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

Problem 6.1.11

Six polarisation states

Determine the polarisations (type, handedness, and the direction of the field or of the major axis) of the waves with the field components below. $β > 0$ , lossless medium of intrinsic impedance $η$ .

$E = (j a_{x} - a_{y}) e^{j 2 z}$
$E = (a_{y} - 2 j a_{z}) e^{- j β x}$
$H = (a_{x} - j a_{z}) e^{j β y}$
$E = (a_{x} + 2 a_{y}) e^{j β z}$
$H = (2 a_{x} + j a_{y}) e^{- j β z}$
$E = (a_{y} + j a_{z}) e^{j β x}$

Adapted from: S. J. Orfanidis, Electromagnetic Waves and Antennas, Problem 2.8.

Try it yourself first.

Strategy

Read $a_{k}$ from the exponent ( $e^{- j β u}$ : towards $+ u$ , $e^{+ j β u}$ : towards $- u$ ), split the amplitude into $E_{r} + j E_{i}$ and apply the handedness test. Since $H = \frac{1}{η} a_{k} \times E$ turns rigidly with $E$ , the test can be applied to $H$ directly in c) and e).

1. Case by case

The field at $t = 0$ is $E_{r}$ , at $t = T /4$ it is $- E_{i}$ . For each wave: $a_{k}$ , the field at $t = 0$ and at $t = T /4$ , the test $(E_{r} \times (- E_{i})) \cdot a_{k}$ and the result.

a) $a_{k} = - a_{z}$ , $- a_{y}$ then $- a_{x}$ : $(a_{y} \times a_{x}) \cdot (- a_{z}) = + 1$ , RHCP.
b) $a_{k} = + a_{x}$ , $a_{y}$ then $2 a_{z}$ : $(a_{y} \times 2 a_{z}) \cdot a_{x} = + 2$ , RH elliptical, AR $= 2$ , major axis $z$ .
c) ( $H$ ) $a_{k} = - a_{y}$ , $a_{x}$ then $a_{z}$ : $(a_{x} \times a_{z}) \cdot (- a_{y}) = + 1$ , RHCP.
d) $a_{k} = - a_{z}$ , $a_{x} + 2 a_{y}$ then $0$ : $E_{i} = 0$ , linear, along $(a_{x} + 2 a_{y}) / 5$ .
e) ( $H$ ) $a_{k} = + a_{z}$ , $2 a_{x}$ then $- a_{y}$ : $(2 a_{x} \times (- a_{y})) \cdot a_{z} = - 2$ , LH elliptical, AR $= 2$ .
f) $a_{k} = - a_{x}$ , $a_{y}$ then $- a_{z}$ : $(a_{y} \times (- a_{z})) \cdot (- a_{x}) = + 1$ , RHCP.

In a), c) and f) the two vectors are perpendicular and of equal length: circular. In d) the field makes the angle $arctan 2 = 63. 4^{\circ}$ with the $x$ -axis.

2. Time domain, two examples

a) E b) E = Re {(j a_{x} - a_{y}) e^{j (ω t + 2 z)}} = - sin (ω t + 2 z) a_{x} - cos (ω t + 2 z) a_{y}, = cos (ω t - β x) a_{y} + 2 sin (ω t - β x) a_{z} .

3. The electric fields of c) and e)

$E = η H \times a_{k}$ :

c) E e) E = η (a_{x} - j a_{z}) \times (- a_{y}) e^{j β y} = - η (a_{z} + j a_{x}) e^{j β y}, = η (2 a_{x} + j a_{y}) \times a_{z} e^{- j β z} = η (j a_{x} - 2 a_{y}) e^{- j β z} .

In e) the major axis of $H$ is along $x$ , that of $E$ along $y$ .

Answer

a) RHCP, b) RH elliptical, AR = 2, c) RHCP, d) linear, e) LH elliptical, AR = 2, f) RHCP

Check

The test on $E$ in c): $E_{r} = - η a_{z}$ , $E_{i} = - η a_{x}$ , and $((- a_{z}) \times a_{x}) \cdot (- a_{y}) = (- a_{y}) \cdot (- a_{y}) = + 1$ : RH, as found from $H$ . In e): $E_{r} = - 2 η a_{y}$ , $E_{i} = η a_{x}$ , and $((- 2 a_{y}) \times (- a_{x})) \cdot a_{z} = 2 (a_{y} \times a_{x}) \cdot a_{z} = - 2$ : LH.

**Figure 6.1.34.** Problem 6.1.11, what an observer facing each wave sees. The transverse plane with the wave coming out of the page, (right axis) $\times$ (up axis) $= a_{k}$ , and the dot at $t = 0$ . Counter-clockwise is RH, clockwise LH. The phasor of a) with $e^{- j 2 z}$ would be LHCP: the hand depends on the field and on the direction of travel, which is why the test contains $a_{k}$ .

Problem 6.1.11: Six polarisation states