1
$\begingroup$

I am trying to understand modes in a cylindrical dielectric waveguide. Does a particular mode only propagate for one particular frequency? Or, does a mode exist simply as long as the signal is above the cut-off frequency?

Basically, if I propagate a signal of a single frequency down the waveguide (a frequency that I know corresponds to the peak resonance of some mode), will I excite only that one mode?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

Transverse (TE and TM) modes in a recangular waveguide typically have dispersion equations $\omega(k_z)$ of the form $$ \omega^2= ak_z^2 + \omega^2_{\rm min}, $$ where $a$ and $\omega_{\rm min}$ depend on the mode and the geometry of the waveguide and $z$ is the direction down the waveguide. So, yes, once the frequency $\omega$ is greater that $\omega_{\rm min}$, there will be some real wavenumber $k_z$ for which the mode will propagate at the given frequency. For frequencies lower than $\omega_{\rm min}$ the wavenumber $k_z$ will be imaginary, and so the mode will be evanescent.

Improve this answer
$\endgroup$
5
  • $\begingroup$ so is it that a mode will propagate at any frequency above that cut-off frequency, but it might be more strongly excited (greater resonance) for some frequencies than others? $\endgroup$
    – zoopnoodle
    Commented Dec 16, 2020 at 13:06
  • $\begingroup$ The extent to which a given mode is excited by a source depends on the impedence matching. In TE and TM dispersive modes the impedence does depend strongly on the frequency near $\omega_{\rm min}$ as you can read here:en.wikipedia.org/wiki/Wave_impedance. The impedence does not depend on the frequency in non-dispersive TEM modes such as in a coaxial cable. $\endgroup$
    – mike stone
    Commented Dec 16, 2020 at 13:27
  • $\begingroup$ Thank you for your answers, I'm afraid I'm not managing to wrap my head around this. Will a signal with a single frequency excite one mode or all modes with a lower cut-off frequency than it? $\endgroup$
    – zoopnoodle
    Commented Dec 16, 2020 at 13:44
  • $\begingroup$ In general a signal of a given frequency will excite all modes, and will radiate down the waveguide those all modes that are above their cutoff frquency. The modes will propagate at different speeds which is why one uses the non-dispersive TEM in coax if you want to send data. If you wish to propagate a single TE mode, for example, it will have to be the one with the lowest cutoff freqency. $\endgroup$
    – mike stone
    Commented Dec 16, 2020 at 13:56
  • $\begingroup$ Okay, I believe I understand - thank you kindly! $\endgroup$
    – zoopnoodle
    Commented Dec 16, 2020 at 14:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.