When defining the dispersion relation in a waveguide we have an equation that also includes a cutoff wave-number. This cutoff wave number depends on the geometry of the waveguide.

$$ k_z = \sqrt{\omega^2/c^2 - (m\pi/a)^2 - (n\pi/b)^2}, $$ where $a,b$ are the dimensions of the waveguide and $n,m$ are the mode numbers.

I understand the mathematics behind this that when $k_z > \omega/c$ then $k_z$ (wavenumber) becomes imaginary and we get an exponential decay of the electric field. But I am not able to understand it intuitively.

What is so special about a waveguide geometry that below certain frequency we get no propagation of power but just an exponential decay in the waveguide?

Can this be explained using the phase relation between reflected and incident waves? Below cutoff frequency will there still be interference between incident and reflected wave?

What exactly is happening between the reflected and incident wave that we get an exponential decay when we excite a waveguide below cutoff frequency?

  • $\begingroup$ Note that we use MathJax to typeset mathematics; you can find a good tutorial here. $\endgroup$ Commented Jul 12, 2020 at 18:16
  • $\begingroup$ Most of the time, you have to trust the mathematics. The cutoff has nothing to do with interference between incident and reflected waves. $\endgroup$ Commented Jul 13, 2020 at 2:35

1 Answer 1


We always talk about the cutoff frequency of a particular mode. Have in mind that, when we talk about modes, we are referring to particular solutions to the Helmholtz Equation in a specific scenario with some boundary conditions that help us to determine these solutions.

Below the cutoff frequency essentially your wave cannot propagate because it is not fulfilling the necessary conditions for propagating inside your structure. As Jerrold pointed out in one of the comments, you have to trust the mathematics

  • 2
    $\begingroup$ A somewhat intuitvely explanation for me is that, below a certain frequency, the wavelength/mode is simply too big to "fit" in the waveguide and therefore doesn't propagate $\endgroup$
    – Gary Allen
    Commented Mar 23, 2023 at 13:04

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