Question 4 of 8: Polarisation of Two Crossed Plane Waves

Question 4: Polarisation of Two Crossed Plane Waves (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 equal-strength 10 GHz plane waves cross at right angles in the horizontal plane, one polarised horizontally and the other vertically, and at the reference point A their sum happens to be linearly polarised.

Find. (i) the rms magnitude of the total electric field at A, and (ii) the position of the nearest point at which the sum becomes circularly polarised.

Approach. Identify the spatial directions of the two electric-field vectors (they turn out to be mutually perpendicular), which makes the rms magnitudes add in quadrature everywhere; then track how the relative phase of the two waves varies with position and find the shortest step that advances it by a quarter cycle.

1. Field magnitude of a single wave. For a plane wave in free space the time-average power density and the rms field are related by \(S=E_{\text{rms}}^{2}/\eta_{0}\), so $$E_{\text{rms}}=\sqrt{S\eta_{0}}=\sqrt{(1)(377)}=19.42\ \text{V/m}$$ Each wave contributes this much on its own.
2. Establish the direction of each electric field. A plane wave’s electric field is perpendicular to its direction of travel. The north-going wave, being horizontally polarised, must therefore have its field along the east–west axis; the east-going wave, being vertically polarised, has its field along the vertical. Those two directions are mutually perpendicular, which is the observation the rest of the question hangs on.
3. Add the two fields at A. Because the vectors are orthogonal, the instantaneous magnitude satisfies \(|\mathbf{E}|^{2}=E_{1}^{2}+E_{2}^{2}\) with no cross term, and averaging over a cycle gives $$E_{\text{tot,rms}}=\sqrt{E_{1,\text{rms}}^{2}+E_{2,\text{rms}}^{2}}=\sqrt{2S\eta_{0}}=\sqrt{2(1)(377)}=\boxed{27.46\ \text{V/m}}$$ The statement that A is linearly polarised tells us the relative phase there is \(0^{\circ}\) or \(180^{\circ}\), but it does not affect this number: two orthogonal polarisations cannot interfere, so the rms total is 27.46 V/m at every point in the region, not just at A.
4. Write the relative phase as a function of position. Take \(x\) due east and \(y\) due north. The north-going wave carries the factor \(e^{-jky}\) and the east-going wave \(e^{-jkx}\), so their phase difference at a general point is $$\Delta\phi=\phi_{1}-\phi_{2}=-ky-(-kx)=k\,(x-y)$$ At A this equals \(0^{\circ}\) or \(180^{\circ}\), which is what makes the sum linear there.
5. State the condition for circular polarisation. Two orthogonal components produce a circle when they are equal in amplitude and in phase quadrature. The amplitudes are already equal (both waves carry 1 W/m²), so the requirement is purely on phase: $$|\Delta\phi|=\frac{\pi}{2}\quad\Longrightarrow\quad k\left|\Delta x-\Delta y\right|=\frac{\pi}{2}\quad\Longrightarrow\quad\left|\Delta x-\Delta y\right|=\frac{\lambda}{4}$$
6. Minimise the distance subject to that condition. The quantity \(x-y\) changes fastest along the direction \((1,-1)/\sqrt{2}\), that is along the south-east/north-west axis, and does not change at all along the north-east/south-west axis. Choosing the fastest direction and moving \(\Delta x=+\lambda/8\), \(\Delta y=-\lambda/8\) satisfies the condition with the shortest step: $$d_{\min}=\sqrt{\left(\frac{\lambda}{8}\right)^{2}+\left(\frac{\lambda}{8}\right)^{2}}=\frac{\lambda}{4\sqrt{2}}$$
7. Insert the wavelength. At 10 GHz, $$\lambda=\frac{c}{f}=\frac{3.00\times10^{8}}{1.00\times10^{10}}=3.00\ \text{cm}\qquad\Longrightarrow\qquad d_{\min}=\frac{0.0300}{4\sqrt{2}}=\boxed{5.30\ \text{mm}}$$ The nearest circularly polarised point therefore lies 5.30 mm from A along the north-west direction, or equally 5.30 mm to the south-east; the two give opposite senses of rotation. Moving a further 5.30 mm the same way returns the field to linear (with the orthogonal tilt), so linear and circular states alternate every 5.30 mm along that axis and repeat every 21.2 mm.