# Waveguides

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## Additional information

To run the model, open main.pro or a file .pro with Gmsh.

## Description of the model

### Classical waveguides

#### General case

Let us consider a hollow cylindrical waveguide of arbitrary cross-sectional shape that has a principal axis in the $z$-direction. The elementary solution of this problem reads \begin{align} {\bf E}(x,y,z,t) &= {\bf E}(x,y) \: e^{i(\pm kz-\omega t)} \\ {\bf H}(x,y,z,t) &= {\bf H}(x,y) \: e^{i(\pm kz-\omega t)} \end{align} where the new unknowns are governed by \begin{align} \left[\nabla_t^2 + (\mu\varepsilon\omega^2 - k^2)\right] \left\{\begin{array}{x}{\bf E}\\{\bf H}\end{array}\right\} = 0 \end{align} where $\nabla_t$ is the transverse part of the Nabla operator.

Parallel and transverse fields

It is useful to separate the fields into components parallel to and transverse to the $z$-direction: \begin{align} {\bf E} &= {\bf E}_z + {\bf E}_t && \text{with } {\bf E}_z = {E}_z \hat{\bf z} \\ {\bf H} &={\bf H}_z + {\bf H}_t && \text{with } {\bf H}_z = {H}_z \hat{\bf z} \end{align}

Some well-known cases:

• Transverse electromagnetic (TEM) waves: if ${E}_z=0$ and ${H}_z=0$ everywhere
• Transverse magnetic (TM) waves: if ${H}_z=0$ everywhere
• Transverse electric (TE) waves: if ${E}_z=0$ everywhere

If both parallel fields are vanishing (TEM case), the transverse fields are the solution of an electrostatic problem in two dimensions.

If at least one parallel field is non-vanishing, the transverse fields are \begin{align} {\bf E}_t &= \frac{i}{\mu\varepsilon\omega^2-k^2} \left[\pm\:k\:\nabla_t{E}_z - \mu\omega\:\hat{\bf z}\times\nabla_t{H}_z\right] \\ {\bf H}_t &= \frac{i}{\mu\varepsilon\omega^2-k^2} \left[\pm\:k\:\nabla_t{H}_z + \varepsilon\omega\:\hat{\bf z}\times\nabla_t{E}_z\right] \end{align}

In TEM, TM and TE cases, the transverse fields are related by $${\bf H}_t = \pm\frac{1}{Z} \hat{\bf z}\times{\bf E}_t$$ where the wave impedance $Z$ is given by $$Z= \left\{\begin{array}{ll} \sqrt{\frac{\mu}{\varepsilon}} &\quad \text{(TEM case)} \\ \frac{k}{k_0} \sqrt{\frac{\mu}{\varepsilon}} &\quad \text{(TM case)} \\ \frac{k_0}{k} \sqrt{\frac{\mu}{\varepsilon}} &\quad \text{(TE case)} \end{array}\right.$$ with $k_0=\omega\sqrt{\mu\varepsilon}$.

Eigenvalue problem

For a waveguide with perfectly conducting borders, the non-vanishing parallel field of TM and TE cases is governed by, respectively, \begin{align} \left[\nabla_t^2 + \gamma^2\right] {E}_z &= 0 \\ \left[\nabla_t^2 + \gamma^2\right] {H}_z &= 0 \end{align} with $\gamma^2 = \mu\varepsilon\omega^2 - k^2$, and is subject to the homogeneous boundary condition ${E}_z=0$ (TM case) or ${\bf n}\cdot\nabla{H}_z = 0$ (TE case).

These equations define eigenvalue problems. There is a spectrum of eigenvalues $\gamma^2_\ell$ and corresponding solutions $\left.E_z\right|_\ell$ or $\left.H_z\right|_\ell$, $\ell=1,2,...$, which form an orthogonal set. For a given frequency $\omega$, the wave number $k$ is determined for each $\ell$: $$k_\ell = \sqrt{\mu\varepsilon\omega^2-\gamma^2_\ell} = \sqrt{\mu\varepsilon} \sqrt{\omega^2-\omega^2_\ell}$$ where $\omega_\ell$ is the cutoff frequency, defined by $$\omega_\ell=\frac{\gamma_\ell}{\sqrt{\mu\varepsilon}}$$ This frequency defines the nature of waves:

• If $\omega>\omega_\ell$, $k_\ell$ is real and the waves are travelling modes.
• If $\omega<\omega_\ell$, $k_\ell$ is imaginary and the waves are evanescent modes.

#### Rectangular waveguide

Let us consider a rectangular waveguide $(x,y)\in[0,a]\times[0,b]$ that has a principal axis in the $z-$direction.

• For TM modes, the solutions for $E_z$ are

\begin{align} \left.E_z\right|_{mn} &= E_0 \sin\left(\frac{m\pi x}{a}\right) \sin\left(\frac{n\pi y}{b}\right) e^{ikz-i\omega t} && \text{with } m,n=0,1,2,... \end{align}

• For TE modes, the solutions for $H_z$ are

\begin{align} \left.H_z\right|_{mn} &= H_0 \cos\left(\frac{m\pi x}{a}\right) \cos\left(\frac{n\pi y}{b}\right) e^{ikz-i\omega t} && \text{with } m,n=0,1,2,... \end{align} In both cases, the eigenvalues and the cutoff frequencies are, respectively, \begin{align} \gamma_{mn}^2 &= \pi^2\left(\frac{m^2}{a^2}+\frac{n^2}{b^2}\right) && \text{with } m,n=0,1,2,... \\ \omega_{mn}^2 &= \frac{\pi}{\sqrt{\mu\varepsilon}}\sqrt{\frac{m^2}{a^2}+\frac{n^2}{b^2}} && \text{with } m,n=0,1,2,... \end{align}

The transverse components can now be solved for. Starting with the TM Modes ($H_z = 0$),

\begin{align} {\bf E}_t =& \frac{i}{\mu\varepsilon\omega^2-k^2} \left[\pm\:k\:\nabla_t{E}_z - \mu\omega\:\hat{\bf z}\times\nabla_t{H}_z\right] \\ =& \frac{i}{\mu\varepsilon\omega^2-k^2} \left[\pm\:k\:\nabla_t{E}_z -0 \right]\\ =& \frac{i}{\mu\varepsilon\omega^2-k^2} \left[\pm\:k\:\nabla_t E_0 \sin\left(\frac{m\pi x}{a}\right) \sin\left(\frac{n\pi y}{b}\right) e^{ikz-i\omega t} \right]\\ =& \frac{i}{\mu\varepsilon\omega^2-k^2} \left[\pm\:k\:E_0\left(\frac{m\pi}{a}\cos\left(\frac{m\pi x}{a}\right) \sin\left(\frac{n\pi y}{b}\right)\hat{\bf x}+ \right.\right. \left.\left. \frac{n\pi}{b}\sin\left(\frac{m\pi x}{a}\right) \cos\left(\frac{n\pi y}{b}\right)\hat{\bf y}\right) e^{ikz-i\omega t} \right] \end{align}

Separating out the factors, then the complete $E$ field descriptions are then:

\begin{align} {\bf E}_x =& \frac{i}{\mu\varepsilon\omega^2-k^2} \left[\pm\:k\:E_0\frac{m\pi}{a}\cos\left(\frac{m\pi x}{a}\right) \sin\left(\frac{n\pi y}{b}\right)\right] e^{ikz-i\omega t}\hat{\bf x} \end{align} \begin{align} {\bf E}_y =& \frac{i}{\mu\varepsilon\omega^2-k^2} \left[\pm\:k\:E_0 \frac{n\pi}{b}\sin\left(\frac{m\pi x}{a}\right) \cos\left(\frac{n\pi y}{b}\right)\right] e^{ikz-i\omega t}\hat{\bf y} \end{align} \begin{align} {\bf E}_z &= E_0 \sin\left(\frac{m\pi x}{a}\right) \sin\left(\frac{n\pi y}{b}\right) e^{ikz-i\omega t} \hat{\bf z} \end{align}

We can then use the relationship between TE and TM modes to solve for the complete $H$ fields.

\begin{align} {\bf H}_t & = \pm\frac{1}{Z} \hat{\bf z}\times{\bf E}_t \\ & = \pm\frac{1}{Z} \left(-{\bf E}_y,{\bf E}_x\right) \end{align} \begin{align} {\bf H}_x =& \mp \frac{k_0}{k} \sqrt{\frac{\varepsilon}{\mu}} \frac{i}{\mu\varepsilon\omega^2-k^2} \left[\pm\:k\:E_0 \frac{n\pi}{b}\sin\left(\frac{m\pi x}{a}\right) \cos\left(\frac{n\pi y}{b}\right)\right] e^{ikz-i\omega t}\hat{\bf x} \end{align} \begin{align} {\bf H}_y =& \pm \frac{k_0}{k} \sqrt{\frac{\varepsilon}{\mu}}\frac{i}{\mu\varepsilon\omega^2-k^2} \left[\pm\:k\:E_0\frac{m\pi}{a}\cos\left(\frac{m\pi x}{a}\right) \sin\left(\frac{n\pi y}{b}\right)\right] e^{ikz-i\omega t}\hat{\bf y} \end{align} \begin{align} {\bf H}_z &= 0 \end{align} Now for the TE, modes, so $E_z=0$, therefore:

\begin{align} {\bf H}_t =& \frac{i}{\mu\varepsilon\omega^2-k^2} \left[\pm\:k\:\nabla_t{H}_z + \varepsilon\omega\:\hat{\bf z}\times\nabla_t{E}_z\right]\\ =& \frac{i}{\mu\varepsilon\omega^2-k^2} \left[\pm\:k\:\nabla_t{H}_z \right]\\ =& \frac{i}{\mu\varepsilon\omega^2-k^2} \left[\pm\:k\:\nabla_t H_0 \cos\left(\frac{m\pi x}{a}\right) \cos\left(\frac{n\pi y}{b}\right) e^{ikz-i\omega t} \right] \\ =& \frac{i}{\mu\varepsilon\omega^2-k^2} \left[\mp\:k\: H_0\left(\frac{m\pi}{a} \sin\left(\frac{m\pi x}{a}\right) \cos\left(\frac{n\pi y}{b}\right)\hat{\bf x}+ \right. \right. \left.\left.\frac{n\pi}{b}\cos\left(\frac{m\pi x}{a}\right) \sin\left(\frac{n\pi y}{b}\right)\hat{\bf y} \right)e^{ikz-i\omega t} \right] \end{align}

Seperating out the factors, then the complete $H$ field descriptions are then:

\begin{align} {\bf H}_x =& \frac{i}{\mu\varepsilon\omega^2-k^2} \left[\mp\:k\: H_0\frac{m\pi}{a} \sin\left(\frac{m\pi x}{a}\right) \cos\left(\frac{n\pi y}{b}\right)e^{ikz-i\omega t} \right]\hat{\bf x} \end{align}

\begin{align} {\bf H}_y =& \frac{i}{\mu\varepsilon\omega^2-k^2} \left[\mp\:k\: H_0\frac{n\pi}{b}\cos\left(\frac{m\pi x}{a}\right) \sin\left(\frac{n\pi y}{b}\right)e^{ikz-i\omega t} \right]\hat{\bf y} \end{align}

\begin{align} {\bf H}_z &= H_0 \cos\left(\frac{m\pi x}{a}\right) \cos\left(\frac{n\pi y}{b}\right) e^{ikz-i\omega t} \hat{\bf z} \end{align}

Once again we can use the relationship between TE and TM modes to solve for the $E$ fields:

\begin{align} {\bf E}_t & = \pm Z \left({\bf H}_y,-{\bf H}_x\right) \end{align}

\begin{align} {\bf E}_x =& \pm \frac{k_0}{k} \sqrt{\frac{\mu}{\varepsilon}} \frac{i}{\mu\varepsilon\omega^2-k^2} \left[\mp\:k\: H_0\frac{n\pi}{b}\cos\left(\frac{m\pi x}{a}\right) \sin\left(\frac{n\pi y}{b}\right)e^{ikz-i\omega t} \right]\hat{\bf x} \end{align}

\begin{align} {\bf E}_y =& \mp \frac{k_0}{k} \sqrt{\frac{\mu}{\varepsilon}} \frac{i}{\mu\varepsilon\omega^2-k^2} \left[\mp\:k\: H_0\frac{m\pi}{a} \sin\left(\frac{m\pi x}{a}\right) \cos\left(\frac{n\pi y}{b}\right)e^{ikz-i\omega t} \right]\hat{\bf y} \end{align}

\begin{align} {\bf E}_z =&0 \end{align}

### Discontinuities and networks

#### Discontinuity in a parallel-plate waveguide

See [1], section 4.6.1.

• Solution for the TE mode

\begin{align} H_z &= H_0 e^{-jk_0 x} + R H_0 e^{jk_0 x} && \text{at } x=x_1 \\ H_z &= T H_0 e^{-jk_0 x} && \text{at } x=x_2 \end{align}

• Boundary conditions

\begin{align} \partial_x H_z &= jk_0 H_z - 2jk_0H_0 e^{-jk_0 x} && \text{at } x=x_1 \\ \partial_x H_z &= -jk_0 H_z && \text{at } x=x_2 \end{align}

• Reflection and transmission coefficients

\begin{align} R &= \left.\frac{H_z - H_0 e^{-jk_0x}}{H_0 e^{ jk_0 x}}\right|_{x=x_1} \\ T &= \left.\frac{H_z}{H_0 e^{-jk_0 x}}\right|_{x=x_2} \end{align}

with $|R|^2+|T|^2=1$.

#### Waveguide with discontinuities

See [1], section 8.5.

• Solution for the TE$_{mn}$ mode

\begin{align} {\bf E}(x,y,z) &= E_0 {\bf e}_{mn}(x,y) e^{-jk_{z_{mn}} z} + R E_0 {\bf e}_{mn}(x,y) e^{jk_{z_{mn}} z} && \text{at } z=z_1 \\ {\bf E}(x,y,z) &= T E_0 {\bf e}_{mn}(x,y) e^{-jk_{z_{mn}} z} && \text{at } z=z_2 \end{align}

• Reflection and transmission coefficients

\begin{align} R &= \left.\frac{{\bf E}(x,y,z) - E_0 {\bf e}_{mn}(x,y) e^{-jk_{z_{10}}z}}{E_0 {\bf e}_{mn}(x,y) e^{jk_{z_{10}}z}}\right|_{z=z_1} \\ T &= \left.\frac{{\bf E}(x,y,z)}{E_0 {\bf e}_{mn}(x,y) e^{-jk_{z_{10}}z}}\right|_{z=z_2} \end{align}

with $|R|^2+|T|^2=1$.
• Reflection and transmission coefficients (improved - the dominant mode $mn$ - using the orthogonality properties of modes)

\begin{align} R &= \frac{e^{-jk_{z_{mn}}z_1}}{E_0} \frac{\int_{S_1}{\bf E}(x,y,z)\cdot{\bf e}_{mn}(x,y) \: dS}{\int_{S_1}{\bf e}_{mn}(x,y)\cdot{\bf e}_{mn}(x,y) \: dS} - e^{-2jk_{z_{mn}}z_1}\\ T &= \frac{e^{jk_{z_{mn}}z_2}}{E_0} \frac{\int_{S_2}{\bf E}(x,y,z)\cdot{\bf e}_{mn}(x,y) \: dS}{\int_{S_2}{\bf e}_{mn}(x,y)\cdot{\bf e}_{mn}(x,y) \: dS} \end{align}

with $|R|^2+|T|^2=1$.

## Some results

Here will be some snapshots and validation cases.

## References

1. J. Jin, The Finite Element Method in Electromagnetics. Second edition. John Wiley & Sons, 2002
 Model developed by A. Modave, B. Klein and C. Geuzaine