Question 8 of 8: Vertical field component from a short element at a new frequency

Question 8: Vertical field component from a short element at a new frequency (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 in free space is characterised by one measured power density, and is then re-examined at a different frequency, a different range and a different elevation with the driving current unchanged.

Find. The rms vertical component of the electric field at the target point.

Approach. Anchor the unknown antenna constants with the one measured field, then scale that single number by the three ratios the question changes, and finally project the result onto the vertical.

1. Convert the reference power density into a field. The maximum on a sphere surrounding a vertical element lies in the equatorial plane, $\theta = 90^\circ$, so $$E_{\text{ref}} = \sqrt{S_1\eta_0} = \sqrt{(10\times10^{-6})(376.8)} = 61.39\ \text{mV/m (rms)}$$ This one number now carries every antenna constant — the element length, the current, and all the numerical factors — so none of them has to be known separately.
2. Write the far field of a short element. For a short current element of length $h$ carrying current $I$, $$E_{\theta} = \frac{\eta_0 k I h\sin\theta}{4\pi r} \;\propto\; \frac{f\,I\,\sin\theta}{r}$$ since $k = 2\pi f/c$. The element length and all constants cancel when two situations of the same antenna are compared, leaving only the three ratios in $f$, $r$ and $\sin\theta$.
3. Locate the target point. On the 15 km sphere, a height of 7.5 km above the horizontal plane through the element means $$\cos\theta = \frac{z}{r_2} = \frac{7.5}{15} = 0.500 \;\Longrightarrow\; \theta = 60.00^\circ, \qquad \sin\theta = 0.8660$$ so the point sits 30° above the horizon as seen from the element, well clear of the axial null.
4. Scale to the new condition. Doubling the frequency at constant current doubles the field, moving from 10 km to 15 km reduces it by $2/3$, and the pattern factor contributes $\sin 60^\circ$: $$E_{\theta} = E_{\text{ref}}\,\frac{f_2}{f_1}\,\frac{r_1}{r_2}\,\sin\theta = 61.39 \times 2 \times \tfrac{2}{3} \times 0.8660 = 70.88\ \text{mV/m}$$ which is the total far field there, all of it in the $\hat{\theta}$ direction.
5. Project onto the vertical. The far field of this antenna has only a $\theta$ component, and $\hat{z}\cdot\hat{\theta} = -\sin\theta$, so the vertical component carries a second factor of $\sin\theta$: $$|E_{z}| = E_{\theta}\sin\theta = 70.88 \times 0.8660 = \boxed{61.39\ \text{mV/m (rms)}}$$ The vertical component therefore varies as $\sin^{2}\theta$, not as $\sin\theta$.
6. Notice why the answer repeats the reference value. The complete scale factor is $$\frac{f_2}{f_1}\cdot\frac{r_1}{r_2}\cdot\sin^{2}\theta = 2 \times \frac{2}{3} \times \frac{3}{4} = 1$$ exactly, so the vertical component at the target equals the total field at the reference point to the last digit. This is an arithmetic accident of the numbers chosen, but it is a useful check: any answer other than 61.39 mV/m means one of the three ratios was applied the wrong way up.

For completeness, the power density at the target point is $S_2 = E_{\theta}^{2}/\eta_0 = 13.33\ \mu\text{W/m}^2$, larger than the reference despite the greater range, because doubling the frequency at fixed current quadruples the radiated power of a short element.