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

The even TE mode

Take $E_{x} = E_{0} cos h y$ inside and match it, at each face, to the decaying exponential outside:

E_{x} (y) = E_{0} {cos h y, cos \frac{h d}{2} e^{- α (∣ y ∣ - d /2)}, ∣ y ∣ \leq \frac{d}{2}, ∣ y ∣ \geq \frac{d}{2} . (7.6.9)

The tangential component $E_{x}$ is continuous at $y = \pm d /2$ by construction. The magnetic field follows from the first set Eq. (7.6.5):

H_{y} = \frac{τ}{ω μ _{0}} E_{x}, H_{z} = - \frac{j}{ω μ _{0}} \frac{d E _{x}}{d y} . (7.6.10)

$H_{z}$ is tangential to the faces, and since both media are nonmagnetic and carry no surface current, it must be continuous as well (the boundary conditions of Sections 3.4.6 and 4.2.3, which hold for time-varying fields too). Continuity of $H_{z}$ means continuity of $d E_{x} / d y$ . At $y = d /2$ , from inside and from outside,

- h E_{0} sin \frac{h d}{2} = - α E_{0} cos \frac{h d}{2} ⟹ α = h tan \frac{h d}{2} . (7.6.11)

The odd TE mode

With $E_{x} = E_{0} sin h y$ inside and $\pm E_{0} sin (h d /2) e^{- α (∣ y ∣ - d /2)}$ outside (upper sign for $y > 0$ ), the same steps give $h E_{0} cos (h d /2) = - α E_{0} sin (h d /2)$ .

Eigenvalue equations of the TE modes

even TE: odd TE: α = h tan \frac{h d}{2}, α = - h cot \frac{h d}{2}, (7.6.12)

to be solved together with $h^{2} + α^{2} = k_{0}^{2} (n_{1}^{2} - n_{2}^{2})$ , Eq. (7.6.8).

In plain words

The field and its slope must both join smoothly at the faces. A cosine inside can join a decaying exponential only at special values of $h$ , where the cosine is already falling at exactly the rate of the exponential. Those values are the modes.

Figure 7.6.3 shows the result for the lowest mode: $E_{x}$ and the slope that defines $H_{z}$ are continuous, and $H_{z}$ changes sign at the centre, where $E_{x}$ is largest.

$The fields of the even TE_0 mode across the slab (computed for V=3.5, slab shaded): E_x (and with it H_y, which has the same profile) and H_z, which is proportional to the slope of E_x and 90^\circ out of phase with it. Both are continuous at the faces y=\pm d/2: this is the condition Eq. (7.6.11) that selects the mode. H_z is normalised to its largest value.$

Figure 7.6.3. The fields of the even TE

_{0}

mode across the slab (computed for

V = 3.5

, slab shaded):

E_{x}

(and with it

H_{y}

, which has the same profile) and

H_{z}

, which is proportional to the slope of

E_{x}

and

9 0^{\circ}

out of phase with it. Both are continuous at the faces

y = \pm d /2

: this is the condition Eq. (7.6.11) that selects the mode.

H_{z}

is normalised to its largest value.

Wave impedance and power

By Eq. (7.6.10) the ratio

Z_{T E} = \frac{E _{x}}{H _{y}} = \frac{ω μ _{0}}{τ} (7.6.13)

is the same inside and outside the slab: $H_{y}$ has the profile of $E_{x}$ . The time-average power density along $z$ is $S_{z} = \frac{1}{2} Re {E_{x} H_{y}^{*}} = ∣ E_{x} ∣^{2} / (2 Z_{T E})$ , and the power carried per metre of width in $x$ is

P = \frac{1}{2 Z _{T E}} \int_{- \infty}^{\infty} ∣ E_{x} (y) ∣^{2} d y [W/m] . (7.6.14)

Across the guide, $S_{y} = - \frac{1}{2} Re {E_{x} H_{z}^{*}} = 0$ because $H_{z}$ is $9 0^{\circ}$ out of phase with $E_{x}$ : no power leaves the slab, as anticipated in Section 7.6.1.

TE modes are made of perpendicularly polarised waves

$E_{x}$ is perpendicular to the plane of incidence of the zig-zagging waves, the $y z$ -plane. A TE mode is therefore built of the perpendicularly (S) polarised plane waves of Section 6.4, and its reflection at the faces is governed by $Γ_{⊥}$ . Section 7.6.8 uses exactly this.

7.6.3 TE modes of the slab

The even TE mode

The odd TE mode

Wave impedance and power