? The 100 watt number seems to account for all of the transfer of energy. I told you, the 33.33W joins the 100W forward wave at the impedance discontinuity and heads toward the load where a new 33.33W is rejected by the mismatched load. Maybe that is the point of your confusion. The 33.33W that makes a round trip to the impedance discontinuity and back to the load is not the same 33.33W that is reflected by the load. TV ghosting proves that to be true. Let's look at the following system with two sources. Each source is a signal generator equipped with a circulator and load. SGCL1 is equipped with a 50 ohm circulator and load and SGCL2 is equipped with a 150 ohm circulator and load. The two signal generators are phase locked but can be turned on and off independently. rho=0.5 100W SGCL1---50 ohm feedline---+---150 ohm feedline---33.33W SGCL2 Pfwd1-- Pfwd2-- --Pref1 --Pref2 Except for the amount of power input. Although you didn't mention it explicitly, I assume SGCL2 inputs 33.33 watts. "33.33W SGCL2" seems pretty explicit to me. Obviously there's that much more power being input to the network in this example that the others we've worked on. We are also taking out more power than we previously were. I note that interference takes place. Please note that the 25W reflected when SGCL1 only is on is 35.36V at zero deg, and 0.707A at 180 deg. When SGCL2 only is on, the 25W not re-reflected is 35.36V at 180 deg, and 0.707A at zero deg. That's why they cancel when both are on. Assume there is a set of feedline lengths that will accomplish the given values. Vref2 must be 180 degrees out of phase with Vfwd1 at the impedance discontinuity. I think 1WL of 50 ohm feedline and 1/2WL of 150 ohm feedline will do that if the voltages from the two sources are max+ at the same time. Well, you won't get a phase reversal from reflection at the 150/50 ohm boundary as you would with the 150/450 ohm boundary in our other problem, so I think the 150 ohm feedline would need to be an odd number of 1/4WL's in order to put Vref1 out of phase with Vfwd1 at the first boundary. But since we're really interested in the relative phase between Vref2 and Vref1, and Vref1 becomes inverted at the 50/150 ohm boundary, you would want to maintain the phase of Vref2 by using an even multiple of 1/4 wave lengths of the 150 ohm line. Nope, you are wrong about that but I'll let you figure our your own error. The only voltage phase reversal in the given example is in the re-reflected wave associated with the 8.33W. Vfwd1 travels 1WL while Vref2 travels 1/2WL. That is enough to put them 180 degrees out of phase. Vfwd1(|rho|^2) undergoes zero phase shift. Vref2(1-|rho|^2) undergoes zero phase shift. -- 73, Cecil http://www.qsl.net/w5dxp "One thing I have learned in a long life: that all our science, measured against reality, is primitive and childlike ..." Albert Einstein -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 80,000 Newsgroups - 16 Different Servers! =----- |
Richard Harrison wrote:
Terman said about the same thing about cancellation of the reflected wave in a waveguide. Terman offered several methods to generate the offsetting wave in a waveguide, and there are several ways to generate a wave which produces cancellation on a transmission line, too. Trouble is, Terman didn't say anything about what happens to the intrinsic energy that existed in the waves before they were canceled. I say the energy in rearward-traveling canceled reflected waves is reflected back toward the load. Jim (and others) disagree even though we know that all the energy winds up incident upon the load. -- 73, Cecil, W5DXP |
Jim Kelley wrote:
W5DXP wrote: The 33.33W that makes a round trip to the impedance discontinuity and back to the load is not the same 33.33W that is reflected by the load. TV ghosting proves that to be true. TV ghosting is primarily the result of multipathing, which is external to the antenna system. :-) I'm talking about TV ghosting due to reflections on the transmission line between a TV test generator and a TV receiver. The only "multipathing" involved is the different paths taken by the reflected waves inside the transmission line. The only voltage phase reversal in the given example is in the re-reflected wave associated with the 8.33W. How are you determining your phase reversals? Don't forget that phase reversal on reflection occurs when a wave encounters a more dense medium (or a higher impedance). Nope, a voltage phase reversal occurs when a wave encounters a *lower* impedance. A voltage wave encountering a higher impedance does not reverse phase. In the earlier example, the only lower impedance encountered is by the rearward-traveling wave flowing back into the impedance discontinuity. I repeat, Vref2(-rho) is the only voltage phase reversal. -- 73, Cecil, W5DXP |
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