Short vertical element radiating at two frequencies: 22-Elec-A7 Electromagnetics, May 2017, Q8

Question 8: Short vertical element radiating at two frequencies (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. One short vertical current element radiating into free space at two frequencies with equal current amplitudes, calibrated by a single measured field at a broadside point.

Given data
Quantity	Symbol	Value
Element length	$h$	$1$ m, vertical
Frequencies	$f_1$, $f_2$	$10$ MHz and $5$ MHz
Current amplitudes	$I$	equal at both frequencies
Reference point P1	$r_1$	$10$ km horizontally from the element
Measured field at P1	$E_1$	$100\ \mu$V/m rms at $10$ MHz
Target point P2	—	$2.0$ km vertically above P1
Medium	—	free space

Find. The rms vertical component of the electric field at P2 at $5$ MHz.

Figure 8.1 — the reference ray is broadside (theta = 90 degrees); the target ray is longer and off broadside, and the vertical projection of E costs one further factor of sin(theta).

Approach. Write the short-element far field as a scaling law in $f$, $I$, $\sin\theta$ and $1/r$; the measured value at P1 calibrates every constant, so only the ratios matter. Then project the field, which is transverse to the ray, onto the vertical.

Confirm the element is electrically short. $\lambda_1 = c/f_1 = 30.0$ m and $\lambda_2 = 60.0$ m, so $h/\lambda = 0.033$ and $0.017$ respectively — both far below $0.1$, so the current may be taken as uniform in phase and the $\sin\theta$ pattern applies at both frequencies.
Far field of a short vertical element. With $k = 2\pi f/c$, $$E_\theta = \frac{\eta_0 k I h \sin\theta}{4\pi r} \;\propto\; \frac{f\,I\,\sin\theta}{r},$$ where $\theta$ is measured from the element's own (vertical) axis. Because the element, its length and its current are common to both frequencies, every constant cancels in a ratio and no antenna parameter needs to be evaluated.
Geometry of the two points. P1 is broadside, so $\theta_1 = 90^\circ$ and $\sin\theta_1 = 1$. For P2, $2.0$ km above P1: $$r_2 = \sqrt{(10\,000)^{2}+(2\,000)^{2}} = 10\,198\ \text{m}, \qquad \sin\theta_2 = \frac{10\,000}{10\,198} = 0.9806,$$ i.e. $\theta_2 = 78.69^\circ$ from the axis, an elevation of $11.31^\circ$ above the horizontal.
Scale the total field to P2 at 5 MHz. $$E_{\theta 2} = E_1 \cdot \frac{f_2}{f_1}\cdot\frac{r_1}{r_2}\cdot \frac{\sin\theta_2}{\sin\theta_1} = 100 \times 0.500 \times 0.9806 \times 0.9806 = 48.1\ \mu\text{V/m}.$$ Halving the frequency at fixed current halves the radiated field — a short element is a worse radiator the lower you drive it.
Take the vertical component. $\mathbf{E}$ is along $\hat{\theta}$, perpendicular to the ray, and $\hat{z}\cdot\hat{\theta} = -\sin\theta$, so the vertical part carries a second factor of $\sin\theta$: $$E_z = E_\theta \sin\theta_2 = 48.08 \times 0.9806 \quad\Longrightarrow\quad \boxed{E_{z,\text{rms}} = 47.1\ \mu\text{V/m}} .$$ Equivalently $E_z \propto f I \sin^{2}\theta / r$, and at the broadside reference point $E_z = E_1$ because $\sin\theta_1 = 1$ there.
Reading the result. The frequency halving costs a full factor of two, while the extra $2$ km of height costs only about $6\%$ (the range grows $2\%$ and each of the two $\sin\theta$ factors costs about $2\%$, so $1 - 0.9806^{3} = 5.7\%$). So the answer sits just under half the reference value, $47.1$ against $100\ \mu$V/m — a useful check that no factor has been misplaced.

Final results
Quantity asked	Result
Wavelengths / short-element check	$30.0$ m and $60.0$ m; $h/\lambda = 0.033$ and $0.017$
Slant range to P2	$r_2 = 10\,198$ m
Polar angle at P2	$\theta_2 = 78.69^\circ$ ($\sin\theta_2 = 0.9806$), elevation $11.31^\circ$
Total field at P2 at 5 MHz	$E_{\theta} = 48.1\ \mu$V/m rms
Vertical component at P2 (answer)	$E_z = 47.1\ \mu$V/m rms
Scaling law used	$E_z \propto f\,I\,\sin^{2}\theta/r$

Topic. Far field of an electrically short current element, and the cost of projecting it onto the vertical.

Key relations. $E_\theta = \eta_0 k I h\sin\theta/(4\pi r)$ with $k = 2\pi f/c$ (uniform current); $E_\theta \propto f I \sin\theta/r$; vertical projection $E_z = E_\theta\sin\theta \propto f I \sin^{2}\theta / r$; short-element validity $h/\lambda \lt 0.1$.

Why this works. In the far field the radiated $\mathbf{E}$ is transverse to the ray and lies in the plane containing the ray and the element axis, so it is purely vertical only in the broadside direction. Because one calibrated measurement already contains $\eta_0$, $I$, $h$ and all the numerical factors, the whole problem reduces to three ratios — frequency, range and pattern — which is why no antenna constant is ever needed. The extra $\sin\theta$ for the vertical component is geometry, not radiation physics: $\hat{\theta}$ tilts away from the vertical by exactly the elevation angle.

Common pitfall. Using the elevation angle in $\sin\theta$ instead of the polar angle from the element axis (here $0.196$ instead of $0.981$, wrong by a factor of five), and quoting $E_\theta$ when the question asks for the vertical component — that is one whole factor of $\sin\theta$. Also remember to scale the range with the slant distance, not the horizontal one.

Source. Balanis, Antenna Theory: Analysis and Design 4th ed. ch. 4 (infinitesimal dipole); Ulaby, Fundamentals of Applied Electromagnetics ch. 9 (short-dipole radiation).

Question 8 of 8: Short vertical element radiating at two frequencies

Question 8: Short vertical element radiating at two frequencies (20 marks)