Question 10 of 10: Radiation resistance, efficiency and far field of a short vertical antenna

Question 10: Radiation resistance, efficiency and far field of a short vertical antenna

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A 15 m vertical wire radiator at 1 MHz carrying a uniform current, made of 2 cm radius steel of conductivity $6.2\times10^6$ S/m, radiating 1 kW.

Find. The radiation resistance, the radiation efficiency, and the electric field intensity 20 km away when 1 kW is radiated.

Approach. Confirm the element is electrically short, use the uniform-current (Hertzian) radiation resistance, compare it with the skin-effect ohmic resistance for the efficiency, and get the far field from the radiated power and the directivity of a short element.

1. Check that the element is electrically short. $$\lambda = \frac{c}{f} = \frac{3\times10^8}{10^6} = 300\ \text{m}, \qquad \frac{l}{\lambda} = \frac{15}{300} = 0.05$$ Since $l/\lambda = 0.05 \lt 0.1$ the element is short, so the sinθ Hertzian pattern and the small-antenna formulas are legitimate. This is what the frequency and length are really for.
2. (1) Radiation resistance. For a uniform current over a short element in free space, integrating the far-field Poynting vector over a sphere and equating to $\tfrac{1}{2}I_{pk}^2R_r$ gives $$R_r = 80\pi^2\left(\frac{l}{\lambda}\right)^2 = 80\pi^2(0.05)^2 = 80(9.8696)(0.0025)$$ $$\boxed{R_r = 1.974\ \Omega}$$ The word “uniform” matters: a short dipole with the usual triangular current distribution has only a quarter of this, $20\pi^2(l/\lambda)^2 = 0.493\ \Omega$, because its effective length is halved.
3. (2) Ohmic loss resistance from the skin depth. At 1 MHz the current crowds into a thin surface layer of depth $$\delta = \frac{1}{\sqrt{\pi f\mu_0\sigma}} = \frac{1}{\sqrt{\pi(10^6)(4\pi\times10^{-7})(6.2\times10^6)}} = 0.2021\ \text{mm}$$ so the surface resistance is $R_s = 1/(\sigma\delta) = 0.7980$ mΩ/square. Spreading that over the wire's circumference, $$R_{loss} = R_s\frac{l}{2\pi a_w} = (7.980\times10^{-4})\frac{15}{2\pi(0.02)} = 0.09525\ \Omega$$ Note $\delta$ is a hundred times smaller than the wire radius, which is what justifies using the surface-impedance form rather than the d.c. resistance (which would be only 3.85 mΩ).
4. Combine into the radiation efficiency. The two resistances sit in series at the terminals, and only the radiation part delivers useful power: $$e_r = \frac{R_r}{R_r + R_{loss}} = \frac{1.974}{1.974 + 0.09525} = \frac{1.974}{2.069}$$ $$\boxed{e_r = 0.9540 = 95.40\ \%}$$ A short radiator can still be efficient provided its radiation resistance is not too small; halve the length and $R_r$ falls fourfold while $R_{loss}$ only halves, so efficiency degrades quickly below $l/\lambda \approx 0.02$.
5. (3) Far field from the radiated power. A short element has directivity $D = 1.5$ with its maximum broadside ($\theta = 90^\circ$), so the peak power density at 20 km is $$S_{max} = \frac{D\,P_{rad}}{4\pi r^2} = \frac{1.5(1000)}{4\pi(2\times10^4)^2} = 2.984\times10^{-7}\ \text{W/m}^2$$ and since $S = E_{rms}^2/\eta_0$ for a plane wave, $$\boxed{E_{rms} = \sqrt{S_{max}\eta_0} = \sqrt{(2.984\times10^{-7})(376.99)} = 10.61\ \text{mV/m}}$$ equivalently a peak amplitude of $\sqrt{2}(10.61) = 15.00$ mV/m.
6. Verify by the independent current route. From $P_{rad} = \tfrac{1}{2}I_{pk}^2R_r$ the required current is $I_{pk} = \sqrt{2000/1.974} = 31.83$ A. Feeding that into the short-element far field $E_\theta = \eta_0 k I l\sin\theta/(4\pi r)$ at $\theta = 90^\circ$ gives 15.00 mV/m peak — identical to the pattern route, which confirms both $R_r$ and the field. The 31.8 A also shows why such antennas need heavy conductors.