Question 7 of 8: Counter-propagating orthogonal waves: distance to circular polarisation
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, May 2015 —
07-Elec-A7 Electromagnetics. Three hours, closed book; one of two approved
calculators (Casio or Sharp) permitted. Eight questions, all of equal value; the
paper’s note 3 states that any five questions constitute a complete paper
and only the first five presented will be marked. Because this set is a study
resource, all eight questions are solved here. The permitted aids
printed on the cover are used throughout:
$\varepsilon_0 = 8.85\times10^{-12}\ \text{F/m}$ and
$\mu_0 = 4\pi\times10^{-7}\ \text{H/m}$, giving
$\eta_0 = \sqrt{\mu_0/\varepsilon_0} = 376.6\ \Omega$ and
$c = 1/\sqrt{\mu_0\varepsilon_0} \approx 3\times10^{8}\ \text{m/s}$.
Reference texts. D. M. Pozar, Microwave Engineering, 4th ed.
(transmission-line transients, matching, waveguides); M. N. O. Sadiku,
Elements of Electromagnetics, 7th ed. (magnetostatics, Faraday’s law,
plane waves); W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed.
(polarisation, boundary conditions); C. A. Balanis, Antenna Theory: Analysis
and Design, 4th ed. (short elements, image theory); F. T. Ulaby and U. Ravaioli,
Fundamentals of Applied Electromagnetics, 8th ed. (waves and power density).
Given. Two 10 GHz plane waves of equal power density 1 $\mu\text{W/m}^2$ travel in opposite directions along one horizontal axis; one is vertically polarised, the other horizontally, and at the reference point their resultant is linearly polarised (i.e. they are in phase there).
Given data
Quantity
Symbol
Value
Frequency
$f$
$10\ \text{GHz}$
Power density of each wave
$S$
$1\ \mu\text{W/m}^2$
Intrinsic impedance of free space
$\eta_0$
$376.6\ \Omega$
Polarisations
—
one vertical, one horizontal (orthogonal)
Propagation directions
—
anti-parallel, along the same horizontal axis
State at the reference point
—
linear (components in phase)
Find. The shortest distance and the direction from that point to a point of circular polarisation, and the amplitude of the field there.
Figure 7.1 — polarisation state of the resultant along the propagation axis. Circular polarisation recurs every λ/4, with the handedness alternating.
Approach. Write both waves with their own propagation signs, form the relative phase as a function of position, set it to 90$^\circ$, and separately convert the given power density into a field amplitude.
Field amplitude of each wave. For a plane wave in free space the time-average Poynting magnitude is $S = E_{\text{rms}}^2/\eta_0$, so $$E_{\text{rms}} = \sqrt{S\,\eta_0} = \sqrt{(1\times10^{-6})(376.6)} = 1.941\times10^{-2}\ \text{V/m} = 19.41\ \text{mV/m}$$$$E_{0} = \sqrt{2}\,E_{\text{rms}} = 27.45\ \text{mV/m}\ \ \text{(peak)}$$ Both waves have the same amplitude because their power densities are equal.
Relative phase along the axis — the key step. Let the axis be $z$, with the vertical wave travelling in $+z$ and the horizontal wave in $-z$: $$\mathbf{E} = \hat{x}\,E_0 e^{-jkz} + \hat{y}\,E_0 e^{+jkz}$$ taking the reference point (where they are in phase, hence linearly polarised at 45$^\circ$) as $z=0$. The phase difference is $$\Delta\phi = (+kz) - (-kz) = 2kz$$ It accumulates at twice the usual rate, because the two waves march in opposite directions.
Condition for circular polarisation. Two equal-amplitude orthogonal components in quadrature trace a circle, so we need $|\Delta\phi| = \pi/2$: $$2kz = \frac{\pi}{2} \;\Longrightarrow\; z = \frac{\pi}{4k} = \frac{\lambda}{8}$$$$z = \frac{0.030}{8} = \boxed{3.75\ \text{mm}}$$
Direction. The move is along the common propagation axis, and either sense works: moving 3.75 mm one way gives $\Delta\phi = +90^\circ$ (say right-hand circular), and 3.75 mm the other way gives $-90^\circ$ (left-hand circular). Both are the same shortest distance; the two points differ only in handedness. Moving perpendicular to the axis changes nothing, since neither wave’s phase depends on the transverse coordinates.
Amplitude of the resultant. At the circular-polarisation point $$\mathbf{E}(t) = \hat{x}\,E_0\cos\omega t + \hat{y}\,E_0\sin\omega t \;\Longrightarrow\; |\mathbf{E}| = E_0 \ \text{(constant in time)}$$ so $$\boxed{|\mathbf{E}| = 27.4\ \text{mV/m}}$$ which is also its rms value, since the magnitude does not vary. Equivalently, $E_{\text{rms,total}} = \sqrt{2}\,E_{\text{rms,component}} = \sqrt{2}(19.41) = 27.4\ \text{mV/m}$.
Two features of this configuration are worth noting. First, because the two waves are orthogonally polarised they never interfere in intensity: each component keeps a constant amplitude everywhere, and only the traced figure changes with position — there is no standing-wave pattern of maxima and nulls. Second, the state repeats every $\lambda/4 = 7.5\ \text{mm}$, cycling linear, circular, linear (orthogonal to the first), circular of the opposite hand, and back. Dividing the resultant by $\sqrt{2}$ a second time to “get the rms” is the classic error here: for a circularly polarised field the magnitude is already time-invariant.
