Antenna Array Questions & Answers for Exams 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} )
Answer: b
Clarification: 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}))
Answer: a
Clarification: 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))
Answer: b
Clarification: 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)
Answer: d
Clarification: 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)
Answer: a
Clarification: 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).)
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°
Answer: b
Clarification: 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
Answer: b
Clarification: 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})
Answer: b
Clarification: 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°
Answer: b
Clarification: The direction of the secondary maxima (minor lobes) occur at θs
(θ_s=cos^{-1} (frac{λ}{2πd} left[-β±frac{(2s+1)}{N} πright]))
⇨ (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
Answer: a
Clarification: 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.)
Global Education & Learning Series – Antennas.
all exam questions on Antenna Array, here is complete set of 1000+ Multiple Choice Questions and Answers .