Question 3 of 8: Spacing and amplitude of the circularly polarised planes formed by two crossed beams
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, December 2017 — 16-Elec-A7 Electromagnetics. Three hours, closed book (one Casio or Sharp approved calculator). Eight questions of equal value; the rubric states that any five questions constitute a complete paper and that only the first five answered will be marked. All eight are solved here, because the set is a study resource rather than a sitting. Aids printed on the paper: $\varepsilon_0 = 8.85 \times 10^{-12}\ \text{F/m}$ and $\mu_0 = 4\pi \times 10^{-7}\ \text{H/m}$.
Reference texts. D. M. Pozar, Microwave Engineering, 4th ed. (transmission-line transients, stub networks, waveguides); W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed. (plane waves, polarisation, Biot–Savart, inductance); M. N. O. Sadiku, Elements of Electromagnetics, 7th ed. (magnetostatics, guided waves); F. T. Ulaby and U. Ravaioli, Fundamentals of Applied Electromagnetics, 8th ed. (wave polarisation, radiation); C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed. (short-element fields).
Constants used throughout. The aid sheet's $\varepsilon_0$ and $\mu_0$ give $c = 2.9986 \times 10^{8}\ \text{m/s}$ and $\eta_0 = 376.82\ \Omega$, while every round figure the paper quotes (10 km at $3 \times 10^{8}\ \text{m/s}$, 300 and 600 MHz, 10 GHz giving a 3 cm wavelength) is built on $c = 3.00 \times 10^{8}\ \text{m/s}$ and $\eta_0 = 120\pi = 376.99\ \Omega$. Those are the values used below; the two sets differ by less than 0.05 % and no answer or conclusion changes.
Question 3: Spacing and amplitude of the circularly polarised planes formed by two crossed beams
Given. Two equal-strength 10 GHz beams cross symmetrically about north, one horizontally and one vertically polarised.
Given data
Quantity
Symbol
Value
Frequency
$f$
$10\ \text{GHz}$
Power density of each wave
$S$
$10\ \text{W/m}^{2}$
Bearing of wave 1 (horizontally polarised)
—
$30^\circ$ east of north
Bearing of wave 2 (vertically polarised)
—
$30^\circ$ west of north
Half-angle between the beams
$\alpha$
$30^\circ$
Free-space impedance
$\eta_0$
$120\pi = 377\ \Omega$
Find. The spacing of the vertical north–south planes on which the resultant is circularly polarised, and the amplitude of that circular field.
Figure 3.1 — plan view of the two beams and the circularly polarised north-south planes. Only the relative phase varies in space, and it varies with easting alone.
Approach. Confirm the two field vectors are orthogonal and equal in amplitude, so circular polarisation only needs a quarter-cycle of relative phase; then find where in space that phase difference occurs, and read the amplitude off the power density.
Get the wavelength and the field strengths. $\lambda = c/f = (3 \times 10^{8})/10^{10} = 0.030\ \text{m} = 3.0\ \text{cm}$, and from $S = E_{rms}^{2}/\eta_0$,$$E_{rms} = \sqrt{S\,\eta_0} = \sqrt{10 \times 376.99} = 61.40\ \text{V/m}, \qquad E_{pk} = \sqrt{2}\,E_{rms} = 86.83\ \text{V/m}$$for each wave separately.
Set up unit vectors. Working in east–north–up coordinates $(\hat{x}, \hat{y}, \hat{z})$, the propagation directions are $\hat{k}_1 = (\sin 30^\circ, \cos 30^\circ, 0)$ and $\hat{k}_2 = (-\sin 30^\circ, \cos 30^\circ, 0)$. Wave 1 is horizontally polarised, so its field lies along $\hat{e}_1 = \hat{z} \times \hat{k}_1 = (-\cos 30^\circ, \sin 30^\circ, 0)$; wave 2 is vertically polarised, so $\hat{e}_2 = \hat{z}$.
Check that circular polarisation is even possible. $\hat{e}_1 \cdot \hat{e}_2 = 0$: the two field vectors are orthogonal, and the two amplitudes are equal because the power densities are equal. Those are exactly the two amplitude conditions for a circular resultant, so the only remaining variable is the relative phase. It also means the beams cannot interfere: there is no intensity standing wave anywhere, and the total power density is $2S = 20\ \text{W/m}^{2}$ everywhere.
Find how the relative phase varies in space. Wave $i$ carries the phase $-k\,\hat{k}_i \cdot \mathbf{r}$, so the phase difference is$$\Delta\phi(\mathbf{r}) = -k\,(\hat{k}_1 - \hat{k}_2) \cdot \mathbf{r} = -k\,(2\sin\alpha)\,x$$because $\hat{k}_1 - \hat{k}_2 = (2\sin\alpha, 0, 0)$ points due east. The relative phase depends on the easting $x$ and on nothing else — which is precisely why the surfaces of constant polarisation state are vertical north–south planes, as the question states.
Locate the circular planes. Circular polarisation requires $|\Delta\phi| = 90^\circ$ modulo $180^\circ$, so $k(2\sin\alpha)x = \pi/2 + n\pi$, giving planes at $x = \lambda/4 + n\lambda/2$ once $2\sin 30^\circ = 1$ is substituted. Successive planes are therefore$$\boxed{\Delta x = \frac{\pi}{k\,(2\sin\alpha)} = \frac{\lambda}{2} = 1.5\ \text{cm apart}}$$with the first one $0.75\ \text{cm}$ from the crossing point. Note that the handedness alternates from plane to plane (the phase is $+90^\circ$ then $-90^\circ$), so planes of the same sense are $3.0\ \text{cm}$ apart.
Get the amplitude of the circular field. Where two equal orthogonal components are in quadrature the tip of the resultant traces a circle whose radius is the peak of one component, and the magnitude is then constant in time:$$|\mathbf{E}_{tot}| = E_{pk} = \sqrt{2 S \eta_0} = \boxed{86.8\ \text{V/m}}$$Because the magnitude never varies, this is simultaneously the peak and the rms value of the resultant — there is no second division by $\sqrt{2}$. As a check, $|\mathbf{E}_{tot}|^{2}/\eta_0 = 86.83^{2}/376.99 = 20.0\ \text{W/m}^{2}$, the sum of the two incident power densities.
Check: the $2\sin\alpha$ slip factor equals exactly 1 at $\alpha = 30^\circ$, which makes the plane spacing coincide numerically with $\lambda/2$. That is an accident of this geometry, not a general result — at $\alpha = 20^\circ$ the factor is 0.684 and the spacing would be $0.731\lambda$. Always carry the $(\hat{k}_1 - \hat{k}_2)$ magnitude explicitly.
