In a reflectarray antenna, each element on the antenna aperture requires a different electrical phase due to different distances from the antenna feed. One way to achieve this phase variation is to vary the size of the elements on the reflectarray. What I do not understand is: why does changing the element dimension change the electrical phase of the element? I am looking for answers with a qualitative explanation in addition to any maths that may be necessary. Thank you!

  • \$\begingroup\$ Due to the fractional wavelength effects \$\endgroup\$ Mar 4, 2018 at 1:11
  • \$\begingroup\$ Due to size issues, most antenna's are 1/4 wavelength- and work just fine. \$\endgroup\$
    – user105652
    Mar 4, 2018 at 1:42

1 Answer 1


why does changing the element dimension change the electrical phase of the element?

At a specific wavelength an antenna will look electrically like a pure resistance. That resistance will be antenna copper losses plus the radiation resistance. A more efficient antenna will have most of the resistance coming from the radiation resistance.

If that ideal antenna were made slightly longer it would change its impedance by introducing inductance. If it were slightly shorter than optimum capacitive reactance would be added to the impedance: -

enter image description here

Dipole antenna picture source

The black line is resistance and the blue lines are reactance. At the optimum (given that the conductor diameter needs to be taken into account) wavelength (about 0.47) the reactance is zero. Notice that you don't need to make the antenna much longer or shorter to introduce inductance or capacitance.

This gives you the opportunity to tune the phase angle of the received signal from each element to coincide because that reactive impedance is driving into a resistive 75 ohms coax line.

Obviously some antennas produce a different impedance and will need a different feed line impedance.


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