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Much noise has been radiated. I speculate that a reminder about what
linearity means might get things back on track. In a linear network (lumped or distributed) superposition (of linear signals) produces correct results. The last statement works in both directions. (The degree to which a network is linear is the same as the degree to which superposition is valid.) (If one supplies a large enough signal to any network, it will become non-linear - as in letting-out-the-smoke-put-in-at-the-factory.) The catch in all of the above is that superposition only applies to linear signals and power (however indicated) is not a linear signal. Power, which could be complex power S = V*I* (the phasor voltage time the conjugate of the phasor current) or the magnitude of S (apparent power) or the real part of S ("real" power), simply does not obey superposition even in a network that is linear. Bottom line: assuming the use of networks (lumped or distributed) that are essentially linear, one is only allowed to combine phasor voltages or phasor currents (but not their product nor the square of such linear signals). Once combined, the resultant voltage and the resultant current may be used to find a measure of power. (The "combined" mentioned must be a linear, additive process.) It seems to me that Roy, and others, have plowed this ground many times. 73 Mac N8TT -- J. Mc Laughlin; Michigan U.S.A. Home: "Roy Lewallen" wrote in message snip I've posted many, many times on this topic and have shown a number of cases where the load is perfectly matched but the power dissipated in the source resistor is less than or greater than the "reverse power", clearly demonstrating that this concept is incorrect. There are several examples at Food for thought.txt available at http://eznec.com/misc/food_for_thought/. Because I've posted so much on the topic I won't do it all again. But I know at least one person on this newsgroup would be glad to have an opportunity to express his views once again. I'll leave this discussion to those who want to revisit it; I don't. But I do want to caution readers that this view of "reflected power" is demonstrably incorrect. Roy Lewallen, W7EL |
On Wed, 29 Jun 2005 19:42:53 -0700, Frank Gilliland
wrote: [snipped in the interest of brevity] The error is even more insignificant when there are a host of variables and confounds between the SWR meter and the transmitted field that can (and frequently do) affect the objective -- field strength. Often, field strength is of zero importance. What do you do when the device under test isn't supposed to radiate? That device probably wouldn't make a very good radio, would it? My "SWR Meter" is one of these: http://users.adelphia.net/~n2pk/VNA/VNAarch.html I have 1 mW to radiate. What kind of FSM should I use? The simplest example of this would be a CATV system, yet VSWR is *extremely* important in cascaded networks. Thank you for making my point. Not even you have made your point. It's much simpler (and just plain logical) to measure the field strength directly instead of measuring an abstract value halfway towards the objective and relying on nothing more than speculation that the rest is working according as expected. More baloney and it isn't even sliced. The word is "blarney". My Webster's says: baloney n (bologna) : pretentious nonsense : BUNKUM --- often used as a generalized expression of disagreement.... I could not be more accurate. And although the syntax of my statement was somewhat 'convoluted', A ray of hope the logic is sound smashed -- you can dyno your engine all day, but the only way to know for sure how fast you can get down the quarter mile is to run the race. Uh huh. By this convoluted "logic" I guess you would avoid any dyno testing at all and just go do hit-and-miss tuning at the drag strip. |
Reading the mail appearing in this thread is more fun than watching Saturday
Night Live! Walt, W2DU |
J. Mc Laughlin wrote:
Bottom line: assuming the use of networks (lumped or distributed) that are essentially linear, one is only allowed to combine phasor voltages or phasor currents (but not their product nor the square of such linear signals). Power can certainly be combined, just not superposed. Here is the irradiance equation from _Optics_, by Hecht. Itot = I1 + I2 + Sqrt(I1*I2)cos(theta) Irradiance is power/unit-area. The last term is known as the "interference" term. Hecht provides separate chapters for interference and superposition, the best treatment of those two subjects of which I am aware. Here is the transmission line forward power equation from Dr. Best's QST transmission line article. Ptot = P1 + P2 + Sqrt(P1*P2)cos(theta) It is virtually identical to the irradiance equation above. The last term is known to be the "interference" term and for a Z0-matched system, THETA EQUALS ZERO, so for a Z0- matched system, a complete analysis can be done using only the forward and reflected power magnitudes. This is something I and others have been saying for years and it has been called "gobbledegook" (sic) by Roy (and worse by others) even though Roy admits that he doesn't care to understand where the power goes. It seems to me that Roy, and others, have plowed this ground many times. Yes, and they are still not 100% correct. Roy has said: 'I personally don't have a compulsion to understand where this power "goes".' Too bad he doesn't have that compulsion because, with a small amount of mental effort, he surely would have figured it out before me - if by no other means, simply by referencing the chapter on EM wave interference in _Optics_. -- 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 =---- |
Frank wrote, "Prove it."
OK, here I am at the track (the bench). I have an SWR meter that I've verified with my HP8653 to behave like a short section of 50 ohm line at the frequency of interest. I put a load on its output that I've also verified to be 50 ohms at the frequency of interest. I've applied power to the load through the SWR meter. The indicated SWR is 1.23:1. I took the SWR meter apart, and located a particular resistor. I changed its value slightly. I re-verified that the meter still looks like a short section of 50 ohm line. I re-ran the experiment of applying power through the meter to the load. The indicated SWR is now 1.05:1. Yes, I really have done that! This particular meter is built, as very many of them are, to sample current and voltage at a point of essentially zero length on the line. The current sample (through a current transformer: line center passes through a toroid; secondary is several turns, loaded by that calibration resistor) is converted to a voltage by dropping it through a resistance, and by changing that resistance, I can change the relative amount the current contributes to the measurement. In other words, if the voltage sample is v(samp)=k*v(line), I want to adjust the current sampling so v(i(samp)) = k*Zo*i(line), where Zo is the impedance to which the meter is calibrated to measure SWR. In some meters, there is a means to adjust the voltage sampling ratio easily with a variable trimmer capacitor. Either way works. The adjustment DOES have a TINY effect on the impedance the meter presents to the line it's in, but that is very minor, compared with the range of adjustment of the impedance calibration value. Yes, I really have adjusted a meter which uses the variable capacitor, too. Cheers, Tom |
On Thu, 30 Jun 2005 09:41:05 -0500, Cecil Moore
wrote: Power can certainly be combined, just not superposed. Here is the irradiance equation from _Optics_, by Hecht. Itot = I1 + I2 + Sqrt(I1*I2)cos(theta) According to this formula, for two laser beams of 100W each, without any phase difference, we can illuminate a target with Itot = 100 + 100+ Sqrt(100*100)cos(0) Itot = 300W Every CBer's dream.... Hecht should sue you for copyright Unfair Use. |
SWR=Strewn With Rumor
"Walter Maxwell" wrote in message ... Reading the mail appearing in this thread is more fun than watching Saturday Night Live! Walt, W2DU |
K7ITM wrote:
Yes, I really have adjusted a meter which uses the variable capacitor, too. In the old Heathkit SWR meter were instructions to install either two 50 ohm resistors for a 50 ohm SWR meter or two 75 ohm resistors for a 75 ohm SWR meter. We used a lot of RG-11 back in those days. -- 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 =---- |
On Thu, 30 Jun 2005 01:30:39 GMT, "Tom Donaly"
wrote: Convolution is a mathematical stunt you can perform with two functions: f(x)* g(x) = (integral from 0 to x) f(t)g(x-t) dt. At least that's how it's explained in Schaum's Outline book _Differential Equations_. It's pretty tough to see how it relates to power in a transmission line. Maybe someone has a use for it there. 73, Tom Donaly, KA6RUH **** Yes Tom Convultion was the wrong term to use. I made a mistake because i type as i think and on occasion hit send before i reread what i have written. I still contend that a sinusoidal wave travelling down a coax is comprised of perpendicular(orthogonal) E and H fields. The these vector fields that induce sinusodial current and voltage potential vectors in and between the shield and center conductors as the wave travels. Both the source and reflected waves are comprised of two vector fields, E and H. Granted this is true only when the load reflection coefficient is not zero. In that case of zero, then there is no reflected power. It is possible to derive from the vector current and vector voltage a magnitude of those vectors and thus a produce two scalar quantities that can be pluged into Ohm's Law and derive an instantaineous power at a given time and position on the coax. That both source and reflected sinusoidal current and voltage can have derived scalar values. These values can be directly added. This all started from an SWR question. I contend that the instantaineous power at any given time and position of the coax can be expressed as the sum of the magnitudes or scalar quantities of the source and reflected powers. If you are wanting just the magnitudes of the power, then this should work. james |
Richard Clark wrote:
On Thu, 30 Jun 2005 09:41:05 -0500, Cecil Moore wrote: Power can certainly be combined, just not superposed. Here is the irradiance equation from _Optics_, by Hecht. Itot = I1 + I2 + Sqrt(I1*I2)cos(theta) According to this formula, for two laser beams of 100W each, without any phase difference, we can illuminate a target with Itot = 100 + 100+ Sqrt(100*100)cos(0) Itot = 300W Sorry, I made a mistake in the equation. Please forgive my omission. Here's the correct equation: Itot = I1 + I2 + 2*Sqrt(I1*I2)cos(theta) So if theta equals zero, Itot would be 400w, not 300w. This is the power of superposition of coherent waves. When you superpose two 100w coherent laser beams, the resultant power is indeed 400w and must be supplied by the sources or supplied by destructive interference from somewhere else. This is all explained in _Optics_, by Hecht. Hows about reading it so I won't have to explain superposition to you? -- 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 =---- |
Cecil Moore wrote:
Itot = I1 + I2 + Sqrt(I1*I2)cos(theta) Sorry, should be: Itot = I1 + I2 +2*SQRT(I1*I2)cos(theta) Ptot = P1 + P2 + Sqrt(P1*P2)cos(theta) Sorry should be: Ptot = P1 + P2 + 2*Sqrt(P1*P2)cos(theta) -- 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 =---- |
On Thu, 30 Jun 2005 11:18:37 -0500, Cecil Moore
wrote: Sorry, I made a mistake in the equation. .... Hows about reading it so I won't have to explain superposition to you? Certainly no one stands any chance of understanding your explanations given the continuing goof-ups. |
On Thu, 30 Jun 2005 11:18:37 -0500, Cecil Moore
wrote: When you superpose two 100w coherent laser beams, the resultant power is indeed 400w Yowza! You, with W's help, can roll your Social Security over into investments in the CB amplifier Market. and must be supplied by the sources [Hecht rolling his eyes] So, this means that Hecht's formula only works for steady state? :-) If both are 100W pulses, and the lasers are off before the target are pulse illuminated -um- 1.) 100W 2.) 200W 3.) 400W 4.) no hundred W or supplied by destructive interference from somewhere else. Maybe two more magic lasers? This is all explained in _Optics_, by Hecht. Somehow, I don't think so. How many errors can our readers count? For N = number of words in orginal posting errors = N! Such is the problem of Xerox research. |
On Thu, 30 Jun 2005 12:08:05 -0400, "Fred W4JLE"
wrote: SWR=Strewn With Rumor SWR = Strife Without Regret |
Richard Clark wrote:
Certainly no one stands any chance of understanding your explanations given the continuing goof-ups. Richard, I cannot recall you and I ever disagreeing upon a technical subject. Your objections are just personal pot shots. -- 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 =---- |
On Thu, 30 Jun 2005 13:08:52 -0500, Cecil Moore
wrote: Richard, I cannot recall you and I ever disagreeing upon a technical subject. Convenient memory. |
When you superpose two 100w coherent laser beams, the resultant
power is indeed 400w This is correct if the two beams are coherent and have the same polarization. Very hard to do at optical frequencies, much easier at lower frequencies. Tor N4OGW |
Richard Clark wrote:
Cecil Moore wrote: When you superpose two 100w coherent laser beams, the resultant power is indeed 400w Yowza! You, with W's help, can roll your Social Security over into investments in the CB amplifier Market. Note the following extremely important qualification. The extra power must come from somewhere, either from the two sources or from destructive interference. So says Hecht. When you phasor add 100v to 100v, what V^2/Z0 do you get? and must be supplied by the sources or or supplied by destructive interference from somewhere else. [Hecht rolling his eyes] So, this means that Hecht's formula only works for steady state? :-) Yep, irradiance is a quantity averaged over time. It is steady-state by definition, an accumulated effect. If both are 100W pulses, and the lasers are off before the target are pulse illuminated -um- 1.) 100W 2.) 200W 3.) 400W 4.) no hundred W You will get interference rings of 400w/unit-area and rings of 0w/unit-area all averaging out to 200w total. All this is covered in _Optics_. Please spare us your ignorance and read the book. If the sources are incapable of supplying the extra power, For every P1+P2+2*SQRT(P1*P2), i.e. constructive interference, there is a P1+P2-2*SQRT(P1*P2), i.e. destructive interference. -- 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 =---- |
Earlier today, I wrote, "...HP8653..." Ooops. Belay that. It's
HP8753E. |
On Thu, 30 Jun 2005 13:28:24 -0500, Cecil Moore
wrote: You will get interference rings [Hecht rolling his eyes] for a target that is smaller than a wavelength? Must be another failure of Hecht (or Hecht pupil). of 400w/unit-area and rings of 0w/unit-area all averaging out to 200w total. Which, of course, cannot be found in the formula: Itot = I1 + I2 + 2*Sqrt(I1*I2)cos(theta) nor its correction: Itot = I1 + I2 +2*SQRT(I1*I2)cos(theta) If the sources are incapable of supplying the extra power, Which, in the end was a non sequitur. For every P1+P2+2*SQRT(P1*P2), i.e. constructive interference, Which, of course, cannot be found in the formula: Itot = I1 + I2 + 2*Sqrt(I1*I2)cos(theta) nor its correction: Itot = I1 + I2 +2*SQRT(I1*I2)cos(theta) there is a P1+P2-2*SQRT(P1*P2), i.e. destructive interference. Which, of course, cannot be found in the formula: Itot = I1 + I2 + 2*Sqrt(I1*I2)cos(theta) nor its correction: Itot = I1 + I2 +2*SQRT(I1*I2)cos(theta) When combined with the adhominem (which, of course, reveals another inaccuracy, one of assignment): Please spare us your ignorance and read the book. is possibly the best advice (once the assignment is corrected), given the continuing goof-ups. |
wrote:
When you superpose two 100w coherent laser beams, the resultant power is indeed 400w This is correct if the two beams are coherent and have the same polarization. Very hard to do at optical frequencies, much easier at lower frequencies. As in coherent RF waves confined to a transmission line, eh? -- 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 =---- |
|
When you superpose two 100w coherent laser beams, the resultant
power is indeed 400w This is correct if the two beams are coherent and have the same polarization. Very hard to do at optical frequencies, much easier at lower frequencies. Hi Tor, We've seen the math pencil-whipped both ways now to cover all the available answers. The devil's in the details that are not found in: Itot = I1 + I2 + 2*Sqrt(I1*I2)cos(theta) not to mention the glaring mistakes of the first posting of this formula. So? Superposition works. With a yagi antenna, through superposition you get an EM wave which has larger intensity in certain directions than for a single dipole with the same power. Someone far away can't tell the difference between switching to a yagi and turning on a linear. What the formula doesn't say is that in any real system, the wave must have a finite extent (not be a infinite plane wave). Then there must be destructive interference in some directions. So there isn't a problem with conservation of energy. Tor N4OGW |
When you superpose two 100w coherent laser beams, the resultant power is indeed 400w This is correct if the two beams are coherent and have the same polarization. Very hard to do at optical frequencies, much easier at lower frequencies. As in coherent RF waves confined to a transmission line, eh? That's nothing new. Why do they call them "waveguides" :)? Jackson has a chapter with all the hairy details. Tor N4OGW |
wrote:
When you superpose two 100w coherent laser beams, the resultant power is indeed 400w This is correct if the two beams are coherent and have the same polarization. Very hard to do at optical frequencies, much easier at lower frequencies. As in coherent RF waves confined to a transmission line, eh? That's nothing new. Then why are some of the "experts" on this newsgroup protesting it so much? Some don't seem to comprehend the energy situation associated with superposition of EM fields. -- 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 =--- |
On Thu, 30 Jun 2005 06:47:07 GMT, Owen wrote:
The only other inference you can make from one instrument with regard to the other will be if one of the instruments shows zero reflected power, then you know the VSWR that the other instrument will indicate. As Cecil has hinted ), you can also infer that the impedance at the point where you have two sets of Pf,Pr readings is that the impedance (R, X) is one of two values (which can be visualised as the intersection of two circles on a Smith Chart), one only in the cases VSWR1*VSWR2=Zo1/Zo2 (or the inverse) (which can be visualised as the kissing of two circles on a Smith Chart). The latter includes the case above where one or other instruments indicates VSWR=1, but is more exact because one of the circles is infinitely small. These cases are not reliably of practical value, the conversion of error in the measurements made with each instrument, into error in the estimated R and X at the point could be huge. So, I "correct" my statement to "The only other inference that you can reasonably reliably make from one instrument with regard to the other will be if one of the instruments shows zero reflected power, then you know R and X, and the VSWR that the other instrument should indicate. Thanks Cecil. Owen -- |
On Thu, 30 Jun 2005 07:36:21 -0500, Cecil Moore
wrote in : Frank Gilliland wrote: Cecil Moore wrote: But I specifically stated above the Z0 environment was different from 50 ohms. The same type of error happens when one uses a 50 ohm SWR meter in a 75 ohm coaxial line. If that were true then the mere existence of standing waves could render any measurements worthless. Regardless, I did the experiment a long time ago -- take a 50 ohm SWR meter and plug it into a 75 ohm line -- it gives you almost the same measurement (in fact, I didn't see -any- difference at all). Please run it again in the following configuration: Xmtr--1/4WL 75 ohm line--SWR meter--1/4WL 75 ohm line--50 ohm load The SWR meter will read 2.25:1 when the actual SWR is 1.5:1 Xmtr--1/2WL 75 ohm line--SWR meter--1/2WL 75 ohm line--50 ohm load The SWR meter will read 1:1 when the actual SWR is 1.5:1 I'm not going to argue this -- either you can play with theory and speculate about the results, or you can do the experiment yourself, observe the empirical evidence, and -then- use theory to explain the results. When you get around to doing the latter give me a holler in rrcb since I'm done cross posting on this topic. And BTW, the best location for the directional coupler is at the feedpoint of the antenna. Barring that, the next best place is at the transmitter. Regardless of it's location, you should -never- leave the coupler floating with the coax or you will end up with results like what you describe above. ----== 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 =---- |
Frank Gilliland wrote:
Regardless of it's location, you should -never- leave the coupler floating with the coax or you will end up with results like what you describe above. The results above obey the laws of physics. What laws do your results obey? -- 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 =---- |
Cecil Moore wrote:
Frank Gilliland wrote: Regardless of it's location, you should -never- leave the coupler floating with the coax or you will end up with results like what you describe above. The results above obey the laws of physics. What laws do your results obey? You guys are just itchin' for a visit from the coax length police. :D |
"Tom Ring" wrote in message . .. Walter Maxwell wrote: Reading the mail appearing in this thread is more fun than watching Saturday Night Live! Walt, W2DU Some of the people involved appear to be listening from inside Faraday Cages! tom K0TAR Faraday used the cages after the monkeys were through with them. The monkeys left them so fouled up that EM waves couldn't penetrate the walls. Walt, W2DU |
But I specifically stated above the Z0 environment was different
from 50 ohms. The same type of error happens when one uses a 50 ohm SWR meter in a 75 ohm coaxial line. If that were true then the mere existence of standing waves could render any measurements worthless. Regardless, I did the experiment a long time ago -- take a 50 ohm SWR meter and plug it into a 75 ohm line -- it gives you almost the same measurement (in fact, I didn't see -any- difference at all). Any small error you might see is, as I said before, insignificant, especially considering the reason you are measuring SWR in the first place. The objective is simply to get the reading as low as practially possible. If you feel the need to quibble about a couple tenths of a point on a ratio then maybe you're spending a little too much time playing with the calculator instead of the antenna. I'm running RG-6 out to my 2 meter antenna. I put my cheap RS HF meter inline to see what I'd read. I got my expected 1.5:1. Wattage read 1/2 of what the radio is rated for. It gets out and I'm not worried about it. |
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