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#51




Dr. Slick wrote:
Do you know of anyone who has mathematically derived the 73 Ohms of a dipole in free space? _Fields_and_Waves_in_Communication_Electronics_, Ramo, Whinnery, & Van Duzer. Pages 647, 648, sections 12.05, 12.06. The feedpoint current of a dipole is caused by the inphase superposition of the forward current and reflected current at the balanced feedpoint of a standing wave antenna. A 1/2WL dipole and an open 1/4WL stub have similarities. The deviation of the feedpoint impedance from zero ohms gives an indication of the losses due to dissipation and radiation.  73, Cecil http://www.qsl.net/w5dxp = Posted via Newsfeeds.Com, Uncensored Usenet News = http://www.newsfeeds.com  The #1 Newsgroup Service in the World! == Over 80,000 Newsgroups  16 Different Servers! = 
#52




Kraus comes up with Z=73 + j42.5. He then goes on to say that an actual
dipole is made a few % shorter, which yields 65 + j0. When I did an EZNEC calculation on a 1/2 wave dipole at 3MHz, I did not quite get that. For a #30 wire in free space I got 76.81 + j43.89 at 3 MHz, and 72.88 + j0.3465 at 2.94 MHz. I let EZNEC tell me what the wavelength was, and used 1/2 of that for the length of the dipole. Tam/WB2TT "W5DXP" wrote in message ... Dr. Slick wrote: Do you know of anyone who has mathematically derived the 73 Ohms of a dipole in free space? _Fields_and_Waves_in_Communication_Electronics_, Ramo, Whinnery, & Van Duzer. Pages 647, 648, sections 12.05, 12.06. The feedpoint current of a dipole is caused by the inphase superposition of the forward current and reflected current at the balanced feedpoint of a standing wave antenna. A 1/2WL dipole and an open 1/4WL stub have similarities. The deviation of the feedpoint impedance from zero ohms gives an indication of the losses due to dissipation and radiation.  73, Cecil http://www.qsl.net/w5dxp 
#53




On Thu, 17 Jul 2003 07:49:40 0400, Jack Smith
wrote: Do we worry about matching 8 ohms of electrical speaker impedance to 413 Rayleighs? C.f. Paul Klipsch and the Horn speaker. Hi Jack, Someone must, or we would see more 600 Ohm speakers. It is facile by half to simply accept the end product of design and anoint it as an example of a general solution. There are many antennas that are NOT 50 Ohms. What they ARE is actually of no consequence except in the sense of efficiency and mission (the antenna cares not a whit about either). I have many examples of 2 Ohm antennas (and lower) and of 600 Ohm antennas (and easily higher) and ALL can be induced to radiate all of the power applied to them. The distinguishing factor across the board is that each hiZ antenna presents similar, physical characteristics to all other hiZ antennas; and each lowZ antenna presents similar, physical characteristics to all other lowZ antennas. All couple power to the same load of the Ã¦ther. Clearly impedance and size are correlated and it is up to the designer to accommodate losses to achieve similar performance. The same statement is equally applicable to speakers of any impedance. Is the antenna transforming its Z to that of the Ã¦ther? Of course it is just as the speaker is. Are they both transducers? Of course they are when transducer is applied loosely (but strictly speaking  no). Injecting this notion that transducers are a class distinct from transformers is simply myopic to force an argument. No sooner is the notion introduced than we find the correlative transducer of the receive antenna introduced to recover the power  now transformed (and very inefficiently one might add). The remainder of that power becomes part of the background noise of the cosmos (far more of it than is ever recovered for actual use). Transducers, as a class, are far more prone to the loss through resistance than transformers  by definition. The speaker is feeding a lossy medium of air, and the sonar is feeding the less lossy medium of water. The difference is in the compression characteristics that turns power into heat. Core loss of the transformer is not due to compression, but is a direct analog (and electrons bumping into each other and atoms does constitute a form of compressive loss). There is no loss in space/Ã¦ther but neither are there any phonons, the classic transport of transducer emission and coupling. If an antenna is to qualify as transducer, it must be with the proviso that it is distinctly different from every other transducer in lacking the common transport mechanism of phonons. This is like say walking is a form of mass transportation if you simply ignore the word mass. To support these specious forms requires enormous exaggerations. Another transducer available as a common example (or perhaps not for less wellheeled equipment) is found in the Collins mechanical filter for interstage coupling. It has both input and output transducers that couple the mechanical (and thus heatprone) energy into nickelsteel resonant disks. Nickelsteel is obviously less compressive than either air or water, and exhibits far higher Q (which is a factor of both antennas and transducers  in their medium) to the advantage of the circuit. To any bench tech working on receivers, they would unhesitatingly call these IF Transformers. Does an antenna "transform" any Z to another Z? The process is obviously performed with concomitant and equivalent issues of efficiency regardless of the term inserted between quotes. Does the term substitution bring any change, or does it correct any error? No. It is a tautology to suggest that "transducer" is appropriate when every presumption finds a corresponding "transducer" necessitated by the force of discussing fields (how does one know these fields exist without the absolute necessity of completing the transformer action?). One may "know" in the purely abstract sense, but such knowledge through the centuries has rarely preceded the actuality of observation in the real transformed world. The distinction between transduction and transformation does not preclude the sense of an antenna serving as a bridge between two system impedances. Neither hiZ nor lowZ structures have a stranglehold on design, except through economy. We commonly employ very lowZ sources (transistors) to feed modestZ loads (a common quarterwave antenna). The economic factor of that load (a quarterwave at 160M) is sometimes unsupportable and yet we find very few short, lowZ antennas designed with direct feed from the same lowZ transistor. Economy again forces some form of transformation (I would hesitate to call a Tuner a transducer) in that the commercial market sees very little sense in building lowZ sources for an incredibly small niche who would refuse to pay the price. Instead, commercial design accommodates to one Z and expects the user to transform it along the way. The same logic extends to, and through the antenna. 73's Richard Clark, KB7QHC 
#54




On Thu, 17 Jul 2003 09:06:32 0500, "William E. Sabin"
[email protected] wrote: Dilon Earl wrote: Where does the loss occur? If you have 3 db of mismatch loss, is it in the coax, tank circuit? The loss in "mismatch loss" refers only to the fact that the power delivered by the generator to the load is less than it would be if the load resistance were the same value as the generator resistance, in other words if the load and generator were "matched". The best way to get a handle on this subject is to draw a diagram of a generator with voltage V=10, an internal resistance of 50 ohms, and a load resistor of R ohms. Let R vary from 1 ohm to 100 ohms and calculate the power dissipated in the generator resistance (50 ohms), the power in the load resistance (R), and the total power. Plot a graph of the three quantities. The load power goes through a maximum when R=50 ohms. The maximum power dissipated in the generator resistance is 10^2/50=2 W, which occurs when R=0 ohms. The minimum power dissipated in the generator resistance is 3.33^2/50=0.22 W which occurs when R=100 ohms. When R=50 ohms, the load power is 5^2/50=0.5 W (the maximum value), the dissipation in the generator resistance is 5^2/50=0.5 W and the total power is 10^2/100=1 W. Bill W0IYH Bill; Thanks, that all makes sense. Can you consider a Transmitter to have an internal resistance like the generator that changes with the plate and tune controls? If I have a 100 watt transmitter and my wattmeter shows 3 watts reflected. Is 3 watts actually being dissipated in the tank and final PA? Sorry to ask such simple questions. I did search through Google on posts on this subject, just never could find the answer I was looking for. 
#55




On Thu, 17 Jul 2003 17:00:55 GMT, Dilon Earl
wrote: If I have a 100 watt transmitter and my wattmeter shows 3 watts reflected. Is 3 watts actually being dissipated in the tank and final PA? Hi Dilon, Does it become 3 watts hotter under the same drive conditions without the reflected power? You would be surprised how few pundits actually discuss this in these terms. Of course everyone would be surprised if anyone attempted to perform this chore. I like to include this jab at those who rave on about the impossibility of knowing the internal resistance of a transmitter and are satisfied to squeak out 100W RF for 250W DC in. 73's Richard Clark, KB7QHC 
#56




"Reg Edwards" wrote in message ...
I didn't say they were unimportant. I said they served only to add to the confusion when considering operation of the usual amateur installation when the generator internal resistance is unknown. Indeed, and not only that, the generator (ham transmitter) is commonly neither a linear system nor time invariant. Also, maximum power (conjugatematched load) from a linear generator is generally not the most efficient case. A great many generators and amplifiers are distincly NOT designed to deliver power to a matched load, but rather to deliver power efficiently to a specific load which is mismatched with respect to the output impedance of the generator/amplifier. There are times when knowing that a generator is a linear 50 ohm source (within some small tolerance) is importantI deal with them all the time in the work I dobut in a typical ham transmitter application, that's very seldom if ever the case. Cheers, Tom 
#57




"Ian White, G3SEK" wrote in message ...
A transducer is any gadget that converts energy from one form into a *different* form. Examples include a loudspeaker (electrical energy to sound/mechanical energy), a microphone (the reverse), a light bulb and a photocell. It's a useful word for a useful idea. If an antenna is not a transformer of some type, then why is it affected by it's surroundings so much? They obviously are, just like the primary's impedance is affected by what the secondary sees in a transformer. That's a perfect example of the trap, because in reality it's not "just like". An antenna also has Efield interactions with its environment that a transformer doesn't have, so any resemblance will literally be only halftrue. Roy has clarified this adequately already. Ok, I was half correct then. Two transducers make up one transformer. Certainly two dipoles very close to one another will affect each other's impedance. And a regular transformer with a core can definitely be affected by a close EMfield. Slick 
#58




Dr. Slick wrote: Do you know of anyone who has mathematically derived the 73 Ohms of a dipole in free space? Antennas, John Kraus, McGraw Hill 1950, Chapter 56, pages 143 to 146 gives a complete derivation for a 1/2 wavelength Antenna. You need some Calculus and infinite series to understand the derivation. Deacon Dave, W1MCE 
#59





#60




Dilon Earl wrote:
Bill; Thanks, that all makes sense. Can you consider a Transmitter to have an internal resistance like the generator that changes with the plate and tune controls? If I have a 100 watt transmitter and my wattmeter shows 3 watts reflected. Is 3 watts actually being dissipated in the tank and final PA? No. If the transmitter output is 100 W and the reflected power is 3 W, then the 100 W is the difference between 100+3=103 W (forward power) and 3 W (reflected power). The question "where does the reflected power go?" never seems to have an acceptable answer. Very strange. A good way to look at is as follows: The junction of the transmitter output jack and the coax to the antenna is a "node", which is just a "point" or "location" where the jack and the coax meet. At this node the voltage is exactly equal to the voltage output of the amplifier (VPA) and also the voltage across the input of the coax (VCOAX). The voltage VCOAX) across the coax is equal to the phasor sum of a forward voltage wave that travels toward the antenna and a reverse voltage wave that is traveling from the antenna backward toward the transmitter. Also, at the node, IPA is the current from the PA and ICOAX is the phasor sum of a current wave that travels to the antenna and a return current wave that travels toward the transmitter. At the node, the IPA current and the ICOAX current are exactly equal and in the same direction (toward the antenna). At the node the IPA current is equal to the ICOAX coax forward current minus the ICOAX reflected current. In other words there is an *EQUILIBRIUM* at the node between VPA voltage and VCOAX voltage, and an *EQUILIBRIUM* between IPA current and ICOAX (forward and reflected) current. This explanation accounts for everything that is going on at the node. The answer to the question "where does the reflected power go?" is the following: "It is a nonsense question that has caused nothing but misery". The reflected power does not actually *GO* anywhere. The correct answer is that forward and reflected coax waves always combine precisely and exactly with the voltage and current that is delivered by the PA. The voltage and current at the junction are correctly accounted for. The basic principles here are Kirchhoff's voltage law and Kirchhoff's current law, as applied to the node. You can study Kirchhoff's laws in the textbooks. If we apply these laws and calculate the 100 W power out of the PA and the 100 W power that is dumped into the coax, they are exactly equal. They cannot possibly be unequal. The power delivered is the real part of the product of VPA and IPA (100 W), which is identical to the real part of the product of VCOAX and ICOAX (100 W). Observe carefully the following: We do not need to know anything about the PA and its circuitry. The PA is nothing more than an anonymous "black box". In other words, any 100 W (output) PA will perform exactly as I have described. Bill W0IYH 
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