Question 8 of 8: Where two quadrature current elements radiate a linearly polarised field

Question 8: Where two quadrature current elements radiate a linearly polarised field

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 at right angles, driven in quadrature with equal current amplitudes.

Find. The locus on the 1 km sphere where the resultant field is linearly polarised, and the rms amplitude of the field there.

Approach. Calibrate each element's field from its stated maximum power density, write the resultant as an in-phase vector plus a quadrature vector, and demand that the two be collinear — the condition for linear polarisation.

1. Calibrate the fields and check the model. $\lambda = c/f = 60\ \text{m}$, so $h/\lambda = 1/60 = 0.017$: both elements are electrically short and each has the elementary $\sin\theta$ pattern about its own axis, $\theta$ measured from that axis. The sphere at $1\ \text{km} = 16.7\lambda$ is comfortably in the far field. Each element's own maximum, at broadside to itself, is$$E_m = \sqrt{S_{max}\,\eta_0} = \sqrt{10^{-7} \times 376.99} = 6.14\ \text{mV/m rms}$$The stated length and frequency serve only to license the short-element pattern; every antenna constant is already folded into $S_{max}$.
2. Write the resultant as two vectors. A short element along $\hat{u}$ radiates a field parallel to the transverse part of its own axis, $\hat{u}_{\perp} = \hat{u} - (\hat{u}\cdot\hat{r})\hat{r}$, whose length is $\sin\theta$. Putting the vertical element along $\hat{z}$, the horizontal one along $\hat{x}$, and the $90^\circ$ current phase into a factor $j$,$$\mathbf{E}_{tot} = E_m\,\hat{z}_{\perp} + jE_m\,\hat{x}_{\perp}$$so the in-phase part points along $\hat{z}_{\perp}$ and the quadrature part along $\hat{x}_{\perp}$.
3. Impose linear polarisation. A phasor of the form $\mathbf{a} + j\mathbf{b}$ describes a straight-line oscillation exactly when $\mathbf{a}$ and $\mathbf{b}$ are collinear, i.e. when $\hat{z}_{\perp} \times \hat{x}_{\perp} = 0$. Expanding the two projections for a general direction $\hat{r}$ gives the compact identity$$\hat{z}_{\perp} \times \hat{x}_{\perp} = (\hat{r}\cdot\hat{y})\,\hat{r}$$which vanishes if and only if $\hat{r}\cdot\hat{y} = 0$.
4. Name the locus. $\hat{y}$ is the direction perpendicular to both element axes, so the condition says $\hat{r}$ must lie in the plane spanned by the two axes:$$\boxed{\text{the great circle in the vertical plane containing both element axes}}$$For a vertical element and a horizontal element running, say, east–west, that is the whole vertical east–west great circle. It is a complete circle, not just the two element axes — those axes are merely the special points on it where one component vanishes altogether.
5. Find the amplitude on that circle. Take $\hat{r} = (\sin\theta, 0, \cos\theta)$ with $\theta$ measured from the vertical. The vertical element contributes $E_m\sin\theta$ and the horizontal element $E_m\cos\theta$ (its own angle from the horizontal axis is $90^\circ - \theta$), and both lie along the same tangent to the circle. Two collinear phasors in quadrature add in quadrature, so$$E_{rms}(\theta) = \sqrt{(E_m\sin\theta)^{2} + (E_m\cos\theta)^{2}} = \boxed{E_m = 6.14\ \text{mV/m rms}}$$independent of $\theta$. The rms field is constant all the way round the circle, and equals each element's own individual maximum. The power density there is $E_m^{2}/\eta_0 = 10^{-7}\ \text{W/m}^{2}$ — the same as one element's maximum, not twice it.
6. Contrast with the rest of the sphere. Broadside to both elements ($\hat{r} = \pm\hat{y}$, due north or south for an east–west horizontal element) the two contributions are equal in magnitude, mutually perpendicular and $90^\circ$ apart in phase: that direction is circularly polarised, with $E_{rms} = \sqrt{2}E_m = 8.68\ \text{mV/m}$ and $2 \times 10^{-7}\ \text{W/m}^{2}$. Everywhere else the state is elliptical. So the linear-polarisation circle is also where the total power density is at its lowest.