1.

The relation between the spacing factor σ and the scaling factor τ is given by ____(a) \(σ=tan^{-1}(\frac{1-τ}{4σ}) \)(b) \(σ=tan^{-1}(\frac{1-τ}{2σ}) \)(c) \(σ=tan^{-1}⁡(\frac{4σ}{1-τ}) \)(d) \(σ=tan^{-1}⁡(\frac{2σ}{1-τ}) \)This question was addressed to me by my school principal while I was bunking the class.My question comes from Frequency Independent Antenna in section Frequency Independent Antenna of Antennas

Answer»

Right option is (a) \(σ=tan^{-1}(\frac{1-τ}{4σ}) \)

Easiest EXPLANATION: In LPDA, the ratio of successive spacing of elements is EQUAL to the ratio of adjacent dipole lengths. The spacing factor is \(σ=\frac{d_n}{2L_n} \,and\, d_n =\) spacing BETWWEN adjacent elements‘n’ and ‘n+1’ and Ln is the length of n^th dipole. The relation between the spacing factor σ and the SCALING factor τ is \(σ=tan^{-1}(\frac{1-τ}{4σ}). \)


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