Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Problem 6.1.3

Interactive: Polarisation Explorer

Problem 6.1.3

Elliptical polarisation

The electric field of a plane wave travelling towards $+ z$ in a lossless medium is

E (z) = E_{0} (a_{x} + e^{j π /3} a_{y}) e^{- j β z} .

Trace $E (t)$ at $z = 0$ and find the sense of rotation.
Find the semi-axes, the axial ratio (also in dB) and the tilt angle of the polarisation ellipse.
Decompose the wave into an RHCP and an LHCP wave. Which fraction of the power does each carry? What would a linear antenna along $x$ receive?

Adapted from: S. J. Orfanidis, Electromagnetic Waves and Antennas, Example 2.5.1(d).

Try it yourself first.

Strategy

Read $E_{1} = E_{2} = E_{0}$ and $δ = + 6 0^{\circ}$ from the phasor, then use the ellipse formulas, Eq. (6.1.58), and the circular components, Eq. (6.1.61).

a) Points in time

At $z = 0$ , $E_{x} = E_{0} cos ω t$ and $E_{y} = E_{0} cos (ω t + 6 0^{\circ})$ :

$ω t$	$E_{x} / E_{0}$	$E_{y} / E_{0}$
$0$	$1$	$0.5$
$3 0^{\circ}$	$0.866$	$0$
$9 0^{\circ}$	$0$	$- 0.866$
$12 0^{\circ}$	$- 0.5$	$- 1$
$18 0^{\circ}$	$- 1$	$- 0.5$
$27 0^{\circ}$	$0$	$0.866$

The tip moves from the first quadrant down through the $+ x$ axis into the fourth quadrant: clockwise seen with $+ z$ pointing towards the viewer. This agrees with the rule, Eq. (6.1.56), $0 < δ < π$ : left-handed.

b) The ellipse

$E_{0}^{4} + E_{0}^{4} + 2 E_{0}^{4} cos 12 0^{\circ} = E_{0}^{2}$ , so

a^{2} = \frac{1}{2} (2 + 1) E_{0}^{2}, b^{2} = \frac{1}{2} (2 - 1) E_{0}^{2} ⟹ a = 1.225 E_{0}, b = 0.707 E_{0}, AR = \frac{a}{b} = 3 = 1.73.

In decibels $20 lo g_{10} 3 = 4.77 dB$ . The tilt: $tan 2 τ = 2 E_{0}^{2} cos 6 0^{\circ} / (E_{0}^{2} - E_{0}^{2}) \to \infty$ , so $2 τ = 9 0^{\circ}$ and $τ = 4 5^{\circ}$ . The major axis lies on the line $y = x$ .

c) Two circles

With $j e^{j 6 0^{\circ}} = e^{j 15 0^{\circ}}$ ,

E_{R} E_{L} = \frac{E _{0}}{2} (1 + e^{j 15 0^{\circ}}) = E_{0} sin 1 5^{\circ} e^{j 7 5^{\circ}} = 0.259 E_{0} e^{j 7 5^{\circ}}, = \frac{E _{0}}{2} (1 - e^{j 15 0^{\circ}}) = E_{0} cos 1 5^{\circ} e^{- j 1 5^{\circ}} = 0.966 E_{0} e^{- j 1 5^{\circ}} .

Both circular waves $E_{R, L} (a_{x} \mp j a_{y})$ have $∣ a_{x} \mp j a_{y} ∣^{2} = 2$ , so the power splits as $∣ E_{R} ∣^{2} : ∣ E_{L} ∣^{2}$ :

\frac{P _{R}}{P} \frac{P _{L}}{P} = \frac{∣ E _{R} ∣ ^{2}}{∣ E _{R} ∣ ^{2} + ∣ E _{L} ∣ ^{2}} = sin^{2} 1 5^{\circ} = 6.7 %, = cos^{2} 1 5^{\circ} = 93.3 %.

An $x$ -directed linear antenna sees only $E_{x}$ : $E_{1}^{2} / (E_{1}^{2} + E_{2}^{2}) = 50 %$ .

Answer

LH elliptical, AR = 3 (4.77 dB), τ = 4 5^{\circ}, 93.3 % LHCP + 6.7 % RHCP

Turn on "RHCP + LHCP parts" below: the big left-handed circle of radius $0.966 E_{0}$ and the small right-handed one of radius $0.259 E_{0}$ add up to the ellipse.

InteractiveLoading simulation…

Check

AR = \frac{∣ E _{L} ∣ + ∣ E _{R} ∣}{∣ E _{L} ∣ - ∣ E _{R} ∣} = \frac{0.966 + 0.259}{0.966 - 0.259} = \frac{1.225}{0.707} = 3,

and $∣ E_{L} ∣ > ∣ E_{R} ∣$ confirms the left hand.

**Figure 6.1.26.** Problem 6.1.3. Left: the trace at $z = 0$ with the values of $ω t$ marked, $+ z$ out of the page. The ellipse, with $a = 1.225 E_{0}$ along $y = x$ and $b = 0.707 E_{0}$ , is the sum of two counter-rotating circles: the large LH one of radius $0.966 E_{0}$ and the small RH one of radius $0.259 E_{0}$ . Where they line up the field is $a = 0.966 + 0.259$ , where they oppose it is $b = 0.966 - 0.259$ . Right: the fraction of the power received by matched antennas. An LHCP antenna loses only $10 lo g_{10} 0.933 = - 0.30 dB$ , an RHCP one $- 11.7 dB$ .

Problem 6.1.3: Elliptical polarisation