Timeline for Frequency and Wavenumbers in Rectangular Waveguide for TE01
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 13 at 3:44 | comment | added | Puk | @hyportnex Yes, I should have written $k_3^2 > 0$, thanks. $k_3^2 < 0$ (for which $k_3$ is imaginary) gives the evanescent modes. | |
| Jan 13 at 3:29 | comment | added | hyportnex | should write: $k_3^2 >0$ for a propagating mode, $k_3 <0$ just propagates in the opposite direction to $k_3 >0$ . Below cutoff, also called evanescent mode, waveguides are used to design frequency selective structures, filters and some such, Regarding both their performance (insertion loss/selectivity) and size they are between lumped element and resonant transmission line structures. | |
| Jan 13 at 2:15 | vote | accept | Poisson Aerohead | ||
| Jan 13 at 2:14 | comment | added | Poisson Aerohead | Yup, got it, it is truly the "cut off" frequency, and not really useful for sending anything down the guide. Anything less is going to decay and anything more will propagate at some rate. The boundary is basically standing. To send anything down the guide at any sort of "reasonable rate" you would need some "safe margin" like you said. Thanks. | |
| Jan 13 at 1:57 | comment | added | Puk | @PoissonAerohead Yes, any value of $k_3 > 0$ yields a valid propagating mode, for which $\omega=c\sqrt{k_1^2+k_2^2+k_3^2}$. $k_3 = 0$ thus yields the minimum possible frequency of traveling waves. $k_3=0$ is a special case that marks the border between propagating and evanescent waves. It isn't really a propagating mode, the Poynting vector has no component down the waveguide in this mode. It is not really useful in in waveguides in practice as far as I know, normally $\omega$ is chosen (or the waveguide is designed) so that $k_3 > 0$ by a safe margin. | |
| Jan 13 at 1:36 | comment | added | Poisson Aerohead | Is this a roundabout way of saying that $k_{3}$ is indeterminate and so the minimum frequency is for $k_{3}=0$ so that $k = k_{2}$ and $\omega = k_{2}c$? If so, are we saying that the wavelength in the $z$ direction is then infinite? If so, are we saying that the wave is not actually propagating down the guide? It is basically a standing wave? Thanks. | |
| Jan 13 at 1:27 | history | answered | Puk | CC BY-SA 4.0 |