For an isotropic antenna P_n(θ, φ) = 1, D = 1, for all θ and φ. The beam area for the isotropic antenna is given by:

Question

Question

This question was previously asked in

UGC NET Paper 2: Electronic Science 29 Oct 2022 Shift 1

Answer 1

Concept:

Isotropic antenna:

where, P_n(θ,ϕ )_max is maximum power density

Pn (θ,ϕ )avg is average power density

Pn (θ,ϕ )avg = \(\frac{4\pi}{{ \iint p_n\left(\theta,\phi\right)d\mathrm{Ω}}}\) = \(\frac{4\pi}{ ΩA}\)

\({ \iint p_n\left(\theta,\phi\right)d\mathrm{Ω}}\) is beam area( ΩA)

Calculation:

given:

Pn (θ,ϕ )max =1

D=1

\(D = \frac{p_n(θ,ϕ)_{max}}{p_n(θ,ϕ)_{avg}}\)

Pn (θ,ϕ )avg =1

Ω_A = \(\frac{4\pi}{Pn (θ,ϕ )avg}\)

= \({4\mathrm{\pi}}\)

so correct option is 2.