Exercise 6.1.10: Faraday rotation as two circular waves

About the further exercises of Section 6.1

Seventeen further exercises in the spirit of the problems, the home exercises and the worked examples, graded from easy to harder within each group. Exercises 6.1.1–6.1.4 are about uniform plane waves, $k$ , $λ$ , $η$ and the triad, Exercises 6.1.5–6.1.6 about power, energy and loss, Exercises 6.1.7–6.1.10 about polarisation, Exercises 6.1.11–6.1.12 about the Doppler effect, and Exercises 6.1.13–6.1.17 mix power, exposure and polarisation. Besides the relations listed at the head of this subsection they use the lossy forms $k = ω μ ε_{c} = β - j α$ , $ε_{c} = ε (1 - j σ / ω ε)$ , $η = μ / ε_{c} = ∣ η ∣ e^{j θ_{η}}$ and $P_{a v} = \frac{∣ E _{0} ∣ ^{2}}{2∣ η ∣} e^{- 2 α a_{k} \cdot r} cos θ_{η} a_{k}$ (Exercise 6.1.6 only), the average energy density $w_{a v} = \frac{1}{4} (ε ∣ E ∣^{2} + μ ∣ H ∣^{2})$ , Faraday's law $E = - j ω Φ$ , and the polarisation match $∣ p_{w} \cdot p_{a}^{*} ∣^{2}$ . A field factor $e^{- α z}$ is $8.686 α z dB$ .

Exercise 6.1.10

Faraday rotation as two circular waves

Along the Earth's magnetic field the ionosphere is circularly birefringent: an RHCP and an LHCP wave travel through it unchanged, but with slightly different phase constants $β_{R}$ and $β_{L}$ . A satellite at the top of a path of length $L$ sends the linear wave $E (0) = E_{0} a_{x}$ , propagating towards $+ z$ . (a) Write $E (0)$ as a sum of an RHCP and an LHCP wave, propagate both to $z$ , and show that the field is still linearly polarised, its plane turned by $Ω = (β_{L} - β_{R}) z /2$ from $x$ towards $y$ . (b) At $400 MHz$ the extra phase of the LHCP wave over the path is $(β_{L} - β_{R}) L = 3.0 rad$ . Find $Ω$ and the power a linear receiving antenna aligned with $x$ collects, in dB. (c) The extra phase scales as $1/ f^{2}$ . Repeat for $1.6 GHz$ . (d) What would an RHCP receiving antenna collect in each case, and what if the satellite sent RHCP?

Try it yourself first.

Strategy

A linear wave is the sum of two equal circular waves of opposite hands. If the medium delays one of them more than the other, their sum is still a line, turned by half the phase difference.

$a_{x} = \frac{1}{2} [(a_{x} - j a_{y}) + (a_{x} + j a_{y})]$ , so with $\overset{ˉ}{β} = \frac{1}{2} (β_{R} + β_{L})$ and $Δ = \frac{1}{2} (β_{L} - β_{R})$ ,

E (z) = \frac{E _{0}}{2} [(a_{x} - j a_{y}) e^{- j β_{R} z} + (a_{x} + j a_{y}) e^{- j β_{L} z}] = \frac{E _{0}}{2} e^{- j \overset{ˉ}{β} z} [(a_{x} - j a_{y}) e^{j Δ z} + (a_{x} + j a_{y}) e^{- j Δ z}] .

Collecting terms with $e^{j Δ z} + e^{- j Δ z} = 2 cos Δ z$ and $- j (e^{j Δ z} - e^{- j Δ z}) = 2 sin Δ z$ :

Answer

E (z) Ω = E_{0} (cos Ω a_{x} + sin Ω a_{y}) e^{- j \overset{ˉ}{β} z}, = Δ z = \frac{( β _{L} - β _{R} ) z}{2}

The vector multiplying $e^{- j \overset{ˉ}{β} z}$ is real: at every $z$ the wave is linear, along a direction that turns steadily with distance.

$Ω = 3.0/2 = 1.5 rad = 85. 9^{\circ}$ . A linear antenna along $x$ receives $cos^{2} Ω$ of the power:

cos^{2} 85. 9^{\circ} = 0.0050 (- 23.0 dB), almost crossed .

At $1.6 GHz$ the extra phase is $3.0 \times (400/1600)^{2} = 0.1875 rad$ , $Ω = 0.094 rad = 5.3 7^{\circ}$ , and $cos^{2} Ω = 0.991$ ( $- 0.04 dB$ ).

A linear wave is half RHCP and half LHCP in power, and the ionosphere changes only the relative phase of the two halves: an RHCP antenna always receives $\frac{1}{2}$ ( $- 3 dB$ ), at both frequencies and at every time of day. If the satellite sends RHCP, there is only the RH wave, which arrives as RHCP with nothing but a phase delay: $0 dB$ into an RHCP antenna.

Answer

400 MHz : Ω = 85. 9^{\circ}, - 23.0 dB 1.6 GHz : Ω = 5. 4^{\circ}, - 0.04 dB CP link: no rotation loss

Check

With $β_{L} = β_{R}$ the formula gives $Ω = 0$ : no rotation in an ordinary medium.

Remark

Because the rotation varies with the electron content of the ionosphere, that is with the time of day, the season and the solar cycle, a linear link at UHF can fade by tens of dB at unpredictable times. At L-band the effect is small but not negligible, and the cure is the one of Application 6.1.2: circular polarisation, whose states are the natural modes of the medium and suffer no rotation at all. The ionosphere itself is a plasma, the subject of Section 6.2.

cf. Problem 6.1.3