Section 6.1 · Problems · 1 of 11

Problem 6.1.1: Three polarisations

Interactive: Polarisation Explorer

Problem 6.1.1

Three polarisations

Find the polarisations of the plane waves with the given field components:

$E (z) = E_{0} (a_{x} - j a_{y}) e^{- j β z}$ ,
$H (x) = H_{0} (a_{y} + j a_{z}) e^{j β x}$ ,
$E (y) = E_{0} (a_{x} - a_{z}) e^{- j β y}$ .

$E_{0}$ and $H_{0}$ are real and positive, and the medium is lossless with intrinsic impedance $η$ . Use the algorithm of Section 6.1.6 and the IEEE convention.

Try it yourself first.

Strategy

Follow the recipe: (1) get $E$ , from $E = - η a_{k} \times H$ if $H$ is given, and read $a_{k}$ from the exponent, (2) go to the time domain, (3) take the point where the space coordinate is zero, (4) follow the tip of $E$ from $ω t = 0$ to $ω t = π /2$ , (5) compare the sense of rotation with $a_{k}$ .

a) 1. Direction

$e^{- j β z}$ : the wave travels towards $+ z$ , $a_{k} = a_{z}$ , and $a_{z} \cdot (a_{x} - j a_{y}) = 0$ , as it must be for a transverse wave.

2. Time domain

With $- j = e^{- j π /2}$ ,

E (z; t) = Re {E_{0} (a_{x} - j a_{y}) e^{- j β z} e^{j ω t}} = E_{0} [cos (ω t - β z) a_{x} + sin (ω t - β z) a_{y}] .

3. At

z = 0

$E (0; t) = E_{0} (cos ω t a_{x} + sin ω t a_{y})$ .

4. The tip

$∣ E ∣ = E_{0}$ at every instant: the tip runs round a circle, from $E_{0} a_{x}$ at $ω t = 0$ to $E_{0} a_{y}$ at $ω t = π /2$ , $- E_{0} a_{x}$ at $π$ and $- E_{0} a_{y}$ at $3 π /2$ .

5. The hand

The rotation axis is $a_{x} \times a_{y} = a_{z} = a_{k}$ : with the right thumb along $+ z$ the fingers curl from $x$ to $y$ .

Answer

(a) right-hand circular polarisation (RHCP), towards + z

b) Direction and

E

$e^{+ j β x}$ : the wave travels towards $- x$ , $a_{k} = - a_{x}$ . With $a_{x} \times a_{y} = a_{z}$ and $a_{x} \times a_{z} = - a_{y}$ ,

E = - η a_{k} \times H = η H_{0} a_{x} \times (a_{y} + j a_{z}) e^{j β x} = η H_{0} (a_{z} - j a_{y}) e^{j β x} .

Time domain and

x = 0

$E (x; t) = η H_{0} [cos (ω t + β x) a_{z} + sin (ω t + β x) a_{y}]$ . At $x = 0$ the tip goes from $η H_{0} a_{z}$ ( $ω t = 0$ ) to $η H_{0} a_{y}$ ( $ω t = π /2$ ) on a circle of radius $η H_{0}$ . The rotation axis $a_{z} \times a_{y} = - a_{x}$ is $a_{k}$ : right hand.

Answer

(b) RHCP, towards - x

Check

$H$ itself turns from $H_{0} a_{y}$ ( $ω t = 0$ ) to $- H_{0} a_{z}$ ( $ω t = π /2$ ), about $a_{y} \times (- a_{z}) = - a_{x}$ : the same answer, as it must be, since $E$ and $H$ always rotate together.

c) Direction

$e^{- j β y}$ : towards $+ y$ , $a_{k} = a_{y}$ , and $(a_{x} - a_{z}) \cdot a_{y} = 0$ . There is no $j$ : both components are in phase,

E (y; t) = E_{0} cos (ω t - β y) (a_{x} - a_{z}) .

The direction $(a_{x} - a_{z}) / 2$ never changes. Only the length does, between $+ 2 E_{0}$ and $- 2 E_{0}$ : the tip moves on a line, at $4 5^{\circ}$ between $+ x$ and $- z$ . The magnetic field completes the triad:

H = \frac{1}{η} a_{y} \times E = - \frac{E _{0}}{η} (a_{x} + a_{z}) e^{- j β y} .

Answer

(c) linear polarisation along \frac{a _{x} - a _{z}}{2}, towards + y

Pick (a), (b) and (c) in the menu below: in (a) and (b) the field turns counter-clockwise for you, facing the wave, so both are RHCP, while in (c) it stays on a line.

InteractiveLoading simulation…

Check

$(a_{x} - a_{z}) \times (- (a_{x} + a_{z})) = 2 a_{y} = 2 a_{k}$ : $E \times H$ points along $a_{k}$ . The handedness test confirms (a) at once: $E_{r} = E_{0} a_{x}$ , $E_{i} = - E_{0} a_{y}$ , $(a_{x} \times a_{y}) \cdot a_{z} = + 1$ , RH.

**Figure 6.1.24.** Problem 6.1.1, what an observer facing each wave sees. Each panel shows the transverse plane with the wave coming out of the page, (right axis) $\times$ (up axis) $= a_{k}$ , and the dot marks $ω t = 0$ . There a right-handed field turns counter-clockwise. The hand is always judged against $a_{k}$ : the phasor of (b) with $e^{- j β x}$ instead of $e^{+ j β x}$ would turn the same way in time but be LHCP.

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