Power Density Radiated by a Short Horizontal Element: 22-Elec-A7 Electromagnetics, December 2016, Q8

Question 8: Power Density Radiated by a Short Horizontal Element (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. A short horizontal radiating element in free space, calibrated by a single measured field strength at a point directly overhead.

Given data
Quantity	Symbol	Value
Operating frequency	f₁	10 MHz (λ = 30 m)
Reference range, directly overhead	r₁	1 km
Measured field at that point	E₁	100 µV/m
Horizontal run to the asked point	—	2 km
Elevation of the asked point	ψ	30°
Reduced frequency for part (ii)	f₂	5 MHz (λ = 60 m)
Intrinsic impedance of free space	η₀	377 Ω

Find. (i) the power density at the elevated point, and (ii) the same quantity when the frequency is halved with everything else unchanged.

Figure 8.1 — The two observation points: the reference point directly overhead and the asked point 2 km out at 30° elevation.

Approach. Recognise the overhead measurement as the pattern maximum of a horizontal element, use it to calibrate the radiation without ever unpacking the element’s length or current, then scale by the inverse-square law for the new range and by $f^{2}$ for the new frequency.

Convert the reference field to a power density. In the far field of any antenna the wave is locally plane, so $$S_{1}=\frac{E_{1}^{2}}{\eta_{0}}=\frac{\left(100\times10^{-6}\right)^{2}}{377}=2.653\times10^{-11}\ \text{W/m}^{2}=26.53\ \text{pW/m}^{2}$$
Identify what the reference point represents. A short element radiates with the pattern $\sin\theta$, where $\theta$ is measured from the element’s own axis. The element here is horizontal, so a point directly above it lies at $\theta=90^{\circ}$: the reference measurement is taken on the pattern maximum. That single number therefore calibrates the entire product of element length, current and frequency, and no antenna constant needs to be evaluated anywhere in this question.
State the general scaling law. For a fixed element and current, $$S(r,\theta)=S_{1}\left(\frac{r_{1}}{r}\right)^{2}\sin^{2}\theta$$ because the field falls as $1/r$ and follows $\sin\theta$ in angle, and power density goes as the square of the field.
Work out the slant range to the asked point. The point is 2 km away horizontally and $30^{\circ}$ above the horizon, so $$r_{2}=\frac{2000}{\cos30^{\circ}}=2309\ \text{m},\qquad \left(\frac{r_{1}}{r_{2}}\right)^{2}=\left(\frac{1000}{2309}\right)^{2}=0.1875$$ The point is therefore 2.31 km from the element, not 2 km, and it sits 1155 m above it.
Fix the pattern factor. The paper specifies the elevation of the second point but not its bearing relative to the element’s axis, so the natural reading — and the one consistent with the reference point — is that it lies in the same vertical plane broadside to the element, where $\theta$ remains $90^{\circ}$ and $\sin\theta=1$. The pattern therefore contributes no reduction and the answer is governed purely by spreading: $$S_{2}=S_{1}\left(\frac{r_{1}}{r_{2}}\right)^{2}=\left(26.53\right)\left(0.1875\right)=\boxed{4.97\ \text{pW/m}^{2}}$$
Part (ii): scale with frequency. For a short element carrying an unchanged current over an unchanged length, the radiated field is proportional to frequency, $E_{\theta}\propto f\,I\,\ell$, so the power density scales as $f^{2}$: $$S_{2}'=S_{2}\left(\frac{f_{2}}{f_{1}}\right)^{2}=\left(4.97\right)\left(\frac{5}{10}\right)^{2}=\boxed{1.24\ \text{pW/m}^{2}}$$ Halving the frequency therefore costs a factor of four, or 6 dB — the well-known penalty for operating a fixed physical antenna at a lower frequency, where it is electrically shorter and radiates less efficiently for the same current.
Confirm the assumptions the method rests on. The wavelength is 30 m at 10 MHz and 60 m at 5 MHz, so an element short at the higher frequency remains short at the lower one and the $\sin\theta$ pattern stays valid throughout. Both observation points also sit many wavelengths away ($r/\lambda$ is 33 and 77 at 10 MHz), comfortably in the far field where $S=E^{2}/\eta_{0}$ applies.

Check: the bearing of the second point is not stated. Because the element is horizontal, its $\sin\theta$ pattern depends on azimuth as well as elevation, and the paper fixes only the elevation. The solution above takes the point in the vertical plane broadside to the element, where $\sin\theta=1$; this is the reading that makes the stated data sufficient and that matches the reference point’s own plane. If instead the point were taken in the vertical plane containing the element, the angle from the axis would be $30^{\circ}$, the factor $\sin^{2}\theta=0.25$ would apply, and the answers would become 1.24 pW/m² for part (i) and 0.311 pW/m² for part (ii). Any point between those two azimuths lies between these bounds. A candidate should state the assumption explicitly, as the paper’s own note 1 invites.

Question 8 — radiated power density
Part	Quantity	Value
—	Reference power density at 1 km overhead	26.53 pW/m²
—	Slant range to the asked point	2309 m (height 1155 m)
—	Spreading factor (r₁/r₂)²	0.1875
(i)	Power density at 10 MHz	4.97 pW/m²
(ii)	Power density at 5 MHz	1.24 pW/m² (6 dB lower)
—	Alternative azimuth (in the element’s plane)	1.24 and 0.311 pW/m² — see callout

Topic. Far-field power density of a short (Hertzian) current element and its frequency scaling.

Key relations. $S=E^{2}/\eta_{0}$; $E_{\theta}=\eta_{0}kI\ell\sin\theta/(4\pi r)$, so $E\propto fI\ell\sin\theta/r$; hence $S\propto f^{2}I^{2}\ell^{2}\sin^{2}\theta/r^{2}$; short-element validity requires $\ell\ll\lambda$.

Why this works. Every constant describing the antenna — its length, its current, the impedance of free space, the factor $4\pi$ — appears in the field as one multiplicative group, so a single calibrated measurement at a known range and known pattern angle determines that group once and for all. Every subsequent question about the same antenna reduces to ratios of range and of the angular factor, which is why the arithmetic here is so short. The frequency dependence survives that cancellation only because part (ii) deliberately changes $f$ while holding the physical element and its current fixed; the element then represents a smaller fraction of a wavelength and radiates a proportionally weaker field.

Common pitfall. Using the 2 km horizontal run directly as the range, which ignores the 1155 m of height and overstates the answer by a third; and measuring the pattern angle from the vertical, which is correct for a vertical element but not for the horizontal one specified here. A third trap is to assume the frequency change cancels because “the element is the same” — it is the same physical element, but not the same electrical one.

Source. Balanis, Antenna Theory: Analysis and Design, 4th ed., §4.1–4.2 (infinitesimal dipole fields and radiated power); Ulaby & Ravaioli, Fundamentals of Applied Electromagnetics, 7th ed., §9.1.

Question 8 of 8: Power Density Radiated by a Short Horizontal Element

Question 8: Power Density Radiated by a Short Horizontal Element (20 marks)