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Right OPTION is (c) 180

To explain I would say: The PROGRESSIVE phase shift of the end-fire array is 180°. It is a linear array whose DIRECTION of radiation is along the AXIS of the array. For a BROADSIDE array it is 0°.

The correct choice is (a) BROAD side

Easy EXPLANATION: The total PHASE difference of the fields is GIVEN by Ѱ=kdcosθ+β

Here β is the progressive phase shift

⇨ β=0, array is a uniform broadside array

⇨ β=180, array is a uniform end-fire array

Yagi-uda antenna, fishbone antenna are end-fire antenna array.

The correct OPTION is (a) 4λ/2

For explanation I would say: The spacing between elements should not equal to integral multiples of λ to avoid grating lobes. The option 4λ/2=2λ

So when d=2λ grating lobes occurs which MEANS maxima are found at other ANGLES also. So this is not a DESIRED spacing.

Correct answer is (b) \(cos^{-1} (±\frac{n}{2})\)

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

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

⇨ \(θ_n=cos^{-1}⁡(\frac{λ}{2π(λ/4)}) [±\frac{2πn}{N}])=cos^{-1} (±\frac{4n}{8})=cos^{-1} (±\frac{n}{2})[n=1,2,3 \,and\, n≠N,2N…]\)

Correct option is (a) \(\frac{SIN(2kdcosθ)}{2kdcosθ} \)

The 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

\(\frac{Nᴪ}{2}=2kdcosθ\)

\(AF=\frac{sin(Nᴪ/2)}{N \frac{ᴪ}{2}}=\frac{sin(2kdcosθ)}{2kdcosθ} \)

Right ANSWER is (a) True

Easiest explanation: For ISOTROPIC ELEMENTS it radiated N all directions. The radiation pattern of the array is the pattern of the individual array element is MULTIPLIED by the array factor. So its total radiation pattern depends on the nulls of the array factor only.

Correct option is (a) \(\frac{Nᴪ}{2}=±1.391 \)

The best I can EXPLAIN: 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 \(\frac{Nᴪ}{2}=±1.391. \)

Right option is (a) N elements PLACED at equidistance and fed CURRENTS of equal magnitude and progressive PHASE shift

The explanation: An array is said to be LINEAR if N elements are spaced equally long the line and is a uniform array if the current is fed with equal magnitude to all elements and progressive phase shift along the line. HIGH directivity can be obtained by antenna array.

The CORRECT option is (a) β=±kd

Easy explanation: 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°

The correct option is (a) \(\frac{5}{4} π\)

To explain I would say: The secondary maxima occur when the NUMERATOR of the array factor equals to 1.

⇨ \(sin(\frac{Nᴪ}{2})=±1\)

⇨ \(\frac{Nᴪ}{2}=±\frac{2s+1}{2} π \)

⇨ \(ᴪ=±\frac{2s+1}{N}π=\frac{2(2)+1}{4} π=\frac{5}{4} π.\)

The CORRECT choice is (a) 0

For explanation I would SAY: 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 a broadside maximum radiation is normal to axis of array so θ=90

=> β=0

The correct choice is (a) d < λ

The explanation is: GRATING lobes are the minor and UNNECESSARY lobes other than the major lobe.

To avoid grating lobes, KD(cosθ-cosθm)≤2π

\(\frac{d}{λ}≤\frac{1}{1+|cosθ_m |} \)

For BROADSIDE to avoid grating lobes (θm=90)

⇨ \(\frac{d}{λ}\) < 1

⇨ d < λ

Right choice is (a) \(\FRAC{d}{λ}≤\frac{1}{1+|cosθ_m |} \)

The explanation: Grating lobes are the MINOR and unnecessary lobes other than the major lobe.

To avoid grating lobes, KD(cosθ-cosθm) ≤ 2π

θm – Direction of MAXIMUM radiation

⇨ \(\frac{2πd}{λ}(cosθ-cosθ_m)≤2π\)

\(\frac{d}{λ}≤\frac{1}{cosθ-cosθ_m} \)

\(\frac{d}{λ}≤\frac{1}{1+|cosθ_m |} \)

Correct choice is (a) \(\FRAC{sin(Nᴪ/2)}{Nᴪ/2}\)

The best 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 \(\frac{sin(Nᴪ/2)}{Nᴪ/2}\)

The correct choice is (c) Same input CURRENT is fed through the ARRAY, but the phase excitation is varies progressively

Easy EXPLANATION: For end-fire array:

Phase excitation difference β=±kd

Maximum radiation occurs along the axis of the array that is at θ=0° or 180°.

Even though same input current s fed to the arrays of equal magnitude, the phase VARY progressively along the line to get the unidirectional pattern.

The correct answer is (a) 0° or 180°

The EXPLANATION is: In an End-fire ARRAY the MAXIMUM radiation is along the axis of the array. So it is at EITHER 0° or 180°. In broad-side array the maximum radiation is perpendicular to the axis of array that is at 90°.

The correct choice is (d) sinc(2cosθ)

The explanation is: 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

\(\frac{Nᴪ}{2}=4kdcosθ=4(\frac{2π}{λ})(\frac{λ}{4})cosθ=2πcosθ\)

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

The CORRECT option is (a) d < λ/2

To explain: GRATING lobes are the minor and unnecessary lobes other than the major lobe.

To avoid grating lobes, kd(cosθ-cosθm)≤2π

\(\frac{d}{λ}≤\frac{1}{1+|cosθ_m|} \)

For broadside to avoid grating lobes (θm=0)

⇨ \(\frac{d}{λ}\) < 1/2

⇨ d < λ/2