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#1
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Cecil Moore wrote:
Over the years, I have challenged anyone on this newsgroup to create a standing wave in a single source system without having the existence of a forward wave and a reflected wave. Nobody has furnished any proof that standing waves are possible in a single source system without the existence of forward and reflected waves. Cecil, Why would anyone try to prove that the basic math of adding sinusoidal functions is incorrect? To the contrary, you are the one who insists that a standing wave and its constituent traveling wave components are somehow different and unique. No one denies the simultaneous existence of standing waves and traveling waves. Isn't superposition wonderful! 73, Gene W4SZ |
#2
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Gene Fuller wrote:
Why would anyone try to prove that the basic math of adding sinusoidal functions is incorrect? To the contrary, you are the one who insists that a standing wave and its constituent traveling wave components are somehow different and unique. Actually, it was you who made that assertion and thanks for the opportunity to quote you once again: Gene Fuller, W4SZ wrote: In a standing wave antenna problem, such as the one you describe, there is no remaining phase information. Any specific phase characteristics of the traveling waves died out when the startup transients died out. So standing waves are "somehow different" from traveling waves according to your own assertions. The traveling wave possesses phase characteristics and the standing wave doesn't. Phase is gone. Kaput. Vanished. Cannot be recovered. Never to be seen again. So it was you who asserted that standing wave current is "somehow different" from traveling wave current and I agree with you. It's obvious they are "somehow different" because they have different mathematical equations. Have you changed your mind since your above quoted posting? No one denies the simultaneous existence of standing waves and traveling waves. Of course they do, Gene, that is the whole point. Here is a quote from K8LV's article: "I wish to emphasize the fact that the forward and reverse waves really do not exist separately ..." That certainly *denies* the separate existence of the underlying traveling waves so your above assertion is false. I believe that W7EL also denies the separate existence of forward and reverse waves and introduced the technical term, "sloshing", to explain what happens to the energy in a transmission line with reflections. -- 73, Cecil http://www.qsl.net/w5dxp |
#3
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Cecil wrote:
This is just one more example of the dumbing down of amateur radio accompanying the dumbing down of the US educational system in general. Unfortunately, it seems to be a trend that cannot be reversed because it is the biased view being pushed by the ARRL and its supporters. Hmmm...not sure I agree that the folks at ARRL are deliberately being dumb (or maybe I just misunderstood you). Seems more unintentional to me. After all, the technical editor of QEX let publish that bizarre article that claimed to prove by math that phasing SSB receivers were not possible. Can''t imagine a political motivation for that though I have to wonder bigtime how that one got by. QEX really really needs for some good peer review. Ah, to have Ham Radio magazine back again. Loved that thing. Learned most of my radio from it. 73, Glenn AC7ZN |
#4
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Cecil Moore wrote:
Gene Fuller wrote: Why would anyone try to prove that the basic math of adding sinusoidal functions is incorrect? To the contrary, you are the one who insists that a standing wave and its constituent traveling wave components are somehow different and unique. Actually, it was you who made that assertion and thanks for the opportunity to quote you once again: Gene Fuller, W4SZ wrote: In a standing wave antenna problem, such as the one you describe, there is no remaining phase information. Any specific phase characteristics of the traveling waves died out when the startup transients died out. So standing waves are "somehow different" from traveling waves according to your own assertions. The traveling wave possesses phase characteristics and the standing wave doesn't. Cecil, You keep making the same mistake. Yes, you can analyze traveling waves instead of standing waves if you so choose. However, there is not one bit of additional physical information in the traveling waves that is not in the standing wave. Any "phase characteristic" is simply a function of the mathematical manipulations you use. Perhaps someday you will actually understand superposition, but I won't hold my breath. 73, Gene W4SZ |
#5
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Gene Fuller wrote:
Isn't superposition wonderful! 73, Gene W4SZ Yup, it's why I find religion so amusing. tom K0TAR |
#6
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En/na Dave ha escrit:
... Come on, just a little fight??? Just a question about coax cables: I know that a coax cable does not radiate (if common mode currents properly suppressed) because both conductors are apparently "in the same place" (wouldn't know how to express it in more technical terms). Now the question is: This is true for big distances from the coax, but is it also true when you get very close to the coax? Imagine a conductor taped to the outside of a coax for some meters. The capacitive coupling to the braid is much higher than the coupling to the inner conductor. Would it pick some of the current in the coax. If not, why not? (apart from fun I'm really interested in the answer as I'm not quite sure if a coax running parallel to unshielded and not twisted computer cables would pick harmonics from it on RX or create interferences on TX) -- Toni |
#7
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no... and yes.... there would be some pickup, but only from leakage through
an imperfect braid. because the current is on the inside of the shield it would not couple to the cable on the outside. the electric and magnetic fields are contained completely inside the shield. "EA3FYA - Toni" wrote in message ... En/na Dave ha escrit: ... Come on, just a little fight??? Just a question about coax cables: I know that a coax cable does not radiate (if common mode currents properly suppressed) because both conductors are apparently "in the same place" (wouldn't know how to express it in more technical terms). Now the question is: This is true for big distances from the coax, but is it also true when you get very close to the coax? Imagine a conductor taped to the outside of a coax for some meters. The capacitive coupling to the braid is much higher than the coupling to the inner conductor. Would it pick some of the current in the coax. If not, why not? (apart from fun I'm really interested in the answer as I'm not quite sure if a coax running parallel to unshielded and not twisted computer cables would pick harmonics from it on RX or create interferences on TX) -- Toni |
#8
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EA3FYA - Toni wrote:
Just a question about coax cables: I know that a coax cable does not radiate (if common mode currents properly suppressed) because both conductors are apparently "in the same place" (wouldn't know how to express it in more technical terms). Here's why it doesn't radiate: In a coaxial cable with a solid shield, the differential mode current is entirely inside the shield. Current and fields penetrate only a very small distance from the inner surface of the shield, and no significant amount ever makes it through to the outside. This is assuming that the shield is at least several skin depths thick, which is a good assumption at HF and above. Common mode current, by contrast, flows on the outside of the shield, and its field radiates outward from there. Now the question is: This is true for big distances from the coax, but is it also true when you get very close to the coax? Imagine a conductor taped to the outside of a coax for some meters. The capacitive coupling to the braid is much higher than the coupling to the inner conductor. Would it pick some of the current in the coax. If not, why not? Again assuming a solid shield -- the center conductor carries a current and therefore creates a field. The inner surface of the shield carries an equal and opposite current and also creates a field. But those fields are equal and opposite, and cancel at all points beyond a thin layer on the inner surface of the shield. Since there's no significant field at any point outside the shield, it doesn't matter where you look, you won't find any, and there isn't any field to couple to anything else. In reality, any shield other than a completely solid one (such as the shield of hard line or semi-rigid coax) will leak some because of gaps or holes. And the field will couple more strongly to wires which are close than those which are far away. Whether the amount of leakage is significant or not depends on the application. (apart from fun I'm really interested in the answer as I'm not quite sure if a coax running parallel to unshielded and not twisted computer cables would pick harmonics from it on RX or create interferences on TX) You might get enough leakage through the shield of ordinary coax to cause problems in both cases. It depends on the transmit power level, the signals in the wires, the length over which they're bundled, the frequencies involved, the quality of the shield, and so forth. Separating them even a small distance would reduce the coupling considerably. But you're likely to have more trouble with common mode current. Roy Lewallen, W7EL |
#9
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En/na Roy Lewallen ha escrit:
EA3FYA - Toni wrote: I know that a coax cable does not radiate (if common mode currents properly suppressed) because both conductors are apparently "in the same place" (wouldn't know how to express it in more technical terms). Here's why it doesn't radiate: In a coaxial cable with a solid shield, the differential mode current is entirely inside the shield. Current and fields penetrate only a very small distance from the inner surface of the shield, and no significant amount ever makes it through to the outside. This is assuming that the shield is at least several skin depths thick, which is a good assumption at HF and above. When you say "Current and fields penetrate only a very small distance...", I agree for the current part, but I'm not so sure for the fields part: As I understand it you can not "stop a field" in no way, though you can certainly nullify it with an identical but opposite field. Then the question is whether the two fields (the one from the current flowing in the shield + the one from the current flowing in the inner conductor) nullify at all points in the immediate vicinity of the shield. I certainly believe it but would like to understand why this is so. I guess the mathematical proof would involve assuming the braid is an infinite number of conductors equally spaced around the center conductor, each having it's infinitesimal share of the shield current, and integrating all of their fields at the point of interest (Would probably be able to do so back when I was at university but now it is too strong math for me). Would this be a good approximation of the problem? -- Toni |
#10
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![]() Toni wrote: En/na Roy Lewallen ha escrit: EA3FYA - Toni wrote: I know that a coax cable does not radiate (if common mode currents properly suppressed) because both conductors are apparently "in the same place" (wouldn't know how to express it in more technical terms). Here's why it doesn't radiate: In a coaxial cable with a solid shield, the differential mode current is entirely inside the shield. Current and fields penetrate only a very small distance from the inner surface of the shield, and no significant amount ever makes it through to the outside. This is assuming that the shield is at least several skin depths thick, which is a good assumption at HF and above. When you say "Current and fields penetrate only a very small distance...", I agree for the current part, but I'm not so sure for the fields part: As I understand it you can not "stop a field" in no way, though you can certainly nullify it with an identical but opposite field. You bet you can stop a field. It can be stopped either by reflection, absorption, or a combination of the two. Inside an anechoic chamber, absorbing materials stop internal fields to prevent reflections. A screen room or metallic shield reflects external fields. Then the question is whether the two fields (the one from the current flowing in the shield + the one from the current flowing in the inner conductor) nullify at all points in the immediate vicinity of the shield. I certainly believe it but would like to understand why this is so. Indeed they do. Look up Ampere's Law. If you draw a boundary through the middle of the shield or outside the shield, you'll find that the sum of currents within that boundary is zero. According to the law, that means that no net field penetrates the boundary. Because of the physical symmetry, no net field means no field at all. I guess the mathematical proof would involve assuming the braid is an infinite number of conductors equally spaced around the center conductor, each having it's infinitesimal share of the shield current, and integrating all of their fields at the point of interest (Would probably be able to do so back when I was at university but now it is too strong math for me). Would this be a good approximation of the problem? No, it's not that complicated, but a path or surface integration is required to use Ampere's law. Roy Lewallen, W7EL |
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