Question

The array factor of 4- isotropic elements of broadside array is given by ____________

(a) \(\frac{sin(2kdcosθ)}{2kdcosθ} \)

(b) \(\frac{sin(kdcosθ)}{2kdcosθ} \)

(c) \(\frac{sin(2kdcosθ)}{kdcosθ} \)

(d) \(\frac{cos(2kdcosθ)}{2kdcosθ} \)

I have been asked this question in an international level competition.

This intriguing question comes from Radiation Pattern for 4-Isotropic Elements in portion Antenna Array of Antennas

Answer 1

Correct option is (a) \(\frac{sin(2kdcosθ)}{2kdcosθ} \)

The explanation: Normalized array factor is given by

\(AF=\frac{sin(Nᴪ/2)}{N \frac{ᴪ}{2}}\)

And ᴪ=kdcosθ+β

Since its given broad side array β=0,

ᴪ=kdcosθ+β=kdcosθ

\(\frac{Nᴪ}{2}=2kdcosθ\)

\(AF=\frac{sin(Nᴪ/2)}{N \frac{ᴪ}{2}}=\frac{sin(2kdcosθ)}{2kdcosθ} \)