Harry wrote:
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. For your "passing-out" test, calculate the radiation pattern of a two-radial groundplane antenna directly from Maxwell's equations (including the two small cross-polarized lobes that nobody else mentions), using pencil and paper only, in a pub. I've seen it done by a math professor from W6. (Nah... it must have been a party trick... he must have rehearsed that a million times... mustn't he?) When he'd finished, Charlie sat back, smiled asked for comments. The rest of us passed out. -- 73 from Ian G/GM3SEK 'In Practice' columnist for RadCom (RSGB) http://www.ifwtech.co.uk/g3sek |
"Harry" wrote about Roy Lewallen:
You may want to write a book titled "Story of the 73 Ohms". ___________ It should include the point that the 73 ohm radiation resistance value applies to a physical 1/2-wave, thin-wire, linear dipole in free space, and that a reactance term of + j42.5 ohms also applies to such a configuration (Kraus 3rd Edition, p. 182). The dipole length needs to be shorted by a few percent in order to zero out the reactance term, at which time the resistance term will be about 65 ohms. RF |
"Roy Lewallen" wrote in message
... 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 Very interesting references. In particular Kraus' 2nd ed. pp 359 - 408. Also Stutzman and Thiele, 1st ed. pp 306 - 374. Pretty much grad level, needs effort, (for me) even though I took Stutzman and Thiele's antenna course in 1998 (NCEE). Regards, Frank |
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. S = E X H |
In message , Cecil Moore
writes 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. Just a quick question. What is the impedance at the centre of an infinitely long dipole (in free space)? Ian. -- |
Hi Cecil,
I find another formula in Balnais book (section 4.3, 1982 edition) for Rr: Rr = 20 * (pi)^2 * ( l / lamda)^2 (4-37) If l = lamda/2, then this formula gives Rr .=. 50 Ohms. (4-93) the one you quoted is Rr = (120*pi) / (4*pi) * Cin( 2*pi ) = 73 Ohms, (for l = lamda/2) where 120*pi is the intrinsic impedance for a free-space medium. Would you please explain the difference between these two formulas (4-37 and 4-93)? Thanks! -- Harry |
Anyway, reading the op carefully, he asked for an exact _explanation_, not
an exact calculation result.. Steve, K;9.D,C'I "Bill Turner" wrote in message ... Reg Edwards wrote: There's no such value as 'exact'. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Nonsense. I have exactly two pencils in my pencil holder. :-) 73, Bill W6WRT |
Cecil Moore wrote: Ian Jackson wrote: Just a quick question. What is the impedance at the centre of an infinitely long dipole (in free space)? Same as a terminated dipole in an anechoic chamber? 600-800 ohms? Not at zero Hz. ac6xg |
Harry wrote: Hi Cecil, I find another formula in Balnais book (section 4.3, 1982 edition) for Rr: Rr = 20 * (pi)^2 * ( l / lamda)^2 (4-37) If l = lamda/2, then this formula gives Rr .=. 50 Ohms. (4-93) the one you quoted is Rr = (120*pi) / (4*pi) * Cin( 2*pi ) = 73 Ohms, (for l = lamda/2) where 120*pi is the intrinsic impedance for a free-space medium. Would you please explain the difference between these two formulas (4-37 and 4-93)? Thanks! -- Harry 4-37 should apparently read: Rr = 30 * (pi)^2 * (1 / lambda)^2. ac6xg |
Ian Jackson wrote:
Just a quick question. What is the impedance at the centre of an infinitely long dipole (in free space)? Same as a terminated dipole in an anechoic chamber? 600-800 ohms? -- 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 =--- |
All times are GMT +1. The time now is 04:24 PM. |
Powered by vBulletin® Copyright ©2000 - 2024, Jelsoft Enterprises Ltd.
RadioBanter.com