Does HFSS account for mutual coupling effect automatically?

I have 5-element λ/2 dipoles spaced 0.1λ apart (they all show in the model...I am not using the Antenna Array Setup wizard). They are all excited with 1W each and the port impedance for each is 73ohms. After simulation, the power accepted is about half less than power incident. When simulating one half-wave dipole alone with 73ohms port impedance, the power accepted is about the same as the power incident which tells me the antenna is well-matched. When in array, the power accepted is about half less than power incident, which tells me that the rest of power incident is lost due to mutual coupling between adjacent antennas. So, is it safe to say that HFSS does automatically account for mutual coupling effects of adjacent excited array elements?

I actually matched each of the antennas using the re and im Active Z values provided by HFSS and got the power accepted to be close to power incident, but when I place a lossy dielectric near the matched array, I don't get as much absorption in the dielectric as when I place the non-matched array near the lossy dielectric? Anyone knows why? I am tempted to just go without the matched array which is actually good since there won't be a need for a matching network.

Thank you all.

Yes, HFSS must take into account the coupling between the elements. You should be able to see that in the values of S12, S13, S14, S15, S23, S24 .. etc. i.e. all values of Snm, where n does not equal m. These will show the coupling between the elements.

If the antennas in the array show only half the incident power is absorbed, then that is because the impedance of the array is no longer close to 73 Ohms, so 50% of the power is reflected.

I have never used the active Z parameters in HFSS. However, I would expect that if an antenna is well matched, then putting a lossy dielectric near it to cause a big change in S11 (I'm considering only a single antenna here). For example, if the return loss starts off at 30 dB, I can imagine that it is quite possible to reduce that to 10 dB by putting a lossy material nearby. So there's a change of 20 dB in S11. Conversely, if it was poorly matched (say return loss 3 dB), then you are not likely to cause a 20 dB change in S11 by putting a lossy dielectric near it in some random location, though you might if the location just happens to be right. However, whilst I would expect that the lossy dielectric to have a major change on S11 if the antenna is matched, I'm not so sure that would result in more or less absorption in the lossy dielectric.

Of course, if you put a dielectric near an antenna, then mathed it with some matching network, S11 would indicate a good return loss. If you then remove that dielectric, S11 would show a poorer return loss.

One would expect even a perfect dielectric to cause a massive change in S11 if close to a well matched antenna. But none of that change would be due to absorption, but simply the change in impedance.

If you post your .hfss file, perhaps someone could help you, but to me at least, nothing you have stated seems odd.

Dave

thank you, drkirkby, your answer really helped me. Is their a way to prove the above statement mathematically, that is, when antenna matched and next to lossy dielectric the S11 could drop by 20dB vs when it is not matched, it will not drop by that much. Although conceptually it makes sense, it would be nice to confirm it with the math behind it.

Thank you.

The return loss could drop substantially if the antenna is placed next to a lossy dielectric. Ultimately, putting it next to a lossy dielectric will cause it to have some impedance R + j X. If your matching network matches your system impedance (probably 50 Ohms) to R + j X, then the return loss can be theoretically infinite. So the lossy dielectric could cause the return loss to change from 3 dB to 30 dB. However, that is unlikely to happen with random placement of the lossy dielectric. If the antenna is well matched without the lossy dielectric, a randomly placed lossy dielectric is far more likely to cause S11 to change for the worst.

I'm not convinced you will be able to find any mathematical proof of this, but I suspect you can show it to be so statistically with modelling software.

One argument might be that if the antenna is perfectly matched, then any introduction of a dielectric must make S11 worst. It can't possibly improve S11, since the antenna was previously perfectly matched. However, if the antenna is poorly matched, then introducing a dielectric can make S11 better of it can make it worst. If fact, if you had the impedance's of an antenna with 1001 different locations of a lossy dielectric, there must be some impedance where 500 locations make S11 worst and 500 make S11 better. Of course, for a different impedance, it might happen that 900 make S11 better and 100 make S11 worst, or the other way around. But if the antenna is perfectly matched to start with, any introduction can only make S11 worst.

I think based on the above, it is reasonable to say that the likelihood of S11 improving with random placement of a dielectric depends on how close to matched it was originally. Clearly if the antenna was perfectly matched, any introduction of a dielectric will make matters worst. If the antenna was very well matched (say return loss 40 dB), then it is highly unlikely that random placement of a dielectric will improve upon the very good 40 dB return loss. It seems far more likely that placing a dielectric near the antenna will make S11 worst.

It looks to me its easy to prove the limiting case (when antenna is perfectly matched, introducing a dielectric can only make matters worst). I can't see how you will get a mathematical proof when not in the limit, but it seems logical. I think the best you could do is some sort of statistical analysis.

Dave