Analytical solutions of the integrals for the mode-matching of symmetric rectanglular wave guide
2017-05-06

# 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; $$k_c = \sqrt{ \left(\frac{m\pi}{a}\right)^2 + \left(\frac{n\pi}{b}\right)^2 }$$ and $$\beta = \sqrt{k^2-k_c^2}$$

## Notes

• This normalization is only valid for both $$m$$ and $$n$$ non-zero. for TE$$_{m\ 0}$$ or TE$$_{0\ n}$$ there is a factor of 2 in power, i.e., $$\sqrt2$$ in amplitude.
• The name individual parts of $$E_x$$, $$E_y$$ are given for the coefficient pairs $$m/a$$, $$n/b$$ etc. for $$x$$, $$y$$ and their signs, respectively. Where as the common part means the rest terms with square roots, $$k_c$$, $$v$$ and eventually the $$\beta$$ with its sign (propogating direction).
• The coefficient of $$E_y$$ in TM mode is chosen to be positive real (or negative imaginary). A side effect is that this makes the junction reciprocal.

# 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

### Form

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

### Solution (mathematica)

0 except $$m = p \ne 0$$ combining with
• $$c\,q \ne b\,n$$ $$-\frac{a b^2 c n \left(\sin \left(\frac{\pi n (c-b)}{2 c}\right)-(-1)^q \sin \left(\frac{\pi n (b+c)}{2 c}\right)\right)}{2 \pi (b n-c q) (b n+c q)}$$
• $$c\,q = b\,n$$ and $$q \ne 0$$ $$\frac{a b \left(\sin \left(\frac{\pi q (b-c)}{2 b}\right)+\sin \left(\frac{\pi q (3 b+c)}{2 b}\right)+2 \pi q \cos \left(\frac{\pi q (b-c)}{2 b}\right)\right)}{8 \pi q}$$
• $$c\,q = b\,n$$ and $$q = 0$$ $$\frac{a b}{2}$$

## Orthogonal Integral 2

### Form

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

### Solution (mathematica)

0 except $$m = p$$ combining with
• $$c\,q = b\,n$$ and $$p = 0$$ (and $$q \ne 0$$) $$-\frac{a b \left(\sin \left(\frac{\pi q (b-c)}{2 b}\right)+\sin \left(\frac{\pi q (3 b+c)}{2 b}\right)-2 \pi q \cos \left(\frac{\pi q (b-c)}{2 b}\right)\right)}{4 \pi q}$$
• $$c\,q = b\,n$$ and $$p \ne 0$$ (and $$q \ne 0$$) $$-\frac{a b \left( \sin \left(\frac{\pi q (b-c)}{2 b}\right) + \sin \left(\frac{\pi q (3 b+c)}{2 b}\right) -2 \pi q \cos \left(\frac{\pi q (b-c)}{2 b}\right) \right)}{8 \pi q}$$
• $$c\,q \ne b\,n$$ and $$p = 0$$ $$-\frac{a b c^2 q \left(\sin \left(\frac{\pi n (c-b)}{2 c}\right)-(-1)^q \sin \left(\frac{\pi n (b+c)}{2 c}\right)\right)}{\pi (b n-c q) (b n+c q)}$$
• $$c\,q \ne b\,n$$ and $$p \ne 0$$ $$-\frac{a b c^2 q \left(\sin \left(\frac{\pi n (c-b)}{2 c}\right)-(-1)^q \sin \left(\frac{\pi n (b+c)}{2 c}\right)\right)}{2 \pi (b n-c q) (b n+c q)}$$