# Antenna Array Questions and Answers – Radiation Pattern for 4-Isotropic Elements

This set of Antennas Multiple Choice Questions & Answers (MCQs) focuses on “Radiation Pattern for 4-Isotropic Elements”.

1. 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) $$\frac{cos(2kdcosθ)}{2kdcosθ}$$

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

2. A 4-isotropic element broadside array separated by a λ/2 distance has nulls occurring at ____________
a) $$cos^{-1} (±\frac{n}{2})$$
b) $$cos^{-1} (±\frac{4n}{2})$$
c) $$sin^{-1} (±\frac{n}{2})$$
d) $$sin^{-1} (±\frac{n}{4})$$

Explanation: The nulls of the N- element array is given by
$$θ_n=cos^{-1}⁡(\frac{λ}{2πd} \left[-β±\frac{2πn}{N}\right])=cos^{-1}⁡(\frac{λ}{2πd} \left[±\frac{2πn}{N}\right])$$
⇨ $$θ_n=cos^{-1}⁡(\frac{λ}{2π(λ/2)}\left[±\frac{2πn}{N}\right])=cos^{-1} (±\frac{2n}{4})=cos^{-1} (±\frac{n}{2})$$

3. A 4-isotropic element broadside array separated by a λ/4 distance has nulls occurring at ____________
a) cos-1 (±n)
b) $$cos^{-1} (±\frac{n}{2})$$
c) $$sin^{-1} (±\frac{n}{2})$$
d) sin-1 (±n)

Explanation: The nulls of the N- element array is given by
$$θ_n=cos^{-1}⁡(\frac{λ}{2πd} \left[-β±\frac{2πn}{N}\right])=cos^{-1}⁡(\frac{λ}{2πd} \left[±\frac{2πn}{N}\right])$$
⇨ $$θ_n=cos^{-1}(⁡\frac{λ}{2π(λ/4)}\left[±\frac{2πn}{N}\right])=cos^{-1} (±\frac{4n}{4})=cos^{-1} (±n) \left[n=1,2,3 \,and \,n≠N,2N…\right]$$

4. The array factor of 4- isotropic elements of broadside array separated by a λ/4 is given by ____________
a) sinc(cosθ)
b) cos(sinθ)
c) sin(sinθ)
d) sin(cosθ)

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
$$\frac{Nᴪ}{2}=2kdcosθ=2(\frac{2π}{λ})(\frac{λ}{4})cosθ=πcosθ$$
$$AF=\frac{sin(Nᴪ/2)}{N \frac{ᴪ}{2}}=\frac{sin(πcosθ)}{πcosθ}=sinc(cosθ).$$

5. The array factor of 4- isotropic elements of broadside array separated by a λ/2 is given by ____________
a) sinc(2cosθ)
b) sin(2πcosθ)
c) sinc(2πsinθ)
d) sin(2sinθ)

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
$$\frac{Nᴪ}{2}=2kdcosθ=2(\frac{2π}{λ})(\frac{λ}{2})cosθ=2πcosθ$$
$$AF=\frac{sin(Nᴪ/2)}{N \frac{ᴪ}{2}}=\frac{sin(2πcosθ)}{2πcosθ}=sinc(2cosθ).$$

6. What is the direction of first null of broadside 4-element isotropic antenna having a separation of λ/2?
a) 60°
b) 30°
c) 180°
d) 0°

Explanation: The nulls of the N- element array is given by
$$θ_n=cos^{-1}⁡(\frac{λ}{2πd} \left[-β±\frac{2πn}{N}\right])=cos^{-1}⁡(\frac{λ}{2πd} \left[±\frac{2πn}{N}\right])$$
⇨ $$θ_n=cos^{-1}⁡(\frac{λ}{2π(λ/2)} \left[±\frac{2πn}{N}\right])=cos^{-1} (±\frac{2n}{4})=cos^{-1}(±\frac{n}{2})$$
⇨ $$n=1 (first \,null) cos^{-1} (±\frac{n}{2})=cos^{-1} (±\frac{1}{2})=60° or 120°.$$

7. What is the direction of first null of broadside 4-element isotropic antenna having a separation of λ/4?
a) 0
b) 60
c) 30
d) 120

Explanation: The nulls of the N- element array is given by
$$θ_n=cos^{-1}⁡(\frac{λ}{2πd} \left[-β±\frac{2πn}{N}\right])=cos^{-1}⁡(\frac{λ}{2πd} \left[±\frac{2πn}{N}\right])$$
$$θ_n=cos^{-1}⁡(\frac{λ}{2π(λ/4)}\left[±\frac{2πn}{N}\right])=cos^{-1} (±\frac{4n}{4})=cos^{-1}(±n)=0$$

8. The necessary condition for maximum of the second side lobe of n element array is __________
a) $$\frac{Nᴪ}{2}=±\frac{5π}{2}$$
b) $$\frac{Nᴪ}{2}=±\frac{3π}{2}$$
c) $$\frac{Nᴪ}{2}=±\frac{π}{2}$$
d) $$\frac{Nᴪ}{2}=±\frac{4π}{2}$$

Explanation: The secondary maxima occur when the numerator of the array factor equals to 1.
⇨ $$sin(\frac{Nᴪ}{2})=±1$$
⇨ $$\frac{Nᴪ}{2} =±\frac{2s+1}{2}π$$
⇨ $$\frac{Nᴪ}{2}=±\frac{5π}{2}$$ [s=2 for second minor lobe].

9. The direction of the first minor lobe of 4 element isotropic broadside array separated by λ/2 is ___________
a) 41.4°
b) 30°
c) 60°
d) 90°

Explanation: The direction of the secondary maxima (minor lobes) occur at θs
$$θ_s=cos^{-1} (\frac{λ}{2πd} \left[-β±\frac{(2s+1)}{N} π\right])$$
⇨ $$θ_s=cos^{-1} (\frac{λ}{2π(λ/2)} \left[±\frac{3}{4} π\right])$$ (s=1 for 1st minor lobe)
⇨ $$θ_s=cos^{-1} (±\frac{3}{4})=41.4°$$

10. 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) 150

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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