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}
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{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}
0 except \(m = p \ne 0\) combining with
-
\(c\,q \ne b\,n\)
\begin{equation}
-\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)}
\end{equation}
-
\(c\,q = b\,n\) and \(q \ne 0\)
\begin{equation}
\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}
\end{equation}
-
\(c\,q = b\,n\) and \(q = 0\)
\begin{equation}
\frac{a b}{2}
\end{equation}
Orthogonal Integral 2
Form
\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}
0 except \(m = p\) combining with
-
\(c\,q = b\,n\) and \(p = 0\) (and \(q \ne 0\))
\begin{equation}
-\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}
\end{equation}
-
\(c\,q = b\,n\) and \(p \ne 0\) (and \(q \ne 0\))
\begin{equation}
-\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}
\end{equation}
-
\(c\,q \ne b\,n\) and \(p = 0\)
\begin{equation}
-\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)}
\end{equation}
-
\(c\,q \ne b\,n\) and \(p \ne 0\)
\begin{equation}
-\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)}
\end{equation}
