Dave Palmer wrote: >> Assuming that what I am actually taking about is dipole gain (I am a bit of an ignoramus I'm afraid) Can anyone give me a basic approximate formula for the variation of gain with frequency for frequencies that are up to a factor of (say) 10 away (above and below) from the resonant frequency of a half wave dipole. Is the maximum gain cyclic (e.g is there a resonance at, say, a dipole length of 1.5, 2.5 etc wavelengths or does the gain just "disappear" when the frequency moves away from a half wave dipole condition?). If the gain is cyclic what would be an approximate formula for the gain at, and around, these cyclic frequencies?(Note that I am not interested in polar diagram directions, merely gain) <<
Dave, Resonance in a wire in space does indeed occur at particular and cyclic lengths; 0.5, 1, 1.5, 2 and so on wavelengths. For a wire connected at one end to ground, it occurs at multiples of a quarter wavelength. A non-resonant, _short_ dipole with 50 percent efficiency has directive "gain" of 1.64 dB (1.76 dB for perfect efficiency) above an isotropic antenna. In the classic _Antennas_, Kraus has its maximum effective aperture as about 0.12 square wavelengths. A half-wave dipole has 2.15 dB directive gain over isotropic and Kraus has its maximum effective aperture as 0.13 square wavelengths. This is the area from which power in the incident wave is delivered to the antenna. (You also need to account for inefficiency in the antenna when calculating the current that flows in it.) Page 13-2 of the ARRL Antenna handbook has a graph showing gain over a dipole of a wire antenna as a function of wire length (and the angle with respect to the wire at which this gain is realized). It is about 2 dB at 2.75 wavelength long, almost exactly 4 dB at 5 wavelengths long and then almost a straight lineup to about 6.5 wavelengths (about 5.2 dB). From this point to 10 wavelengths (about 7.4 dB) another line segment can be drawn that is not far from the actual plot. Johnson and Jasik have a formula (page 11-5, Antenna Engineering Handbook, Second Edition). I can try to enter it understandably from my ASCII account. E field in volts per meter = the product of: 1. The magnitude of field strength at a distance due to antinode current. (60 * current in amps at maxima)/distance from the antenna) 2. The envelope of the pattern lobes (inverse of the sine of the angle from the wire axis) (This is actually the maximum value of any lobe in the radiation pattern) and 3. Information about the angles of zeros of the pattern (this is either +1 or -1) (sin OR cos of the angle *(pi * length of the wire / wavelength) * cos of the angle) NOTE: cosine or sine depending on whether length is an odd or even number of half wavelengths, respectively. You can rearrange this to calculate the maximum current in a wire at any frequency at which it is resonant. (Kraus shows how the formula is gotten, taking the fields of all the infinitesimal dipoles comprising a dipole antenna of any length.) For a physical antenna, gain over an isotropic source is four times pi times the actual effective aperture, divided by the square of the wavelength. The maximum voltage on an antenna is determined by efficiency and radiation resistance (the antenna always re-radiates). If you can calculate the current flowing as a function of the incident field, you can calculate the voltage along the antenna. Kraus points out that because antennas are not infinitely thin, the current at minima does not actually go to zero. Therefore the impedance and voltage at those points does not go to infinity. However, for a thin wire the impedance at current minima such as the ends can be in the thousands of ohms. The radiation resistance of a dipole in free space is 72 ohms. If antenna current due to the incident field is 1 amp, this gives a power in the antenna of 72 watts. At a point on the antenna where impedance is 1000 ohms, this will give about 270 volts. I believe this is the significance of a dipole in an explosive atmosphere, as you have to protect from creating ignition sources. And at sufficiently high impedance, or high power, given small diameter or pointed structures, corona will develop even when another conducting object is not near enough to flash over. The ARRL Antenna book points out that a long-wire antenna's radiation resistance at current maxima is higher than 72 ohms. Also, you need to consider that the radiation resistance is a function not only of the resonant wire, but also of its exposure to reflected fields from ground and nearby objects. The graph on page 3-11 shows that a dipole's radiation resistance varies from zero when on a perfectly conducting ground, to almost 100 ohms at about 0.4 wavelengths above ground, down to about 58 ohms at 0.6 wavelengths above ground and oscillates around the 72 ohms value as height above ground is increase (though the excursions from 72 ohms become less and less). This has significance if the dipole is located close to a large conducting object. For example, a 5 cm long wire one cm away from a chassis will have a radiation resistance of just 20 ohms at 3 GHz. This increases the current flowing in the antenna, and therefore, the voltage. Against this, is the fact that it is likely to be very different from a thin wire at these frequencies and the maximum impedance much lower than a thousand ohms, reducing the voltage. I hope this helps. Cortland ------------------------------------------- This message is from the IEEE EMC Society Product Safety Technical Committee emc-pstc discussion list. Visit our web site at: http://www.ewh.ieee.org/soc/emcs/pstc/ To cancel your subscription, send mail to: [email protected] with the single line: unsubscribe emc-pstc For help, send mail to the list administrators: Ron Pickard: [email protected] Dave Heald: [email protected] For policy questions, send mail to: Richard Nute: [email protected] Jim Bacher: [email protected] All emc-pstc postings are archived and searchable on the web at: http://ieeepstc.mindcruiser.com/ Click on "browse" and then "emc-pstc mailing list"
