Exercise 6.1.9: How circular is a “circular” antenna?

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 6.1.9

How circular is a “circular” antenna?

A satellite-navigation antenna is designed for RHCP but is in fact slightly elliptical: right-handed with axial ratio $r = AR$ , which data sheets give in dB. (a) Write its polarisation vector with circular components and show that it receives the fractions

\frac{( r + 1 ) ^{2}}{2 ( r ^{2} + 1 )} of an RHCP wave, \frac{( r - 1 ) ^{2}}{2 ( r ^{2} + 1 )} of an LHCP wave .

(b) Evaluate the co-polar loss and the response to an LHCP echo for $AR = 1 dB$ (towards the zenith) and $6 dB$ (near the horizon). (c) Which axial ratio is needed for a $20 dB$ rejection of LHCP relative to RHCP?

Try it yourself first.

Strategy

Any polarisation is $E_{R} (a_{x} - j a_{y}) + E_{L} (a_{x} + j a_{y})$ , and $AR = (∣ E_{R} ∣ + ∣ E_{L} ∣) / (∣ E_{R} ∣ - ∣ E_{L} ∣)$ for an RH state. Use $∣ p_{w} \cdot p_{a}^{*} ∣^{2}$ with unit vectors.

Inverting the axial ratio, $∣ E_{L} ∣/∣ E_{R} ∣ = (r - 1) / (r + 1)$ . Take $∣ E_{R} ∣ = r + 1$ , $∣ E_{L} ∣ = r - 1$ and an arbitrary phase $ψ$ between them, which only sets the tilt:

p_{a} N^{2} = \frac{( r + 1 ) ( a _{x} - j a _{y} ) + ( r - 1 ) e ^{j ψ} ( a _{x} + j a _{y} )}{N}, = 2 (r + 1)^{2} + 2 (r - 1)^{2} = 4 (r^{2} + 1) .

For an RHCP wave $p_{w} = (a_{x} - j a_{y}) / 2$ . Since $(a_{x} - j a_{y}) \cdot (a_{x} - j a_{y})^{*} = 2$ and $(a_{x} - j a_{y}) \cdot (a_{x} + j a_{y})^{*} = 0$ ,

∣ p_{w} \cdot p_{a}^{*} ∣^{2} = \frac{∣2 ( r + 1 ) ∣ ^{2}}{2 N ^{2}} = \frac{( r + 1 ) ^{2}}{2 ( r ^{2} + 1 )} .

For an LHCP wave only the $(r - 1)$ term survives: $(r - 1)^{2} /2 (r^{2} + 1)$ . The two add up to $1$ , as they must, and neither depends on $ψ$ .

$r = 1 0^{A R_{dB} /20}$ :

AR	$r$	RHCP wave (co-polar)	LHCP echo (cross-polar)
$1$ dB	$1.122$	$0.9967$ ( $- 0.014$ dB)	$0.0033$ ( $- 24.8$ dB)
$6$ dB	$1.995$	$0.901$ ( $- 0.45$ dB)	$0.099$ ( $- 10.0$ dB)

A rejection of $((r + 1) / (r - 1))^{2} = 100$ requires $(r - 1) / (r + 1) = 0.1$ , that is $r = 1.1/0.9 = 1.222$ :

Answer

AR \leq 20 lo g_{10} 1.222 = 1.74 dB for 20 dB rejection of LHCP

Check

For $r = 1$ (a perfect RHCP antenna) the fractions are $1$ and $0$ , and for $r \to \infty$ (a linear antenna) both tend to $\frac{1}{2}$ , the entries of Table 6.1.1.

Remark

The co-polar loss is negligible even for a poor antenna, $0.45 dB$ at $6 dB$ AR. What the axial ratio really controls is the rejection of the opposite hand. That is why GPS antenna data sheets specify the AR so carefully, and why the " $- \infty$ " of an ideal RHCP antenna for LHCP becomes $- 10$ to $- 25 dB$ in practice (Application 6.1.2): an antenna with a $6 dB$ AR at low elevation lets a ground echo through only $10 dB$ down.

cf. Problem 6.1.3