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Next: Exercises Up: Resonant Cavities and Waveguides Previous: Waveguides

Dielectric Waveguides

We have seen that it is possible to propagate electromagnetic waves down a hollow conductor. However, other types of guiding structures are also possible. The general requirement for a guide of electromagnetic waves is that there be a flow of energy along the axis of the guiding structure, but not perpendicular to the axis. This implies that the electromagnetic fields are appreciable only in the immediate neighborhood of the guiding structure.

Consider a uniform cylinder of arbitrary cross-section made of some dielectric material, and surrounded by a vacuum. This structure can serve as a waveguide provided the dielectric constant of the material is sufficiently large. Note, however, that the boundary conditions satisfied by the electromagnetic fields are significantly different to those of a conventional waveguide. The transverse fields are governed by two equations: one for the region inside the dielectric, and the other for the vacuum region. Inside the dielectric, we have

$\displaystyle \left[ \nabla_s^{\,2} +\left(\epsilon_1 \,\frac{\omega^{\,2}}{c^{\,2}} - k_g^{\,2}\right) \right]\psi = 0.$ (1372)

In the vacuum region, we have

$\displaystyle \left[ \nabla_s^{\,2} +\left( \frac{\omega^{\,2}}{c^{\,2}} - k_g^{\,2}\right) \right]\psi = 0.$ (1373)

Here, $ \psi(x,y)\, {\rm e}^{\,{\rm i}\, k_g\, z}$ stands for either $ E_z$ or $ H_z$ , $ \epsilon_1$ is the relative permittivity of the dielectric material, and $ k_g$ is the guide propagation constant. The propagation constant must be the same both inside and outside the dielectric in order to allow the electromagnetic boundary conditions to be satisfied at all points on the surface of the cylinder.

Inside the dielectric, the transverse Laplacian must be negative, so that the constant

$\displaystyle k_s^{\,2} = \epsilon_1\, \frac{\omega^{\,2}}{c^{\,2}} -k_g^{\,2}$ (1374)

is positive. Outside the cylinder the requirement of no transverse flow of energy can only be satisfied if the fields fall off exponentially (instead of oscillating). Thus,

$\displaystyle k_t^{\,2} = k_g^{\,2} - \frac{\omega^{\,2}}{c^{\,2}}$ (1375)

must be positive.

The oscillatory solutions (inside) must be matched to the exponentiating solutions (outside). The boundary conditions are the continuity of normal $ {\bf B}$ and $ {\bf D}$ and tangential $ {\bf E}$ and $ {\bf H}$ on the surface of the tube. These boundary conditions are far more complicated than those in a conventional waveguide. For this reason, the normal modes cannot usually be classified as either pure TE or pure TM modes. In general, the normal modes possess both electric and magnetic field components in the transverse plane. However, for the special case of a dielectric cylinder tube of circular cross-section, the normal modes can have either pure TE or pure TM characteristics. Let us examine this case in detail.

Consider a dielectric cylinder of dielectric constant $ \epsilon_1$ whose transverse cross-section is a circle of radius $ a$ . For the sake of simplicity, let us only search for normal modes whose electromagnetic fields have no azimuthal variation. Equations (1374) and (1376) yield

$\displaystyle \left(r^{\,2}\,\frac{d^2}{d r^{2}} + r\,\frac{d}{dr} + r^{\,2}\, k_s^{\,2}\right) \psi = 0$ (1376)

for $ r<a$ . The general solution to this equation is some linear combination of the Bessel functions $ J_0(k_s\, r)$ and $ Y_0(k_s \,r)$ . However, because $ Y_0(k_s \,r)$ is badly behaved at the origin ($ r=0$ ), the physical solution is $ \psi(r)\propto J_0(k_s\, r)$ .

Equations (1375) and (1377) yield

$\displaystyle \left(r^{\,2}\,\frac{d^{\,2}}{d r^{\,2}} + r\,\frac{d}{dr} - r^{\,2}\, k_t^{\,2}\right) \psi = 0.$ (1377)

which can be rewritten

$\displaystyle \left(z^{\,2}\,\frac{d^{\,2}}{d z^{\,2}} + z\,\frac{d}{dz} - z^{\,2}\right) \psi = 0,$ (1378)

where $ z=k_t\, r$ . The above can be recognized as a type of modified Bessel equation, whose most general form is

$\displaystyle \left[z^{\,2}\,\frac{d^2}{d z^2} + z\,\frac{d}{dz} - (z^{\,2}+m^{\,2})\right] \psi = 0.$ (1379)

The two linearly independent solutions of the previous equation are denoted $ I_m(z)$ and $ K_m(z)$ . Moreover, $ I_m(z)\rightarrow\infty$ as $ \vert z\vert\rightarrow \infty$ , whereas $ K_m(z)\rightarrow 0$ . Thus, it is clear that the physical solution to Equation (1379) (i.e., the one that decays as $ \vert r\vert\rightarrow \infty$ ) is $ \psi(r)\propto K_0(k_t\, r)$ .

The physical solution is then

$\displaystyle \psi(r) = J_0(k_s \,r)$ (1380)

for $ r\leq a$ , and

$\displaystyle \psi (r)= A\,K_0(k_t \,r)$ (1381)

for $ r>a$ . Here, $ A$ is an arbitrary constant, and $ \psi(r) \,
{\rm e}^{\,{\rm i}\,k_g\, z}$ stands for either $ E_z$ or $ H_z$ . It follows from Equations (1335)-(1336) (using $ \partial/\partial\theta = 0$ ) that

$\displaystyle H_r$ $\displaystyle = {\rm i}\,\frac{k_g}{k_s^{\,2}} \frac{\partial H_z}{\partial r},$ (1382)
$\displaystyle E_\theta$ $\displaystyle =- \frac{\omega\,\mu_0}{k_g} \,H_r,$ (1383)
$\displaystyle H_\theta$ $\displaystyle = {\rm i}\, \frac{\omega\,\epsilon_0\,\epsilon_1}{k_s^{\,2}} \frac{\partial E_z}{\partial r},$ (1384)
$\displaystyle E_r$ $\displaystyle = \frac{k_g}{\omega\, \epsilon_0\,\epsilon_1} \,H_\theta$ (1385)

for $ r\leq a$ . There are an analogous set of relations for $ r>a$ . The fact that the field components form two groups--that is, ($ H_r$ , $ E_\theta$ ), which depend on $ H_z$ , and ($ H_\theta$ , $ E_r$ ), which depend on $ E_z$ --implies that the normal modes take the form of either pure TE modes or pure TM modes.

For a TE mode ($ E_z=0$ ) we find that

$\displaystyle H_z(r)$ $\displaystyle = J_0(k_s\, r),$ (1386)
$\displaystyle H_r (r)$ $\displaystyle =-{\rm i}\,\frac{k_g}{k_s} \,J_1(k_s\, r),$ (1387)
$\displaystyle E_\theta(r)$ $\displaystyle = {\rm i}\, \frac{\omega\,\mu_0}{k_s}\, J_1(k_s\, r)$ (1388)

for $ r\leq a$ , and

$\displaystyle H_z(r)$ $\displaystyle = A \,K_0(k_t \,r),$ (1389)
$\displaystyle H_r (r)$ $\displaystyle = {\rm i}\,A\,\frac{ k_g }{k_t}\, K_1(k_t \,r),$ (1390)
$\displaystyle E_\theta(r)$ $\displaystyle = -{\rm i}\,A\, \frac{\omega\,\mu_0}{k_t} \,K_1(k_t\, r)$ (1391)

for $ r>a$ . Here we have used the identities

$\displaystyle J_0'(z)$ $\displaystyle \equiv -J_1(z),$ (1392)
$\displaystyle K_0'(z)$ $\displaystyle \equiv -K_1(z),$ (1393)

where $ '$ denotes differentiation with respect to $ z$ . The boundary conditions require $ H_z(r)$ , $ H_r(r)$ , and $ E_\theta(r)$ to be continuous across $ r=a$ . Thus, it follows that

$\displaystyle A\,K_0(k_t\, a)$ $\displaystyle =J_0(k_s \,a),$ (1394)
$\displaystyle -A\, \frac{K_1(k_t \,r) }{k_t}$ $\displaystyle = \frac{J_1(k_s \,a)}{k_s}.$ (1395)

Eliminating the arbitrary constant $ A$ between the above two equations yields the dispersion relation

$\displaystyle \frac{J_1(k_s \,a)}{k_s\,J_0(k_s \,a)} + \frac{K_1(k_t \,a)}{k_t\,K_0(k_t \,a)}= 0,$ (1396)


$\displaystyle k_t^{\,2} + k_s^{\,2} = (\epsilon_1 -1)\,\frac{\omega^{\,2}}{c^{\,2}}.$ (1397)

Figure: Graphical solution of the dispersion relation (1398). The curve $ A$ represents $ -J_1(k_s/a)/k_s \,J_0(k_s\, a)$ . The curve $ B$ represents $ K_1(k_t \,a) / k_t\, K_0(k_t\, a)$ .
\epsfysize =4in

Figure 26 shows a graphical solution of the above dispersion relation. The roots correspond to the crossing points of the two curves; $ -J_1(k_s \,a)/k_s \,J_0(k_s\, a)$ and $ K_1(k_t \,a) / k_t\, K_0(k_t\, a)$ . The vertical asymptotes of the first curve are given by the roots of $ J_0(k_s \,a)=0$ . The vertical asymptote of the second curve occurs when $ k_t=0$ : that is, when $ k_s^{\,2} \,a^{\,2} = (\epsilon_1-1)\,\omega^{\,2}\,a^{\,2}/c^{\,2}$ . Note, from Equation (1399), that $ k_t$ decreases as $ k_s$ increases. In Figure 26, there are two crossing points, corresponding to two distinct propagating modes of the system. It is evident that if the point $ k_t=0$ corresponds to a value of $ k_s \,a$ that is less than the first root of $ J_0(k_s \,a)=0$ then there is no crossing of the two curves, and, hence, there are no propagating modes. Because the first root of $ J_0(z)=0$ occurs at $ z=2.4048$ (see Table 2), the condition for the existence of propagating modes can be written

$\displaystyle \omega > \omega_{01} = \frac{2.4048\,c}{\sqrt{\epsilon_1 -1}\, a}.$ (1398)

In other words, the mode frequency must lie above the cutoff frequency $ \omega_{01}$ for the $ {\rm TE}_{\,01}$ mode [here, the 0 corresponds to the number of nodes in the azimuthal direction, and the 1 refers to the first root of $ J_0(z)=0$ ]. It is also evident that, as the mode frequency is gradually increased, the point $ k_t=0$ eventually crosses the second vertical asymptote of $ -J_1(k_s/a)/k_s \,J_0(k_s\, a)$ , at which point the $ {\rm TE}_{\,02}$ mode can propagate. As $ \omega$ is further increased, more and more TE modes can propagate. The cutoff frequency for the $ {\rm TE}_{\,0l}$ mode is given by

$\displaystyle \omega_{0l} = \frac{j_{0l}\,c}{\sqrt{\epsilon_1 -1}\, a},$ (1399)

where $ j_{0l}$ is $ l$ th root of $ J_0(z)=0$ (in order of increasing $ z$ ).

At the cutoff frequency for a particular mode, $ k_t=0$ , which implies from Equation (1377) that $ k_g = \omega/c$ . In other words, the mode propagates along the guide at the velocity of light in vacuum. At frequencies below this cutoff frequency, the system no longer acts as a guide, but rather as an antenna, with energy being radiated radially. For frequencies well above the cutoff, $ k_t$ and $ k_g$ are of the same order of magnitude, and are large compared to $ k_s$ . This implies that the fields do not extend appreciably outside the dielectric cylinder.

For a TM mode ($ H_z=0$ ) we find that

$\displaystyle E_z(r)$ $\displaystyle = J_0(k_s r),$ (1400)
$\displaystyle H_\theta(r)$ $\displaystyle =-{\rm i}\,\frac{\omega\,\epsilon_0\,\epsilon_1}{k_s} \,J_1(k_s \,r),$ (1401)
$\displaystyle E_r(r)$ $\displaystyle = -{\rm i}\, \frac{k_g}{k_s}\, J_1(k_s\, r)$ (1402)

for $ r\leq a$ , and

$\displaystyle E_z(r)$ $\displaystyle = A \,K_0(k_t \,r),$ (1403)
$\displaystyle H_\theta(r)$ $\displaystyle ={\rm i}\,A\,\frac{\omega\,\epsilon_0}{k_t}\, K_1(k_t\, r),$ (1404)
$\displaystyle E_r(r)$ $\displaystyle = {\rm i}\,A\, \frac{k_g}{k_t} \,K_1(k_t \,r)$ (1405)

for $ r>a$ . The boundary conditions require $ E_z(r)$ , $ H_\theta(r)$ , and $ D_r(r)$ to be continuous across $ r=a$ . Thus, it follows that

$\displaystyle A\,K_0(k_t\, a)$ $\displaystyle =J_0(k_s \,a),$ (1406)
$\displaystyle -A\, \frac{K_1(k_t \,r) }{k_t}$ $\displaystyle = \epsilon_1\, \frac{J_1(k_s\, a)}{k_s}.$ (1407)

Eliminating the arbitrary constant $ A$ between the above two equations yields the dispersion relation

$\displaystyle \frac{\epsilon_1\,J_1(k_s \,a)}{k_s\,J_0(k_s\, a)} + \frac{K_1(k_t \,a)}{k_t\,K_0(k_t\, a)}= 0.$ (1408)

It is clear, from this dispersion relation, that the cutoff frequency for the $ {\rm TM}_{\,0l}$ mode is exactly the same as that for the $ {\rm TE}_{\,0l}$ mode. It is also clear that, in the limit $ \epsilon_1\gg 1$ , the propagation constants are determined by the roots of $ J_1(k_s\, a)\simeq 0$ . However, this is exactly the same as the determining equation for TE modes in a metallic waveguide of circular cross-section (filled with dielectric of relative permittivity $ \epsilon_1$ ).

Modes with azimuthal dependence (i.e., $ m>0$ ) have longitudinal components of both $ {\bf E}$ and $ {\bf H}$ . This makes the mathematics somewhat more complicated. However, the basic results are the same as for $ m=0$ modes: that is, for frequencies well above the cutoff frequency the modes are localized in the immediate vicinity of the cylinder.

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Next: Exercises Up: Resonant Cavities and Waveguides Previous: Waveguides
Richard Fitzpatrick 2014-06-27