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The Extreme Failure of Poor Concepts in Discussing Thin Layer Reflections
Now we venture to new materials, and in this case a solar cell, described in text as: 1w | 1/4WL | laser-----air-----|---thin-film---|---Germanium---... 1st medium | 2nd medium | 3rd medium n = 1.0 n = 2 n = 4.04 where the second medium might be Arsenic trisulfide glass or Lanthanum flint glass. When you take the intensity times the area for both the reflected and refracted beams, the total energy flux must equal that in the incident beam. That equation appears as: (r² + (t² · n2² · cos(theta-t) / n1² · cos(theta-i))) = 1 It stands to reason that this can be quickly reduced without need to use transcendentals for an angle of incidence of 0° (which results in a refractive angle of 0°). All that needs to be known are the coefficients which for that same angle simplify to r = 0.667 a value that is the limit of an asymptote; it is also invested with either a + or - sign depending upon the polarization (another issue that was discarded in the original discussion as more unknown than immaterial) t = 0.667. a value that is the limit of an asymptote; here, too, there are polarization issues we will discard as before. All this discarding comes only by virtue of squaring: r² = 0.445 t² = 0.445 I presume that the remainder of the math can be agreed to exhibit: that part of the energy reflected amounts to 11% or otherwise expressed as: 110mW and that part of the energy transmitted amounts to 89% or otherwise expressed as: 890mW It follows that at the second boundary there is less power available due to the conservation of power observed at the first boundary. They exhibit these results, by percentages: that part of the energy reflected amounts to 11% or otherwise expressed as: 98mW and that part of the energy transmitted amounts to 89% or otherwise expressed as: 792mW Needless to say, that same first interface is going to conserve energy by the total of refraction and reflection being equal to the energy incident upon it. I will skip that to allow ALL of this second reflection to "try" to totally cancel the first reflection: 110mW - 92mW or 18mW But how does this "perfect" result fare in the real world where this 1W laser has an un-cancelled reflection remainder? How does it fare in an application to reduce reflections that the Solar Cell is sensitive to? A laser of this power will have a beam size on the order of 1mm². I typically describe the available power from the sun as being 1000W/M². This laser then would, in terms of W/M², be quite powerful at 1,000,000W/M². This is more than 1000 times more power than the sun's exposure to the same target. However, the sun does not radiate one wavelength of energy. So, if we were to reduce the amount of sunlight confined to that in the Lased BW at its operating wavelength; then let's consider that the sun's power is in a BW of 2000nM and we are talking about (and I will be MOST generous to offer an absurdly wide) Lased BW of 20nM. It follows that the Lased power is thus 100,000 times brighter than the sun in the same BW for the same wavelength. But this still neglects that the sun's power is not evenly distributed throughout this 2000nM BW. I am not going to pencil whip this further. Let's simply return to that un-cancelled power and look at it instead: 18mW in that same 1mm², which if we cast to the same terms of comparison to sunlight it becomes 18000W/M² which is 18 TIMES THE POWER OF THE SUN's total BW emission. As you may guess I am going to use the same BW correction to find that the un-cancelled reflection products have 1800 TIMES MORE POWER THAN THE SUN! 73's Richard Clark, KB7QHC |
Richard Clark wrote:
Richard's following example is similar to a Z0-matched lossless transmission line example. 1/4WL 1w XMTR---50 ohm ---+---100 ohm---+---200 ohm---+-200 ohm load feedline feedline feedline Pfor=3D1w Pfor=3D1.125w Pfor=3D1w Pref=3D0w Pref=3D0.125w Pref=3D0w Anyone capable of solving transmission line problems can verify the above values which are the same as the values in Richard's example although he doesn't realize it yet. Now we venture to new materials, and in this case a solar cell, described in text as: Richard, you failed to admit your errors on the previous example and failed to appologize to me for all your insults. Now you present yet another example before the first one was resolved. And this example contains many of the errors that you made in the earlier one. It's probably time to shut off your output and engage in a little input before you embarrass yourself any farther. 1w | 1/4WL | laser-----air-----|---thin-film---|---Germanium---... 1st medium | 2nd medium | 3rd medium n =3D 1.0 n =3D 2 n =3D 4.04 where the second medium might be Arsenic trisulfide glass or Lanthanum flint glass. When you take the intensity times the area for both the reflected and refracted beams, the total energy flux must equal that in the incident beam. That equation appears as: (r=B2 + (t=B2 =B7 n2=B2 =B7 cos(theta-t) / n1=B2 =B7 cos(theta-i)= )) =3D 1 It stands to reason that this can be quickly reduced without need to use transcendentals for an angle of incidence of 0=B0 (which results in a refractive angle of 0=B0). All that needs to be known are the coefficients which for that same angle simplify to r =3D 0.667 a value that is the limit of an asymptote; it is also invested with either a + or - sign depending upon the polarization (another issue that was discarded in the original discussion as more unknown than immateria= l) Uhhhhh Richard, the amplitude reflection coefficient, r, is: r =3D (nt-ni)/(nt+ni) =3D (2-1)/(2+1) =3D 0.3333 =3D (4-2)/(4+2) You are never going to get it right as long as you cannot even calculate the reflection coefficient. r=B2 =3D 0.445 Wrong again! r^2 =3D Reflectance, R =3D (0.3333)^2 =3D 0.1111 or 11.11% t=B2 =3D 0.445 t^2 doesn't matter. The Transmittance, T =3D (1-R) =3D 0.8889 or 88.89% I presume that the remainder of the math can be agreed to exhibit: that part of the energy reflected amounts to 11% or otherwise expressed as: 110mW That's correct but doesn't agree with your r^2 value. Let's say 11.11% or 111.1mW for four digit accuracy. and that part of the energy transmitted amounts to 89% or otherwise expressed as: 890mW Let's say 88.89% or 888.9mW for four digit accuracy. It follows that at the second boundary there is less power available due to the conservation of power observed at the first boundary. They exhibit these results, by percentages: that part of the energy reflected amounts to 11% or otherwise expressed as: 98mW and that part of the energy transmitted amounts to 89% or otherwise expressed as: 792mW You are ignoring the wave cancellation energy and getting the wrong answers. You have proved my point better than I ever could have. The first internal reflection is 888.9mW*0.1111 =3D 98.76mW. When that first internal reflection encounters Medium 1, 87.78mW will be transmitted through to Medium 1. It is 180 degrees out of phase with the first internal reflection so the two waves will start to cancel. Thus 2*87.78 =3D 175.56mW will join the forward wave in Medium 2. Even after this first re-reflection event, the forward power in Medium 2 will be 1064.5mW on its way to 1125mW during steady-state. It is readily apparent that you simply don't know how to perform this analysis either during the transient state or steady-state. The steady-state values are easily known and yours are wrong. Have you never solved a problem like this before? Needless to say, that same first interface is going to conserve energy by the total of refraction and reflection being equal to the energy incident upon it. I will skip that to allow ALL of this second reflection to "try" to totally cancel the first reflection: 110mW - 92mW or 18mW You calculations are once again wrong. When you add up all the infinite number of reflections and re-reflections, you will find that the sum of all the internal transmissions into Medium 1 indeed does equal 111.1mw thus totally canceling all of the reflections. Assuming a lossless system, there will be one watt delivered into medium 3. The incident power necessary for that event is 1w/0.8889 =3D 1.125w. All reflections from the thin-film are canceled. Here are all the steady-state powers calculated for you: 1w | 1/4WL | laser-----air-----|---thin-film---|---Germanium---... 1st medium | 2nd medium | 3rd medium n =3D 1.0 n =3D 2.0 n =3D 4.0 Pfor=3D1w Pfor=3D1.125w Pfor=3D1w Pref=3D0w Pref=3D0.125w Pref=3D0w Note that T*Pref2 =3D 0.8889*0.125 =3D 111.1mW, exactly the amount of energy required to cancel the initial external reflection of 111.1mW. In the 2nd medium, P1=3D888.9mW, P2=3DPref2*R =3D 125mW*0.1111 =3D 13.89mW P1 + P2 + 2*sqrt(P1*P2) =3D 888.9 + 13.89 + 222.2 =3D 1125mW Hecht's equations are correct after all. What you have proven is that when you ignore the interference energy, you will get the wrong answer. -- 73, Cecil http://www.qsl.net/w5dxp |
Richard Clark wrote:
Needless to say, that same first interface is going to conserve energy by the total of refraction and reflection being equal to the energy incident upon it. I will skip that to allow ALL of this second reflection to "try" to totally cancel the first reflection: 110mW - 92mW or 18mW Richard, if you follow the leads of other gurus on this newsgroup, you will 'ploink' me, never admit a mistake, and dupe yourself into believing that you were right all along. Your "Extreme Failures of Poor Concepts" could have been avoided if you guys had just read and understood the following two quotes that I have posted multiple times. What is it about, "... all 'lost' reflected intensity will appear as enhanced intensity in the transmitted beam.", that you guys don't understand? www.mellesgriot.com/products/optics/oc_2_1.htm "Clearly, if the wavelength of the incident light and the thickness of the film are such that a phase difference exists between reflections of p, then reflected wavefronts interfere destructively, and overall reflected intensity is a minimum. If the two reflections are of equal amplitude, then this amplitude (and hence intensity) minimum will be zero." "In the absence of absorption or scatter, the principle of conservation of energy indicates all 'lost' reflected intensity will appear as enhanced intensity in the transmitted beam. The sum of the reflected and transmitted beam intensities is always equal to the incident intensity. This important fact has been confirmed experimentally." http://micro.magnet.fsu.edu/primer/j...ons/index.html "... when two waves of equal amplitude and wavelength that are 180- degrees out of phase with each other meet, they are not actually annihilated. All of the photon energy present in these waves must somehow be recovered or redistributed in a new direction, according to the law of energy conservation ... Instead, upon meeting, the photons are redistributed to regions that permit constructive interference, so the effect should be considered as a redistribution of light waves and photon energy rather than the spontaneous construction or destruction of light." -- 73, Cecil http://www.qsl.net/w5dxp |
Richard Clark wrote:
Now we venture to new materials, and in this case a solar cell, described in text as: 1w | 1/4WL | laser-----air-----|---thin-film---|---Germanium---... 1st medium | 2nd medium | 3rd medium n = 1.0 A n = 2 B n = 4.04 I have added points 'A' and 'B' to Richard's diagram. Let's make the math simple by having the index of refraction, n, of the 3rd medium be equal to 4.0. The power reflection coefficient is 0.1111 and the power transmission coefficient is 0.8889. All powers will be stated in mW. Since n2 = sqrt(n1*n4), the system will be reflectionless in the 1st medium. Proof of that assertion will be left to the reader but that is a necessary and sufficient condition for a lossless system. Here is the transient state of the system an instant after the first internal reflection from point B arrives back at point A. 1st medium 2nd medium 3rd medium A B 1000mW----| | |----888.9mW----| 111.1mW----| |----790.1mW |----98.76mW----| 87.79mW----| | |----10.97mW Since the 2nd medium is 1/4WL, in the 1st medium, the 111.1mW external reflection will be 180 degrees out of phase with the 87.79mW internal transmission leaving very little of the external reflection uncanceled. A quick calculation indicates 1.37mW left uncanceled. When the second internal reflection from point B arrives back at point A, more wave cancellation will occur. After very few iterations of reflections, the 111.1mW external reflection will be canceled. As in the earlier example, the result is flat black for reflections in the 1st medium. Richard Clark sez: As you may guess I am going to use the same BW correction to find that the un-cancelled reflection products have 1800 TIMES MORE POWER THAN THE SUN! 1800 TIMES MORE POWER THAN THE SUN from flat black???? Quick, Richard, get a patent on that process. Someone sent me an email wondering if Richard C. has a death wish. :-) -- 73, Cecil http://www.qsl.net/w5dxp |
Here is the transient state of the system an instant
after the first internal reflection from point B arrives back at point A. 1st medium 2nd medium 3rd medium A B 1000mW----| | |----888.9mW----| 111.1mW----| |----790.1mW |----98.76mW----| 87.79mW----| | |----10.97mW You are correct that in this system (1/4-wave matching layer), the total reflectance goes to zero. However, your understanding of superposition is wrong. You CANNOT superimpose POWERS, or even talk about the "power" of various reflections in the same media. You can only add the wave amplitudes (electric fields). THEN you take the total amplitude and square that to get the power. Tor N4OGW |
but of course ground isn't ground. there is no such thing as ground, or a
ground potential, or a perfect ground, or anything like that. and yes, currents flowing in the ground can be a bad thing, but only if you don't understand them and design to handle them. "Zombie Wolf" wrote in message ... Ah but you must avoid upsetting GOD (oops I mean Richard.). A simple answer about grounding given to a guy a while back put him into apoplectic fits, necessitating the doubling of his meds. You see, "ground" ISN'T "ground". That isn't "clever" enough for Richard. He promptly claimed that rf currents flowing in a ground system were a "bad thing' and proceeded to "explain" it all to us. I wish my elmer 30 years ago had been as "clever" as Richard. All these books on the shelves simply are irrelevant, because nothing can be simple, you see... It must be complicated, even if it IS wrong........ "W5DXP" wrote in message oups.com... Richard Clark wrote: Needless to say, that same first interface is going to conserve energy by the total of refraction and reflection being equal to the energy incident upon it. I will skip that to allow ALL of this second reflection to "try" to totally cancel the first reflection: 110mW - 92mW or 18mW Richard, if you follow the leads of other gurus on this newsgroup, you will 'ploink' me, never admit a mistake, and dupe yourself into believing that you were right all along. Your "Extreme Failures of Poor Concepts" could have been avoided if you guys had just read and understood the following two quotes that I have posted multiple times. What is it about, "... all 'lost' reflected intensity will appear as enhanced intensity in the transmitted beam.", that you guys don't understand? www.mellesgriot.com/products/optics/oc_2_1.htm "Clearly, if the wavelength of the incident light and the thickness of the film are such that a phase difference exists between reflections of p, then reflected wavefronts interfere destructively, and overall reflected intensity is a minimum. If the two reflections are of equal amplitude, then this amplitude (and hence intensity) minimum will be zero." "In the absence of absorption or scatter, the principle of conservation of energy indicates all 'lost' reflected intensity will appear as enhanced intensity in the transmitted beam. The sum of the reflected and transmitted beam intensities is always equal to the incident intensity. This important fact has been confirmed experimentally." http://micro.magnet.fsu.edu/primer/j...ons/index.html "... when two waves of equal amplitude and wavelength that are 180- degrees out of phase with each other meet, they are not actually annihilated. All of the photon energy present in these waves must somehow be recovered or redistributed in a new direction, according to the law of energy conservation ... Instead, upon meeting, the photons are redistributed to regions that permit constructive interference, so the effect should be considered as a redistribution of light waves and photon energy rather than the spontaneous construction or destruction of light." -- 73, Cecil http://www.qsl.net/w5dxp |
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Hi Tor,
This is all taken care of in the equation for energy conservation that I posted. The solution is valid barring a simple error of subtraction that should read: 110mW - 98mW Not sure which indices you are now referring to: n_3=4.0 or 4.04? With 4.0, there is zero reflected light. With 4.04 there is a small amount. Tor N4OGW |
Richard Clark wrote:
wrote: However, your understanding of superposition is wrong. You CANNOT superimpose POWERS, or even talk about the "power" of various reflections in the same media. You can only add the wave amplitudes (electric fields). THEN you take the total amplitude and square that to get the power. This is all taken care of in the equation for energy conservation that I posted. Tor, by misunderstanding my posting, you have just fed the monster. :-) -- 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 =---- |
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A B
1000mW----| | |----888.9mW----| 111.1mW----| |----790.1mW |----98.76mW----| 87.79mW----| | |----10.97mW ^^^^^ I would not even dream of superposing powers. Perhaps you don't understand the role of the interference term in the Then don't draw diagrams like above labeled with powers of multiple reflections bouncing around in the same media :) Eugene Hecht would be surprised to hear you say such since he talked about the powers (irradiance) of various reflections in the same media. Only when dealing with the interference of TWO waves. He works with electric field when discussing anti-reflection layers. See 9.7.1 I'm not saying the power equation you give is wrong, but it is correct for just two waves. With more than two waves, it obviously becomes more complicated Tor N4OGW |
Richard Clark wrote:
On 26 Jul 2005 14:54:49 -0500, wrote: Not sure which indices you are now referring to: n_3=4.0 or 4.04? With 4.0, there is zero reflected light. With 4.04 there is a small amount. It doesn't matter one iota. Of course it matters. The 1,2,4 combination meets the necessary and sufficient conditions for the elimination of reflections in medium 1 assuming a steady-state lossless system, i.e. perfectly matched. The point of the matter is with such design characteristics fulfilled, then each interface reflects/transmits in the same proportion. With each reflecting/transmitting in the same proportion, the second interface, by the actions of the first, must have less incident upon it. That is simply not true, Richard. You obviously have not read my Melles-Griot web posting and/or simply don't understand it. The forward power in medium 2 is greater than the forward power in either medium 1 or medium 3. You keep making that same mistake over and over and over. Why don't you simply take time to understand the truth? What is even more obvious is that following the second interface, you have lost roughly 20% of that incident upon the first interface. But you have gained the energy involved in the wave cancellation in medium 1. The forward power in medium 2 is higher than it is in medium 1. You don't actually believe the BS by some gurus on this newsgroup that standing waves don't contain any energy, do you? There is not enough energy at the second interface, reflected back, to "Totally Cancel" the energy in the reflection of the first interface. There can be no other outcome. Only in your mind, Richard, only in your mind. The outcome in reality is complete cancellation of the reflections in medium 1 assuming 1,2,4 indices of refraction. That difference yields first interface reflection products that have: 1200 TIMES MORE POWER THAN THE SUN! Somebody get the net! -- 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 =---- |
Tor, by misunderstanding my posting, you have just fed the monster. :-) I see :O Well, I can't get everyone to understand basic physics. Tor N4OGW |
Ah but you must avoid upsetting GOD (oops I mean Richard.). A simple answer
about grounding given to a guy a while back put him into apoplectic fits, necessitating the doubling of his meds. You see, "ground" ISN'T "ground". That isn't "clever" enough for Richard. He promptly claimed that rf currents flowing in a ground system were a "bad thing' and proceeded to "explain" it all to us. I wish my elmer 30 years ago had been as "clever" as Richard. All these books on the shelves simply are irrelevant, because nothing can be simple, you see... It must be complicated, even if it IS wrong........ "W5DXP" wrote in message oups.com... Richard Clark wrote: Needless to say, that same first interface is going to conserve energy by the total of refraction and reflection being equal to the energy incident upon it. I will skip that to allow ALL of this second reflection to "try" to totally cancel the first reflection: 110mW - 92mW or 18mW Richard, if you follow the leads of other gurus on this newsgroup, you will 'ploink' me, never admit a mistake, and dupe yourself into believing that you were right all along. Your "Extreme Failures of Poor Concepts" could have been avoided if you guys had just read and understood the following two quotes that I have posted multiple times. What is it about, "... all 'lost' reflected intensity will appear as enhanced intensity in the transmitted beam.", that you guys don't understand? www.mellesgriot.com/products/optics/oc_2_1.htm "Clearly, if the wavelength of the incident light and the thickness of the film are such that a phase difference exists between reflections of p, then reflected wavefronts interfere destructively, and overall reflected intensity is a minimum. If the two reflections are of equal amplitude, then this amplitude (and hence intensity) minimum will be zero." "In the absence of absorption or scatter, the principle of conservation of energy indicates all 'lost' reflected intensity will appear as enhanced intensity in the transmitted beam. The sum of the reflected and transmitted beam intensities is always equal to the incident intensity. This important fact has been confirmed experimentally." http://micro.magnet.fsu.edu/primer/j...ons/index.html "... when two waves of equal amplitude and wavelength that are 180- degrees out of phase with each other meet, they are not actually annihilated. All of the photon energy present in these waves must somehow be recovered or redistributed in a new direction, according to the law of energy conservation ... Instead, upon meeting, the photons are redistributed to regions that permit constructive interference, so the effect should be considered as a redistribution of light waves and photon energy rather than the spontaneous construction or destruction of light." -- 73, Cecil http://www.qsl.net/w5dxp |
Cecil Moore wrote: Powers obey the equation P1 + P2 + 2*(P1*P2)cos(theta) I thought you said nature doesn't obey math models. :-) P1 + P2 would be superposing powers. I didn't do that. P1 + P2 + 2*sqrt(P1*P2)cos(theta) is NOT superposing powers. You do understand that interference is the result of the superposition of waves traveling in the same direction, right? Eugene Hecht would be surprised to hear you say such since he talked about the powers (irradiance) of various reflections in the same media. Again Cecil, you're really in no position to speak on his behalf. He might not think that highly of your doing so in fact. Perhaps you should obtain his permission first. I did not superpose powers. I used Hecht's equations for powers (irradiance). Actually, the fact that those equations appear in his book (as well as just about any other optics book) does not convey any rights of ownership or invention to the author. They were around a long time before Dr. Hecht. The only difference in irradiance and power is the unit-area which, for our example, is arbitrary. You should show him your equation with all the power terms (P1-P4) in it! You had some of the powers going one way, and the others going the other way. You needed them all to add up to some particular number, so you just added them. - or maybe subtracted some of them. I don't recall exactly. |
wrote:
W5DXP wrote: Eugene Hecht would be surprised to hear you say such since he talked about the powers (irradiance) of various reflections in the same media. Only when dealing with the interference of TWO waves. He works with electric field when discussing anti-reflection layers. See 9.7.1 I'm not saying the power equation you give is wrong, but it is correct for just two waves. With more than two waves, it obviously becomes more complicated medium 1 medium 2 medium 3 A B 1000mW----| | |----888.9mW----| 111.1mW----| |----790.1mW |----98.76mW----| 87.79mW----| | |----10.97mW I appologize if the diagram wasn't clear. It is extremely difficult to draw graphics in ASCII. But I only ever deal with TWO waves. Above in medium 1, the two waves are the 111.1 wave and the 87.79 wave. There are only TWO rearward-traveling waves there. Hecht's irradiance equation works for those TWO waves. The TWO waves in medium 2 are the 888.9 wave and the 10.97 wave. There are only TWO forward-traveling waves there. Hecht's irradiance equation works for those TWO waves. Since I am not trying to calculate the power in the standing waves, there are only ever TWO values of power to deal with. Sorry for the confusion. -- 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 =---- |
Jim Kelley wrote:
You do understand that interference is the result of the superposition of waves traveling in the same direction, right? Of course, there are two superposition terms in each direction represented by the two S-parameter equations. b1 = s11*a1 + s12*a2 toward the source b2 = s21*a1 + s22*a2 toward the load The two terms in the first equation interfere with each other toward the source. The two terms in the second equation interfere with each other toward the load. When b1=0, the system is matched, and total destructive interference exists toward the source. In that case, total constructive interference exists toward the load. Actually, the fact that those equations appear in his book (as well as just about any other optics book) does not convey any rights of ownership or invention to the author. They were around a long time before Dr. Hecht. Didn't mean to imply that he owns them. "Hecht's equations" simply means they appear in his book. They were around a long time before anyone on this newsgroup called them "gobblegook". The interference equations also appear in Chipman's 1968 "Transmission Lines". You should show him your equation with all the power terms (P1-P4) in it! You had some of the powers going one way, and the others going the other way. Nope, you are confused. All powers are going the same direction for any individual equation in my article. Destructive interference energy in one direction equals constructive interference energy in the opposite direction. If you will wade through the transient buildup to steady-state of the recent examples, you won't be confused anymore by direction. During the transient buildup phase, when the first destructive interference occurs toward the source, an equal amount of constructive interference energy flows toward the load - same energy, just flowing in the opposite direction. If destructive interference energy in one direction equals P3+P4, constructive interference energy in the opposite direction equals P3+P4 to satisfy the conservation of energy principle. Perhaps you are remembering the equation (Pload = Pfor - Pref) where Pfor and Pref are flowing in different directions. But that equation doesn't appear in my article. -- 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 =---- |
wrote:
You CANNOT superimpose POWERS, ... Hi Tor, I appologize that the earlier energy flow chart was not drawn as well as it might have been. I've expanded that chart in the following and provided the power management equations associated with the superposition of EM fields. The 'X' event is the initial reflection event at the first discontinuity, point A. The 'Y' event is the initial reflection event at the second discontinuity, point B. The 'Z' event is the initial re-reflection event back at point A. 100W 1/4WL XMTR---50 ohm---+--------100 ohm---------+---200 ohm---200 load feedline A feedline B feedline Just before the first reflection arrives back at A Pfor(50) | | 100W----X | X----+ | 11.11W----X \ Pfor(100) | Pref(50) | +----88.89W----Y | Y----79.01W | +----9.876W----Y Pfor(200) | / Pref(100) | | --+ | | | Just after the first reflection arrives back at A Pfor(50) | Pfor(100) | 100W----X 109.67W | X----+ | 11.11W----X \ P1 | P3 | +----88.89W----Y | Y----79.01W Pref(50) | +----9.876W----Y Pfor(200) 0.139W | / Pref(100) | Z----+ | 8.779W----Z | P4 Z----1.097W | | P2 | P1 = Pfor(50)*(1-rho^2), P2 = Pref(100)*rho^2 P3 = Pref(100)*(1-rho^2), P4 = Pfor(50)*rho^2 How much destructive interference between wave(P3) and wave(P4)? Pref(50) = P3 + P4 - 2*sqrt(P3*P4) = 11.11W + 8.779W - 19.75W = 0.139W 19.75W of destructive interference, negative since cos(180)=-1 How much constructive interference between wave(P1) and wave(P2)? Pfor(100) = P1 + P2 + 2*sqrt(P1*P2) = 88.89W + 1.097W + 19.75W = 109.67W 19.75W of constructive interference, positive since cos(0)=+1 Destructive interference energy = constructive interference energy (toward the source) (toward the load) It follows that |P1*P2| = |P3*P4| in order to satisfy the conservation of energy principle. -- 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 =--- |
Tor, N4OGW wrote:
"You CANNOT superimpose POWERS, or even talk about the "power" of various reflections in the same media." It is done all the time. P=Esquared / R = Isquared x R. On page 99 of the 1955 edition of "Electronic and Radio Engineering" Terman writes: "The standing-wave ratio S is one means of expressing the magnitude of the reflectiom coefficient; the exact relation between the two is: S=1+absolute value of the reflection coefficient / 1-absolute value of the reflection coefficient or absolute value of the reflection coefficient= S-1/S+1 Standing-wave ratio=S=Emax/Emin This definition of standing-wave ratio is sometimes called voltage standing-wave ratio (VSWR) to distinguish it from the standing-wave ratio expressed as a power ratio, which is (Emax/Emin)squared." Best regards, Richard Harrison, KB5WZI |
Richard Harrison wrote: Tor, N4OGW wrote: "You CANNOT superimpose POWERS, or even talk about the "power" of various reflections in the same media." It is done all the time. P=Esquared / R = Isquared x R. On page 99 of the 1955 edition of "Electronic and Radio Engineering" Terman writes: "The standing-wave ratio S is one means of expressing the magnitude of the reflectiom coefficient; the exact relation between the two is: S=1+absolute value of the reflection coefficient / 1-absolute value of the reflection coefficient or absolute value of the reflection coefficient= S-1/S+1 Standing-wave ratio=S=Emax/Emin This definition of standing-wave ratio is sometimes called voltage standing-wave ratio (VSWR) to distinguish it from the standing-wave ratio expressed as a power ratio, which is (Emax/Emin)squared." I think he means that it does not occur in nature, and it therefore shouldn't be done mathematically. That's because power doesn't propagate, and I hasten to add - neither do Poynting vectors. Nor does power reflect, refract, or diffract. Power is a rate at which energy is transferred, absorbed, or dissipated. It's not a wave which propagates. It's a mathematical product of two of the characteristics of a wave that propagates. Numbers on the other hand can indeed be multiplied, divided, squared, added, and subtracted. We can often even get the right answer when we do that with power numbers. But that fact can't necessarily be extrapolated to mean that the given mathematical operation also takes place in a transmission line. An that is what the author in question is attempting to have us believe. 73, ac6xg |
Richard Harrison wrote:
. . . Standing-wave ratio=S=Emax/Emin This definition of standing-wave ratio is sometimes called voltage standing-wave ratio (VSWR) to distinguish it from the standing-wave ratio expressed as a power ratio, which is (Emax/Emin)squared." Now, that's an interesting concept. The VSWR is the ratio of the maximum voltage anywhere along the line to the minimum voltage anywhere along the line (presuming the line is sufficiently long for a full maximum and minimum to occur). We can use a voltage probe to actually measure this voltage in a slotted line. If the voltage is sinusoidal, so is the power, just at twice the frequency. (I'm sure you can find that in Terman somewhere, but it's simple to derive with a little trigonometry.) Since you say that power superposes, we should expect power waves to add and cancel just like voltage waves. And this should result in dips and peaks in the power as we move along the line, just like voltage, right? I'd expect the dips and peaks to be twice as closely spaced, however, because the wavelength of the power waves would be half the wavelength of the voltage waves. So is the ratio of the maximum to minimum power along the line due to this superposition of power waves equal to the PSWR? Roy Lewallen, W7EL |
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Richard Clark wrote: It is quite evident that through the actions of the first interface, that there is less energy incident upon the second interface. Further, given that both interfaces operate with identical reflective and transmissive properties, it follows the second interface could not reflect enough to totally negate the reflections of the first. True for any one reflection. But as an optical engineer I'm sure you're aware that, even in a lossy medium, a given wave reflects back and forth multiple times before it's amplitude is reduced to insignificance. As you know, the measured amplitude at a surface would then be the superposition of multiple successively reflected waves. So says JM Vaughn in his book "The Fabry-Perot Interfermeter". He cites Born and Wolf a lot. Also Kuhn, Steel, Liddell, Mcleod, Meissner, Tolansky, Jacquinot, and of course, Fabry and Perot. Strangely, no mention of Hecht. For those who are interested, a Fabry-Perot interferometer is an optical instrument comprising two partially reflective parallel mirror surfaces separated by some fixed or variable distance. It's basically a narrow-bandpass filter at light wavelengths. 73, ac6xg |
Roy Lewallen wrote:
Since you say that power superposes, we should expect power waves to add and cancel just like voltage waves. Strawman Alert!!! Richard H. did NOT say power superposes. It is obvious that what he was disagreeing with was the statement: "You CANNOT ... even talk about the "power" of various reflections in the same media." -- 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 =---- |
Richard Clark wrote:
It is quite evident that through the actions of the first interface, that there is less energy incident upon the second interface. On the contrary, it is quite evident that through the actions of the first interface, namely reflection and wave cancellation, that there is *MORE* energy incident upon the second interface. Here's the example from my earlier posting. Those who can, please calculate the forward power in the 1/4WL 100 ohm section. Who believes, like Richard C., that it cannot possibly be 100 watts or more? 100W 1/4WL XMTR---50 ohm---+--------100 ohm---------+---200 ohm---200 load feedline A feedline B feedline Torr had a question in this regard, but as he is a casual correspondent we have not seen any further comment from him - nor would I expect him to have amplified on your observation above. Why don't you send Tor an email and allow him to teach you how to handle problems like these? -- 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 =---- |
Jim Kelley wrote:
That's because power doesn't propagate, and I hasten to add - neither do Poynting vectors. Nor does power reflect, refract, or diffract. Power is a rate at which energy is transferred, absorbed, or dissipated. It's not a wave which propagates. It's a mathematical product of two of the characteristics of a wave that propagates. I'm not arguing with you but as with all words, it depends upon how one defines "power". You have your own personal narrow definition. The power industry has a different definition. Even the field of RF engineering has a definition of power different from yours. The IEEE Dictionary has 11 pages of definitions dealing with power including the "power reflection coefficient", "power density of a traveling wave", "radiated power", "power-flow vector", "power transfer", "power 'carried' by a waveguide" ... Why don't you visit your local power company, call a meeting of their engineers, and inform them that there is no power flowing in their transmission lines? :-) -- 73, Cecil http://www.qsl.net/w5dxp |
Jim Kelley wrote:
"That`s because power doesn`t propagate, and hasten to add - neither do Poynting vectors." Words are not objects. They merely represent objects. We use abstractions for brevity and clarity. Even the best say: "--the Poynting vector or power density (watts per square meter)." See Kraus` 3rd edition of "Antennas" page 73, under "Power Patterns". Best regards, Richard Harrison, KB5WZI |
On Fri, 29 Jul 2005 16:58:29 -0700, Jim Kelley
wrote: It is quite evident that through the actions of the first interface, that there is less energy incident upon the second interface. Further, given that both interfaces operate with identical reflective and transmissive properties, it follows the second interface could not reflect enough to totally negate the reflections of the first. True for any one reflection. Hi Jim, And true for ALL accumulated reflections there after. Reflections do not add any energy to the cup when the first interface is draining it more quickly. But as an optical engineer I'm sure you're aware that, even in a lossy medium, a given wave reflects back and forth multiple times before it's amplitude is reduced to insignificance. As you know, the measured amplitude at a surface would then be the superposition of multiple successively reflected waves. My analysis allowed ALL of the energy in the reflection from the second interface ( 0.098X) to combine with the first reflection (0.11X). This total superposition was both more than generous, and at the same time very unlikely; and yet with this generous allowance there is still excess reflection from the first interface. Hence for something less than total superposition of ALL energies, it hardly bodes a better yield in total cancellation - the energy just isn't there in the first place. 0.098X 0.11X is the simple economics of the balance. As an optical engineer, I've dealt with the harsh reality of this myth of total reflection cancellation. I've designed systems with 9 orders of dynamic range and the couple of percent dashed off as being invisible by academics was distinctly and overwhelmingly present. Basically I had the advantage in fluorescence measurement in being able to turn off my detector during the initial flash to suppress the reflection products in time rather than in these shenanigans (I did them too - and certainly much more - because even the detector can be blinded in its "off" state). Basically these claims are for first year students where demanding too much inquiry would push them into switching majors to Business school. Simple optics with simple, ordinary glasses exhibit quite useful results, but they do not embody a proof. To anyone following the math of my presentation, it is quite obvious what WOULD tend towards a more complete cancellation - and such a subtle shift in the formula diverges only slightly from the choir book hymn. It's not that hard when the interface ratios drive the answer. 73's Richard Clark, KB7QHC |
Richard Harrison wrote:
Jim Kelley wrote: "That`s because power doesn`t propagate, and hasten to add - neither do Poynting vectors." Words are not objects. They merely represent objects. We use abstractions for brevity and clarity. Even the best say: "--the Poynting vector or power density (watts per square meter)." See Kraus` 3rd edition of "Antennas" page 73, under "Power Patterns". If we send a pulse of energy down a transmission line, one wonders where the power in the pulse is if not in the pulse. -- 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 =---- |
Richard Clark wrote:
And true for ALL accumulated reflections there after. Reflections do not add any energy to the cup when the first interface is draining it more quickly. If the first interface is a Z0-match, it is not "draining it" at all. That's what you are missing. The examples have been for matched systems. If you change your medium 3 to an index of refraction of 4.0 instead of 4.04, your example is *perfectly matched* and the first interface is NOT draining the system because all reflections are canceled. 100% of the source power is being delivered through the first interface because of that wave cancellation. At that point it is being joined by the reflected power so the forward power in the second medium is greater than the source power. In the second medium, forward power equals source power plus reflected power. In your example, source power is one watt, forward power in the second medium is 1.125 watt, and reflected power in the second medium is 0.125 watt. My analysis allowed ALL of the energy in the reflection from the second interface ( 0.098X) to combine with the first reflection (0.11X). This total superposition ... Power cannot be superposed. If you are going to deal with power, you need to use the interference equations for power calculations. Or alternately, convert the powers to voltages, superpose the voltages, and then calculate the power from the superposed voltage. That's what most RF engineers do and it works well. As an optical engineer, I've dealt with the harsh reality of this myth of total reflection cancellation. Since you don't understand the physics, it is no wonder that you failed to accomplish total reflection cancellation. And total reflection cancellation is easier to accomplish in a transmission line than it is with light in air. -- 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 =---- |
Richard Harrison wrote: Jim Kelley wrote: "That`s because power doesn`t propagate, and hasten to add - neither do Poynting vectors." Words are not objects. They merely represent objects. We use abstractions for brevity and clarity. Even the best say: "--the Poynting vector or power density (watts per square meter)." See Kraus` 3rd edition of "Antennas" page 73, under "Power Patterns". So I guess that must be where he talks about power and Poynting vectors that pretend to be electromagnetic fields and propagate along waveguides? 73, ac6xg |
Richard Clark wrote: On Fri, 29 Jul 2005 16:58:29 -0700, Jim Kelley wrote: It is quite evident that through the actions of the first interface, that there is less energy incident upon the second interface. Further, given that both interfaces operate with identical reflective and transmissive properties, it follows the second interface could not reflect enough to totally negate the reflections of the first. True for any one reflection. Hi Jim, And true for ALL accumulated reflections there after. Reflections do not add any energy to the cup when the first interface is draining it more quickly. Yes, that was my point. You seemed to have momentarily overlooked that detail. My analysis allowed ALL of the energy in the reflection from the second interface ( 0.098X) to combine with the first reflection (0.11X). Single (first) reflection only. This total superposition was both more than generous, and at the same time very unlikely; and yet with this generous allowance there is still excess reflection from the first interface. Your generosity is hardly the issue. Previously reflections are more to the point. Hence for something less than total superposition of ALL energies, it hardly bodes a better yield in total cancellation - the energy just isn't there in the first place. 0.098X 0.11X is the simple economics of the balance. Sounds good. It's wrong, but the sound of it is good nevertheless. As an optical engineer, I've dealt with the harsh reality of this myth of total reflection cancellation. Failure can tend to make some people bitter. ;-) Basically these claims are for first year students where demanding too much inquiry would push them into switching majors to Business school. Simple optics with simple, ordinary glasses exhibit quite useful results, but they do not embody a proof. To anyone following the math of my presentation, it is quite obvious what WOULD tend towards a more complete cancellation - and such a subtle shift in the formula diverges only slightly from the choir book hymn. It's not that hard when the interface ratios drive the answer. Another good sounding thing. 73, ac6xg |
Cecil Moore wrote: Roy Lewallen wrote: Since you say that power superposes, we should expect power waves to add and cancel just like voltage waves. Strawman Alert!!! Richard H. did NOT say power superposes. It is obvious that what he was disagreeing with was the statement: "You CANNOT ... even talk about the "power" of various reflections in the same media." I think you should be allowed to talk about whatever you want, Cecil. That sort of thing can be highly therapeutic. ;-) 73, ac6xg |
W5DXP wrote: Why don't you visit your local power company, call a meeting of their engineers, and inform them that there is no power flowing in their transmission lines? :-) Are you talking about the guys with the clipboards, or the guys with the hard hats? 73, ac6xg |
On Sat, 30 Jul 2005 13:08:40 -0700, Jim Kelley
wrote: Hence for something less than total superposition of ALL energies, it hardly bodes a better yield in total cancellation - the energy just isn't there in the first place. 0.098X 0.11X is the simple economics of the balance. Sounds good. It's wrong, but the sound of it is good nevertheless. Hi Jim, As the total energy returning from the second interface is 0.098X, not 0.11X, and not even all of that passes through the back of the first interface, where do you come up with the remaining difference to accomplish a complete cancellation? Even if an infinite summation of ALL reflections kept any energy from transiting the second interface (a generous allowance not likely to be observed anywhere), then it cannot exceed what energy was initially introduced into the system. With that generous allowance, the infinite summation can only equal 0.098X which remains less than the reflection from the first interface of 0.11X. Without that generous allowance, it must be something less, which can only increase the total, uncancelled reflection product. Either way, "total cancellation" is not total. 73's Richard Clark, KB7QHC |
Jim Kelley wrote:
W5DXP wrote: Why don't you visit your local power company, call a meeting of their engineers, and inform them that there is no power flowing in their transmission lines? :-) Are you talking about the guys with the clipboards, or the guys with the hard hats? I've seen power engineers with clipboards and hardhats. Their title is "Power Engineer". What will they think when you tell them that power generators don't generate power, power transmission lines don't transfer power, and power meters don't meter power? They may all end up in a white coat because of you. The rest of the world simply does not share your definition of "power". Did you see my question? If the power associated with an EM pulse is not in the pulse, where is it? A power meter registers power during the pulse but nothing before and after the pulse. The power was there during the pulse. Where was it before the pulse arrived? Where did it go after the pulse left? -- 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 =---- |
As the total energy returning from the second interface is 0.098X, not
0.11X, ... The total energy returning from the second interface is 0.125 watts. Your solution is incorrect. Why don't you take a poll to see who agrees with you? Even if an infinite summation of ALL reflections kept any energy from transiting the second interface (a generous allowance not likely to be observed anywhere), then it cannot exceed what energy was initially introduced into the system. It cannot exceed the energy. It can certainly exceed the steady-state source power (joules/sec) and it does. Some energy is sourced during the transient stage that does not reach the load. It is stored in the second medium as standing waves. The forward power in the first medium is one watt with no reflections. The forward power in the second medium is 1.125 watts. The reflected power in the second medium is 0.125 watts. The forward power in the third medium is one watt with no reflections. Have you never seen the forward power in a transmission line exceed the source power? If the system is Z0-matched and the SWR on a 1/4WL transformer is 5.83:1, the steady-state forward power on the 1/4WL transformer is *double* the steady-state source power. Either way, "total cancellation" is not total. In your n=1,2,4 example, reflected wave cancellation is total in medium 1 (assuming a lossless system). If n2=sqrt(n1*n3), the reflected wave cancellation is total. -- 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 =---- |
Jim Kelley wrote:
"That`s because power doesn`t propagate, I hasten to add - neither do Poynting vectors.: Some world-class experts disagree with Jim. Here is a sample quotation using the words "power flow". From E.M. Purcell writing about "Antenna Gain and Receiving Cross Section" on page 19 of "Radar System Engineering" edited by Louis M. Ridenour: "If the transmitting antenna were to radiate energy isotropically-that is, uniformly in all directions-the power flow through unit area at a distance R, from the antenna could be found by dividing P, the total radiated power, by 4piRsquared." Best regards, Richard Harrison, KB5WZI |
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