7.6.9 Surface waves along a dielectric-coated conductor

A dielectric layer of index $n_{1}$ and thickness $t$ on a perfectly conducting plane, with air (or any medium of index $n_{2}$ ) above it, is a guide for surface waves: waves bound to the surface and decaying away from it. Its modes follow from the slab by image theory. The perfect conductor requires $E_{tan} = 0$ on its surface. A slab mode of thickness $d = 2 t$ that already has $E_{tan} = 0$ on its midplane $y = 0$ is undisturbed if a metal sheet is placed there, and conversely the metal and its mirror image of the coating reproduce such a mode (Figure 7.6.8), much as the image charges of Section 3.4.8 reproduce the field above a grounded plane. Which slab modes qualify?

TE: $E_{x} (0) = 0$ requires $E_{x} \propto sin h y$ , the odd TE $_{m}$ ( $m = 1, 3, \dots$ ).
TM: $E_{z} \propto d H_{x} / d y$ (Eq. (7.6.15)) vanishes at $y = 0$ for $H_{x} \propto cos h y$ , the even TM $_{m}$ ( $m = 0, 2, \dots$ ).

The normalised frequency is that of the full slab, now with $t = d /2$ :

Surface waves on a coated conductor

V TM_{0} TE_{1} TM_{2} = k_{0} t n_{1}^{2} - n_{2}^{2}, : no cutoff, : f_{c} = \frac{c}{4 t n _{1}^{2} - n _{2}^{2}}, : 2 f_{c}, \dots (7.6.32)

The modes appear in the order TM $_{0}$ , TE $_{1}$ , TM $_{2}$ , TE $_{3}$ , …, with cutoffs at $V = mπ /2$ .

$A dielectric layer of thickness t on a perfect conductor (PEC) behaves like the upper half of a slab of thickness 2t: the metal can replace the midplane of every slab mode that already has E_{\rm tan}=0 there. The profile on the right is H_x of the TM_0 surface wave, a cosine with its maximum at the metal and an exponential tail in the air.$

Figure 7.6.8. A dielectric layer of thickness

t

on a perfect conductor (PEC) behaves like the upper half of a slab of thickness

2 t

: the metal can replace the midplane of every slab mode that already has

E_{tan} = 0

there. The profile on the right is

H_{x}

of the TM

_{0}

surface wave, a cosine with its maximum at the metal and an exponential tail in the air.

In plain words

A metal plane with a dielectric coat is half of a slab. It keeps the half of the slab modes whose electric field is perpendicular to the metal, and it guides the lowest TM wave at every frequency.

Surface impedance

Seen from the air, the coated surface presents a reactance. For a TM wave, $H_{x} \propto e^{- α s}$ at the height $s$ above the coating and, by Eq. (7.6.15), $E_{z} = (j / ω ε_{0}) (- α) H_{x}$ . The ratio of the tangential fields that points into the surface (power flowing in the $- y$ direction) is

Z_{s} Z_{s} = - \frac{E _{z}}{H _{x}} = j \frac{α}{ω ε _{0}} (TM, inductive), = \frac{E _{x}}{H _{z}} = - j \frac{ω μ _{0}}{α} (TE, capacitive) . (7.6.33)

A purely reactive surface absorbs no power, and the wave travels along it without loss. Cheng derives the same two impedances for the slab, his Eqs. (10-261) and (10-279), and concludes that an inductive surface supports TM surface waves and a capacitive one TE surface waves.

Where surface waves matter

A wire with a dielectric coat guides the TM $_{0}$ surface wave (the Goubau line, used for low-loss feeds and for broadband links over power lines). Every microstrip substrate (Section 7.7) is a coated conductor: it guides TM $_{0}$ from d.c. and TE $_{1}$ above $f_{c}$ , and circuits and antennas printed on it leak part of their energy into these surface waves. The TE $_{1}$ cutoff Eq. (7.6.32) is one of the upper frequency limits of a microstrip line.

Quick check

In which order do the surface waves of a coated conductor appear as the frequency rises, and why is there no TE $_{0}$ surface wave?

TM $_{0}$ (from d.c.), TE $_{1}$ , TM $_{2}$ , TE $_{3}$ , …, at $V = 0, π /2, π, 3 π /2, \dots$ TE $_{0}$ of the full slab has $E_{x} \propto cos h y$ , which is largest on the midplane, so it cannot coexist with a metal sheet there.