



(these GIFs are compressed to fit the upload limit and I'll share the full-quality versions soon)

Here are the high-quality versions of these:
https://www.icloud.com/iclouddrive/05fbbMTPgdNvm_Sj1NQXOHuFg#Cantenna
Apologies, that first link needed sign-in. Here are direct download ones:
https://www.icloud.com/iclouddrive/08e7LiE9o-GiXvpmf7qsDXG2w#cantenna2%5Fwhy%5Fprobe
https://www.icloud.com/iclouddrive/036c0REMzBqvMCoR2UKf3hkfw#cantenna2%5Fstandingwave
https://www.icloud.com/iclouddrive/0275TPqLPDyQs68LyOSwnVuRQ#cantenna2%5Frx
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Hi Wolfgang,
Thanks, yes, I'm trying to create an ideal-dimension reference cantenna model to get a simulated pattern I can compare against your measured one. The length of the cantenna that I'm using for this model (which is the same as the new one I made) is 320 mm with some non-ideal dimensions/probe placement, so that overlength may be part of it. It can also be a bug in my model, I'll double-check everything and share my findings.
Still getting the hang of openEMS, so I really appreciate the feedback!
Thanks.
Hi Andrew,
Yes, exactly! The wave reflects off the shorting back wall, which forces a field null (zero) right at the wall. A quarter-wavelength (guide wavelength) away from the wall, the incoming and reflected waves line up in phase and add together, giving an antinode (field peak). The probe sits at that antinode to grab the strongest field.
Good thinking regarding the next peak. I was wondering the same thing last week. From what I read, in principle, yes: the next peak is ~3Lg/4 up. But the standard cantenna length is about 0.75 Lg (= 3Lg/4) so that peak is near the open mouth where the open aperture disturbs the standing-wave pattern, making it less predictable than the first peak near the back wall. The Lg/4 peak near the back sits in the cleanest and most predictable part of the standing-wave pattern, so it's the better, standard spot.
For longer cantennas, this would be interesting to test. Has anyone experimented with placing the probe near the 3Lg/4 peak in a longer waveguide and compared the results?
Thanks.

Note: Lg here is the guide wavelength (the wavelength inside the waveguide), which is longer than the free-space wavelength.
https://www.changpuak.ch/electronics/cantenna.php
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Hi Robert,
"E-plane"/"H-plane" are just labels for the two cut-planes, not separate field components, both are the power pattern. In the far field, E and H are perpendicular and in phase, with H = E/η₀, so the Poynting vector reduces to:
S = ½ Re(E × H*) = |E|² / (2η₀)
So power density ∝ |E_far|². We compute directivity directly from that:
D(θ,φ) = ( |E(θ,φ)|² / |E|²ₘₐₓ ) · D_max (plotted in dBi)
where:
S = Poynting vector — power flow per unit area (W/m²)
E = electric field (V/m); H = magnetic field (A/m)
η₀ = impedance of free space ≈ 377 Ω
Re = real part; × = cross product; * = complex conjugate
θ (theta) = polar angle measured from the antenna's +z axis (boresight in this model)
φ (phi) = azimuth angle (around the axis)
D(θ,φ) = directivity (radiation concentration vs an isotropic source); D_max = its peak value
|E|²ₘₐₓ = peak of |E|² over all directions; dBi = dB relative to isotropic
So the polar plots are the power (directivity) pattern, and E×H is fully captured by |E|²/η₀.
I'm still refining the model and fixing a couple of issues Wolfgang flagged.
Thanks!
Hi Wolfgang,
I redid the cantenna pattern for an ideal D = 150 mm can (single-mode TE-11 at 1420 MHz) after fixing a few bugs in my simulation, you can find it attached. This one's computed from the aperture physics directly, the standard circular-waveguide aperture formula for the main beam plus rim-edge diffraction for the back/side lobes.
The E-plane is a bit narrower than the H-plane.
Thanks.