Field of a Short Vertical Element Raised Above Its Ground Plane: 22-Elec-B9 Electromagnetic Field, Transmission Lines, Antennas, and Radiation, May 2013, Q6

22-Elec-B9 Electromagnetic Field, Transmission Lines, Antennas, and Radiation · May 2013

Question 6: Field of a Short Vertical Element Raised Above Its Ground Plane (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 vertical current element above a perfectly conducting horizontal ground plane, observed at a fixed point on that plane 1 km from the original site.

Given data
Quantity	Case A (original)	Case B (moved)
Element length $l$	0.50 m	0.25 m
Frequency $f$	10 MHz	15 MHz
Wavelength $\lambda$	30 m	20 m
Height above the plane $h$	0 (on the plane)	1000 m
Total radiated power $P$	$P_A$	$2P_A$
Horizontal range to the observer $\rho$	1000 m	1000 m
Measured field	100 µV/m RMS	required

Find. The RMS electric field intensity at the same point on the ground plane after the element is raised and re-tuned.

Figure 6.1 — Side view for Case B. The image element below the plane is co-directional and equidistant from any point on the plane, so the two rays always add in phase there.

Approach. Replace the ground plane by the image element, observe that at a point on the plane the direct and image paths are exactly equal, express the resulting field in terms of the element's moment, and then use the supplied radiation-resistance formula to convert the moment into radiated power — remembering that the image also changes how much power a given current radiates, and that this effect is different on the plane and 50 wavelengths above it.

Apply image theory at a point on the plane. A vertical element at height $h$ has an in-phase image at depth $h$. For an observer on the plane at horizontal range $\rho$ both rays travel $R = \sqrt{\rho^2 + h^2}$, the horizontal field components cancel and the vertical ones add, giving $$|E| = \frac{2K\sin^{2}\theta}{R} = \frac{2K\rho^{2}}{R^{3}}, \qquad K = \frac{\eta I l}{2\lambda},$$ with $\sin\theta = \rho/R$. Setting $h = 0$ recovers the familiar monopole-on-a-plane result $|E| = 2K/\rho$, which is the correct check on the formula.
Eliminate the current using the supplied aid. The aid $R_{\text{rad}} = (2\pi/3)Z_0 (l/\lambda)^2$ is the free-space value, so the power a current $I$ radiates into the half space above the plane must account for the image:
• Case A, on the plane: the element and its contiguous in-phase image act as one element of length $2l$, so the upper half space receives $P_A = 2\cdot\tfrac12 I_A^{2}R_{\text{rad}}$ (the familiar result that a short monopole has twice the free-space radiation resistance of its own length);
• Case B, $h = 50\lambda$: the element–image coupling averages to zero over the hemisphere, so $P_B = \tfrac12 I_B^{2}R_{\text{rad}}$.
Solving each for the moment $Il/\lambda$, $$K_A = \frac{\eta}{2}\sqrt{\frac{3P_A}{2\pi Z_0}}, \qquad K_B = \frac{\eta}{2}\sqrt{\frac{3P_B}{\pi Z_0}} \quad\Longrightarrow\quad \frac{K_B}{K_A} = \sqrt{\frac{2P_B}{P_A}} = \sqrt{4} = 2 .$$ The length and the wavelength cancel identically, so the stated changes of length and frequency do not enter. (Numerically the effective radiation resistance is 0.439 Ω on the plane and 0.123 Ω when raised, and the current adjusts accordingly.) A direct integration of $|E|^2$ over the upper hemisphere confirms the factor of two between the two heights.
Evaluate the geometry factor for each case. $$\text{Case A: } R_A = \rho = 1000\ \text{m}, \qquad \frac{\rho^{2}}{R_A^{3}} = 1.000\times 10^{-3}\ \text{m}^{-1},$$ $$\text{Case B: } R_B = \sqrt{1000^2 + 1000^2} = 1414.2\ \text{m}, \qquad \frac{\rho^{2}}{R_B^{3}} = 3.536\times 10^{-4}\ \text{m}^{-1}.$$ The observation direction in Case B is $45^\circ$ below the horizontal, so $\sin\theta = 0.7071$ and the element pattern itself costs a factor of two in power.
Form the ratio. $$\frac{E_B}{E_A} = \frac{K_B}{K_A}\cdot \frac{\rho^{2}/R_B^{3}}{\rho^{2}/R_A^{3}} = 2\times 0.35355 = 0.7071 .$$ The doubled power buys $\sqrt{2}$, releasing the element from the plane buys a further $\sqrt{2}$ (the same power no longer has to be shared with a coherent image), and the extra range and off-broadside direction together cost $2\sqrt{2}$.
Report the field. $$\boxed{\ E_B = 0.7071 \times 100\ \mu\text{V/m} = 70.7\ \mu\text{V/m (RMS)}\ }$$

The changes in frequency and element length are deliberate distractors. For an electrically short element the far-field strength in a given direction is set by the radiated power, the geometry and the ground-plane configuration, because shortening the element raises the current needed to radiate that power in exactly the inverse proportion. Both configurations remain electrically short ($l/\lambda$ = 0.0167 and 0.0125), so the short-element pattern $\sin\theta$ applies throughout and the comparison is legitimate.

Final results — Question 6
Quantity	Case A	Case B
Effective radiation resistance ($P = \tfrac12 I^2 R$)	0.439 Ω (2 × 0.219)	0.123 Ω
Moment ratio $K_B/K_A$	1	2
Slant range to the observer	1000 m	1414.2 m
$\sin\theta$ at the observer	1.000	0.7071
Geometry factor $\rho^{2}/R^{3}$	$1.000\times 10^{-3}$	$3.536\times 10^{-4}$
Field strength (RMS)	100 µV/m	70.7 µV/m

Check: assumptions behind the comparison. The ground plane is taken as a perfect conductor of unlimited extent, so image theory is exact and the observation point stays in the far field of both configurations. “Total power radiated” is read as the power actually launched into the half space above the plane in both cases. The raised element's coupling to its image is negligible at $h = 1000\ \text{m} = 50\lambda$ at 15 MHz, but on the plane the coupling is total — which is why the element on the plane radiates twice the power for the same current. A candidate who applies the free-space $R_{\text{rad}}$ unchanged to both positions obtains $\sqrt{2}\times 0.3536\times 100 = 50\ \mu\text{V/m}$; that reading ignores the image's effect on radiated power in Case A and is not adopted here.

Topic. Image theory over a conducting plane and the power-only dependence of a short element's far field.

Key relations. $E_\theta = j\eta I l \beta\sin\theta\,e^{-j\beta R}/(4\pi R)$; vertical element over a PEC plane has an in-phase image; $R_{\text{rad}} = (2\pi/3)Z_0(l/\lambda)^2$; $P = \tfrac12 I^{2}R_{\text{rad}}$ far above the plane, $P = I^{2}R_{\text{rad}}$ on it.

Why this works. Image theory replaces the boundary condition on an infinite conducting plane with a fictitious source, and for a vertical electric element that image points the same way as the original. At an observation point lying on the plane the two sources are equidistant, so their contributions arrive with no relative delay whatever the height — there is no interference lobing to worry about, only a clean factor of two on the normal component. Once the field is written in terms of the element moment, the supplied radiation-resistance formula converts moment to radiated power, and the $(l/\lambda)$ factors cancel between the two expressions. The image acts twice: it doubles the field at the observer in both positions, but it doubles the radiated power for a given current only when it is contiguous with the element. Keeping those two effects separate is the real content of the question.

Common pitfall. Scaling the field with $l/\lambda$ as if the current were held constant, which produces a spurious factor of $(0.25/20)/(0.50/30) = 0.75$. The question fixes the power, not the current. A second trap is using $1/R$ alone and forgetting the $\sin^2\theta$ that image addition contributes at a point on the plane (that alone would give 141 µV/m). A third is applying the free-space radiation resistance to the element sitting on the plane: there the element and its image radiate as one element of twice the length, so the same current puts twice the power into the upper half space, and ignoring it understates the answer as 50 µV/m.

Source. Balanis, Antenna Theory 4th ed., §4.2 (infinitesimal dipole) and §4.7 (ground effects, vertical element above an infinite plane); Sadiku, Elements of Electromagnetics 7th ed., §13.2.

Question 6 of 8: Field of a Short Vertical Element Raised Above Its Ground Plane

Question 6: Field of a Short Vertical Element Raised Above Its Ground Plane (20 marks)