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Wave Guide

Parallel Plate Waveguide

TE Waves

\[ \begin{gather} \vec E = E_0 \sin{(k_x x)}\sin{(\omega t - k_z z)} \,\hat{a_y} \\\\ \vec H = - \frac{E_0}{\eta}\sin{\theta} \sin{(k_x x)}\sin{(\omega t - k_z z)} \,\hat{a_x} +\frac{E_0}{\eta}\cos{\theta} \sin{(k_x x)}\sin{(\omega t - k_z z)} \,\hat{a_z} \end{gather} \]
  • cutoff wavelength
\[ \begin{gather} a = m \, \frac{\lambda_x}{2}=\frac{m\,\displaystyle \frac{\lambda}{\cos\theta}}{2} \\\\ \implies \lambda =\frac{2a}{m}\cos\theta \\\\ \implies \lambda_c = \big[\lambda\big]_{\cos\theta=1} =\frac{2a}{m} \\\\ \implies \cos\theta = \frac{\lambda }{\lambda_c} \end{gather} \]
  • guide wavelength (in \(z\) direction)
\[ \begin{gather} \lambda_g = \frac{2\pi}{k_z}=\frac{2\pi}{k\sin\theta}=\frac{\lambda}{\sin\theta} \\\\ \implies \sin\theta=\frac{\lambda}{\lambda_g} \end{gather} \]

by the \(\lambda_c\) and \(\lambda_g\) we can also write the equations as

\[ \begin{gather} \vec E = E_0 \sin{(k_x x)}\sin{(\omega t - k_z z)} \,\hat{a_y} \\\\ \vec H = - \frac{E_0}{\eta}\frac{\lambda}{\lambda_g} \sin{(k_x x)}\sin{(\omega t - k_z z)} \,\hat{a_x} +\frac{E_0}{\eta}\frac{\lambda}{\lambda_c} \sin{(k_x x)}\sin{(\omega t - k_z z)} \,\hat{a_z} \end{gather} \]

TM Waves

\[ \begin{gather} \vec H = H_0 \cos{(k_x x)}\sin{(\omega t - k_z z)} \,\hat{a_y} \\ \end{gather} \]
\[ \begin{align} \vec E &= \eta H_0\sin{\theta} \sin{(k_x x)}\sin{(\omega t - k_z z)} \,\hat{a_x} + \eta H_0\cos{\theta} \sin{(k_x x)}\sin{(\omega t - k_z z)} \,\hat{a_z} \\\\ &= \eta H_0\frac{\lambda}{\lambda_g} \sin{(k_x x)}\sin{(\omega t - k_z z)} \,\hat{a_x} + \eta H_0\frac{\lambda }{\lambda_c} \sin{(k_x x)}\sin{(\omega t - k_z z)} \,\hat{a_z} \end{align} \]

Rectangular Waveguide

  • \(\gamma\)
\[ \begin{gather} k^{2} = k_x^{2} + k_y^{2} + k_z^{2} = k_x^{2} + k_y^{2} + (-j\gamma)^{2} \\\\ \implies \gamma^{2} = k_t^{2} - k^{2} \end{gather} \]

TM mode

tip

notice that we assuming perfect conductors.

thus,

\[ \begin{gather} \vec E_n = 0 \\\\ \implies E_{zs}=0, \qquad\text{at boundary} \end{gather} \]
\[ \begin{gather} E_{zs} = E_0\sin{\left(\frac{m\pi x}{a}\right)}\sin{\left(\frac{n\pi y}{b}\right)}\,e^{-\gamma z} \end{gather} \]

TE mode

\[ \begin{gather} H_{zs} = H_0 \cos{\left(\frac{m\pi x}{a}\right)}\cos{\left(\frac{n\pi y}{b}\right)}\,e^{-\gamma z} \end{gather} \]

Resonators

  • TM
\[ \begin{align} E_{zs} &= \frac{E_0}{2}\sin{\left(\frac{m\pi x}{a}\right)}\sin{\left(\frac{n\pi y}{b}\right)}\left(e^{-jk_z z}+e^{jk_z z}\right) \\\\ &= E_0\sin{\left(\frac{m\pi x}{a}\right)}\sin{\left(\frac{n\pi y}{b}\right)}\cos{\left(\frac{p \pi z}{d}\right)} \end{align} \]
  • TE
\[ \begin{align} H_{zs} &= H_0\cos{\left(\frac{m\pi x}{a}\right)}\cos{\left(\frac{n\pi y}{b}\right)}\sin{\left(\frac{p\pi z}{d}\right)} \end{align} \]

Q Factors

\[ \begin{gather} Q = 2\pi\frac{W}{P_L\,T}= \frac{\omega W}{P_L} \end{gather} \]

recall

\[ \begin{gather} Q = \frac{X}{R} = \frac{\omega L}{R} = \frac{\omega I^{2}L}{I^{2}R} = \frac{\omega W}{P} \end{gather} \]

Dominant Mode

  • the mode with lowest \(f_c\) (i.e. largest \(\lambda_c\)).

Cylindrical

  • \(\beta_c\)
\[ \begin{gather} k^{2} = \beta_c^{2} + \beta^{2} \end{gather} \]

TM mode

\[ \begin{gather} E_{zs} = J_n\big[\beta_c \rho\big] \,(A_n \cos{n\phi} + B_n\sin{n\phi})e^{\mp j\beta z} \end{gather} \]

TE mode

\[ \begin{gather} H_{zs} = J_n\big[\beta_c \rho\big](A_n\cos {n\phi} + B_n \sin{n\phi})e^{\mp j \beta z} \end{gather} \]

Dielectric Slab Waveguide

  • cutoff wavelength
\[ \begin{gather} \lambda_c = \frac{2d\sqrt{\varepsilon_{r1} - \varepsilon_{r2}}}{m}\,\qquad m=0, 1, 2,\dots \end{gather} \]

thinking

\[ \begin{gather} k\cdot 2d \cdot \cos{\theta_i} - 2\angle{\Gamma} = 2m\pi \\\\ \implies \frac{2\pi\sqrt{\varepsilon_{r1}}}{\lambda_0}\cdot d \cdot \cos{\theta_i} - \angle{\Gamma} = m\pi \\\\ \implies \angle{\Gamma} = k\cdot d \cdot \cos{\theta_i} - m\pi \end{gather} \]

when cutoff \(\angle{\Gamma} = 0\)

\[ \begin{gather} k_c \,d\cos{\theta_i} = m\pi \\\\ \implies \frac{2\pi\sqrt{\varepsilon_{r1}}}{\lambda_c}\,d\cos{\theta_i} = m\pi \\\\ \frac{2\pi\sqrt{\varepsilon_{r1}}}{\lambda_c}\,d\sqrt{1-\frac{\varepsilon_2}{\varepsilon_1}} = m\pi \\\\ \implies \lambda_c = \frac{2d\sqrt{\varepsilon_{r1} - \varepsilon_{r2}}}{m} \end{gather} \]