What does excitation frequency mean for $TE_{mn}$ modes?

I am working on a problem in which there is a rectangular, perfectly conducting waveguide with side lengths $2D$ along the x-axis and $3D$ along the y-axis. We are supposed to find the values of $m$ and $n$ for which the $TE_{mn}$ modes can be excited if the cutoff frequency is $\frac{7}{2} \omega_{10}$, where $\omega_{mn}$ is the cutoff frequency for the mode.

My main problem with this is I'm unsure what it means for the $TE_{mn}$ modes to be excited, nor what the excitation frequency is. The magnetic field corresponding to the $TE_{mn}$ mode is:

$$B_z = B_0 cos(\frac{m \pi x}{3D})cos(\frac{n \pi y}{2D})$$

The cutoff frequency is:

$$\omega_{mn} = c \pi \sqrt{(m/3D)^2+(n/2D)^2}$$ making the minimum cutoff frequency: $$\omega_{10} = \frac{c \pi}{3D}$$ so that the excitation frequency is: $$\omega = \frac{7 c \pi}{6D}$$ As mentioned before, I'm not really sure where to go from here, but I chose to look for all of the modes where the cutoff frequency was less than $\omega$. This led to:

$$4m^2 + 9n^2 < 49$$

I then concluded that the modes were:

$$TE_{10}, TE_{01}, TE_{11}, TE_{20},TE_{02},TE_{12},TE_{21},TE_{30},TE_{31}$$

• Yeah, that's pretty much what you needed to do. – Emilio Pisanty Nov 27 '16 at 22:43

My main problem with this is I'm unsure what it means for the $TE_{mn}$ modes to be excited

This means, you applied a signal to the input end of the waveguide that has non-zero overlap with the mode in question.

For example, this can be done with a little antenna projected into the waveguide and fed by a coax or other transmission line. We are supposed to find the values of $m$ and $n$ for which the $TE_{mn}$ modes can be excited
In my opinion, this is a poor way to word the question. You can excite any mode you like, however that mode may produce evanescent rather than travelling waves at the excitation frequency, and thus not be useful for transmitting power or signals along the waveguide. A better wording would be "Find the values of m and n for all the $TE_{mn}$ modes that are propagating modes [at some frequency]".
• @infinitylord, an evanescent wave is a solution to the wave equation with the form $\exp(\gamma_z z)$ instead of $\exp(i k_z z)$ (or some analogous form in different coordinate systems). It's not the same as a standing wave. A standing wave is a superposition of two travelling waves. – The Photon Nov 28 '16 at 2:43
• For example, in a rectangular waveguide if you excite a mode below its cut-off frequency, the corresponding solution in 3 D would be: $A\exp(i k_x x)\exp(i k_y y)\exp(-\gamma_z z)$, an evanescent wave. You'll run into similar results in QM for the wavefunction at the edges of a finite energy well, in EM in a dilectric waveguide in the cladding region, etc., etc. – The Photon Nov 28 '16 at 2:49