Question 10 of 10: Radiation Resistance, Efficiency and Field Intensity of a Steel Wire Antenna

Question 10: Radiation Resistance, Efficiency and Field Intensity of a Steel Wire Antenna (20 marks)

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.

Find. $R_r$; the radiation efficiency $\xi$; $|E|$ at $r=20$ km.

Approach. $\lambda=c/f=300$ m gives $l/\lambda=0.05\ll1$, so this is an electrically-short antenna; "uniform current" means the idealized (Hertzian) short-dipole radiation-resistance formula applies directly (not the triangular-current form used for an open-circuited linear dipole). The loss resistance comes from skin-effect surface resistance of the steel wire; efficiency is $R_r/(R_r+R_{loss})$. The field at a distance uses the short dipole's directivity $D=1.5$ with the GIVEN radiated power (independent of parts 1–2).

1. Part (a)/1 — Radiation resistance. $\lambda=c/f=(3\times10^8)/(10^6)=300$ m, $l/\lambda=15/300=0.0500$: $$R_r=80\pi^2\left(\frac{l}{\lambda}\right)^2=80\pi^2(0.0500)^2=\boxed{1.98\ \Omega.}$$
2. Part (b)/2 — Loss resistance (skin effect) and efficiency. Surface resistance of the steel wire: $$R_s=\sqrt{\frac{\pi f\mu_0}{\sigma}}=\sqrt{\frac{\pi(10^6)(4\pi\times10^{-7})}{6.2\times10^6}}=7.98\times10^{-4}\ \Omega/\text{sq}.$$ The high-frequency (skin-effect) loss resistance of a round wire of length $l$, radius $b$ is $R_{loss}=R_sl/(2\pi b)$: $$R_{loss}=\frac{(7.98\times10^{-4})(15)}{2\pi(0.02)}=\boxed{0.0953\ \Omega.}$$ $$\xi=\frac{R_r}{R_r+R_{loss}}=\frac{1.98}{1.98+0.0953}=\boxed{95.4\%.}$$
3. Part (c)/3 — Field intensity at 20 km for 1 kW radiated. A short dipole has directivity $D=1.5$ in its maximum (broadside) direction, so the maximum time-average power density at distance $r$ is $$S_{max}=\frac{DP_{rad}}{4\pi r^2}=\frac{(1.5)(1000)}{4\pi(20000)^2}=2.984\times10^{-7}\ \text{W/m}^2.$$ Relating this to the peak field via $S_{avg}=|E_{max}|^2/(2\eta_0)$: $$|E_{max}|=\sqrt{2\eta_0S_{max}}=\sqrt{2(376.7)(2.984\times10^{-7})}=\boxed{0.01499\ \text{V/m}=15.0\ \text{mV/m}.}$$