Question 5 of 8: Circular Polarization from Two Crossed Current Elements

Question 5: Circular Polarization from Two Crossed Current Elements (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 co-located short current elements — one vertical, one horizontal along north–south — driven in quadrature with the vertical element carrying twice the current.

Find. At least one direction of propagation in which the two radiated fields combine into a circularly polarised wave.

Approach. Circular polarization needs two orthogonal field components of equal amplitude in quadrature. The quadrature is already supplied by the currents, so the task reduces to finding a direction where the two $\sin\theta$ patterns equalise the amplitudes while the field directions stay perpendicular.

1. Check the short-element approximation. At 10 MHz the wavelength is $\lambda = c/f = 30$ m, so each 1 m element is $\lambda/30$ long and the supplied formula $E = Z_0 I l k \sin\theta\,e^{-jkr}/(4\pi r)$, with $k = 2\pi/\lambda = 0.2094$ rad/m, applies. Both elements share the same $l$, $k$ and $r$, so only the products $I\sin\theta$ differ.
2. Radiation patterns of the two elements. Each element radiates as $\sin\theta$ measured from its own axis, with the field lying along the local $\hat{\boldsymbol{\theta}}$ direction. Writing $\theta_v$ for the angle from the vertical and $\theta_h$ for the angle from the north–south axis,$$|E_v| \propto 2I\sin\theta_v,\qquad |E_h| \propto I\sin\theta_h .$$
3. Equal-amplitude condition. Circular polarization demands the two components be equal in magnitude:$$2\sin\theta_v = \sin\theta_h .$$The right-hand side can be at most 1, so $\sin\theta_v \leq 1/2$: the direction must lie well away from the horizon, where the stronger vertical element would otherwise dominate.
4. Choose the convenient solution. Take the equality case $\sin\theta_h = 1$, which means the direction is perpendicular to the north–south element, i.e. it lies in the vertical east–west plane. Then$$\sin\theta_v = \tfrac{1}{2}\;\Longrightarrow\;\theta_v = 30^{\circ}\ \text{from the vertical},$$so the ray sits 60° above the horizon, pointing east (or west).
5. Confirm the two fields are orthogonal. Along that ray $\hat{\mathbf{r}} = (\sin 30^{\circ},\,0,\,\cos 30^{\circ})$ in (east, north, up) coordinates. The vertical element radiates a field in the plane of $\hat{\mathbf{z}}$ and $\hat{\mathbf{r}}$ — the vertical east–west plane — while the north–south element, being perpendicular to $\hat{\mathbf{r}}$, radiates a field along $\hat{\mathbf{y}}$ (due north). Those two directions are at right angles, so the components are genuinely orthogonal:$$\mathbf{E}_v \cdot \mathbf{E}_h = 0 .$$
6. Combine. Equal magnitudes ($2\sin 30^{\circ} = 1\cdot\sin 90^{\circ} = 1$ in units of $Z_0 I l k/4\pi r$), perpendicular directions, and the $90^{\circ}$ current phase difference carried straight through to the fields: the tip of the total E vector traces a circle. Hence$$\boxed{\theta = 30^{\circ}\ \text{from the vertical, in the vertical E--W plane}}$$By symmetry the same holds due west, and for the two mirror directions 30° from the downward vertical, so there are four such rays in all. The sense (right- or left-hand) is set by the sign of the 90° phase and reverses between the eastward and westward rays.