Question 3 of 8: Circular polarisation from two counter-propagating waves
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2014 — 07-Elec-A7 Electromagnetics. Three hours, closed book; one of two approved calculators permitted. Eight questions, all of equal value (20 marks each). The rubric states that any five questions constitute a complete paper and that only the first five presented are marked — all eight are worked here, because the set is a study resource rather than a sat examination. Aids printed on the paper: $\varepsilon_0 = 8.85\times10^{-12}\ \text{F/m}$ and $\mu_0 = 4\pi\times10^{-7}\ \text{H/m}$, from which $\eta_0=\sqrt{\mu_0/\varepsilon_0}\approx 377\ \Omega$ and $c = 3\times10^{8}\ \text{m/s}$.
Reference texts. D. M. Pozar, Microwave Engineering, 4th ed. (transmission lines, waveguides); M. N. O. Sadiku, Elements of Electromagnetics, 7th ed.; W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed. (statics, Faraday's law, plane waves); F. T. Ulaby, Fundamentals of Applied Electromagnetics, 8th ed. (polarisation, radiation); C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed. (short elements, image theory).
Question 3: Circular polarisation from two counter-propagating waves (20 marks)
Given. Two equal-strength plane waves of the same frequency travel in opposite directions along one horizontal axis with mutually orthogonal linear polarisations, and they are known to be in phase at one reference point.
Given data
Quantity
Symbol
Value
Frequency
$f$
$10\ \text{GHz}$
Power density of each wave
$S$
$10\ \text{kW/m}^{2}$
Medium
$\eta_0$
free space, $377\ \Omega$
Polarisation of wave 1
—
vertical, travels $+z$
Polarisation of wave 2
—
horizontal, travels $-z$
Reference point
$z=0$
the two fields are in phase
Find. The shortest distance from the in-phase point to a point where the resultant field is circularly polarised, and the rms amplitude of the resultant there.
Because the two waves travel in opposite directions, their relative phase advances at $2k$ per unit distance, so the polarisation state cycles linear → circular → linear every $\lambda/8$.
Approach. Write the two fields as orthogonal components with opposite propagation signs, note that their relative phase is $-2kz$, and set that phase to $90^{\circ}$. The amplitudes are already equal from the equal power densities, so the two circular-polarisation conditions are satisfied together.
Wavelength and phase constant. $$\lambda=\frac{c}{f}=\frac{3\times10^{8}}{10^{10}}=0.03\ \text{m}=3\ \text{cm},\qquad k=\frac{2\pi}{\lambda}=209.4\ \text{rad/m}.$$
Write the two fields. Taking $\hat{x}$ vertical and $\hat{y}$ horizontal, $$\mathbf{E}_1=\hat{x}E_0e^{-jkz},\qquad \mathbf{E}_2=\hat{y}E_0e^{+jkz}.$$ The relative phase of the two orthogonal components is therefore $\phi(z)=-2kz$, which is zero at the stated in-phase point $z=0$.
Impose the circular-polarisation conditions. Circular polarisation requires equal component amplitudes (automatic here, since the power densities are equal) and a phase difference of $\pm 90^{\circ}$: $$2kz=\frac{\pi}{2}\;\Rightarrow\;z=\frac{\pi}{4k}=\frac{\lambda}{8}\;\Rightarrow\;\boxed{d=\frac{\lambda}{8}=3.75\ \text{mm}} .$$
Amplitude of each wave from its power density. $$S=\frac{E_{rms}^{2}}{\eta_0}\;\Rightarrow\;E_{rms}=\sqrt{S\eta_0}=\sqrt{(10^{4})(377)}=1942\ \text{V/m}$$ per wave, whose peak value is $\sqrt{2}(1942)=2746\ \text{V/m}$.
RMS amplitude of the circular resultant. For a circularly polarised field the instantaneous magnitude is constant — it equals the peak of either component — so its rms value is that same constant. Equivalently, adding the two orthogonal components in quadrature, $$E_{rms,tot}=\sqrt{E_{rms,1}^{2}+E_{rms,2}^{2}}=\sqrt{2}\,(1942)$$ $$\Rightarrow\;\boxed{E_{rms,tot}=2746\ \text{V/m}\approx 2.75\ \text{kV/m}} .$$
Confirm the polarisation pattern. Sampling $E_x(t)=E_{pk}\cos\omega t$ and $E_y(t)=E_{pk}\cos(\omega t+90^{\circ})$ over a cycle gives a magnitude that never varies, confirming a true circle; at $z=0$ the same sampling collapses onto a $45^{\circ}$ straight line, i.e. linear polarisation.
An important feature of this problem is that no conventional standing wave forms. Two counter-propagating waves of the same polarisation would interfere and produce nulls, but these two are orthogonal, so each keeps its own constant amplitude everywhere and only the relative phase varies with position. The consequence is that the resultant amplitude is the same $2.75\ \text{kV/m}$ at every point; what changes with $z$ is the shape traced by the field vector, cycling linear at $45^{\circ}$, circular, linear at $135^{\circ}$, circular of the opposite hand, and back again with period $\lambda/4=7.5$ mm.
