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

Problem 7.6.1

TE modes of a dielectric sheet

A lossless dielectric sheet with $ε_{r} = 5$ , $μ = μ_{0}$ , of thickness $d = 2 cm$ in air guides a wave of $f = 10 GHz$ along $z$ .

Find NA, $V$ , the cutoff frequencies, and the TE and TM modes that propagate.
Solve the eigenvalue equations of the TE modes (graphically, then numerically). For TE $_{0}$ find $h$ , $α$ , $τ$ , $n_{eff}$ , $λ_{g}$ and the ray angle $θ$ , and compare with $θ_{c}$ .
For TE $_{0}$ : the field at the surface relative to the centre, the depth of the tail, and the fraction of the power inside the sheet.
In which frequency band, or below which thickness at $10 GHz$ , is the sheet single-mode?

Try it yourself first.

Strategy

Everything is in normalised form: compute $V$ once, count the branches it reaches, and find each root $(u, w)$ on the circle $u^{2} + w^{2} = V^{2}$ . The physical quantities then follow from $h = 2 u / d$ , $α = 2 w / d$ and $τ = k_{0}^{2} n_{1}^{2} - h^{2}$ .

1. Bookkeeping

$n_{1} = 5 = 2.236$ , $n_{2} = 1$ , $d = 0.02 m$ . At $10 GHz$ , $λ_{0} = 3 cm$ and $k_{0} = 2 π / λ_{0} = 209.4 rad/m$ , so $k_{0}^{2} n_{1}^{2} = 2.193 \times 1 0^{5} (rad/m)^{2}$ .

$NA = 5 - 1 = 2$ and

V = \frac{π d}{λ _{0}} NA = \frac{π \times 2}{3} \times 2 = \frac{4 π}{3} = 4.189, \frac{2 V}{π} = 2.67.

The TE and TM branches with $mπ /2 < V$ , that is $m = 0, 1, 2$ , are reached. The cutoffs are

f_{c, m} = \frac{m c}{2 d NA} = \frac{m \times 3 \times 1 0 ^{8}}{2 \times 0.02 \times 2} = 3.75 m GHz : 0, 3.75, 7.5 and 11.25 GHz .

At $10 GHz$ TE $_{0}$ , TE $_{1}$ , TE $_{2}$ and TM $_{0}$ , TM $_{1}$ , TM $_{2}$ propagate. TE $_{3}$ and TM $_{3}$ need $11.25 GHz$ .

Graphically the quarter circle of radius $4.189$ cuts the branch $u tan u$ on $(0, π /2)$ , the branch $- u cot u$ on $(π /2, π)$ and again $u tan u$ on $(π, 3 π /2)$ (Figure 7.6.14a). Bisection on each branch gives

	$u$	$w$
TE $_{0}$	$1.264$	$3.994$
TE $_{1}$	$2.502$	$3.360$
TE $_{2}$	$3.653$	$2.050$

	$τ$ [rad/m]	$n_{eff}$	$θ$
TE $_{0}$	$450.9$	$2.153$	$74. 3^{\circ}$
TE $_{1}$	$395.9$	$1.890$	$57. 7^{\circ}$
TE $_{2}$	$293.1$	$1.399$	$38. 7^{\circ}$

For TE $_{0}$ :

h α τ n_{eff} = \frac{2 u}{d} = \frac{2 \times 1.264}{0.02} = 126.4 rad/m, = \frac{2 w}{d} = 399.3 Np/m, = 2.193 \times 1 0^{5} - 126. 4^{2} = 450.9 rad/m, = \frac{τ}{k _{0}} = 2.153, λ_{g} = \frac{2 π}{τ} = 13.9 mm .

The ray angle follows from $τ = k_{0} n_{1} sin θ$ : $sin θ = 2.153/2.236 = 0.963$ , $θ = 74. 3^{\circ}$ , far beyond the critical angle $θ_{c} = arcsin (1/ 5) = 26. 6^{\circ}$ . TE $_{1}$ and TE $_{2}$ (at $57. 7^{\circ}$ and $38. 7^{\circ}$ ) are also totally reflected, as every guided mode must be.

Inside, $E_{x} = E_{0} cos h y$ , so at the face $∣ E_{x} (d /2) ∣/∣ E_{x} (0) ∣ = cos u = 0.302$ . Outside the field falls as $e^{- α (∣ y ∣ - d /2)}$ : it drops to $1/ e$ in $1/ α = 2.50 mm$ and by $20 dB$ ( $e^{- α s} = 0.1$ ) in $s = ln 10/ α = 5.8 mm$ . With $sin 2 u / (2 u) = 0.2276$ and $cos^{2} u / w = 0.0228$ ,

Γ = \frac{1.2276}{1.2276 + 0.0228} = 0.982.

Only TE $_{0}$ and TM $_{0}$ remain for $V < π /2$ , that is for $f < 3.75 GHz$ at $d = 2 cm$ , or, at $10 GHz$ , for

d < \frac{λ _{0}}{2 NA} = \frac{3 cm}{4} = 7.5 mm .

Answer

V = 4.189 : TE_{0, 1, 2}, TM_{0, 1, 2} single-mode for d < 7.5 mm or f < 3.75 GHz TE_{0} : τ = 450.9 rad/m, n_{eff} = 2.153, θ = 74. 3^{\circ}, Γ = 98 %

Check

$u tan u = 1.264 \times 3.159 = 3.994 = w$ and $u^{2} + w^{2} = 1.598 + 15.95 = 17.55 = V^{2}$ . Each $n_{eff}$ lies between $n_{2} = 1$ and $n_{1} = 2.236$ , and the higher the order, the closer to $n_{2}$ .

(a) Roots on the circle V=4.189. — (a) Roots on the circle $V = 4.189$ .

(b) E_x(y) of the three TE modes. — (a) Roots on the circle $V = 4.189$ .

Reading the graphs

The TM roots of the same circle are $u = 1.495$ , $2.946$ and $4.055$ , each to the right of its TE partner: a larger $h$ and therefore a smaller $τ$ (TM $_{0}$ : $n_{eff} = 2.119$ against $2.153$ ). TE $_{2}$ works close to its cutoff $V = π$ : its tail, $1/ α = 4.9 mm$ , is twice that of TE $_{0}$ , and its $n_{eff} = 1.40$ is already close to $1$ .

Problem 7.6.1: TE modes of a dielectric sheet