# 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.

Sanfoundry Global Education & Learning Series – Antennas.

To practice all areas of Antennas, here is complete set of 1000+ Multiple Choice Questions and Answers.

Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!