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Rectangular Waveguides

Rectangular waveguides are important for two reasons. First of all, the Laplacian operator separates nicely in Cartesian coordinates, so that the boundary value problem that must be solved is both familiar and straightforward. Second, they are extremely common in actual application in physics laboratories for piping e.g. microwaves around as experimental probes.

In Cartesian coordinates, the wave equation becomes: ( x + y + (&mu#mu;&epsi#epsilon;&omega#omega;^2 - k^2)) &psi#psi;= 0

This wave equation separates and solutions are products of sin, cos or exponential functions in each variable separately. To determine which combination to use it suffices to look at the BC's being satisfied. For TM waves, one solves for $ \psi = E_z$ subject to $ E_z\vert _S = 0$ , which is automatically true if: E_z(x,y) = &psi#psi;_mn(x,y) = E_0(m&pi#pi; xa)(n &pi#pi;yb) where $ a$ and $ b$ are the dimensions of the $ x$ and $ y$ sides of the boundary rectangle and where in principle $ m,n = 0,1,2...$ .

However, the wavenumber of any given mode (given the frequency) is determined from: k^2 = &mu#mu;&epsi#epsilon;&omega#omega;^2 - &pi#pi;^2(m^2a^2 + n^2b^2) + where $ k^2 > 0$ for a ``wave'' to exist to propagate at all. If either index $ m$ or $ n$ is zero, there is no wave, so the first mode that can propagate has a dispersion relation of: k_11^2 = &mu#mu;&epsi#epsilon;&omega#omega;^2 - &pi#pi;^2(1a^2 + 1b^2) so that: &omega#omega;&ge#ge;&pi#pi;&mu#mu;&epsi#epsilon; 1a^2 + 1b^2 = &omega#omega;_c,TM(11) Each combination of permitted $ m$ and $ n$ is associated with a cutoff of this sort - waves with frequencies greater than or equal to the cutoff can support propogation in all the modes with lower cutoff frequencies.

If we repeat the argument above for TE waves (as is done in Jackson, which is why I did TM here so you could see them both) you will be led by nearly identical arguments to the conclusion that the lowest frequency mode cutoff occurs for $ a > b$ , $ m = 1$ and $ n = 0$ to produce the $ H_z(x,y) = \psi(x,y)$ solution to the wave equation above. The cutoff in this case is: &omega#omega;&ge#ge;&pi#pi;&mu#mu;&epsi#epsilon; 1a = &omega#omega;_c, TE(10) < &omega#omega;_c,TM(11) There exists, therefore, a range of frequencies in between where only one TE mode is supported with dispersion: k^2 = k_10^2 = &mu#mu;&epsi#epsilon;&omega#omega;^2 - &pi#pi;^2a^2.

Note well that this mode and cutoff corresponds to exactly one-half a free-space wavelength across the long dimension of the waveguide. The wave solution for the right-propagating TE mode is: H_z & = & H_0 (&pi#pi;xa) e^ikz - i&omega#omega;t
H_x & = & ik&mu#mu;&epsi#epsilon;&omega#omega;^2 - k^2 H_zx = -ika&pi#pi; H_0 (&pi#pi;xa) e^ikz - i&omega#omega;t
E_y & = & &mu#mu;&omega#omega;k H_x = i&mu#mu;&omega#omega;a&pi#pi; H_0 (&pi#pi;xa) e^ikz - i&omega#omega;t We used $ \gamma^2 = \mu\epsilon\omega^2 - k^2 = \pi^2/a^2$ and $ \Vec{E}_\perp = ik/\gamma^2 \del_\perp \psi$ to get the second of these, and $ \Vec{H}_\perp = \frac{k}{\omega\mu}(\Hat{z} \times
\Vec{E}_\perp)$ ) to get the last one.

There is a lot more one can study in Jackson associated with waveguides, but we must move on at this time to a brief look at resonant cavities (another important topic) and multipoles.

next up previous contents
Next: Resonant Cavities Up: Wave Guides Previous: Summary of TE/TM waves   Contents
Robert G. Brown 2017-07-11