Question 8 of 8: Vertical elements over a ground plane: the 30 MHz vertical field

Question 8: Vertical elements over a ground plane: the 30 MHz vertical field (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. Two short vertical current elements share a conducting ground plane: a 1 m, 10 MHz element sitting in the plane (which produces a measured 1 $\mu\text{V/m}$ rms at 5 km along the plane), and a 0.2 m, 30 MHz element 2.89 km directly above it carrying one tenth of the current.

Find. The vertical component of the 30 MHz electric field at the same 5 km point on the ground plane.

Approach. Use the short-element far field $E_\theta \propto f I h \sin\theta / r$, extract the vertical component by projecting $\hat{\theta}$ onto $\hat{z}$ (which supplies a second factor of $\sin\theta$), and then take the ratio to the measured reference so that every antenna constant, and the ground-plane image factor, cancels.

1. Far field of a short vertical element over a ground plane. For an element of length $h \ll \lambda$ carrying current $I$, $$E_\theta = \frac{\eta_0 k I h \sin\theta}{4\pi r}\times 2 = K\,\frac{f\,I\,h\,\sin\theta}{r}$$ where the factor 2 is the in-phase image supplied by the conducting plane and $k = 2\pi f/c$ has been absorbed into the constant $K$. Only ratios of $f$, $I$, $h$, $\sin\theta$ and $r$ will be needed, so $K$ never has to be evaluated.
2. The vertical component carries a second $\sin\theta$. The far field points along $\hat{\theta}$, and $\hat{z}\cdot\hat{\theta} = -\sin\theta$. Hence $$|E_z| = |E_\theta|\sin\theta = K\,\frac{f\,I\,h\,\sin^{2}\theta}{r}$$ This is the marked step: for the reference element the point lies on the plane, i.e. at $\theta = 90^\circ$ where $\sin\theta = 1$, so its measured 1 $\mu\text{V/m}$ is the vertical component.
3. Calibrate on the reference element. With $\theta_1 = 90^\circ$ and $r_1 = 5.0\ \text{km}$, $$E_1 = K\,\frac{f_1 I_1 h_1}{r_1} = 1.0\ \mu\text{V/m}$$ Everything unknown about the antennas and the excitation is now bundled into this one measured number.
4. Geometry for the elevated element. The observation point is 5.0 km along the plane and the element is 2.89 km above it, so $$r_2 = \sqrt{(5000)^2 + (2890)^2} = 5775\ \text{m} = 5.775\ \text{km}$$$$\sin\theta_2 = \frac{5000}{5775} = 0.8658, \qquad \theta_2 = 120.0^\circ \ \text{(measured from the element axis, }+\hat{z})$$
5. Form the ratio. Dividing the two vertical-component expressions, every constant — including the image factor of 2, which applies to both elements because the observation point lies on the plane — cancels: $$\frac{E_{z,2}}{E_1} = \frac{f_2}{f_1}\cdot\frac{I_2}{I_1}\cdot\frac{h_2}{h_1}\cdot\frac{\sin^{2}\theta_2}{\sin^{2}\theta_1}\cdot\frac{r_1}{r_2}$$$$= (3)(0.1)(0.2)\,(0.8658)^{2}\left(\frac{5000}{5775}\right) = 0.03894$$
6. Evaluate. $$E_{z,2} = 0.03894 \times 1.0\ \mu\text{V/m} = \boxed{3.89\times10^{-8}\ \text{V/m} = 38.9\ \text{nV/m rms}}$$

The five factors pull in opposite directions and it is instructive to see which wins. Tripling the frequency helps (a short element radiates in proportion to its length in wavelengths), but the current is down by ten and the length by five, so the source is 16.7 times weaker overall; the obliquity $\sin^2\theta$ costs a further 25 % and the longer slant range another 13 %. The net result is about a 26-fold reduction from the 10 MHz reference. Note finally that at a point on the conducting plane the direct and image contributions of the elevated element travel equal distances and arrive in phase, and their horizontal components cancel exactly — so the total field there is purely vertical, and the “vertical component” asked for is in fact the whole field.