This question is about reflectarray antennas. In the paper below, it is mentioned that:

"A second useful property of aperiodic arrays is the possibility to reduce the number of elements in one assigned aperture without major impact on the beamwidth."

Viganó, M. C., Toso, G., Caille, G., Mangenot, C., & Lager, I. E. (2009). Sunflower Array Antenna with Adjustable Density Taper. International Journal of Antennas and Propagation, 2009, 1-10.

I understand that aperiodic lattices are able to reduce sidelobe levels by suppressing grating lobe formation. However, why does the periodicity of the lattice affect the beamwidth of the antenna? I am hoping to get an answer on this that includes a qualitative explanation in addition to any math that is required. Thank you!

  • \$\begingroup\$ Please don't repost the same question under a different title, edit your original question. \$\endgroup\$
    – jramsay42
    Mar 4, 2018 at 4:21
  • \$\begingroup\$ This is a different question. This question is on beamwidth while the previous one was on bandwidth \$\endgroup\$
    – Fromage
    Mar 4, 2018 at 6:17
  • \$\begingroup\$ My apologies, much of the wording and formatting was the same. \$\endgroup\$
    – jramsay42
    Mar 4, 2018 at 7:19
  • \$\begingroup\$ Yupz I have a science fair next week and I realised I still don't understand some concepts that I used in my report... that's why so many questions asked by me on reflectarray antennas \$\endgroup\$
    – Fromage
    Mar 4, 2018 at 12:51

1 Answer 1


According to Linear arrays with variable interelement spacings:

The beam width of the main lobe is rather insensitive to changes of the element positions and to the total number of elements; it depends primarily on the total length of the array. Unless the excitations of the elements are strongly tapered, and the inter element spacing vary wildly, the following formula can be used with reasonable accuracy for determining the total length \$L\$ of a broadside array required to produce a 3-dB beam width, \$\Delta\theta\$, where it is assumed that \$\Delta\theta\ll1\$ radian $$\frac{L}{\lambda}\approx\frac{1}{\Delta\theta}$$

Here \$\lambda\$ is the the wavelength and the expression can be solved for the beam width: $$\Delta\theta \approx \frac{\lambda}{L}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.