E-field at a distance from an open ended waveguide with known dB input? - Page 1
Forum section: electronics / rf-microwave

author: Plasmateur
author_rank: Regular Contributor
timestamp: October 03, 2019, 02:31:42 pm
content: I've done a lot of Google searching on this one but I keep coming up with how to calculate the electric field inside a waveguide.I'm looking for information on how to calculate the electric field exiting an open ended waveguide - no horn.I know what my input is in terms of dBm.Is there an easy calculation for this? Is it broken up into near field, far field, and in between?Thanks!

author: radiolistener
author_rank: Super Contributor
timestamp: October 03, 2019, 02:55:04 pm
content: I'm not sure, but probably you can calculate it from your dBm power with take into account that the air has impedance very close to the free space - about 376 Ohm.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 03, 2019, 03:25:41 pm
content: Would it not be equivalent to a short dipole of the same voltage, over a ground plane (the open flange face)?Tim

author: radiolistener
author_rank: Super Contributor
timestamp: October 03, 2019, 03:31:37 pm
content: Quote from: T3sl4co1l on October 03, 2019, 03:25:41 pmFor waveguide output all power is radiated in single direction. But there is also loss in the waveguide, which will depends on waveguide shape, dimensions and surface resistance...

author: EEEnthusiast
author_rank: Frequent Contributor
timestamp: October 03, 2019, 03:43:09 pm
content: A simple estimate would be P = |E|^2/Zo where Zo is the intrinsic impedance of free space (377 Ohms).Thus, |E| = sqrt (P * Zo)

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 03, 2019, 03:54:27 pm
content: Quote from: radiolistener on October 03, 2019, 03:31:37 pmWhich direction is that?Quote from: EEEnthusiast on October 03, 2019, 03:43:09 pmOr perhaps even more to-the-point, you could take the ratio of waveguide impedance to free-space impedance, and calculate the reflection coefficient; the remainder should be radiated. (Just adding the reciprocal step on, taking Zo_waveguide to find E then finding free-space P from E.)This is complicated by the differing aperture sizes: the waveguide contains the wave stacked on top of itself (confined by the boundary), so I suspect the reflection loss is actually higher than the impedance ratio would suggest. (Also, is the waveguide impedance high to begin with, on account of its geometry? I forget.)Of course this also assumes a matched source, so the reflected wave is absorbed. If not, the waveguide will have SWR > 1 and the radiated power depends on waveguide length and source impedance.Tim

author: radiolistener
author_rank: Super Contributor
timestamp: October 03, 2019, 04:18:52 pm
content: Quote from: T3sl4co1l on October 03, 2019, 03:54:27 pmThe direction of waveguide output

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 03, 2019, 04:40:12 pm
content: So it makes a perfect plane wave? From a sub-wavelength aperture?Tim

author: Plasmateur
author_rank: Regular Contributor
timestamp: October 03, 2019, 05:07:40 pm
content: This discussion has generated a little confusion for me.So my 50Ohm coaxial line is terminated to the waveguide adapter. Because the impedance of free space isn't 50Ohms, no matter what there reflection back into the source of the time varying signal?Or if I go from coaxial cable, to waveguide, to waveguide, back to coaxial cable, there will be reflections at each transition?

author: fourfathom
author_rank: Super Contributor
timestamp: October 03, 2019, 05:28:42 pm
content: Quote from: Plasmateur on October 03, 2019, 05:07:40 pmNo, there is a matching section that couples the 50 Ohm port to the waveguide. This minimizes reflections and maximizes power transfer. Here is a brief description of coax/waveguide coupling:https://www.microwaves101.com/encyclopedias/waveguide-to-coax-transitions

author: hagster
author_rank: Frequent Contributor
timestamp: October 03, 2019, 05:35:17 pm
content: So assuming good VSWR(or accounting for reflected power if not) all you input energy will be present at the waveguide aperture. You can calculate the filed strength here by dividing your input power by the waveguide aperture area. You can then use the 377ohms of free space to calculate the E field strength.To calculate the same thing at a distance is harder as their are lots of disruptions due to fringing fields around the aperture. You could do some computational modeling to work this out or you could just ignore it and go for a worst case by using the real aperture as the effective aperture and calculating the antenna gain. Then subtract the free space pathloss and calculate the field strength at the required distance using the reverse process into an imagined aperture.

author: TheUnnamedNewbie
author_rank: Super Contributor
timestamp: October 03, 2019, 05:54:08 pm
content: The reflection at a open metal waveguide is somewhere around -15 dB, I seem to recall. I could be totally wrong.

author: hagster
author_rank: Frequent Contributor
timestamp: October 03, 2019, 06:33:40 pm
content: Quote from: TheUnnamedNewbie on October 03, 2019, 05:54:08 pmIn terms of accepted power that's very close to 100%.

author: rfeecs
author_rank: Frequent Contributor
timestamp: October 03, 2019, 06:51:31 pm
content: Basically whathagstersaid. Most people only care about far field gain, which can be calculated approximately:https://www.govinfo.gov/content/pkg/GOVPUB-C13-60518677c4755c38f2b65eaf1a46f264/pdf/GOVPUB-C13-60518677c4755c38f2b65eaf1a46f264.pdfIf you need near field you may have to resort to computer simulation.You can look at the plots of gain. It looks like the peak gain is approximately 6dB to 8dB.You can then plug in your power, gain of 7dB and distance into a calculator like this to get your field strength:http://www.giangrandi.ch/electronics/anttool/tx-field.shtmlThis is assuming reflections and other losses are minimal.

author: Plasmateur
author_rank: Regular Contributor
timestamp: October 04, 2019, 01:09:28 am
content: Thiar calculator is exactly what I found with my searching but I wasn't sure if there were reflections from going coaxial to open ended.Thank you everyone for the links!