[CST MWS] ""farfield is inaccurate!!" means .

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Insang Woo

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i simulated the simple dipole and examined the farfield.

But, with the farfield pattern picture, there is warning.

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"warning : farfield is inaccurate"
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anybody know this meaning? now i attach the very picture.



 

Re: [CST MWS] ""farfield is inaccurate!!" mea

that's because the simulation boundary is to close to the dipole with respect to the wavelength. it is better to use the open + add space boundary when simulating antennas in free space. cst will set a proper simulation boundary for you

best regards
 
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    Ibso

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Re: [CST MWS] ""farfield is inaccurate!!" mea

kleinesmurf69 said:
that's because the simulation boundary is to close to the dipole with respect to the wavelength. it is better to use the open + add space boundary when simulating antennas in free space. cst will set a proper simulation boundary for you

best regards
Click to expand...

thank you, mr. kleinsmurf69!

But, as you can check my picture, i want to see the radiaton of dipole [which is loacated at the boundary between uppper air layer and lower substrate (my case is lossless GaAS)].
So, to give a infinite concept to all directions (=no reflected wave from any boundary) , i attached the all sideplanes to open boundary plane.
So, if i would use the open (add), that is out of my original intention.

could examine my structre closer. plz.

thank you again your comment!
 

Re: [CST MWS] ""farfield is inaccurate!!" mea

if you do not want to use the open(add) function (all it does is pads the volume with λ/8 free space on all sides, by default), then you could make the substrate bigger - it also implies that your substrate is large (approx. infinite) in comparison to the dipole. of course it will take more time, but at least it will give you a useful result. maye uploading a pic with the boundary conditions will help.
 

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