# Antenna Array Questions and Answers – Radiation Pattern of 8-Isotropic Elements

This set of Antenna Array Questions & Answers for Exams focuses on “Radiation Pattern of 8-Isotropic Elements”.

1. 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) $$\frac{cos(2kdcosθ)}{2kdcosθ}$$

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 arrayβ=0,
ᴪ=kdcosθ+β=kdcosθ
$$\frac{Nᴪ}{2}=4kdcosθ$$
$$AF=\frac{sin(Nᴪ/2)}{N \frac{ᴪ}{2}}=\frac{sin(4kdcosθ)}{4kdcosθ}$$

2. An 8-isotropic element broadside array separated by a λ/2 distance has nulls occurring at ____
a) $$cos^{-1} (±\frac{n}{4})$$
b) $$cos^{-1} (±\frac{n}{2})$$
c) $$sin^{-1} (±\frac{n}{2})$$
d) $$sin^{-1} (±\frac{n}{4})$$

Explanation: The nulls of the N- element array is given by
$$θ_n=cos^{-1}⁡(\frac{λ}{2πd}[-β±\frac{2πn}{N}])=cos^{-1}⁡(\frac{λ}{2πd} [±\frac{2πn}{N}])$$
⇨ $$θ_n=cos^{-1}⁡(\frac{λ}{2π(λ/2)} [±\frac{2πn}{N}])=cos^{-1} (±\frac{2n}{8})=cos^{-1} (±\frac{n}{4})$$

3. An 8-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
$$θ_n=cos^{-1}⁡(\frac{λ}{2πd}\left[-β±\frac{2πn}{N}\right])=cos^{-1}⁡(\frac{λ}{2πd} [±\frac{2πn}{N}])$$
⇨ $$θ_n=cos^{-1}⁡(\frac{λ}{2π(λ/4)}) [±\frac{2πn}{N}])=cos^{-1} (±\frac{4n}{8})=cos^{-1} (±\frac{n}{2}) [n=1,2,3 \,and\, n≠N,2N…]$$

4. The array factor of 8- isotropic elements of broadside array separated by a λ/4 is given by ____
a) sinc(cosθ)
b) cos(sinθ)
c) sin(sinθ)
d) sinc(2cosθ)

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

5. The array factor of 8 – isotropic elements of broadside array separated by a λ/2 is given by ____
a) sinc(4cosθ)
b) sin(2πcosθ)
c) sinc(4πsinθ)
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
$$\frac{Nᴪ}{2}=4kdcosθ=4(\frac{2π}{λ})(\frac{λ}{2})cosθ=4πcosθ$$
$$AF=\frac{sin(Nᴪ/2)}{N \frac{ᴪ}{2}}=\frac{sin(4πcosθ)}{4πcosθ}=sinc(4cosθ).$$
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6. What is the direction of first null of broadside 8-element isotropic antenna having a separation of λ/2?
a) 60°
b) 75.5°
c) 37.5°
d) 57.5°

Explanation: The nulls of the N- element array is given by
$$θ_n=cos^{-1}⁡(\frac{λ}{2πd} [-β±\frac{2πn}{N}])=cos^{-1}⁡(\frac{λ}{2πd} [±\frac{2πn}{N}])$$
⇨ $$θ_n=cos^{-1}⁡(\frac{λ}{2π(λ/2)} [±\frac{2πn}{N}])=cos^{-1} (±\frac{2n}{8})=cos^{-1} (±\frac{n}{4})$$
⇨ $$n=1 (first \,null) cos^{-1} (±\frac{n}{4})=cos^{-1} (±\frac{1}{4})=75.5°.$$

7. What is the direction of first null of broadside 8-element isotropic antenna having a separation of \frac{λ}{4}?
a) 0
b) 60
c) 30
d) 120

Explanation: The nulls of the N- element array is given by
$$θ_n=cos^{-1}⁡(\frac{λ}{2πd} [-β±\frac{2πn}{N}])=cos^{-1}⁡(\frac{λ}{2πd} [±\frac{2πn}{N}])$$
$$θ_n=cos^{-1}⁡(\frac{λ}{2π(\frac{λ}{4})}\left[±\frac{2πn}{N}\right])=cos^{-1} (±\frac{4n}{8})=cos^{-1} (±\frac{n}{2})=cos^{-1} (±1/2)=60$$

8. The necessary condition for maximum of the first 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.
⇨ $$sin(\frac{Nᴪ}{2})=±1$$
⇨ $$\frac{Nᴪ}{2}=±\frac{2s+1}{2} π$$
⇨ $$\frac{Nᴪ}{2}=±\frac{3π}{2}$$ [s=1 for first minor lobe].

9. The direction of the first minor lobe of 8 element isotropic broadside array separated by λ/2 is ___
a) 41.4°
b) 76.6°
c) 67.7°
d) 90°

Explanation: The direction of the secondary maxima (minor lobes) occur at θs
$$θ_s=cos^{-1} (\frac{λ}{2πd} \left[-β±\frac{(2s+1)}{N} π\right])$$
⇨ $$θ_s=cos^{-1} (\frac{λ}{2π(λ/2)} [±\frac{3}{8}π])$$ (s=1 for 1st minor lobe)
⇨ $$θ_s=cos^{-1} (±\frac{3}{8})=67.7°$$

10. An 8-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 $$θ_n=cos^{-1}⁡(\frac{λ}{2πd} [-β±\frac{2πn}{N}])$$
Since its given broad side array $$β=±kd=±\frac{2πd}{λ}=±\frac{π}{2},$$
$$θ_n=cos^{-1}⁡(\frac{2}{π} [∓\frac{π}{2}±\frac{2πn}{8}])$$
$$=cos^{-1}⁡([∓1±\frac{n}{2}])$$
First null at n=1; $$θ_n= =cos^{-1}⁡([1±\frac{1}{2}]) (considering \,β=-\frac{π}{2})$$
$$θ_n =cos^{-1} (\frac{1}{2})\,or \,cos^{-1} (3/2)$$
$$θ_n =cos^{-1} (\frac{1}{2})=60.$$

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