Analytical solutions of the integrals for the mode-matching of symmetric rectanglular wave guide

Total power \(P\) over the cross-section is normalized

TE modes

\begin{aligned} E_z &= 0 \\ H_z &= \frac{j\,k_c}{\sqrt{ab|\beta|\omega\mu}}\; A \cos\frac{m\pi x}{a} \cos\frac{n\pi y}{b} \\ E_x &= \frac{-1}{k_c\sqrt{ab}}\; \sqrt{\frac{\omega\mu}{|\beta|}} \frac{n}{b} A \cos\frac{m\pi x}{a} \sin\frac{n\pi y}{b} \\ E_y &= \frac{1}{k_c\sqrt{ab}}\; \sqrt{\frac{\omega\mu}{|\beta|}} \frac{m}{a} A \sin\frac{m\pi x}{a} \cos\frac{n\pi y}{b} \\ H_x &= \frac{-\beta}{k_c\sqrt{ab}}\; \sqrt{\frac{1}{\omega\mu|\beta|}} \frac{m}{a} A \sin\frac{m\pi x}{a} \cos\frac{n\pi y}{b} \\ H_y &= \frac{-\beta}{k_c\sqrt{ab}}\; \sqrt{\frac{1}{\omega\mu|\beta|}} \frac{n}{b} A \cos\frac{m\pi x}{a} \sin\frac{n\pi y}{b} \end{aligned}

TM modes

\begin{aligned} E_z &= \frac{v\,j\,k_c}{\sqrt{ab|\beta|\omega\epsilon}}\; B \sin\frac{m\pi x}{a} \sin\frac{n\pi y}{b} \\ H_z &= 0 \\ E_x &= \frac{v\,\beta}{k_c\sqrt{ab}}\; \sqrt{\frac{1}{\omega\epsilon|\beta|}} \frac{m}{a} B \cos\frac{m\pi x}{a} \sin\frac{n\pi y}{b} \\ E_y &= \frac{v\,\beta}{k_c\sqrt{ab}}\; \sqrt{\frac{1}{\omega\epsilon|\beta|}} \frac{n}{b} B \sin\frac{m\pi x}{a} \cos\frac{n\pi y}{b} \\ H_x &= \frac{-v}{k_c\sqrt{ab}}\; \sqrt{\frac{\omega\epsilon}{|\beta|}} \frac{n}{b} B \sin\frac{m\pi x}{a} \cos\frac{n\pi y}{b} \\ H_y &= \frac{v}{k_c\sqrt{ab}}\; \sqrt{\frac{\omega\epsilon}{|\beta|}} \frac{m}{a} B \cos\frac{m\pi x}{a} \sin\frac{n\pi y}{b} \\ \end{aligned}
where \(v=1\) if forwards, \(-1\) backwards; \begin{equation} k_c = \sqrt{ \left(\frac{m\pi}{a}\right)^2 + \left(\frac{n\pi}{b}\right)^2 } \end{equation} and \begin{equation} \beta = \sqrt{k^2-k_c^2} \end{equation}


Orthogonal Integration of Fields for an E-Plane Discontinuity

For E-plane discontinuities, \(\int H\mathrm{d}s\) on the small aperture for the continuation of the field whereas \(\int E\mathrm{d}s\) on the large aperture, in order to include the \(E_\perp=0\) metal boundary.

Note: Weight (Base) functions should be conjugated to the transversal fields, however since we put \(j\) into the common coefficiency, the conjugation can be omited.

Orthogonal Integral 1


\begin{equation} \begin{split} \int_{-\frac{a}{2}}^{\frac{a}{2}} \; \int_{-\frac{b}{2}}^{\frac{b}{2}} \quad &\sin\frac{m\pi\,(x+a/2)}{a} \cos\frac{n\pi\,(y+c/2)}{c} \\ &\sin\frac{p\pi\,(x+a/2)}{a} \cos\frac{q\pi\,(y+b/2)}{b} \quad \mathrm{d}y\; \mathrm{d}x \end{split} \end{equation}

Solution (mathematica)

0 except \(m = p \ne 0\) combining with

Orthogonal Integral 2


\begin{equation} \begin{split} \int_{-\frac{a}{2}}^{\frac{a}{2}} \; \int_{-\frac{b}{2}}^{\frac{b}{2}} \quad &\cos\frac{m\pi\,(x+a/2)}{a} \sin\frac{n\pi\,(y+c/2)}{c} \\ &\cos\frac{p\pi\,(x+a/2)}{a} \sin\frac{q\pi\,(y+b/2)}{b} \quad \mathrm{d}y\; \mathrm{d}x \end{split} \end{equation}

Solution (mathematica)

0 except \(m = p\) combining with