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The array factor of 8 – isotropic elements of broadside array is given by ____(a) \(\frac{sin(2kdcosθ)}{2kdcosθ} \)(b) \(\frac{sin(4kdcosθ)}{4kdcosθ} \)(c) \(\frac{sin(2kdcosθ)}{kdcosθ} \)(d) \(\frac{cos(2kdcosθ)}{2kdcosθ} \)The question was asked in an internship interview.The question is from Radiation Pattern of 8-Isotropic Elements topic in chapter Antenna Array of Antennas

Answer» CORRECT choice is (b) \(\frac{SIN(4kdcosθ)}{4kdcosθ} \)

The best I can explain: Normalized array factor is given by \(AF=\frac{sin(Nᴪ/2)}{N \frac{ᴪ}{2}}\)

And ᴪ=kdcosθ+β

Since its given broad SIDE arrayβ=0,

ᴪ=kdcosθ+β=kdcosθ

\(\frac{Nᴪ}{2}=4kdcosθ\)

\(AF=\frac{sin(Nᴪ/2)}{N \frac{ᴪ}{2}}=\frac{sin(4kdcosθ)}{4kdcosθ} \)


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