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) \(\frac{cos(2kdcosθ)}{2kdcosθ} \)
View Answer
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) \(sin^{-1} (±\frac{n}{4}) \)
View Answer
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)
View Answer
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)
View Answer
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)
View Answer
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°
View Answer
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
View Answer
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) \( \frac{Nᴪ}{2}=±\frac{4π}{2}\)
View Answer
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°
View Answer
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
View Answer
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.
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