Antennas Multiple Choice Questions 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} )
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}=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}) )
Answer: a
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} 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)
Answer: a
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} 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)
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}=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)
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}=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°
Answer: a
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} 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
Answer: a
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} 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})
Answer: a
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{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°
Answer: a
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)} left[±frac{3}{4} πright])) (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
Answer: c
Clarification: 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.