You should have made your patch shorter. Try that and see if it is still unstable (shorter isn't about stability, it is just what you should do to get a higher frequency).

It isn't the resonance that is unstable, it is the solution in CST.

Sorry, my bad. I made my patch shorter yes. But there comes the instability.

What do you mean ? the solution in CST would be unstable ?

I have read more of the article you linked to. I don't agree that the behaviour there is unstable. The picture shown just means there are two reflections. If the signal eventually dies away to nothing then CST is stable and the solution will be (nearly) right. In some problems CST keeps calculating for so long that the errors build up and the output eventually starts getting bigger and bigger again. The energy goes up instead of down. That is an unstable result.

hi,

it's not necessary to truncate, u can try to check your solver setting. click Home > Setup Solver > under solver setting u can choose desired Accuracy (normally -30 to -60 dB). also tick the Adaptive mesh refinement. then u'll see the depth of the resonance frequency. optimize your design by varying the patch dimensions.

Thank you both for your answers.

How can I get a CP without cutting edges or using a double feed ?

Could you tell me more about the Adaptative mesh refinement ? I've been using a -60dB accuracy since the beginning. How to see the depth of the resonance frequency and what does it give me to improve my design ?

60 dB seems a bit too much. I usually use 40 dB. You can see the ripples (on the parts of the S11 that should be flat) getting smaller as the energy drops - look at the energy plot during the run, it is this that determines when the prediction finishes and tells you what level it has got down to, and is what the code uses to predict the time left. When the ripples are small enough you can stop it yourself using the abort button - it allows you to save the results at that point, even if you abort.

To get circular polarization one way is to have two linearly polarized radiating modes with 90 degrees phase difference. Cutting the edges or having an oval patch results in the two modes having different resonant frequencies (because the dimensions are different when the shape is rotated 90 degrees) so one is operating below and the other above resonance which results in one being 45 degrees phase shifted and the other -45 degrees phase shifted, giving a total of 90 degrees difference. The other way is to have two feeds 90 degrees out of phase.

If the resonant frequency goes down to -30 dB you don't need any more than this, -20 is usually good enough. antennas bandwidth is usually quoted at the -10 dB level, or perhaps -14 dB sometimes, almost never below -14 dB.

Adaptive mesh refinement can be good, but often the first mesh CST generates is good enough. Try the default, then see how the results change with the adaptive mesh. Too fine a mesh takes for ever to run, and can be unstable. Even if it works, what is the point of a result that is only 0.1% away from perfect when you can't get circuit boards with dielectric constant that accurate?

any how u can try and see the outcome in which design an antenna with truncated edges to generate CP waves. (actually i not sure your aim, because CP is good for microwave transmission). double feed also considerable for CP.

Adaptive mesh refinement in CST is to refined a mesh when u run the simulation design. thus accurate and perfect value obtained. try to change the accuracy to -40 dB perhaps instead of -60 dB. the depth of the resonance frequency can be obtained from s-parameter magnitude in dB result, from this parameter u'll know how well your design or dimensions line are matched. a low return loss of s-parameter (usually less than -10dB) is desirable and it results in a lower insertion loss.

Very resonant structures such as patch antennas are better simulated with Finite Element Method or MoM, time domain sucks for this. Try using the F solver instead.

Marcello

It depends on the Q - as you say, very resonant structures take a very long time in time domain - but patch antennas tend not to be that high Q and can be efficiently solved in the time domain.

The reason time domain takes a long time for high Q structures is that the signal has to bounce about in the resonant structure about Q times before you get the right answer. On a simple resonator like a short transmission line you can see the energy drop every time the pulse transits the resonator and some of the pulse escapes back to the input port. You can also see the pulses arriving back at the port, in the port signal.

Long track lengths in feed networks can also take a long time in the time domain - the calculation that matters goes along the length of the track, not straight across the board, and has to travel one calculation step at a time.

Frequency domain can be quicker for one run, but has to re-run for every frequency, whereas time domain give all frequencies in one run, so the choice depends on the structure and what answer you want. Frequency domain can also give results for different ports or excitations at the same frequency without much further calculation, whereas time domain has to do each excitation separately.

hi, one easy way for producing CP polarization is to cut a slant (45 degree) rectangular slot from the middle of the patch.

regarding the unstable resonant frequency, I think that it relates to correct adjusting the CST parameters.