Rectangular cavity resonator

A rectangular cavity resonator is a piece of rectangular waveguide of dimension a, b, and c.
In figure a rectangular cavity resonator of sides a, b, c is shown.

TE waves in a rectangular cavity resonator

Here we will consider about two transverse electric waves, out of two one moves in +z direction and another in -z direction.
Let the transverse component of electric field for TE waves propagating along +z direction be

$\bar{E}_{1t}(x,y,z,t)=\bar{E}_{1t}(x,y)exp[j(kz-\omega t)]$

Here E_1t(x, y) is the part of E_1t which is independent of z and t, which can be derived from the z-component of H_1z(x,y) by using the equation

$\bar{E}_{1t}(x,y)=-\frac{j\omega\mu}{k'^{2}}\hat{k}\times\left(\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}\right)H_{1z}(x,y)\;\;\;\;\;[k'^{2}=\omega^{2}\mu\varepsilon-k^{2}]$

The value of H_1t (x, y) can be obtained by using

$\left(\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}\right)H_{1z}(x,y)+k'^{2}H_{1z}(x,y)=0$

Boundary conditions

$\left.\begin{matrix}(\partial H_{1z}/\partial x)\end{matrix}\right|_{x=0}=0\;\;\;\;and\;\;\;\;\left.\begin{matrix}(\partial H_{1z}/\partial x)\end{matrix}\right|_{x=a}=0$

$\left.\begin{matrix}(\partial H_{1z}/\partial y)\end{matrix}\right|_{y=0}=0\;\;\;\;and\;\;\;\;\left.\begin{matrix}(\partial H_{1z}/\partial y)\end{matrix}\right|_{y=b}=0$

On solving the above equation we get

$H_{1z}(x,y)=H_{0}\;cos\frac{m\pi x}{a}cos\frac{n\pi y}{b}$

$Here\;\;\;\;\;k'=\pi\sqrt{\frac{m^{2}}{a^{2}}+\frac{n^{2}}{b^{2}}}$

But

$\bar{E}_{1t}(x,y)=-\frac{j\omega\mu}{k'^{2}}\hat{k}\times\left(\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}\right)H_{1z}(x,y)$

$\therefore\;\;\;\;\;\bar{E}_{1t}(x,y)=-\frac{j\omega\mu}{k'^{2}}\left(\hat{j}\frac{\partial}{\partial x}-\hat{i}\frac{\partial}{\partial y}\right)H_{0}\;cos\frac{m\pi x}{a}cos\frac{n\pi y}{b}$

$or\;\;\;\;\;\bar{E}_{1t}=\frac{j\omega\mu H_{0}}{k'^{2}}\left(\hat{j}\frac{m\pi}{a}cos\frac{n\pi y}{b}sin\frac{m\pi x}{a}-\hat{i}\frac{n\pi}{b}sin\frac{n\pi y}{b}cos\frac{m\pi x}{a}\right)$

The transverse component of electric field for TE waves propagating along -z direction will be

$\bar{E}_{2t}(x,y,z,t)=\bar{E}_{2t}(x,y)exp[-j(kz+\omega t)]$

On superposition of waves having electric field E_1t(x, y, z, t) and E_2t(x, y, z, t), we get the electric field

$\bar{E}_{t}(x,y,z,t)=\bar{E}_{1t}(x,y,z,t)+\bar{E}_{2t}(x,y,z,t)$
$or\;\;\;\;\;\bar{E}_{t}(x,y,z,t)=[\bar{E}_{1t}(x,y)e^{jkz}+\bar{E}_{2t}(x,y)e^{-jkz}]e^{-j\omega t}$

Since the tangential component of electric field intensity vanishes at the bounding surface of the cavity.

$\therefore\;\;\;\;\;\bar{E}_{t}(x,y,z,t)|_{z=0}=0\;\;\;and\;\;\;\bar{E}_{t}(x,y,z,t)|_{z=c}=0$

On applying boundary condition E_t (x, y, z, t) |_{z = 0} = 0, we get

$[\bar{E}_{1t}(x,y)+\bar{E}_{2t}(x,y)]e^{-j\omega t}=0$

$or\;\;\;\;\;\bar{E}_{2t}(x,y)=-\bar{E}_{1t}(x,y)$

$\therefore\;\;\;\;\;\bar{E}_{t}(x,y,z,t)=[\bar{E}_{1t}(x,y)e^{jkz}-\bar{E}_{1t}(x,y)e^{-jkz}]e^{-j\omega t}$

$or\;\;\;\;\;\bar{E}_{t}(x,y,z,t)=\bar{E}_{1t}(x,y)[e^{jkz}-e^{-jkz}]e^{-j\omega t}$

$or\;\;\;\;\;\bar{E}_{t}(x,y,z,t)=2j\bar{E}_{1t}(x,y)sinkze^{-j\omega t}$

On applying boundary condition E_t (x, y, z, t) |_{z = c} = 0, we get

$2j\bar{E}_{1t}(x,y)sinkc\;e^{-j\omega t}=0$

or sin kc = 0

or kc = pπ ; where p = 0, 1, 2, 3, …

or k = pπ / c

$\therefore\;\;\;\;\;\bar{E}_{t}(x,y,z,t)=2j\bar{E}_{1t}(x,y)sin\frac{p\pi z}{c}e^{-j\omega t}$

$But\;\;\;\;\;\bar{E}_{1t}=\frac{j\omega\mu H_{0}}{k'^{2}}\left(\hat{j}\frac{m\pi}{a}cos\frac{n\pi y}{b}sin\frac{m\pi x}{a}-\hat{i}\frac{n\pi}{b}sin\frac{n\pi y}{b}cos\frac{m\pi x}{a}\right)$

$\therefore\;\;\;\;\;\bar{E}_{t}(x,y,z,t)=-\frac{2\omega\mu H_{0}}{k'^{2}}\left(\hat{j}\frac{m\pi}{a}sin\frac{m\pi x}{a}cos\frac{n\pi y}{b}-\hat{i}\frac{n\pi}{b}cos\frac{m\pi x}{a}sin\frac{n\pi y}{b}\right)sin\frac{p\pi z}{c}e^{-j\omega t}$

Here p = 0, 1, 2, 3, …

$\therefore\;\;\;\;\;{E}_{x}(x,y,z,t)=\frac{2\omega\mu H_{0}}{k'^{2}}\times\frac{n\pi}{b}cos\frac{m\pi x}{a}sin\frac{n\pi y}{b}sin\frac{p\pi z}{c}e^{-j\omega t}$

${E}_{y}(x,y,z,t)=-\frac{2\omega\mu H_{0}}{k'^{2}}\times\frac{m\pi}{a}sin\frac{m\pi x}{a}cos\frac{n\pi y}{b}sin\frac{p\pi z}{c}e^{-j\omega t}$

$and\;\;\;\;\;{E}_{z}(x,y,z,t)=0$

The components of magnetic field intensity can be calculated by Maxwell’s equation.
From Maxwell’s third equation

$\triangledown\times\vec{E}(x,y,z,t)=-\mu\frac{\partial}{\partial t}\vec{H}(x,y,z,t)$

$\because\;\;\;\;\;{E}_{z}(x,y,z,t)=0$

$\therefore\;\;\;\;\;\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\E_{x}&E_{y}&0\\\end{vmatrix}=-\mu(-j\omega)\vec{H}=j\omega\mu(\hat{i}H_{x}+\hat{j}H_{y}+\hat{k}H_{z})$

$\because\;\;\;\;\;{E}_{y}(x,y,z,t)=-\frac{2\omega\mu H_{0}}{k'^{2}}\times\frac{m\pi}{a}sin\frac{m\pi x}{a}cos\frac{n\pi y}{b}sin\frac{p\pi z}{c}e^{-j\omega t}$

$\therefore\;\;\;\;\;H_{x}=-\frac{1}{j\omega\mu}\frac{\partial E_{y}}{\partial z}=-\frac{1}{j\omega\mu}\left(-\frac{2\omega\mu H_{0}}{k'^{2}}\right)\times\frac{m\pi}{a}\times\frac{p\pi}{c}sin\frac{m\pi x}{a}cos\frac{n\pi y}{b}cos\frac{p\pi z}{c}e^{-j\omega t}$

$or\;\;\;\;\;H_{x}=-\frac{2jH_{0}}{k'^{2}}\times\frac{m\pi}{a}\times\frac{p\pi}{c}sin\frac{m\pi x}{a}cos\frac{n\pi y}{b}cos\frac{p\pi z}{c}e^{-j\omega t}$

$\because\;\;\;\;\;{E}_{x}(x,y,z,t)=\frac{2\omega\mu H_{0}}{k'^{2}}\times\frac{n\pi}{b}cos\frac{m\pi x}{a}sin\frac{n\pi y}{b}sin\frac{p\pi z}{c}e^{-j\omega t}$

$\therefore\;\;\;\;\;H_{y}=\frac{1}{j\omega\mu}\frac{\partial E_{x}}{\partial z}=\frac{1}{j\omega\mu}\left(\frac{2\omega\mu H_{0}}{k'^{2}}\right)\times\frac{n\pi}{b}\times\frac{p\pi}{c}cos\frac{m\pi x}{a}sin\frac{n\pi y}{b}cos\frac{p\pi z}{c}e^{-j\omega t}$

$or\;\;\;\;\;H_{y}=-\frac{2jH_{0}}{k'^{2}}\times\frac{n\pi}{b}\times\frac{p\pi}{c}cos\frac{m\pi x}{a}sin\frac{n\pi y}{b}cos\frac{p\pi z}{c}e^{-j\omega t}$

$and\;\;\;\;\;H_{z}=\frac{1}{j\omega\mu}\left[\frac{\partial E_{y}}{\partial x}-\frac{\partial E_{x}}{\partial y}\right]=2jH_{0}cos\frac{m\pi x}{a}cos\frac{n\pi y}{b}sin\frac{p\pi z}{c}e^{-j\omega t}$

$\because\;\;\;\;\;k'=\pi\sqrt{\frac{m^{2}}{a^{2}}+\frac{n^{2}}{b^{2}}}\;\;\;and\;\;\;k=p\pi /c$

$Now\;\;\;\;\;k'^{2}=\omega^{2}\mu\varepsilon-k^{2}$

$or\;\;\;\;\;\pi^{2}\left(\frac{m^2}{a^2}+\frac{n^2}{b^2}\right)=\omega^{2}\mu\varepsilon-\left(\frac{p\pi}{c}\right)^{2}$

$or\;\;\;\;\;\omega^{2}\mu\varepsilon=\left(\frac{m\pi}{a}\right)^{2}+\left(\frac{n\pi}{b}\right)^{2}+\left(\frac{p\pi}{c}\right)^{2}$

$or\;\;\;\;\;\omega=\frac{\pi}{\sqrt{\mu\varepsilon}}\left[\frac{m^2}{a^2}+\frac{n^2}{b^2}+\frac{p^2}{c^2}\right]^{1/2}$

$or\;\;\;\;\;f=\frac{1}{2\sqrt{\mu\varepsilon}}\left[\frac{m^2}{a^2}+\frac{n^2}{b^2}+\frac{p^2}{c^2}\right]^{1/2}$

This equation give the resonant frequencies for different sets of m, n, p for different cavity modes.
If two or more independent cavity modes have the same resonant frequencies then it is known as degenerate frequency.
It is found that the modes TE₀₀₀, TE₀₀₁, TE₀₁₀ and TE₁₀₀ do not exist in the cavity.
The lowest modes (dominant mode), which are possible in TE mode are TE₁₀₁, TE₀₁₁ etc.
The allowed frequencies corresponding to dominant modes is fundamental frequency.

To know in detail about this lecture click here

TM waves in a rectangular cavity resonator

Here we will consider about two transverse electric waves, out of two one moves in +z direction and another in -z direction.
Let the transverse component of electric field for TM waves propagating along +z direction be

$\bar{E}_{1t}(x,y,z,t)=\bar{E}_{1t}(x,y)exp[j(kz-\omega t)]$

Here E_1t(x, y) is the part of E_1t which is independent of z and t, which can be derived from the z-component of E_1z(x,y) by using the equation

$\bar{E}_{1t}(x,y)=\frac{jk}{k'^{2}}\left(\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}\right)E_{1z}(x,y)\;\;\;\;\;[k'^{2}=\omega^{2}\mu\varepsilon-k^{2}]$

The value of E_1t (x, y) can be obtained by using

$\left(\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}\right)E_{1z}(x,y)+k'^{2}E_{1z}(x,y)=0$

Boundary conditions

$\left.\begin{matrix}(\partial E_{1z}/\partial x)\end{matrix}\right|_{x=0}=0\;\;\;\;and\;\;\;\;\left.\begin{matrix}(\partial E_{1z}/\partial x)\end{matrix}\right|_{x=a}=0$

$\left.\begin{matrix}(\partial E_{1z}/\partial y)\end{matrix}\right|_{y=0}=0\;\;\;\;and\;\;\;\;\left.\begin{matrix}(\partial E_{1z}/\partial y)\end{matrix}\right|_{y=b}=0$

On solving the above equation we get

The transverse component of electric field for TM waves propagating along –z direction will be

The value of E_z (x, y, z, t) can be obtained as

The components of magnetic field intensity can be calculated by Maxwell’s equation.
From Maxwell’s third equation

This equation give the resonant frequencies for different sets of m, n, p for different cavity modes.
If two or more independent cavity modes have the same resonant frequencies then it is known as degenerate frequency.
It is found that the modes TM₀₀₀, TM₀₀₁, TM₀₁₀ and TM₁₀₀, TM₁₀₁, TM₀₁₁ do not exist in the cavity.
The lowest possible mode (dominant mode) in TM wave is TM₁₁₀.
The allowed frequency corresponding to dominant modes is fundamental frequency