Scaling a Short Monopole's Field to a New Frequency and a New Direction: 22-Elec-A7 Electromagnetics, May 2014, Q7

Question 7: Scaling a Short Monopole's Field to a New Frequency and a New Direction (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 data
Quantity	Symbol	Value
Element 1	—	10 MHz, current $I$, length $L$
Element 2	—	5 MHz, current $I/2$, same length $L$
Reference field (10 MHz)	$E_{ref}$	$2.2\times10^{-3}$ V/m rms
Reference point	—	10 km away, on the ground plane
Field point, horizontal offset	$\rho$	3.5 km
Field point, height	$z$	2 km
Ground plane	—	perfectly conducting, horizontal

Find. The rms magnitude of the vertical component of the 5 MHz field at the elevated point.

Figure 7.1 — Elevation view. The two elements are co-located at the origin, so a single geometry serves both frequencies; the inset shows how $E_\theta$ resolves into vertical and horizontal parts.

Approach. Write the far field of a short vertical element over a perfect ground, use the 10 MHz datum to fix the constant, scale it to 5 MHz by frequency and current, apply the new geometry, and finally project onto the vertical.

Far field of a short element over a perfect ground. Image theory replaces the ground plane with a mirror element, doubling the effective moment in the upper half space. For a Hertzian element of length $L$ carrying current $I$,$$E_\theta = \frac{\eta_0 k I L\sin\theta}{2\pi r} = \frac{\eta_0 L}{c}\,\frac{f I\sin\theta}{r},$$with $\theta$ measured from the vertical element axis. Everything fixed by the hardware collects into one constant $K = \eta_0 L/c$.
Calibrate with the 10 MHz measurement. On the ground plane $\theta = 90^{\circ}$ and $\sin\theta = 1$, so$$K f_{10} I_{10} = E_{ref}\,r_{ref} = (2.2\times10^{-3})(10\,000) = 22.0\ \text{V}.$$This one product carries all the unknown hardware constants and never needs to be unpacked.
Scale to the 5 MHz element. Its frequency is halved and its current is halved, and the length is the same:$$K f_{5} I_{5} = 22.0\left(\tfrac12\right)\left(\tfrac12\right) = 5.50\ \text{V}.$$The field of the low-frequency element is therefore only a quarter as strong as its partner at the same range and angle.
Geometry of the elevated point. With a horizontal offset of 3.5 km and a height of 2 km,$$r = \sqrt{3.5^{2} + 2.0^{2}} = 4.031\ \text{km},\qquad \sin\theta = \frac{\rho}{r} = \frac{3.5}{4.031} = 0.868,$$so $\theta = 60.3^{\circ}$ from the vertical, i.e. the point sits about $29.7^{\circ}$ above the horizon as seen from the antenna.
Total field at that point. Substituting,$$E_\theta = \frac{(5.50)(0.868)}{4031} = 1.185\times10^{-3}\ \text{V/m rms}.$$This is the magnitude of the whole far field, which points along $\hat{\theta}$ — perpendicular to the line of sight, not vertical.
Project onto the vertical. Since $\hat{z}\cdot\hat{\theta} = -\sin\theta$, the vertical component picks up a second factor of $\sin\theta$:$$E_z = E_\theta\sin\theta = K f_5 I_5\,\frac{\sin^{2}\theta}{r} = (1.185\times10^{-3})(0.868),$$$$\boxed{E_z = 1.03\times10^{-3}\ \text{V/m rms} = 1.03\ \text{mV/m}.}$$

Check: the question asks for the vertical component, so the second factor of $\sin\theta$ is deliberate and is where most of the marks sit. The total field at that point is 1.185 mV/m and the horizontal (radially outward) component is 0.588 mV/m; the three recombine correctly, $\sqrt{1.029^{2}+0.588^{2}} = 1.185$ mV/m.

A sanity check confirms the scaling: at the original 10 km ground-plane point the 5 MHz element would produce 0.55 mV/m, one quarter of the 2.2 mV/m measured for its partner. Moving the observer to 4.03 km more than doubles the field through the $1/r$ factor, and the two projections onto $\hat{\theta}$ and then onto the vertical claw most of that gain back.

Final Results
Quantity	Symbol	Value
Calibration product, 10 MHz	$K f_{10} I_{10}$	22.0 V
Calibration product, 5 MHz	$K f_{5} I_{5}$	5.50 V
Slant range to the field point	$r$	4.031 km
Polar angle from the vertical	$\theta$	$60.3^{\circ}$ ($\sin\theta = 0.868$)
Total far field there	$E_\theta$	1.185 mV/m rms
Vertical component	$E_z$	1.03 mV/m rms
Horizontal component	$E_\rho$	0.588 mV/m rms

Topic. Far field of a short vertical element over a perfect ground plane, and the resolution of $E_\theta$ into Cartesian components.

Key relations. $E_\theta = \eta_0 kIL\sin\theta/(2\pi r)$ for an element with its image; $E_\theta \propto fI\sin\theta/r$; $\hat{z}\cdot\hat{\theta} = -\sin\theta$, $\hat{\rho}\cdot\hat{\theta} = \cos\theta$.

Why this works. A vertical element standing on a perfect conductor produces, in the upper half space, exactly the field of the element plus its mirror image, which is why the usual dipole result appears with $2\pi$ rather than $4\pi$ in the denominator. Because every hardware detail enters only through the product $KfI$, one measured field value calibrates the whole system and the rest is scaling. The far field is transverse to the line of sight, so its vertical component is not the field itself but its projection, and that projection is what a vertical receiving whip would actually pick up.

Common pitfall. By far the most common error is stopping at $E_\theta$ and reporting 1.185 mV/m as the vertical component, which drops the second $\sin\theta$. A close second is measuring $\theta$ from the ground plane rather than from the element axis, which swaps sine for cosine in both factors and gives 0.34 mV/m. Scaling by the current ratio while forgetting the frequency ratio is the third.

Source. Balanis, Antenna Theory: Analysis and Design, 4th ed., §4.2 and §4.7 (infinitesimal dipole; ground-plane images); Sadiku, Elements of Electromagnetics, 7th ed., §13.2.

Question 7 of 8: Scaling a Short Monopole's Field to a New Frequency and a New Direction

Question 7: Scaling a Short Monopole's Field to a New Frequency and a New Direction (20 marks)