Short Tutorial on Waveguides

Here I present the main theory of guided electromagnetic structures. The first section deals with a generic system. The main result of this section is represented by equation (8), which is an eigenvalue problem. This was made possible by the direct decomposition of the wave equation operator (i.e, the d’Almbertian) in time and then in the direction of propagation. Hence we end up with a simpler equation for the magnetic field, where now the 2D Laplacian operator operates on the magnetic field vector.

The eigenvalue problem can be simplified further in homogeneous media to transverse electric (TE) and transverse magnetic (TM) modes. This simplification allows the equation to be turned to an eigenvalue problem on a scalar field which is the longitudinal component of the magnetic in case of TE or electric in case of TM. I then show basic results of how tangential fields can be calculated and introduce the important concept of wave impedance.

The solution of the TE eigenvalue problem for cylindrical dielectric-coated circular wave guides is discussed in detail. Next, I spend some time discussing the concepts of reflections and transmissions in waveguides. An important takeaway point is the use of the transmission line model as a platform to describe interactions between waves in cascaded waveguides.

This tutorial will be updated regularly and may be split into two or more tutorials to make it easier for readers to digest and appreciate the theory.

An arbitrary cylindrical waveguide

In this section we will derive the wave equation, fields expressions and the mode impedance of an arbitrary cylindrical waveguide. In this case the waveguide cross section can be of an arbitrary shape (not necessarily circular), and is uniform in the longitudinal direction.

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Starting from the Curl equations and assuming no sources are present in the structure (i.e, $\mathbf{J}=0$ , $\rho=0$ )

(1) $\begin{equation*} \nabla\times \mathbf{E}=-\frac{\partial \mathbf{B}}{\partial t}=-\mu\frac{\partial \mathbf{H}}{\partial t} \end{equation*}$

and

(2) $\begin{equation*} \nabla\times \mathbf{H}=\frac{\partial \mathbf{D}}{\partial t}=\epsilon\frac{\partial \mathbf{E}}{\partial t} \end{equation*}$

Taking the curl of (2) and substituting in (1)

(3) $\begin{equation*} \nabla \times \nabla\times\mathbf{H}=\nabla(\nabla \cdot\mathbf{H})-\nabla^2\mathbf{H}=\frac{\partial\nabla\times\epsilon\mathbf{E}}{\partial t}. \end{equation*}$

In regions where $\epsilon$ and $\mu$ are constant, the above equation become

(4) $\begin{equation*} -\nabla^2\mathbf{H}=-\mu\epsilon\frac{\partial^2\mathbf{H}}{\partial t^2}=-\frac{1}{\nu^2}\frac{\partial^2\mathbf{H}}{\partial t^2}, \end{equation*}$

where $\nu\triangleq 1/\sqrt{\mu\epsilon}$ is the speed of the wave in the medium. Therefore

(5) $\begin{equation*} \Box^2\mathbf{H}(\mathbf{r},t)=0, \end{equation*}$

where

$\Box^2\triangleq\nabla^2-\frac{1}{\nu^2}\frac{\partial^2}{\partial t^2}$

is the d’Almbertian (4-D Laplacian operator). Mathematically, $\mathbf{H}$ in equation (??) can be interpreted as the vector fields that belong to the null space of $\Box^2$ . To find possible $\mathbf{H}(\mathbf{r},t)$ , we decompose $\Box^2$ as

(6) $\begin{equation*} \Box^2=\nabla_t^2+\frac{\partial^2}{\partial x^2}-\frac{1}{\nu^2}\frac{\partial^2}{\partial t^2}, \end{equation*}$

where $\nabla_t^2$ is the 2D transverse Laplacian ( $yz$ plane in the above figure). We then seek solutions that are the eigenfunctions of the different components of $\Box^2$ (This approach is the same as separation of variables, but presented from a different perspective.) For instance $\partial^2/\partial t^2$ has eigenfunctions that can be written as $\mathbf{H}_1(\mathbf{r})\exp(i\omega t)$ , where the corresponding eigenvalue is $-\omega^2$ . Plugging the eigenfunction of the second time derivative in Eq. (??) results in the time independent wave equation

(7) $\begin{equation*} \left(\nabla_t^2+\frac{\partial^2}{\partial x^2}+k_0^2\right)\mathbf{H_1}(\mathbf{r})=0, \end{equation*}$

where $k_0'\triangleq \omega/\nu$ is the wavenumber inside the material, which is related to the free space wavenumber $k_0$ by $k_0'=\sqrt{\epsilon_r}k_0$ . (This is the wavenumber of an unbounded wave; it is not to be confused with the propagation wavenumber $k$ that will be shown later to represent propagation in the $x$ direction.) Similarly $\mathbf{H}_1(\mathbf{r})=\mathbf{H}_2(y,z)\exp(ikx)$ is an eigenfunction of $\partial^2/\partial x^2$ . Therefore,

(8) $\begin{equation*} \left(\nabla_t^2+k_t^2\right)\mathbf{H}_2(y,z)=0, \end{equation*}$

where $k_t^2\triangleq k_0'^2-k^2$ . Note that (8) is a vector wave equation, where all the three components $H_x$ , $H_y$ and $H_z$ satisfy it.

Transverse Electric (TE) Waves

Two important classes of solutions to (8) are the Transverse Electric ( $E_x=0$ ) and transverse magnetic ( $H_x=0$ ). For TE waves it is sufficient to solve Eqn. (8) for $H_x$ and subsequently calculate the tangential component $\mathbf{H}_t=H_y\hat{\mathbf{y}}+H_z\hat{\mathbf{z}}$ as (Classical Electromagnetic Theory, J. Vanderlinde, 2006)

(9) $\begin{equation*} \mathbf{H}_t=\frac{ik}{k_t^2}\nabla_t H_x \end{equation*}$

and the $\mathbf{E}$ fields as

(10) $\begin{equation*} \mathbf{E}_t=\frac{-i\omega\mu}{k_t^2}\hat{x}\times \nabla_t H_x. \end{equation*}$

Additionally the wave impedance $Z_c$ is

(11) $\begin{equation*} Z_c=\frac{|E_t|}{|H_t|}=\frac{\omega\mu}{k}=\omega\mu\frac{\lambda_g}{2\pi}. \end{equation*}$

Here $\lambda_g=2\pi/k$ is the guided wavelength (i.e, wavelength in the $x$ direction).

Note that:

The previous analysis is valid for an arbitrary cross section.
Cut-off is defined as the frequency where a wave starts to propagate; for lower frequencies the fields exponentially decay.
$Z_c$ is imaginary below cut-off.
For a wave traveling in the $-x$ direction, $\mathbf{E}_t$ stays the same, while $\mathbf{H}_t$ reverses direction ( $k\rightarrow -k$ ). Hence, the complex Poynting vector $\mathbf{S}=\mathbf{E}_t\times\mathbf{H}_t^*$ changes direction.

Example: Dielectric Coated Waveguide

Please refer to the first figure where we now consider the cross section to be circular. The core is a dielectric material with a dielectric constant $\epsilon_{r1}$ and a radius $b$ . The core is coated by another dielectric material with a dielectric constant $\epsilon_{r2}$ and has a thickness $a-b$ . Due to Azimuthal symmery, it is convenient to represent $\nabla_t^2$ in polar coordinates,

(12) $\begin{equation*} \nabla_t^2=\frac{1}{\rho^2}\frac{\partial^2}{\partial\phi^2}+\frac{1}{\rho}\frac{\partial}{\partial \rho}\left(\rho\frac{\partial}{\partial \rho}\right). \end{equation*}$

Therefore, Eq. (8) becomes

(13) $\begin{equation*} \left(\frac{1}{\rho^2}\frac{\partial^2}{\partial\phi^2}+\frac{1}{\rho}\frac{\partial}{\partial \rho}\left(\rho\frac{\partial}{\partial \rho}\right)+k_t^2\right)\mathbf{H}_2(\rho,\phi)=0. \end{equation*}$

We follow our decomposition procedure and decompose $\mathbf{H}_2(\rho,\phi)$ into the eigenfunctions of $\partial^2/\partial\phi^2$ , which are nothing but of the form $\exp(im\phi)$ . Since $\phi$ and $\phi+2n\pi$ represent the same point, $m$ must be an integer. Therefore letting $\mathbf{H}_2(\rho,\phi)=\mathbf{H}_3(\rho)\exp(im\phi)$ one gets

(14) $\begin{equation*} \left(\frac{1}{\rho}\frac{d}{d \rho}\left(\rho\frac{d}{d \rho}\right)-\frac{m^2}{\rho^2}+k_t^2\right)\mathbf{H}_3(\rho)=0. \end{equation*}$

As previously mentioned, it is sufficient to solve for the $x$ component, $H_{3x}(\rho)$ . Therefore for the $x$ component, Eq. (14) can be re-written as

(15) $\begin{equation*} \left(\rho^2\frac{d^2}{d\rho^2}+\rho\frac{d}{d\rho}+k_t^2\rho^2-m^2\right)H_{3x}=0. \end{equation*}$

Letting $\zeta=k_t\rho$ , Eq. (15) reduces to

(16) $\begin{equation*} \zeta^2\frac{d^2H_{3x}}{d\zeta^2}+\zeta\frac{dH_{3x}}{d\zeta}+\left(\zeta^2-m^2\right)H_{3x}=0, \end{equation*}$

which is nothing but Bessel’s equation. Eqn. (16) has the general solution

(17) $\begin{equation*} H_{3x}(\rho)=AJ_m(\zeta)+BN_m(\zeta), \end{equation*}$

or in terms of $\rho$ as

(18) $\begin{equation*} H_{3x}(\rho)=AJ_m(k_t\rho)+BN_m(k_t\rho). \end{equation*}$

Here $J_m(\cdot)$ and $N_m(\cdot)$ are Bessel and Neumann functions, respectively.

The most general solution of $H_{x}(\rho,\phi,x,t)$ assumes the form

(19) $\begin{equation*} H_x(\rho,\phi,x,t)=\int_{-\infty}^\infty \sum_{m=-\infty}^\infty \sum_k\left(A_mJ_m(k_t\rho)+B_mN_m(k_t\rho)\right)e^{i(m\phi+\omega t-kx})d\omega. \end{equation*}$

Note that $k$ and consequently $k_t$ will be functions of the frequency $\omega$ and the angular modal number $m$ as will be determined from the dispersion relation as shown next. For azimuthally symmetric modes $m=0$ , (19) becomes

(20) $\begin{equation*} H_x(\rho,\phi,x,t)=\int_{-\infty}^\infty \sum_k\left(AJ_0(k_t\rho)+BN_0(k_t\rho)\right)e^{i(\omega t-kx)}d\omega. \end{equation*}$

For region 1, where $\rho\leq b$ and $\epsilon_r=1$ , $B=0$ because $N_0(k_t\rho)$ is singular at the origin. Therefore at a given frequency $\omega$ ,

(21) $\begin{equation*} H_{x}^{(1)}=A_1J_0(k_{t1}\rho)e^{i(\omega t-k_1x)}, \end{equation*}$

where $k_{t1}=\sqrt{k_0^2-k_1^2}$ and in regions 2, where $b\leq\rho\leq a$

(22) $\begin{equation*} H_{x}^{(2)}=\left(A_2J_0(k_{t2}\rho)+B_2N_0(k_{t2}\rho)\right)e^{i(\omega t-k_2x)}, \end{equation*}$

where $k_{t2}=\sqrt{\epsilon_rk_0^2-k_2^2}$ . The tangential electric and magnetic fields can be calculated from Eqs. (10) and (9), respectively, where

(23) $\begin{equation*} \nabla_tH_x=\frac{\partial H_x}{\partial \rho}\hat{\mathbf{\rho}}= \begin{cases} k_{t1}A_1J_0'(k_{t1}\rho)e^{i(\omega t-k_1x)}\hat{\mathbf{\rho}} ,~~ \rho\leq b\\ k_{t2}\left(A_2J_0'(k_{t2}\rho)+B_2N_0'(k_{t2}\rho)\right)e^{i(\omega t-k_2x)}\hat{\mathbf{\rho}} , ~~b\leq\rho\leq a\\ \end{cases} \end{equation*}$

(24) $\begin{equation*} \mathbf{E}_t=-i\omega\mu \begin{cases} \frac{A_1J_0'(k_{t1}\rho)}{k_{t1}}e^{i(\omega t-k_1x)}\mathbf{\hat{\phi}},~~\rho\leq b\\ \frac{A_2J_0'(k_{t2}\rho)+B_2N_0'(k_{t2}\rho)}{k_{t2}}e^{i(\omega t-k_2x)}\mathbf{\hat{\phi}},~~b\leq\rho\leq a \end{cases} \end{equation*}$

(25) $\begin{equation*} \mathbf{H}_t= \begin{cases} ik_1\frac{A_1J_0'(k_{t1}\rho)}{k_{t1}}e^{i(\omega t-k_1x)}\mathbf{\hat{\rho}},~~\rho\leq b\\ ik_2\frac{A_2J_0'(k_{t2}\rho)+B_2N_0'(k_{t2}\rho)}{k_{t2}}e^{i(\omega t-k_2x)}\mathbf{\hat{\rho}},~~b\leq\rho\leq a \end{cases} \end{equation*}$

Boundary Conditions

At the interface $\rho=b$ , $H_x$ , $\mathbf{E}_t$ and $\mathbf{H}_t$ are continuous. Enforcing the continuity of $H_x$ , i.e, $H_{x}^{(1)}(b)=H_{x}^{(2)}(b)$ dictates that $k_1=k_2=k$ . This can be understood by noting that for different values of $x$ , $e^{-ik_jx}$ form coefficients of a system of homogeneous linear equations. Since $x$ is a real number, these coefficients, if $k_1\neq k_2$ can be chosen to be independent, resulting in a trivial solution where $A_1$ , $A_2$ and $B_2$ vanish. Therefore, it must be the case that for a given $x$ , $e^{-ik_1x}=Ce^{-ik_2x}$ , where $C$ is a constant (i.e, the two columns of coefficient matrix are dependent). The constant $C$ can be absorbed in $A_2$ and $B_2$ . Hence $k_1=k_2$ . Additionally,

(26) $\begin{equation*} A_1J_0(k_{t1}b)=A_2J_0(k_{t2}b)+B_2N_0(k_{t2}b). \end{equation*}$

The continuity of $\mathbf{H}_t$ (or $\mathbf{E}_t$ ) leads to

(27) $\begin{equation*} k_{t2}A_1J_0'(k_{t1}b)=k_{t1}\left( A_2J_0'(k_{t2}b)+B_2N_0'(k_{t2}b) \right). \end{equation*}$

Additionally at $\rho=a$ , $\mathbf{E}_t=\mathbf{0}$ . Therefore,

(28) $\begin{equation*} A_2J_0'(k_{t2}a)+B_2N_0'(k_{t2}a)=0. \end{equation*}$

Note that $A_1$ can be set arbitrary. With no loss of generality, we will set $A_1$ to unity.

From (28)

(29) $\begin{equation*} B_2=-\frac{J_0'(k_{t2}a)}{N_0'(k_{t2}a)}A_2. \end{equation*}$

Substituting (29) in (26) and (28) results in

(30) $\begin{equation*} \left[ J_0(k_{t2}b)N_0'(k_{t2}a)-J_0'(k_{t2}a)N_0(k_{t2}b) \right]A_2=J_0(k_{t1}b)N_0'(k_{t2}a) \end{equation*}$

and

(31) $\begin{equation*} k_{t1}\left[J_0'(k_{t2}b)N_0'(k_{t2}a)-J_0'(k_{t2}a)N_0'(k_{t2}b) \right]A_2=k_{t2}J_0'(k_{t1}b)N_0'(k_{t2}a). \end{equation*}$

Therefore after dividing (30) by (31),

(32) $\begin{equation*} k_{t1}J_0(k_{t1}b)\left[J_0'(k_{t2}b)N_0'(k_{t2}a)-J_0'(k_{t2}a)N_0'(k_{t2}b)\right]=k_{t2}J_0'(k_{t1}b)\left[ J_0(k_{t2}b)N_0'(k_{t2}a)-J_0'(k_{t2}a)N_0(k_{t2}b) \right]. \end{equation*}$

Note that for each frequency $\omega$ , the characteristic equation (32) is ultimately a function of $k$ \footnote{This is because $k_{ti}=\sqrt{\epsilon_{ri}k_0^2-k_2}$ .}. Therefore, the characteristic function can be written as $F(\omega,k)=0$ . For a given frequency $\omega$ , Eq. (32) can be satisfied for an infinite number of $k$ . A $k$ value along with $m$ (here is zero), identify the waveguidfe mode. Beyond a cut-off frequency $\omega_{co}$ , $k$ will be real. We will label the different values of $k$ , $k_1,k_2,\cdots,k_n,\cdots$ by the order of their corresponding $\omega_{co}$ , such that $\omega_{co}^{(1)}<\omega_{co}^{(2)}<\omega_{co}^{(3)}<\cdots <\omega_{co}^{(n)}<\cdots$ . This means that mode 1 will be denoted by $TE_{01}$ , mode 2 by $TE_{02}$ , etc.

When $b\rightarrow a$ , the waveguide approaches the conventional air-core circular waveguide. Eq. (32) is reduced to

(33) $\begin{equation*} J_0'(k_{t1}a)=0 \end{equation*}$

and $k_{t1}=p_{1n}/a$ , where $p_{1n}$ is the $n^\textnormal{th}$ root of $J_1(\cdot)$ \footnote{Note that $J_1(z)=-J_0'(z)$ .}. Furthermore, if the inner core is filled with the same dieletric constant as the coating (i,e, the waveguide is completely filled), $k_{t1}=k_{t2}=k_t$ and (33) also follows. Note that in both cases, $k_{ti}$ does not depend on frequency. Hence when two waveguides (completely) filled with two different dielectric materials $\epsilon_{r1}$ and $\epsilon_{r2}$ are cascaded, the $(m,n)$ modes will have the same radial profile. This implies that the tangential fields at their common interface match and the cascade can be described using transmission lines.

Cascade of Two waveguides

The figure below shows two cascaded waveguides that, in general, have different $\epsilon_r$ values. The dispersion relation $(k,k_0)$ of the waveguides and consequently the transverse wavenumbers $k_{ti}$ . Hence in the general case, the continuity of $\mathbf{E}_t$ and $\mathbf{H}_t$ at the common interface $SS'$ cannot be satisfied for a single $TE_{0n}$ mode. In the following discussion, we will deterimne the condition at which $k_{ti}^{(1)}$ and $k_{ti}^{(2)}$ are equal; thus allowing the application of transmission line theory.

Cascade of two waveguides

In the most general setting, we allow the different $\epsilon_{ri}$ values to be arbitrary. Enforcing $k_{ti}^{(1)}$ and $k_{ti}^{(2)}$ to be equal implies that

(34) $\begin{align*} k_{t1}^{(1)}&=k_{t1}^{(2)}\\ \epsilon_{r1}k_0^2-k_1^2&=\epsilon_{r3}k_0^2-k_2^2 \end{align*}$

and

(35) $\begin{align*} k_{t2}^{(1)}&=k_{t2}^{(2)}\\ \epsilon_{r2}k_0^2-k_1^2&=\epsilon_{r4}k_0^2-k_2^2 \end{align*}$

must be simultaneously satisfied. Therefore

(36) $\begin{equation*} \epsilon_{r2}-\epsilon_{r1}=\epsilon_{r4}-\epsilon_{r3}, \end{equation*}$

which means that

(37) $\begin{equation*} k_2=\sqrt{(\epsilon_{r3}-\epsilon_{r1})k_0^2+k_1^2}. \end{equation*}$

In the subsequent analysis, we are interested in the special case where $\epsilon_{r1}=1$ and WG1 is operating at cut-off (i.e, $k_1=0,~k_0=k_{0co}$ ). Therefore