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The correct option is (a) \(ᴪ=\frac{2.782}{N}\)

To EXPLAIN: Normalized ARRAY factor is given by \(AF=\frac{sin(Nᴪ/2)}{N \frac{ᴪ}{2}}=\frac{1}{√2} \)

⇨ \(\frac{Nᴪ}{2}=1.391\)

⇨ \(ᴪ=\frac{2.782}{N} \)

Right option is (a) The single element field is multiplied by the array factor

Easy explanation: TOTAL array field is the field generated by the sum of the individual elements in array and is given SIMPLE by multiplying the field DUE to single element by the array factor. Array factor is the FUNCTION of antenna POSITIONS in the array and its weights. Multiplying the normalized field with the normalized array factor gives the pattern multiplication.

The correct option is (a) \(\frac{SIN(Nᴪ/2)}{Nᴪ/2} \)

To explain: 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 \(\frac{sin(Nᴪ/2)}{Nᴪ/2}. \)

Right choice is (b) DIRECTIVITY DECREASES

For explanation: A single antenna provides low gain and less directivity. To INCREASE the directivity antenna arrays are used. With the antenna arrays, directivity and gain INCREASES and beam width decreases.

The correct choice is (d) π

The best I can explain: The NORMALIZED array FACTOR is given by \(AF_n=\frac{AF}{2}=COS(\frac{kdcosθ+β}{2})\)

⇨ AFn=0

⇨ \(cos⁡(\frac{kdcosθ+β}{2})=0\)

⇨ \(\frac{(\frac{2π}{λ})(\frac{λ}{4})cosθ-\frac{π}{2}}{2}=±\frac{π}{2}\)

⇨ Cosθ=-1

⇨ θ=180 or π

So Nulls occur at 180°.

The correct answer is (a) Pattern multiplication

For EXPLANATION I WOULD SAY: 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.

The correct choice is (a) Vector SUPERPOSITION of INDIVIDUAL FIELD from the element

Best explanation: The total resultant field is obtained by adding all the fields obtained by the individual sources in the array. An Array containing N elements has the resultant field equal to the vector superposition of individual field from the elements.

Right option is (a) High directivity

The explanation is: Long DISTANCE communication REQUIRES antenna with high directivity. To increase the directivity antenna arrays are used. With the antenna arrays, directivity and GAIN increases and BEAM width DECREASES.

The CORRECT choice is (a) \(-\frac{2π}{λ} d \)

EASIEST explanation: In end-fire array the PHASE excitation difference is –kd for θ=0°.

kdcosθ+β=0

β=-kd

\(β=-\frac{2π}{λ} d \)

The CORRECT answer is (a) \(COS^{-1}([±\frac{λ}{Nd}])\)

Explanation: The NULLS of the N- element array is given by \(θ_n=cos^{-1}⁡(\frac{λ}{2πd} [-β±\frac{2πn}{N}])\)

Given it’s a broadside array so β=0 and n=1 for FIRST NULL

\(θ_n=cos^{-1}⁡(\frac{λ}{2πd} [±\frac{2πn}{N}])= cos^{-1}⁡([±\frac{nλ}{Nd}])=cos^{-1}⁡([±\frac{λ}{Nd}])\)

Right choice is (a) 90°

Explanation: In a Broadside array the MAXIMUM radiated is directed TOWARDS the normal to the AXIS of the array. So it is at ANGLE 90°. In the end-fire array maximum radiation is along the axis of the array.

The correct ANSWER is (a) 60°

To elaborate: The nulls of the N- ELEMENT array is given by

\(θ_n=cos^{-1}⁡(\frac{λ}{2πd} \LEFT[-β±\frac{2πn}{N}\right])=cos^{-1}⁡(\frac{λ}{2πd} \left[±\frac{2πn}{N}\right])\)

⇨ \(θ_n=cos^{-1}⁡(\frac{λ}{2π(λ/2)} \left[±\frac{2πn}{N}\right])=cos^{-1} (±\frac{2n}{4})=cos^{-1}(±\frac{n}{2})\)

⇨ \(n=1 (first \,NULL) cos^{-1} (±\frac{n}{2})=cos^{-1} (±\frac{1}{2})=60° or 120°. \)

Right answer is (a) PATTERN multiplication

The explanation is: The radiation pattern of the single array ANTENNA is multiplied by the antenna factor then it is called pattern multiplication. So multiplying the normalized FIELD with the normalized array factor gives the pattern multiplication. Array factor is the function of antenna POSITIONS in the array and its weights. Nulls are known by EQUATING array factor to zero.

Right option is (d) ±kd

Best EXPLANATION: The MAXIMUM array factor occurs when \(\FRAC{sin \frac{Nφ}{2}}{\frac{Nφ}{2}}\) maximum that is \(\frac{Nφ}{2}=0. \)

And φ=kdcosθ+β

=> kdcosθ+β=0 For an end-fire array maximum RADIATION is along the axis of array so

 θ=0° or 180°

=> β=kd when θ=180°

=> β=-kd when θ=0°

Right OPTION is (a) True

The best EXPLANATION: The MAXIMUM of 1st minor LOBE occurs at \(\frac{Nᴪ}{2}=±3π/2\)

⇨ \(AF = \frac{sin(\frac{Nᴪ}{2})}{\frac{Nᴪ}{2}}=0.212= -13.46dB.\)

The correct ANSWER is (a) \(θ_n=cos^{-1}⁡(\frac{λ}{2πd}[-β±\frac{2πn}{N}])\)

To explain: To determine NULL points the array factor is set equal to zero.

\(\frac{SIN(\frac{Nᴪ}{2})}{\frac{Nᴪ}{2}}=0\)

⇨ \(sin(\frac{Nᴪ}{2})=0\)

⇨ \(\frac{Nᴪ}{2}=±nπ \)

⇨ \(kdcosθ+β=±\frac{2nπ}{N} \)

⇨ \(θ_n=cos^{-1}⁡(\frac{λ}{2πd}[-β±\frac{2πn}{N}]).\)

Right answer is (a) 60

To explain: 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; \(θ_n= =cos^{-1}⁡([1±\frac{1}{2}]) (considering \,β=-\frac{π}{2}) \)

\(θ_n =cos^{-1} (\frac{1}{2})\,or \,cos^{-1} (3/2) \)

\(θ_n =cos^{-1} (\frac{1}{2})=60.\)

Correct option is (a) \(cos^{-1}⁡\FRAC{λn}{Nd}\)

Easiest explanation: Nulls OCCURS when array factor \(AF=\frac{sin \frac{Nφ}{2}}{\frac{Nφ}{2}} = 0\)

⇨ \(sin\frac{Nφ}{2} = 0=>\frac{Nφ}{2}=±nπ \,and\, φ=kdcosθ+β=kdcosθ_n=\frac{2π}{λ} dcosθ_n\)

⇨ NULL occurs at \(θ_n=cos^{-1}⁡\frac{λn}{Nd}\)

The CORRECT ANSWER is (a) \(θ_n=cos^{-1}⁡(\left[±\frac{4n}{N}\right])\)

EASY explanation: The nulls of the N- element array is given by \(θ_n=cos^{-1}⁡(\frac{λ}{2πd} \left[-β±\frac{2πn}{N}\right])\)

⇨ \(θ_n=cos^{-1}⁡(\frac{λ}{2πλ/4} [0±\frac{2πn}{N}])\)

⇨ \(θ_n=cos^{-1}⁡([±\frac{4n}{N}])\)

The correct choice is (a) \(θ_n=cos^{-1}⁡(\left[±\frac{N}{2}\RIGHT])\)

The best explanation: The nulls of the N- element ARRAY is given by \(θ_n=cos^{-1}⁡(\frac{λ}{2πd} \left[-β±\frac{2πn}{N}\right])\)

⇨ \(θ_n=cos^{-1}⁡(\frac{λ}{2πλ/4}\left[0±\frac{2πn}{N}\right])\)

⇨ \(θ_n=cos^{-1}⁡(\left[±\frac{4n}{8}\right])\)

⇨ \(θ_n=cos^{-1}⁡(\left[±\frac{n}{2}\right]).\)

Correct answer is (B) 75.5°

For explanation I would say: The nulls of the N- element array is GIVEN by

\(θ_n=COS^{-1}⁡(\frac{λ}{2πd} [-β±\frac{2πn}{N}])=cos^{-1}⁡(\frac{λ}{2πd} [±\frac{2πn}{N}])\)

⇨ \(θ_n=cos^{-1}⁡(\frac{λ}{2π(λ/2)} [±\frac{2πn}{N}])=cos^{-1} (±\frac{2N}{8})=cos^{-1} (±\frac{n}{4}) \)

⇨ \(n=1 (FIRST \,null) cos^{-1} (±\frac{n}{4})=cos^{-1} (±\frac{1}{4})=75.5°. \)

The CORRECT OPTION is (b) No side lobes

Explanation: High directive antennas are required for the long distance communications. The array of antennas is used to INCREASE the DIRECTIVITY. The directivity can be increased by increasing the DIMENSIONS of antenna but it creates side lobes.

Right choice is (a) \(ᴪ=±\frac{2s+1}{N} π \)

The explanation: The SECONDARY MAXIMA occur when the NUMERATOR of the array factor equals to 1.

⇨ \(SIN(\frac{Nᴪ}{2})=±1\)

⇨ \(\frac{Nᴪ}{2}=±\frac{2s+1}{2} π \)

⇨ \(ᴪ=±\frac{2s+1}{N} π .\)

Correct option is (a) 0

Easiest EXPLANATION: The nulls of the N- element ARRAY is GIVEN by

\(θ_n=COS^{-1}⁡(\frac{λ}{2πd} \left[-β±\frac{2πn}{N}\right])=cos^{-1}⁡(\frac{λ}{2πd} \left[±\frac{2πn}{N}\right])\)

\(θ_n=cos^{-1}⁡(\frac{λ}{2π(λ/4)}\left[±\frac{2πn}{N}\right])=cos^{-1} (±\frac{4n}{4})=cos^{-1}(±n)=0\)

The correct choice is (a) cos^-1 (±n)

Best explanation: The NULLS of the N- ELEMENT ARRAY is given by

\(θ_n=cos^{-1}⁡(\frac{λ}{2πd} \left[-β±\frac{2πn}{N}\right])=cos^{-1}⁡(\frac{λ}{2πd} \left[±\frac{2πn}{N}\right])\)

⇨ \(θ_n=cos^{-1}(⁡\frac{λ}{2π(λ/4)}\left[±\frac{2πn}{N}\right])=cos^{-1} (±\frac{4n}{4})=cos^{-1} (±n)\left[n=1,2,3 \,and \,n≠N,2N…\right]\)

Correct OPTION is (a) 0

For explanation: The NORMALIZED array factor is given by \(AF_n=\frac{AF}{2}=cos⁡(\frac{kdcosθ+β}{2})\)

⇨ AFn=0

⇨ \(cos(\frac{kdcosθ+β}{2})=0\)

⇨ ⁡\(\frac{(\frac{2π}{λ})(λ/4)cosθ+\frac{π}{2}}{2}=\frac{π}{2}\)

⇨ Cosθ=1

⇨ θ=0

So Nulls OCCUR at 0°.

Right OPTION is (a) \(\frac{3}{8} π\)

For explanation I would SAY: The secondary maxima occur when the NUMERATOR of the array factor EQUALS to 1.

⇨ \(sin(\frac{Nᴪ}{2})=±1\)

⇨ \(\frac{Nᴪ}{2}=±\frac{2s+1}{2} π \)

⇨ \(ᴪ=±\frac{2s+1}{N}π=\frac{2+1}{8} π=\frac{3}{8} π.\)

Right choice is (c) 90

For 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.

The correct ANSWER is (a) As the number of elements increase in ARRAY it becomes more directive

The explanation is: 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.

Correct answer is (b) False

Easiest explanation: In end-fire ARRAY the phase EXCITATION difference is ±kd and their phase vary progressively and get unidirectional MAXIMUM radiation finally. In broadside SIDE array the phase excitation difference is ZERO.

Correct answer is (a) \(COS^{-1} (±\frac{N}{4})\)

To elaborate: The nulls of the N- element array is GIVEN by

\(θ_n=cos^{-1}⁡(\frac{λ}{2πd}[-β±\frac{2πn}{N}])=cos^{-1}⁡(\frac{λ}{2πd} [±\frac{2πn}{N}])\)

⇨ \(θ_n=cos^{-1}⁡(\frac{λ}{2π(λ/2)} [±\frac{2πn}{N}])=cos^{-1} (±\frac{2n}{8})=cos^{-1} (±\frac{n}{4}) \)

Right ANSWER is (b) Field pattern is the SUM of INDIVIDUAL elements in array

To elaborate: The total resultant field is obtained by adding all the FIELDS obtained by the individual sources in the array. Radiation pattern is obtained by multiplying the individual pattern of the element. Field pattern is the product of individual elements in array. Antenna arrays are USED to get high directivity with less side lobes.

Right answer is (a) \(θ_n=cos^{-1}⁡([∓1±\FRAC{4n}{N}])\)

Explanation: 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 \(β=±kd=±\frac{2πd}{λ}=±\frac{π}{2},\)

\(θ_n=cos^{-1}⁡(\frac{2}{π} [∓\frac{π}{2}±\frac{2πn}{N}])\)

\(θ_n=cos^{-1}⁡([∓1±\frac{4n}{N}])\)

The CORRECT OPTION is (a) 41.4°

To explain I would say: 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 1^st minor lobe)

⇨ \(θ_s=cos^{-1} (±\frac{3}{4})=41.4°\)

Right choice is (B) 60

Easy explanation: The nulls of the N- element array is GIVEN by

 \(θ_n=cos^{-1}⁡(\frac{λ}{2πd} [-β±\frac{2πn}{N}])=cos^{-1}⁡(\frac{λ}{2πd} [±\frac{2πn}{N}])\)

\(θ_n=cos^{-1}⁡(\frac{λ}{2π(\frac{λ}{4})}\LEFT[±\frac{2πn}{N}\right])=cos^{-1} (±\frac{4n}{8})=cos^{-1} (±\frac{n}{2})=cos^{-1} (±1/2)=60\)

The CORRECT option is (a) True

To explain I would say: INCREASING the dimensions of antennas may lead to the appearance of the SIDE lobes. So by placing a group of antennas together the electrical size of ANTENNA can be increased. With the antenna arrays, directivity and gain increases and beam width decreases.

The correct choice is (a) \(θ_m=cos^{-1}⁡(\frac{λ}{2πd}[-β±2πm])\)

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

⇨ \(θ_m=cos^{-1}⁡(\frac{λ}{2πd}[-β±2πm]).\)

Right CHOICE is (a) sinc(2cosθ)

Best 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

\(\frac{Nᴪ}{2}=2kdcosθ=2(\frac{2π}{λ})(\frac{λ}{2})cosθ=2πcosθ \)

\(AF=\frac{sin(Nᴪ/2)}{N \frac{ᴪ}{2}}=\frac{sin(2πcosθ)}{2πcosθ}=sinc(2cosθ).\)

The CORRECT answer is (a) \(\frac{9λ}{4}\)

The explanation: For an N-element end-fire ARRAY, the overall LENGTH of the array is given by ρ=(N-1)d

⇨ ρ=(N-1)d=(10-1)(λ/4)=9 λ/4

Right CHOICE is (a) \(cos^{-1}⁡([±\frac{nλ}{Nd}])\)

The explanation is: The nulls of the N- ELEMENT ARRAY is GIVEN by \(θ_n=cos^{-1}⁡(\frac{λ}{2πd} [-β±\frac{2πn}{N}])\)

Given it’s a BROADSIDE array so β=0

\(θ_n=cos^{-1}⁡(\frac{λ}{2πd} [±\frac{2πn}{N}])= cos^{-1}⁡([±\frac{nλ}{Nd}]).\)

The correct choice is (a) θm=cos^-1⁡(4m)

To EXPLAIN I WOULD say: 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).