##### Rectangular Waveguides
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Assume a waveguide with rectangular cross-section of width \(a\) along \(\hat{\boldsymbol x}\) and \(b\) along \(\hat{\boldsymbol y}\). Without loss of generality (and to follow common historical convention) we will assume \(a \geq b\).

In view of this geometry, we will use separation of variables in Cartesian coordinates. We thus put \[ B_z (x,y) = X(x) Y(y) \] so \[ Y \frac{d^2 X}{dx^2} + X \frac{d^2 Y}{dy^2} + \left[(\omega/c)^2 - k^2 \right] XY = 0. \] Invoking separation of variables, we obtain \[ \frac{d^2 X}{dx^2} = -k_x^2 X, \hspace{5mm} \frac{d^2 Y}{dy^2} = -k_y^2 Y, \hspace{5mm} k_x^2 + k_y^2 + k^2 = \frac{\omega^2}{c^2}. \] The general solution for \(X\) is \[ X(x) = A \sin (k_x x) + B \cos (k_x x) \] and to fit the boundary conditions we must have that \(\partial B_x/\partial x\) must vanish at \(x=0, a\), so \(dX/dx\) must vanish at \(x = 0, a\) and thus \(A = 0\) and \[ k_x = m \pi/a, \hspace{5mm} m = 0, 1, \cdots \] The same reasoning applies to \(Y\) so we get \[ B_z (x, y) = B_0 \cos \frac{m\pi x}{a} \cos \frac{n \pi y}{b} \] which is called the TE\(_{mn}\) mode. Its wavenumber is \[ k = \sqrt{\left(\frac{\omega}{c}\right)^2 - \pi^2 \left[\left(\frac{m}{a}\right)^2 + \left(\frac{n}{b}\right)^2 \right]}. \] In view of this, if the frequency is low enough, namely \[ \omega < c\pi \sqrt{\left(\frac{m}{a}\right)^2 + \left(\frac{n}{b}\right)^2} \equiv \omega_{mn} \] then the wavenumber is imaginary and the travelling wave is exponentially attenuated. This frequency is called the cutoff frequency for this mode.

In terms of the cutoff frequency, we have wavevector \[ k = \frac{1}{c} \sqrt{\omega^2 - \omega_{mn}^2} \] and wave velocity \[ v = \frac{\omega}{k} = \frac{c}{\sqrt{1 - \left(\frac{\omega_{mn}}{\omega}\right)^2}} \] which is greater than \(c\). The energy of the wave however propagates at the group velocity \[ v_g = \frac{d\omega}{dk} = c \sqrt{1 - \left(\frac{\omega_{mn}}{\omega}\right)^2}. \]

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Author: Jean-Sébastien Caux

Created: 2022-03-24 Thu 08:42