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80-m-Antennas on ARDF Download of article with all pictures This is a revised contribution of the in magazine FUNKAMATEUR 1983 H.3 S.138 appearance article, meanwhile reprinted in FUNKAMATEUR-Digest, Theuberger Verlag GmbH, Berlin 1998, S.156. With this article, I want to start a discussion about this theme. Opinions and supplements are welcome at the adress of. Antenna systems, sensitivity and accurately of 3.5MHz direction finding antennas for ARDF Dipl.-Ing. Schade DL7VDB, K. Smet ON4CHE (translation) Corrected version dated July 2005 There are a lot of unclearness about the working, dimensionning, the reachable sensitivity, the range and direction accurately of 3.5 MHz direction finding antennas. But the two parameters: sensitivity and direction accurately are very important parameters of the receivers and have a big influence of the competition results in ARDF contests.
The next paragraphs explain the problems, show the calculations and compare it with the praxis. Ferrite Antennas A loop antenna generates in a electrical field E (in fact we have to use the magnetic field H, but in the far field, it is allowed to use the electrical field) with a wavelength lambda a maximum voltage of: (1) U = 2 * pi * w * A * E / lambda When we push the ferrite in the loop (coil), the field will be grow with a multiplication factor of the relative permeability µw. This permeability is lower than the static permeability and is depend of the ferrite and coil surface. (2) UFA = 2 * pi * µw * w * A * E / lambda This is the reason, why the surface of the induction with ferrite is smaller as without a ferrite core for the same voltage.
We become a higher signal when we put the induction in a resonance circuit tuned at the signal frequency. Matching of the resonance circuit to the low impedance receiver input will be done by a second coil. Figure 1 Figure 1 shows the replacement circuit of a real ferrite antenna. The voltage UFA1 is bigger as the induced voltage UFA of the electromagnetic field because we want to reckon with the quality factor of the resonance circuit.
(3) UFA1 = 2 * pi * µw * w1 * A * Q * E / lambda Figure 2 Transforming the replacement circuit (Figure 1) in a equivalent circuit (Figure 2) gives us the formula of the quality factor Q of the resonance circuit. (4) Q = 1 / (2 * pi * f * L1 / Re * (w2 / w1) 2 + (R1 / (2 * pi * f * L1))) The voltage at receiver input UFA2, will be: UFA2 = w2 * UFA1 / w1 (5) UFA2 = 2 * pi * µw * w2 * A * Q * E / lambda The quotient U/E called also the effective height of antenna hW. The effective height of the ferrite antenna (secondary side) for a signal at the receiver input: (6) hwFA2 = 2 * pi * µw * w2 * A * Q / lambda For example: A receiver FPE80 has approximate values: µw= 45, Q= 65, dL= 12mm, w2= 3. The effective height of the ferrite antenna becomes hwFA2= 0.075m.
Vertical Antenna The received voltage UHA at the vertical antenna of the RDF equipment: (7) UHA = hwHA * E We know that for a antenna (hHA lambda/4) just above the earth surface the effective antenna height hwa= 0.5 * hHA is. Figure 12 Because the vertical antenna is assembled at the receiver, and this receiver is used at a height of 1 to 2m, increased the effective antenna height (see [1]) to: (8) hwHA = c * (hH + hHA) with c = 0.5.0.82 The factor c is dependent of the quotient hH / hHA. In a practical case for a height hH= 1.2m, we can use c=0.8 [1]. Klyuch aktivacii scandoc.
The voltage delivered by the vertical antenna becomes: (9) UHA = 0.8 (hH + hHA) * E Example: How tall is the effective height of the vertical antenna if the user hold the receiver on different levels (example FPE80): • hw = 0.8 * (1.4 + 0.026) m = 1.14 m. Normal uses (1.4 m) • hw = 0.8 * (2.2 + 0.026) m = 1.78 m. With stretched arm (2.2 m) • hw = 0.5 * (0 + 0.026) m = 0.013 m. Just above the earth (theoretical) When we want to calculate the input voltage of the receiver, we have to know the impedance of the vertical antenna at the place of connection with the receiver. The antenna theory tells us that a short vertical antenna (hHA lambda/4) in the feet point pure capacitve is. With a equivalent capacity of: (10) Ca = hHA / 1.8 / ln (1.15 * hHA / dHA)) Ca in pF, hHA and dHA in cm Example: A vertical antenne of 26 cm and a diameter of 2 mm calculate us a feedpoint capacity of Ca= 2.9pF.
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