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A 4-isotropic element end-fire array separated by a λ/4 distance has first null occurring at ____________(a) 60(b) 30(c) 90(d) 150This question was posed to me during an online exam.Origin of the question is Radiation Pattern for 4-Isotropic Elements topic in division Antenna Array of Antennas

Answer»

Right choice is (c) 90

For explanation: The NULLS of the N- ELEMENT array is given by \(θ_n=cos^{-1}⁡(\frac{λ}{2πd} \LEFT[-β±\frac{2πn}{N}\right])\)

Since its given broad side array \(β=±kd=±\frac{2πd}{λ}=±\frac{π}{2},\)

\(θ_n=cos^{-1}⁡(\frac{2}{π} \left[∓\frac{π}{2}±\frac{2πn}{4}\right])\)

=cos^-1([∓1±n])

First null at n=1; θn=cos^-1⁡([1±1) (considering β=\(-\frac{π}{2}) \)

θn = cos^-1⁡ (0) or cos^-1⁡(2)

θn = cos^-1⁡ (1)=90.



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