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Tom Donaly wrote:
Kraus was talking about distribution over _length_. Of course, Kraus is talking about the cos(KX) term where 'X' is _length_ and 'K' is the constant that converts that _length_ into degrees. Given 'K', _length_ and degrees are perfectly correlated. -- 73, Cecil http://www.qsl.net/w5dxp |
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"Richard Clark" wrote in message ... On 19 May 2006 17:09:50 -0700, wrote: Maybe someone can help us here. Hi Glenn, In the half dozen postings that followed this, did you find anything discussed that helped you? Perhaps orthogonal was the wrong word, transverse concepts? 73's Richard Clark, KB7QHC whatever it is, KEEP IT GOING! just got off the tower after pulling up the lifting rope and caught up while my legs are recovering... boy its chilly up there, had to wear my winter coat and hat and gloves! but there are more showers forecast for this afternoon so i'll need some more fresh reading material! FIGHT! FIGHT! FIGHT! |
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Cecil Moore wrote:
Tom Donaly wrote: Cecil Moore wrote: Seems the easiest measurement of nonlinearity would be the harmonics (if any) generated by the antenna that do not appear in the source signal. Which wouldn't tell you a single thing about the current distribution along the length of the dipole. Yes it would. It would be proof that the current distribution along the length of the dipole is sinusoidal no matter what your illusionary perceptions are telling you. For standing wave antennas, if the source is a pure single frequency sine wave and if no harmonics are generated by the antenna system: 1. The forward wave is sinusoidal. 2. The reflected wave is sinusoidal and coherent with the forward wave. 3. Their superposition results in a sinusoidal standing wave with the same angular velocity. Any non-linearity would introduce harmonics. The purpose of most antennas is to radiate electromagnetic waves. That means there is loss. It also means that the current envelope is affected. That's one of the reasons we use EZNEC. I suppose, Cecil, that if you keep repeating the same old tired line, over and over again, you might find someone who will agree with you. Certainly, no antenna measurement would. 73, Tom Donaly, KA6RUH |
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Tom Donaly wrote:
The purpose of most antennas is to radiate electromagnetic waves. That means there is loss. It also means that the current envelope is affected. That's one of the reasons we use EZNEC. The current envelope is affected but remains a linear system function since it is the result of superposition which itself is a linear system function. I suppose, Cecil, that if you keep repeating the same old tired line, over and over again, you might find someone who will agree with you. Certainly, no antenna measurement would. The current envelope is a linear system function. I am repeating the rules and laws of mathematics. Sounds like you need to review the definition of linear systems. You can do that at: http://www.cns.nyu.edu/~david/linear...r-systems.html In particular, quoting: "Systems that satisfy both homogeneity and additivity are considered to be linear systems. These two rules, taken together, are often referred to as the principle of superposition." In general, antennas are linear systems that satisfy the principle of superposition. If they were non-linear, they would not satisfy the principle of superposition. Two linear system functions, like forward waves and reflected waves, cannot superpose to a non-linear function. Therefore, standing waves are linear, not non-linear, functions. To argue otherwise exhibits a certain degree of ignorance. Until the obvious mathematical misconception is corrected, no rational discussion is possible. To the best of my knowledge, Maxwell's equations are also linear system functions so claims of non-linearity also contradict Maxwell's equations. -- 73, Cecil http://www.qsl.net/w5dxp |
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I suppose, Cecil,
that if you keep repeating the same old tired line, over and over again, you might find someone who will agree with you. ========================================= I agreed with Cecil the first time he said it. But I'm only a foreigner. So whatever I say doesn't carry any weight. Or does it? ---- Reg. |
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Reg Edwards wrote:
I suppose, Cecil, that if you keep repeating the same old tired line, over and over again, you might find someone who will agree with you. I agreed with Cecil the first time he said it. But I'm only a foreigner. So whatever I say doesn't carry any weight. Or does it? I dug out my linear network theory book and would like to present a few quotes and comments: "The real world is inherently non-linear." Lightning hitting an antenna can cause arcing and melted wires. "Although nature is non-linear, linear approximations over defined ranges of validity are valid representations of non-linear phenomena." Amateur radio antennas are usually confined to that limited linear range. "The necessary and sufficient conditions for a linear system a (1) validity of the principle of superposition; (2) preservation of scale factor. Does doubling the power input to the antenna ~double the radiated power? Does it ~double the non-radiated losses? "Fortunately for the engineer, however, linear systems are frequently excellent approximations to reality and have a wide range of validity in the real world." Maxwell's equations in particular. Textbook equations for traveling waves and standing waves assume linearity. -- 73, Cecil http://www.qsl.net/w5dxp |
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Reg Edwards wrote:
I suppose, Cecil, that if you keep repeating the same old tired line, over and over again, you might find someone who will agree with you. ========================================= I agreed with Cecil the first time he said it. But I'm only a foreigner. So whatever I say doesn't carry any weight. Or does it? ---- Reg. You're the master of simple approximation, Reg. Cecil thinks your simplified ideas are received wisdom. Knowing you, I find it hard to believe you'd ever agree with anyone. 73, Tom Donaly, KA6RUH |
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On Sat, 20 May 2006 12:14:29 -0000, "Dave" wrote:
KEEP IT GOING! Dave, your trolling effort is rather a poor substitute for the sense of accomplishment. Too many do it far better, with more flair, and offer more entertainment than this pallid use of the CAPS KEYS. |
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Tom Donaly wrote:
Knowing you, I find it hard to believe you'd ever agree with anyone. Reg and I are in perfect agreement on the benefits of a good Cabernet. -- 73, Cecil http://www.qsl.net/w5dxp |
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On Sat, 20 May 2006 17:39:44 GMT, "Tom Donaly"
wrote: Cecil thinks your simplified ideas are received wisdom. Hi Tom, Is there some suggestion of smoldering bush in this parable? Commandments that are unzipped and ready for immediate entablature? 73's Richard Clark, KB7QHC |
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Cecil Moore wrote:
Reg Edwards wrote: I suppose, Cecil, that if you keep repeating the same old tired line, over and over again, you might find someone who will agree with you. I agreed with Cecil the first time he said it. But I'm only a foreigner. So whatever I say doesn't carry any weight. Or does it? I dug out my linear network theory book and would like to present a few quotes and comments: "The real world is inherently non-linear." Lightning hitting an antenna can cause arcing and melted wires. "Although nature is non-linear, linear approximations over defined ranges of validity are valid representations of non-linear phenomena." Amateur radio antennas are usually confined to that limited linear range. "The necessary and sufficient conditions for a linear system a (1) validity of the principle of superposition; (2) preservation of scale factor. Does doubling the power input to the antenna ~double the radiated power? Does it ~double the non-radiated losses? "Fortunately for the engineer, however, linear systems are frequently excellent approximations to reality and have a wide range of validity in the real world." Maxwell's equations in particular. Textbook equations for traveling waves and standing waves assume linearity. You can still pretend a dipole is a "linear system," as you call it, and still understand that the current envelope is not a simple sine function. The Achilles heel of all your reflection mechanics ideas is the assumption that everything is lossless. (Not to mention the fact that it's supposed to exist in outer space.) You and Reg like to think of a dipole as a transmission line, and Reg can even tell you its characteristic impedance (average). What neither he nor you ever mention is the alpha part of the propagation constant. That's the important part, though, since it signifies radiation, the very thing the antenna was designed to do. By the way, why are you quoting from a network theory book when not too long ago you were ranting and raving about the invalidity of the lumped constant model? 73, Tom Donaly, KA6RUH |
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"Richard Clark" wrote in message ... On Sat, 20 May 2006 12:14:29 -0000, "Dave" wrote: KEEP IT GOING! Dave, your trolling effort is rather a poor substitute for the sense of accomplishment. Too many do it far better, with more flair, and offer more entertainment than this pallid use of the CAPS KEYS. I'm just cheering you on... besides if I'm going to troll I might as well abandon some other parts of decency in the process. And it doesn't seem like anyone cares, even with the obvious thread title it took off all on its own. |
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Tom Donaly wrote:
You can still pretend a dipole is a "linear system," as you call it, and still understand that the current envelope is not a simple sine function. Diverting to a "simple" sine function in the same spirit as diverting to a "small" loading coil? If you were always talking about a perfect sine wave, you should have said so long before now and nobody would have disagreed with you. The Achilles heel of all your reflection mechanics ideas is the assumption that everything is lossless. That's NOT the assumption. The assumption is that lossless systems are easiest to understand so let's understand them first before we move on to something more complex. You guys have proven that you don't even understand the simple lossless condition. (Not to mention the fact that it's supposed to exist in outer space.) You and Reg like to think of a dipole as a transmission line, and Reg can even tell you its characteristic impedance (average). What neither he nor you ever mention is the alpha part of the propagation constant. That's the important part, though, since it signifies radiation, the very thing the antenna was designed to do. Only about 1 dB of the steady-state energy stored in a 1/2WL dipole is radiated so radiation is not the largest effect. The radiation from an antenna can be simulated by using resistance wire to simulate a 1 dB loss in a transmission line. The reason that I have rarely mentioned such is that you guys don't understand enough of the basics to proceed to those more complex examples. By the way, why are you quoting from a network theory book when not too long ago you were ranting and raving about the invalidity of the lumped constant model? The lumped constant model is valid under certain conditions. What I object to is its use under known invalid conditions. The lumped constant model and distributed network model are both *linear systems*. -- 73, Cecil http://www.qsl.net/w5dxp |
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Richard Clark wrote:
On Sat, 20 May 2006 17:39:44 GMT, "Tom Donaly" wrote: Cecil thinks your simplified ideas are received wisdom. Hi Tom, Is there some suggestion of smoldering bush in this parable? Commandments that are unzipped and ready for immediate entablature? 73's Richard Clark, KB7QHC Hi Richard, Yes, but there's some evidence Cecil is about to apostatize. I hope he doesn't end up wandering in the wilderness. 73, Tom Donaly, KA6RUH |
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Richard Harrison wrote:
Richard Clark, KB7QHC wrote: "Who. in your estimation, does qualify to discuss it?" If it`s about antennas, I nominate Kraus. If it`s about mathematics, many marhematicians qualify. In algebra, y = mx + b, (the point slope formula), is called linear because it is the graph of a straight line. . . . But of course you realize that the function y = mx + b doesn't meet the requirements of a linear function when applied to network theory. Roy Lewallen, W7EL |
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Richard Harrison wrote:
Cecil, W5DXP wrote: "Assuming the source signal is a pure sine wave, if the standing wave current "isn`t in general sinusoidally shaped (as Roy said)", then the antenna would have to be introducing harmonic radiation that doesn`t exist in the source signal." . . . Either Cecil is misquoting me, or Richard is misquoting Cecil. Cecil calls the total current the "standing wave current". I never said that the standing wave current isn't sinusoidally shaped. I said that the *envelope* of the standing wave -- that is, a graph of the magnitude of the current on a transmission line as a function of position along the line -- is not sinusoidally shaped except in the special case of a complete reflection. This isn't a personal theory, but a very well established fact which can easily be derived from fundamental equations. (Or it can even be found clearly stated in texts for those unable to understand the derivation.) On antennas, the current distribution (magnitude of current vs position) is generally not sinusoidal either, although it's approximately so on thin wire antennas. Assuming that the transmission line is driven with a pure sine wave, the forward, reverse, and total currents as a function of *time* will be sinusoidal and consequently no harmonics will be generated. (I'm of course neglecting nonlinear effects which might occur in a real transmission line or antenna such as magnetic conductors or nonlinear dielectrics. But in practical terms these will be virtually unmeasurable with ordinary coax or twinlead and amateur power levels.) It's hard to determine whether Cecil's sustained confusion between a time waveform and a graph of amplitude vs distance is intentional or if the concepts are really mixed up for him. It has, in either case, provided a convenient diversion from his basic strange theories at the time it was getting particularly hard for him to continue supporting them. Roy Lewallen, W7EL |
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Roy Lewallen wrote:
Richard Harrison wrote: In algebra, y = mx + b, (the point slope formula), is called linear because it is the graph of a straight line. But of course you realize that the function y = mx + b doesn't meet the requirements of a linear function when applied to network theory. I knew what Richard meant. Quoting "Linear Network Theory", by Ferris: "In the functional relationship, h(t)=kf(t), no matter what h(t) and f(t) represent, the relation h(t)=kf(t) must be linear. ... An elementary concept, then, is that h(t) and f(t) are related by a straight line of the form h(t) = mt + b." -- 73, Cecil http://www.qsl.net/w5dxp |
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Roy Lewallen wrote:
Richard Harrison wrote: Cecil, W5DXP wrote: "Assuming the source signal is a pure sine wave, if the standing wave current "isn`t in general sinusoidally shaped (as Roy said)", then the antenna would have to be introducing harmonic radiation that doesn`t exist in the source signal." Either Cecil is misquoting me, or Richard is misquoting Cecil. Richard quoted me correctly. I did *NOT* quote you. I merely stated what I thought you said. Cecil calls the total current the "standing wave current". That is absolutely false. Total current equals standing wave current plus traveling wave current. Or standing wave current equals total current minus traveling wave current. Subtract out the traveling wave component and a pure standing wave component is left. Note that you did not quote me. You merely stated what you thought I said and you were mistaken. I never said that the standing wave current isn't sinusoidally shaped. I said that the *envelope* of the standing wave -- that is, a graph of the magnitude of the current on a transmission line as a function of position along the line -- is not sinusoidally shaped except in the special case of a complete reflection. The envelope of the standing wave current is the same whether reflection is complete or not since the traveling wave current has been subtracted from the total current. Seems you should have said the total current envelope is not sinusoidal. The standing wave current envelope is obviously sinusoidal since the traveling wave has been subtracted. This isn't a personal theory, but a very well established fact which can easily be derived from fundamental equations. Please provide a reference that says after the traveling wave current has been subtracted from the total current, the resulting standing wave current envelope is not sinusoidal. (Or it can even be found clearly stated in texts for those unable to understand the derivation.) On antennas, the current distribution (magnitude of current vs position) is generally not sinusoidal either, although it's approximately so on thin wire antennas. Just because the scale of the position axis changes with VF doesn't mean things become non-sinusoidal. Assuming that the transmission line is driven with a pure sine wave, the forward, reverse, and total currents as a function of *time* will be sinusoidal and consequently no harmonics will be generated. This statement contradicts what I thought you said before. Seems we are now in agreement. -- 73, Cecil http://www.qsl.net/w5dxp |
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Roy Lewallen, W7EL wrote:
"But of course you realize that the function y = mx + b doesn`t meet the requirements of a linear function when applied to network theory." Works for me. Linear means the graph of the function is a straight line. f(x) = y = mx + b is called linear because its graph is a straight line. A straight line is the shortest distance between two points. In y = mx + b, m is a constant determining the slope of the line. x is is the independent variable. b is the offset or point along the x-axis where the line crosses. y then is a linear function of x because its slope is always mx, but displaced in the x-direction by a constant value, namely b. y is linear the same as IR is linear, or by substitution, E is linear in Ohm`s law where E=IR. For any value of I, voltage = IR and the graph of I versus E is a straight line with a slope equal to R. Resistance is a common factor in network theory. Best regards, Richard Harrison, KB5WZI |
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Roy Lewallen, W7EL wrote:
"Either Cecil is misquoting me, or Richard is misquoting Cecil." I apologize if I misquoted anyone. It was unintentional, On page 239 of Kraus` 1950 edition of "Antennas" he wrote: "It is generally assumed that the current distribution of an infinitesimally thin antenna (l/a=infinity) is sinusoidal, and that the phase is constant over a 1/2-wavelength interval, changing abruptly by 180-degrees between intervals." Cosine has the same shape as sine. They are identical except for a 90-degree phase displacement. Sine starts at zero. Cosine starts at one. Sinusoidal covers both sine and cosine shapes. Best regards, Richard Harrison, KB5WZI |
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Cecil Moore wrote:
Roy Lewallen wrote: Richard Harrison wrote: Cecil, W5DXP wrote: "Assuming the source signal is a pure sine wave, if the standing wave current "isn`t in general sinusoidally shaped (as Roy said)", then the antenna would have to be introducing harmonic radiation that doesn`t exist in the source signal." Either Cecil is misquoting me, or Richard is misquoting Cecil. Richard quoted me correctly. I did *NOT* quote you. I merely stated what I thought you said. Cecil calls the total current the "standing wave current". That is absolutely false. Total current equals standing wave current plus traveling wave current. Or standing wave current equals total current minus traveling wave current. Subtract out the traveling wave component and a pure standing wave component is left. I thought I was following this somewhat, but now it appears that I am not. I thought there were two traveling waves, forward and reverse, with the reverse wave being a reflection of the forward wave from the end of the antenna. I thought that the total current was the sum of the two waves, could be measured with an ammeter, and that it could be referred to as the standing wave. Are you now saying that the standing wave and the reverse wave are the same wave? Confusedly, John |
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John - KD5YI wrote:
I thought there were two traveling waves, forward and reverse, with the reverse wave being a reflection of the forward wave from the end of the antenna. I thought that the total current was the sum of the two waves, could be measured with an ammeter, and that it could be referred to as the standing wave. If the antenna were lossless, that would be true. However, the antenna loses RF energy to radiation, ground, I^2*R, etc. Since the same thing is true of a transmission line, let's first look at a transmission line. Let's say we measure (at the tuner) a forward power of 100 watts and a reflected power of 80 watts. We can assume that the losses in the line and the power delivered to the load add up to 20 watts. The 20 watts can be associated with a pure forward traveling wave. The other 80 watts of the forward power and the 80 watts of reflected power can be associated with a pure standing wave. The equations for such would look something like this. Itot = Ifor*cos(kx+wt) + Iref*cos(kx-wt) = Itot = I1*cos(kx+wt) + I2*cos(kx)*cos(wt) I1 would be considered to be the part of the traveling wave that is delivering net power to the load. I2 would be considered to be the standing wave current delivering no net power to the load. It's just another way of mathematically partitioning the currents. Let's assume for the sake of discussion that the forward wave at the antenna feedpoint is 100 watts and the reflected wave at the antenna feedpoint is 80 watts. That means there are 20 watts of total losses. We can partition the currents in the same way that we did in the transmission line. Incidentally, if we assume the antenna is 1/2 wavelength with a Z0 of 600 ohms, we can calculate the feedpoint impedance just as if it were a piece of transmission line with a resistive termination. Anybody want to try that exercise? What impedance is seen looking into a 600 ohm 1/4WL open stub when the forward power is 100 watts and the reflected power is 80 watts? Are you now saying that the standing wave and the reverse wave are the same wave? The reverse wave is half of the standing wave. The other half of the standing wave is part of the forward wave. One might say that the energy in the reverse wave neutralizes the ability of the forward wave to deliver that same amount of energy to the load. And indeed, we know the power delivered to the load is the forward power minus the reflected power. -- 73, Cecil http://www.qsl.net/w5dxp |
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Cecil Moore wrote:
Roy Lewallen wrote: . . . Richard quoted me correctly. I did *NOT* quote you. I merely stated what I thought you said. Cecil calls the total current the "standing wave current". That is absolutely false. Total current equals standing wave current plus traveling wave current. Or standing wave current equals total current minus traveling wave current. Subtract out the traveling wave component and a pure standing wave component is left. I could have sworn Cecil had used it in that sense before. But Ok, I'll go along with the new definition. Any basic analysis of transmission line operation shows that the total current is simply the sum of the traveling wave currents. Consequently the "standing wave current" as defined by Cecil is zero. So by "standing wave current", Cecil means zero. This is getting stranger by the day. Roy Lewallen, W7EL |
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I'm sure that somewhere in one of your texts you can find the definition
of linear as applied to networks. Once you do, though, a little thought is required to discover that y = mx + b doesn't satisfy the criteria for network linearity. To be linear, a network has to satisfy superposition. This means that: If y1 is the response to excitation x1 and y2 is the response to excitation x2, then the response to x1 + x2 must be y1 + y2. Let's try that with your function. The response to x1 is: y(x1) = mx1 + b The response to x2 is: y(x2) = mx2 + b The sum of y(x1) and y(x2) is: y(x1) + y(x2) = m(x1 + x2) + 2b But response to x1 + x2 is: y(x1 + x2) = m(x1 + x2) + b These are not equal as they must be to satisfy superposition and therefore the requirements for linearity. Roy Lewallen, W7EL Richard Harrison wrote: Roy Lewallen, W7EL wrote: "But of course you realize that the function y = mx + b doesn`t meet the requirements of a linear function when applied to network theory." Works for me. Linear means the graph of the function is a straight line. f(x) = y = mx + b is called linear because its graph is a straight line. A straight line is the shortest distance between two points. In y = mx + b, m is a constant determining the slope of the line. x is is the independent variable. b is the offset or point along the x-axis where the line crosses. y then is a linear function of x because its slope is always mx, but displaced in the x-direction by a constant value, namely b. y is linear the same as IR is linear, or by substitution, E is linear in Ohm`s law where E=IR. For any value of I, voltage = IR and the graph of I versus E is a straight line with a slope equal to R. Resistance is a common factor in network theory. Best regards, Richard Harrison, KB5WZI |
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Roy Lewallen wrote:
Any basic analysis of transmission line operation shows that the total current is simply the sum of the traveling wave currents. Consequently the "standing wave current" as defined by Cecil is zero. So by "standing wave current", Cecil means zero. Your last two statements are false. Please stop trying to tell others what I mean when it is obvious that you don't know what I mean. This is getting stranger by the day. Why can't you and I just have a reasonable technical discussion? Sooner or later, the readers who care at all about the truth are going to see through your attempts to falsify what I have said. Since you obviously have not understood what I said in the past I guess I need to simplify it for you so here goes: Let A be the forward traveling current. Let B be the rearward traveling current. Let C be the part of the forward traveling current that flows through the load. Let D be the part of the forward traveling current that doesn't flow through the load and becomes B after being reflected. A = C + D The total forward current is divided into two components, one accepted by the load and one rejected by the load. The following is phasor addition Total Current = A + B = (C + D) + B = C + (D + B) (D + B) is the pure standing wave current component, a part of the forward current equal to the reflected current. The total current is indeed the phasor sum of the forward and reflected current. The total current is also the sum of the load current plus the pure standing wave current. The pure standing wave current envelope is sinusoidal. -- 73, Cecil http://www.qsl.net/w5dxp The day a Guru starts believing that He already knows everything is the day He becomes hopelessly ignorant. |
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Roy Lewallen wrote:
I'm sure that somewhere in one of your texts you can find the definition of linear as applied to networks. Once you do, though, a little thought is required to discover that y = mx + b doesn't satisfy the criteria for network linearity. You have missed the point, Roy. It is the relationship between functions that has to be linear. If f(x) is the input and y is the output where y = h(x) = k*f(x), a plot of the output Vs input will yield a straight line of the form mx + b. -- 73, Cecil http://www.qsl.net/w5dxp |
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Cecil Moore wrote:
John - KD5YI wrote: I thought there were two traveling waves, forward and reverse, with the reverse wave being a reflection of the forward wave from the end of the antenna. I thought that the total current was the sum of the two waves, could be measured with an ammeter, and that it could be referred to as the standing wave. If the antenna were lossless, that would be true. However, the antenna loses RF energy to radiation, ground, I^2*R, etc. Since the same thing is true of a transmission line, let's first look at a transmission line. Let's say we measure (at the tuner) a forward power of 100 watts and a reflected power of 80 watts. We can assume that the losses in the line and the power delivered to the load add up to 20 watts. The 20 watts can be associated with a pure forward traveling wave. The other 80 watts of the forward power and the 80 watts of reflected power can be associated with a pure standing wave. The equations for such would look something like this. Itot = Ifor*cos(kx+wt) + Iref*cos(kx-wt) = Itot = I1*cos(kx+wt) + I2*cos(kx)*cos(wt) I1 would be considered to be the part of the traveling wave that is delivering net power to the load. I2 would be considered to be the standing wave current delivering no net power to the load. It's just another way of mathematically partitioning the currents. Let's assume for the sake of discussion that the forward wave at the antenna feedpoint is 100 watts and the reflected wave at the antenna feedpoint is 80 watts. That means there are 20 watts of total losses. We can partition the currents in the same way that we did in the transmission line. Incidentally, if we assume the antenna is 1/2 wavelength with a Z0 of 600 ohms, we can calculate the feedpoint impedance just as if it were a piece of transmission line with a resistive termination. Anybody want to try that exercise? What impedance is seen looking into a 600 ohm 1/4WL open stub when the forward power is 100 watts and the reflected power is 80 watts? Are you now saying that the standing wave and the reverse wave are the same wave? The reverse wave is half of the standing wave. The other half of the standing wave is part of the forward wave. One might say that the energy in the reverse wave neutralizes the ability of the forward wave to deliver that same amount of energy to the load. And indeed, we know the power delivered to the load is the forward power minus the reflected power. Never mind. |
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Roy, W7EL wrote:
"I`m sure that somewhere in one of your texts you can find the definition of linear as applied to networks." Right. From "The Penguin Desk Encyclopedia of Science and Mathematics": "Linear (algebra) An equation or function of the form ax + by + c = 0 or f(x) = mx +b is called linear because its graph is a line. This has been generalized to a concept called linear combination, which is the sum of 2 or more entities with each multiplied by some number (with not all numbers being 0). Linear combinations of vectors, equations, and functions are commonly employed." I think that definition is broad enough to cover the representations I made in my postings. Best regards, Richard Harrison, KB5WZI |
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I can only hope that some of the readers understand.
Roy Lewallen, W7EL Richard Harrison wrote: Roy, W7EL wrote: "I`m sure that somewhere in one of your texts you can find the definition of linear as applied to networks." Right. From "The Penguin Desk Encyclopedia of Science and Mathematics": "Linear (algebra) An equation or function of the form ax + by + c = 0 or f(x) = mx +b is called linear because its graph is a line. This has been generalized to a concept called linear combination, which is the sum of 2 or more entities with each multiplied by some number (with not all numbers being 0). Linear combinations of vectors, equations, and functions are commonly employed." I think that definition is broad enough to cover the representations I made in my postings. Best regards, Richard Harrison, KB5WZI |
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Richard Harrison wrote: Roy, W7EL wrote: "I`m sure that somewhere in one of your texts you can find the definition of linear as applied to networks." Right. From "The Penguin Desk Encyclopedia of Science and Mathematics": "Linear (algebra) An equation or function of the form ax + by + c = 0 or f(x) = mx +b is called linear because its graph is a line. This has been generalized to a concept called linear combination, which is the sum of 2 or more entities with each multiplied by some number (with not all numbers being 0). Linear combinations of vectors, equations, and functions are commonly employed." I think that definition is broad enough to cover the representations I made in my postings. Best regards, Richard Harrison, KB5WZI You're right in a certain mathematical sense, but that definition isn't correct, as Roy pointed out, in defining superposition. 73, Tom Donaly, KA6RUH |
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OK, you guys can stop now... wx is improving and i found something more fun
to play with anyway.... try this site out: The links don't mail well, but if you go to this link... http://local.live.com/?v=2&sp=aN.42.....022505_k1ttt_ and then do this.... After the page above displays the road view, in the control pad at the top left click on 'bird's eye'... then wait for it to redraw. Then click on the 'W' on the compass rose to view toward the west which seems to be the closest view of my place. Then click on the big set of buildings in that control panel to zoom in. by this point you won't see the station at all, but in the 'scratch pad' control on the right side click on the 'k1ttt' link and it should swing it around and show you a view of my station looking toward the west. the photo is at least 3 to 4 years old now as you can see the 6 element 20m and 8 element 15m yagi's still up that have been replaced. David Robbins K1TTT web: http://www.k1ttt.net AR-Cluster node: 145.69MHz or telnet://dxc.k1ttt.net |
neat aerial photo site... was FIGHT! FIGHT! FIGHT!
OK, you guys can stop now... wx is improving and i found something more
fun to play with anyway.... try this site out: The links don't mail well, but if you go to this link... http://local.live.com/?v=2&sp=aN.42.....022505_k1ttt_ and then do this.... After the page above displays the road view, in the control pad at the top left click on 'bird's eye'... then wait for it to redraw. Then click on the 'W' on the compass rose to view toward the west which seems to be the closest view of my place. Then click on the big set of buildings in that control panel to zoom in. by this point you won't see the station at all, but in the 'scratch pad' control on the right side click on the 'k1ttt' link and it should swing it around and show you a view of my station looking toward the west. the photo is at least 3 to 4 years old now as you can see the 6 element 20m and 8 element 15m yagi's still up that have been replaced. David Robbins K1TTT web: http://www.k1ttt.net AR-Cluster node: 145.69MHz or telnet://dxc.k1ttt.net |
FIGHT! FIGHT! FIGHT!
Oh my goodness! The critical qualification and definition has been provided
by Roy. Roy and Richard then proceeded to nail the subject to the proverbial wall. It is time to pause in grading exams to remind that analytic geometry and circuit theory are not the same thing. In circuit theory (and practice) a circuit/network/or-whatever is linear if, and only if, superposition is satisfied. If superposition is satisfied, then the circuit in question is linear. This concept also has utility in some mechanical structures. In English, it is not uncommon for a word to have quite different meanings depending on where it is used. We all know this. Even when discussing active devices such as BJTs and FETs one finds the word "saturation" used differently depending on the device. Obviously, large enough signals applied to any network that appears to be linear with smaller signals will melt same - letting out the smoke that was placed therein at the factory. Obviously, there are degrees of linearity. A footnote is always understood to say: such-and-such network is acceptably linear for signals larger than A and smaller than B. ... thus ends the lesson. 73, Mac N8TT -- J. Mc Laughlin; Michigan U.S.A. Home: "Roy Lewallen" wrote in message ... I can only hope that some of the readers understand. Roy Lewallen, W7EL |
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On Tue, 23 May 2006 19:11:00 -0400, "J. Mc Laughlin"
wrote: Dear Richard: I have had occasion to note to students that some of what they call music would not be noticeably modified by being amplified by a amplifier having a great deal of distortion. Your point is right on target. Hi Mac, My point may have been on target, but the specifics left an escape for those quick enough to pick up on it. In fact, I would speculate that most audio (as do RF) amplifiers exhibit the gain characteristic of f(x) = y = mx + b and this is called class AB. When you build an amp employing two of them in a push-pull configuration the constants b negate each other and the 2mx remains. Of course, the push-pull configuration is built to nullify the distortion of this "linear" curve. The single power supply Op Amp also suffers from f(x) = y = mx + b with the output floating at half the supply voltage - this has got to be an application killer if it goes straight to the speakers without removing the b with a capacitor. 73's Richard Clark, KB7QHC |
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Tom, KA6RUH wrote:
"You`re right in a certain mathematical sense, but that definition isn`t correct as Roy pointed out in defining superposition." Or in conjugate matching for that matter. The discussion was: Is an antenna a linear device, meaning does it produce amplitude distortion? Amplitude distortion comes from a nonlinear relation between imput and output. It has nothing to do with has the amplitude grown larger or smaller in the antenna? I think the majority of antennas have an output which is a reasonable approximation in form to their input. That goes for frequencies too. The response won`t be the same for all frequencies, but you`re not likely to find new frequencies on the output of the antenna that aren`t on its input. Superposition says that when a number of voltages (distributed in any manner throughout a linear network) are applied to the network simultaneously, the current that flows is the sum of the component currents that would flow if the same currents had acted individually. As Cecil said, nonlinearity produces new frequencies. Good antennas don`t do that, superposition or not. Best regards, Richard Harrison, KB5WZI |
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