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In problems determining the polarization of an antenna array I've looked at, the polarization of the total array is determined by finding the polarization of a single element and then extrapolating that polarization to the whole array?

I'm not sure why this can be done. Is it because the elements in an array are all parallel to each other so the polarization would be the same?

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    \$\begingroup\$ so, ask yourself: how is polarization of a wavefront defined (i.e. what aspect of the electromagnetic field defines that)? What does the antenna have to do with that aspect (to make it simpler: start not with "antenna in general", but with a dipole)? I think that answers your question! \$\endgroup\$ Apr 16, 2020 at 8:10
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    \$\begingroup\$ (also, you seem to assume that in an antenna array, all antennas are identically oriented – that's not necessarily the case, but might be a sensible definition of an array) \$\endgroup\$ Apr 16, 2020 at 8:11
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    \$\begingroup\$ @MarcusMüller The polarization is defined by the direction of the electric field component. So since all the elements in the array would have the same electric field, the polarization for one element would be the polarization for all elements? \$\endgroup\$ Apr 16, 2020 at 8:13
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    \$\begingroup\$ jep, remember that electromagnetic fields do linear superposition, so if my E-field varies only in one direction, and I add a second E-field in the same direction, the direction stays the same. \$\endgroup\$ Apr 16, 2020 at 8:24
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    \$\begingroup\$ By the way, that was an interesting question, and I'd recommend you write an answer to your own question! \$\endgroup\$ Apr 16, 2020 at 8:26

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The polarization is defined by the direction of the electric field component. So since all the elements in the array would have the same electric field, the polarization for one element would be the polarization for all the elements. Electromagnetic fields abide by linear superposition so by varying the electric field in one direction and adding more electric fields in the same direction, the array's electric field direction stays the same.

Thanks to @MarcusMüller for the help!

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