Reg Edwards wrote:
Antenna conductors ARE transmission lines and the same sort of calculations apply. You're right, Reg. Without reflections from the ends of a dipole, the feedpoint impedance would be hundreds of ohms. A standing wave antenna is like a lossy transmission line where the loss is to radiation. And the SWR on a 1/2WL dipole standing wave antenna is quite high - in the neighborhood of 20:1. -- 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 =--- |
It is equally correct to say that the radiation resistance of a dipole
is 2*73 = 146 ohms, since this is the value when the radiation resistance is assumed to be uniformly distributed along its length as is the conductor resistance. For calculating purposes the radiation and conductor resistances can simply be added together. And when referred to the end of a dipole, at HF the radiation resistance is of the order of Q-squared * 73/2 = 3000 ohms. Although this is somewhat indeterminate because of the difficulty of feeding an isolated dipole at one end. The conductor diameter also plays a significant part. ---- Reg. |
Just about any antenna textbook will show you the calculation of a half
wavelength, infinitely thin dipole in free space. For that special case, the answer ends up being simply 30 * Cin(2 * pi), where Cin is a modification of the cosine integral Ci[*]. That's where the "magic number" comes from. Kraus' _Antennas_ is just one of the many textbooks which give the derivation for this. It takes about 3 pages and 23 equations for Kraus to derive. One assumption made in calculation of this value is that the current distribution is sinusoidal, an assumption that's true only for an infinitely thin antenna. For finite thickness wire, the calculation becomes much more difficult. The radiation resistance changes only slowly with wire diameter, however, so sinusoidal distribution is a reasonably good approximation provided that the antenna is thin. Feedpoint reactance, though, varies much more dramatically with both antenna length and diameter. Calculating its value exactly requires solution of a triple integral equation which can't be solved in closed form. That's why computer programs are used to solve it numerically. To include the effect of ground, you need to calculate the mutual impedance between the antenna and its "image". If the antenna is about 0.2 wavelength above ground or higher (for a half wave antenna -- the height must be greater if the antenna is longer), you can assume that the ground is perfect and get a pretty good result. Below that height, the calculation again becomes much more complicated because the quality of the ground becomes a factor. If you're interested in the numerical methods used, locate the NEC-2 manual (available on the web), which describes it. If you're satisfied with approximate results, the work by S.I. Shelkunoff provides formulas for free-space input impedance of antennas with finite diameter wire which can be solved with a programmable calculator or computer. They're detailed in "Theory of Antennas of Arbitrary Size and Shape", in Sept. 1941 Proceedings of the I.R.E. The formulas for R and X contain many terms involving sine and cosine integrals, which can be approximated with numerical series. You'll find additional information in his book _Advanced Antenna Theory_. For approximate calculations of mutual impedance of thin linear antennas, see "Coupled Antennas" by C.T. Tai, in April 1948 Proceedings of the I.R.E. Those also involve multiple terms of sine and cosine integrals. Before numerical calculations became possible, many very good mathematicians and engineers devised a number of approximation methods of varying complexity and accuracy. You'll find their works in various journals primarily in the 1940s - 1960s. The complexity and difficulty of the problem is why virtually all antenna calculations are done today with computers, using numerical methods such as the moment method. In summary, here are your choices: 1. You can calculate the approximate radiation resistance but not reactance of a thin, free-space antenna by assuming a sinusoidal current distribution and using the method Reg described. To include the effect of ground, you have to calculate or look up from a table the mutual impedance between the antenna and its "image", and modify the feedpoint impedance accordingly by applying the mesh equations for two coupled antennas. This method of including the effect of ground becomes inaccurate below around 0.2 wavelength, if the antenna is over typical earth. 2. You can use various approximation methods to calculate reactance, and resistance with better accuracy. But for the effect of ground, you're still limited to being greater than about 0.2 wavelength high. 3. To accurately include the effect of real ground with low antennas, and/or to get resistance and reactance values with arbitrarily good accuracy requires numerical methods. A computer program is the only practical way to do this. A very good basic description of the moment method can be found in the second and later editions of Kraus' _Antennas_. [*] Cin(x) = ln(gamma * x) - Ci(x), where gamma = Euler's constant, 0.577. . . Ci(x) = the integral from -infinity to x of [cos(v)/v dv] = ln(gamma * x) - (x^2)/(2!2) + (x^4)/(4!4) - (x^6)/(6!6). . . Roy Lewallen, W7EL Harry wrote: Hi Tim and Reg, Thank you for your valuable information. Is there any website or textbook that actually shows the step-by-step calculation of this magic number which has been quoted so often in the cable industry? You know most video cables and connectors have characteristic impedance, 75 Ohms. I am not afraid of math. I just like to understand the details of its derivation. -- Harry |
"Cecil Moore" wrote Prad is found by integrating the Poynting Vector over a certain radius. ==================================== Cec, what's the Poynting Vector? --- Reg. |
Reg Edwards wrote:
There's no such value as 'exact'. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Nonsense. I have exactly two pencils in my pencil holder. :-) 73, Bill W6WRT |
Reg Edwards wrote:
"Cecil Moore" wrote Prad is found by integrating the Poynting Vector over a certain radius. Cec, what's the Poynting Vector? Same thing as a power flow vector. The dimensions are energy per second per unit area. If the power flow vector is integrated over the entire surface of a sphere of radius r, the result is total radiated power. I thought maybe you and Mr. Poynting might have worked together at some time in the 20th century. :-) Poynting, John Henry (1852-1914) English physicist, mathematician, and inventor. He devised an equation by which the rate of flow of electromagnetic energy (now called the Poynting vector) can be determined. http://www.cartage.org.lb/en/themes/...oynting/1.html -- 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 =--- |
"Reg Edwards" wrote It is equally correct to say that the radiation resistance of a dipole is 2*73 = 146 ohms, since this is the value when the radiation resistance is assumed to be uniformly distributed along its length as is the conductor resistance. For calculating purposes the radiation and conductor resistances can simply be added together. ===================================== Hints and tips - By the same arithmetical reasoning - The centre feedpoint input resistance of a 1/2-wave dipole is 73 ohms plus HALF of the conductor end-to-end HF resistance, which to be precise sometimes matters. This is exact insofar as 73 is exact - not an approximation. The HF resistance of a copper wire at 20 degrees C is - Rhf = Sqrt( F ) / 12 / d ohms per metre, Where F is the frequency in MHz and d is the wire diameter in millimetres. For example, the end-to-end loss resistance of a 1/2-wave dipole at 1.9 MHz using 16-gauge copper wire is 7.07 ohms, which increases the feedpoint resistance to 73 + 3.53 = 76.53 ohms, to give a radiating efficiency of 95.4 percent. Some people would consider that's enough to lose a contest! Why not start a little notebook to record useful, simple, little formulae such as above which don't appear in Terman et al? Or if they do appear then you can never find the right page in the right volume. ---- Reg. |
Hi Cecil,
From "Antenna Theory" by Balanis: Rr = 2*Prad/|Io^2| = 73 ohms (4-93) I got Balanis' book and found the above formula. Thanks a lot! (Tom also mentioned the same book.) 4-94 (I have the 1982 edition) is from formula 4-70. The math is OK to me. I happen to have a PH.D. degree in math. -- Harry |
Hi Roy,
Thanks for all the valuable information. Are you a professor? You may want to write a book titled "Story of the 73 Ohms". -- Harry |
Harry wrote:
Hi Roy, Thanks for all the valuable information. Are you a professor? Not by a long shot. I'm way too poor a student to ever be considered for a job as a teacher. You may want to write a book titled "Story of the 73 Ohms". It's already been written, by many people who know a great deal more than I. Roy Lewallen, W7EL |
All times are GMT +1. The time now is 09:03 AM. |
Powered by vBulletin® Copyright ©2000 - 2024, Jelsoft Enterprises Ltd.
RadioBanter.com