This set of Antennas Multiple Choice Questions & Answers (MCQs) focuses on “N-element Linear Array”.
1. Which of the following statement is true?
a) As the number of elements increase in array it becomes more directive
b) As the number of elements increase in array it becomes less directive
c) Point to point communication is not possible with more number of array elements
d) There is no uniform progressive phase shift in linear uniform array
View Answer
Explanation: To get a single beam for the point to point communication more number of array elements is used. It increases the directivity of the antenna. An array is said to be uniform if the elements are excited equally and there is a uniform progressive phase shift.
2. Normalized array factor of N –element linear array is ________
a) \(\frac{sin(Nᴪ/2)}{Nᴪ/2} \)
b) \(\frac{cos(Nᴪ/2)}{Nᴪ/2} \)
c) \(N\frac{sin(ᴪ/2)}{ᴪ/2} \)
d) \(N\frac{cos(Nᴪ/2)}{Nᴪ/2} \)
View Answer
Explanation: The N-element linear uniform array, having a constant phase difference will have the array factor \(AF = ∑_{n=1}^Ne^{j(n-1)ᴪ}\)
Normalized array factor is given by \(\frac{sin(Nᴪ/2)}{Nᴪ/2}. \)
3. Which of the following expression gives the nulls for the N- element linear array?
a) \(θ_n=cos^{-1}(\frac{λ}{2πd}[-β±\frac{2πn}{N}])\)
b) \(θ_n=sin^{-1}(\frac{λ}{2πd}[-β±\frac{2πn}{N}])\)
c) \(θ_n=cos^{-1}(\frac{λ}{2πd}[-β±\frac{πn}{N}])\)
d) \(θ_n=cos^{-1}(\frac{λ}{2πd}[β±\frac{2πn}{N}])\)
View Answer
Explanation: To determine Null points the array factor is set equal to zero.
\(\frac{sin(\frac{Nᴪ}{2})}{\frac{Nᴪ}{2}}=0\)
⇨ \(sin(\frac{Nᴪ}{2})=0 \)
⇨ \(\frac{Nᴪ}{2}=±nπ \)
⇨ \(kdcosθ+β=±\frac{2nπ}{N} \)
⇨ \(θ_n=cos^{-1}(\frac{λ}{2πd}[-β±\frac{2πn}{N}]).\)
4. Maximum value of array factor for N-element linear array occurs at ______
a) \(θ_m=cos^{-1}(\frac{λ}{2πd}[-β±2πm])\)
b) \(θ_m=cos^{-1}(\frac{λ}{2πd}[β±2πm])\)
c) \(θ_m=sin^{-1}(\frac{λ}{2πd}[-β±2πm])\)
d) \(θ_m=sin^{-1}(\frac{λ}{2πd}[-β±2πm])\)
View Answer
Explanation: Normalized array factor is given by \(\frac{sin(Nᴪ/2)}{Nsin(\frac{ᴪ}{2})}.\)
To get maximum denominator is equal to zero.
⇨ \(sin(\frac{ᴪ}{2})=0 \)
⇨ \(\frac{ᴪ}{2}=±nπ \)
⇨ kdcosθ+β=±2πm
⇨ \(θ_m=cos^{-1}(\frac{λ}{2πd}[-β±2πm]).\)
5. Find the maximum value of array factor when elements are separated by a λ/4 and phase difference is 0?
a) θm=cos-1(4m)
b) θm=sin-1(4πm)
c) θm=cos-1(4πm)
d) θm=sin-1(2m)
View Answer
Explanation: The maximum value of array factor is \(θ_m=cos^{-1}(\frac{λ}{2πd}[-β±2πm]).\)
\(θ_m=cos^{-1}(\frac{λ}{2πλ/d}[0±2πm]).\)
θm=cos-1(4m).
6. Find the Nulls of the N-element array in which elements are separated by λ/4 and phase difference is 0?
a) \(θ_n=cos^{-1}(\left[±\frac{4n}{N}\right])\)
b) \(θ_n=cos^{-1}(\left[±\frac{2n}{N}\right])\)
c) \(θ_n=sin(\left[±\frac{4n}{N}\right])\)
d) \(θ_n=sin^{-1}(\left[±\frac{4n}{N}\right])\)
View Answer
Explanation: The nulls of the N- element array is given by \(θ_n=cos^{-1}(\frac{λ}{2πd} \left[-β±\frac{2πn}{N}\right])\)
⇨ \(θ_n=cos^{-1}(\frac{λ}{2πλ/4} [0±\frac{2πn}{N}])\)
⇨ \(θ_n=cos^{-1}([±\frac{4n}{N}])\)
7. Find the Nulls of the 8-element array in which elements are separated by λ/4 and phase difference is 0?
a) \(θ_n=cos^{-1}(\left[±\frac{n}{2}\right])\)
b) \(θ_n=cos^{-1}(\left[±\frac{n}{4}\right])\)
c) \(θ_n=sin^{-1}(\left[±\frac{n}{2}\right])\)
d) \(θ_n=sin^{-1}(\left[±\frac{n}{4}\right])\)
View Answer
Explanation: The nulls of the N- element array is given by \(θ_n=cos^{-1}(\frac{λ}{2πd} \left[-β±\frac{2πn}{N}\right])\)
⇨ \(θ_n=cos^{-1}(\frac{λ}{2πλ/4}\left[0±\frac{2πn}{N}\right])\)
⇨ \(θ_n=cos^{-1}(\left[±\frac{4n}{8}\right])\)
⇨ \(θ_n=cos^{-1}(\left[±\frac{n}{2}\right]).\)
8. The radiating pattern of single element multiplied by the array factor simply gives the ___________
a) Pattern multiplication
b) Normalized array factor
c) Beamwidth of the array
d) Field strength of the array
View Answer
Explanation: The radiation pattern of the single array antenna is multiplied by the antenna factor then it is called pattern multiplication. Array factor is the function of antenna positions in the array and its weights. Total array field is the field generated by the sum of the individual elements in array.
9. Condition for the half power width of the Array factor is given by ___________
a) \(\frac{Nᴪ}{2}=±1.391 \)
b) \(\frac{Nᴪ}{2}=±3\)
c) \(\frac{Nᴪ}{2}=±0.5\)
d) \(\frac{Nᴪ}{2}=±1\)
View Answer
Explanation: Array factor is the function of antenna positions in the array and its weights. Half power beamwidth is also known as the 3 decibel points. The half power beam widths of the array factor will occur at \(\frac{Nᴪ}{2}=±1.391. \)
10. The maximum of the first minor lobe of array factor occurs at 13.46 dB down the maximum major lobe.
a) True
b) False
View Answer
Explanation: The maximum of 1st minor lobe occurs at \(\frac{Nᴪ}{2}=±3π/2\)
⇨ \(AF = \frac{sin(\frac{Nᴪ}{2})}{\frac{Nᴪ}{2}}=0.212= -13.46dB.\)
Sanfoundry Global Education & Learning Series – Antennas.
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