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{{metamodel|waveguides}}
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{{metamodelGetDP|waveguides}}
  
 
{{mytexdefs}}
 
{{mytexdefs}}
  
== Introduction ==
+
== Additional information ==
  
This academic example introduces the numerical solution of eigenvalue problems.
+
To run the model, open '''main.pro''' or a file '''.pro''' with Gmsh.
The Helmholtz equation (scalar and vector versions) is considered with homogeneous Dirichlet boundary conditions,
 
for different basic geometries (i.e. linear, squared, circular and cuboid domain).
 
These problems have a family of solutions. The eigenvalue solver used for this example
 
provide the first eigenvalues and the associated eigenfunctions (i.e. the first possible solutions).
 
  
$\rightarrow$ To run the example, open '''main.pro''' with Gmsh.
+
== Description of the model ==
  
== Description of the problem ==
+
=== Classical waveguides ===
  
==== Mathematical and numerical formulations ====
+
==== General case ====
  
==== Waveguides ====
+
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
 +
\begin{equation}
 +
  {\bf H}_t = \pm\frac{1}{Z} \hat{\bf z}\times{\bf E}_t
 +
\end{equation}
 +
where the wave impedance $Z$ is given by
 +
\begin{equation}
 +
  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.
 +
\end{equation}
 +
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 $\ell$:
 +
\begin{equation}
 +
  k_\ell
 +
    = \sqrt{\mu\varepsilon\omega^2-\gamma^2_\ell}
 +
    = \sqrt{\mu\varepsilon} \sqrt{\omega^2-\omega^2_\ell}
 +
\end{equation}
 +
where $\omega_\ell$ is the cutoff frequency, defined by
 +
\begin{equation}
 +
  \omega_\ell=\frac{\gamma_\ell}{\sqrt{\mu\varepsilon}}
 +
\end{equation}
 +
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}
 +
 
 +
 
 +
 
 +
 
 +
 
 +
<!--
 +
 
 +
The complete solution for the TM<sub>10</sub> and TE<sub>10</sub> modes are, respectively,
 +
\begin{align}
 +
  \begin{cases}
 +
    \displaystyle E_z = E_0 \sin\left(\frac{\pi x}{a}\right) e^{ikz-i\omega t} \\
 +
    \displaystyle E_x = i\frac{ka}{\pi} E_0 \cos\left(\frac{\pi x}{a}\right) e^{ikz-i\omega t} \\
 +
    \displaystyle H_y = i\frac{\varepsilon\omega a}{\pi} E_0 \cos\left(\frac{\pi x}{a}\right) e^{ikz-i\omega t}
 +
  \end{cases}
 +
  \quad\quad\text{and}\quad\quad
 +
  \begin{cases}
 +
    \displaystyle H_z = H_0 \cos\left(\frac{\pi x}{a}\right) e^{ikz-i\omega t} \\
 +
    \displaystyle H_x = -i\frac{ka}{\pi} H_0 \sin\left(\frac{\pi x}{a}\right) e^{ikz-i\omega t} \\
 +
    \displaystyle E_y =i\frac{\mu\omega a}{\pi} H_0 \sin\left(\frac{\pi x}{a}\right) e^{ikz-i\omega t}
 +
  \end{cases}
 +
\end{align}
 +
 
 +
 
 +
-->
 +
 
 +
=== Discontinuities and networks ===
 +
 
 +
==== Discontinuity in a parallel-plate waveguide ====
 +
 
 +
See <ref name=Jin2002 />, 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 <ref name=Jin2002 />, 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 ==
 
== Some results ==
  
Here are some snapshots.
+
Here will be some snapshots and validation cases.
  
 
== References ==
 
== References ==
Line 27: Line 286:
 
<references>
 
<references>
  
<ref name=Evans2010> L.C. Evans, ''Partial Differential Equations''. Second edition. American Mathematical Society, 2010</ref>
 
 
<ref name=Jin2002> J. Jin, ''The Finite Element Method in Electromagnetics''. Second edition. John Wiley & Sons, 2002</ref>
 
<ref name=Jin2002> J. Jin, ''The Finite Element Method in Electromagnetics''. Second edition. John Wiley & Sons, 2002</ref>
  
 
</references>
 
</references>
  
 
+
{{metamodelfooter|waveguides}}
{{metamodelfooter|academic_eigenvalues}}
 

Revision as of 08:58, 25 July 2015

2D and 3D models of metallic waveguides

Download model archive (waveguides.zip)
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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 \begin{equation} {\bf H}_t = \pm\frac{1}{Z} \hat{\bf z}\times{\bf E}_t \end{equation} where the wave impedance $Z$ is given by \begin{equation} 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. \end{equation} 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 $\ell$: \begin{equation} k_\ell = \sqrt{\mu\varepsilon\omega^2-\gamma^2_\ell} = \sqrt{\mu\varepsilon} \sqrt{\omega^2-\omega^2_\ell} \end{equation} where $\omega_\ell$ is the cutoff frequency, defined by \begin{equation} \omega_\ell=\frac{\gamma_\ell}{\sqrt{\mu\varepsilon}} \end{equation} 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. 1.0 1.1 J. Jin, The Finite Element Method in Electromagnetics. Second edition. John Wiley & Sons, 2002

Model developed by A. Modave, B. Klein and C. Geuzaine