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The CORRECT choice is (a) 70λ/D

Explanation: The area the POWER is RADIATED is given by BEAMWIDTH. Half power Beamwidth is the area at which the power is radiated 50% of peak power. The half-power beamwidth is given by70λ/D.

Correct answer is (a) \(D = \FRAC{4\pi}{\lambda^2}A_{EFF}\)

The EXPLANATION: For a UNIFORM APERTURE antenna, the physical and effective apertures are equal.

The relation between directivity and the effective aperture of the antenna is given by

\(D = \frac{4\pi}{\lambda^2}A_{eff}\).

Right choice is (a) Horn antenna

The explanation is: The aperture of antenna at the END DETERMINES its shape. Horn antenna has a rectangular aperture. Helical, Parabolic, CONICAL antenna have circular APERTURES. Helical antenna belongs to frequency INDEPENDENT antenna.

Right choice is (a) True

Easiest EXPLANATION: Equivalence PRINCIPLE follows the uniqueness theorem. It provides a unique solution for the boundary conditions. So this is used in the far FIELD analysis of the aperture ANTENNAS in the OUTER regions.

Right choice is (a) Equivalence principle

For explanation I would say: Equivalence principle follows the uniqueness THEOREM. It provides a UNIQUE solution for the boundary conditions. So this is used in the analysis of the APERTURE antennas. Friss transmission is used to relate the distance and power RADIATION between the antennas. Reflectivity and diffraction are not the PRINCIPLES used for analysis of aperture antenna.

Correct choice is (a) f(x, y)=f(x) f(y)

The explanation is: The RADIATION pattern for the RECTANGULAR aperture is likely RELATABLE to the line source distributions. If the functions f(x) and f(y) are SEPARABLE, then total pattern will be the PRODUCT of the two functions. f(x, y)=f(x)f(y).

The correct ANSWER is (a) True

To explain I WOULD say: The uniform APERTURE produces the high SLL under a constant phase amplitude excitation. To reduce this SLL effect, tapering is done. Tapering is done maximum at center and reduces to zero at the edges for an equivalent source distribution.

Correct CHOICE is (a) Js=n×H=0;Ms=-n×E

The explanation is: The EQUIVALENT surface CURRENTS Js, Ms radiates fields H, E respectively. The conditions for the electric CONDUCTOR equivalent is Js=n×H=0;Ms=-n×E.

Right option is (B) Js=n×H;Ms=-n×E=0

To explain: The equivalent surface currents Js, Ms RADIATES fields H, E. The conditions for the ELECTRIC conductor equivalent is Js=n×H;Ms=-n×E=0. This is one of the equivalence principle MODES.

Right choice is (a) Tapering

Easiest explanation: The uniform aperture produces the high SLL under a CONSTANT phase amplitude excitation. To reduce this SLL effect, tapering is DONE. Tapering is done maximum at CENTER and reduces to ZERO at the edges for anequivalent source distribution.

Right answer is (a) 140λ/D

The EXPLANATION is: The AREA the POWER is radiated is given by BEAM-width. The beam-width between the FIRST nulls is the FNBW. For circular aperture the FNBW is given by 140λ/D. The half-power beam width is given by 70λ/D.

Correct answer is (a) True

Explanation: The equivalence principle replaces the aperture ANTENNA with surface CURRENTS and thereby fields. It is DERIVED from the uniqueness theorem and provides a unique solution for the boundary conditions for the fields. This is mainly used for the far field analysis of the aperture antenna.

The CORRECT OPTION is (a) True

For explanation I would say: The radiation PATTERN for the rectangular distributions is similar to the line source distributions. In this the patterns is calculated by INTEGRATION in two directions as rectangle have two different lengths in different direction.

Correct option is (a) 0.886λ/a

Explanation: By equating the field in H-plane to half POWER point

\(\frac{sin⁡(0.5kasin\THETA)}{0.5kasin\theta} = \frac{1}{\sqrt{2}} => \theta =arcsin⁡(\frac{0.443\lambda}{a})\)

Now HPBW = \(2 arcsin⁡(\frac{0.443\lambda}{a}) \approx 0.886\lambda/a.\)

Right option is (a) sin^-1⁡(λ/b)

To elaborate: The area the POWER is radiated is GIVEN by Beam-width. The beam-width between the FIRST nulls is the FNBW. \(\FRAC{kb}{2}\) sinθ=nπ

⇨ θ= sin^-1⁡(nλ/b)

⇨ For first null n=1θ= sin^-1⁡(λ/b).

The correct option is (c) Principle PATTERNS in E-plane is same and H-plane is different

The EXPLANATION is: Tapering is done to reduce the SLL EFFECT. SINCE the E-plane is not tapered, its principle pattern is same but in H-plane as the aperture is tapered principle pattern will be different from the uniform aperture antenna.

The correct option is (a) True

Easiest EXPLANATION: The FNBW in E-plane is GIVEN by FNBW=2 θ = sin^-1⁡\(⁡(\frac{n\lambda}{b})\). As the dimension of the antenna aperture increases, the FNBW will decrease. Thereby, beam-width becomes narrower.

The correct OPTION is (a) Half-Dipole

Easiest EXPLANATION: ANTENNA with an APERTURE at the end is known as aperture antenna. Horn antenna, Waveguide antenna and slot antenna are examples of aperture antenna. Half-dipole is a WIRE antenna.

The CORRECT CHOICE is (c) 70

The best I can explain: The first null beam-width is GIVEN by 140λ/D.

⇨ \(\frac{140\lambda}{D} = \frac{140}{2}=70.\)

Correct answer is (a) UNIQUENESS Theorem

The best explanation: Equivalence principle follows the uniqueness theorem. It provides a unique solution for the BOUNDARY conditions. Uniqueness theorem is defined from the pointing theorem. FRISS transmission is USED to relate the distance and power radiation between the ANTENNAS. Gauss theorem states that total electric flux enclosed by charge is equal to net positive charge.

Right OPTION is (a) 35

To EXPLAIN I would say: The half-power beamwidth is given by70λ/D.

⇨ \(\frac{70\lambda}{D}=\frac{70}{2}=35.\)

Correct choice is (a) \(\FRAC{sin⁡(0.5kbsin\theta)}{0.5kbsin\theta}\)

Explanation: The principle plane PATTERNS for the uniform rectangular APERTURE antenna is given by

F(θ) = \(\frac{sin⁡(0.5kbsin\theta)}{0.5kbsin\theta}\) and main LOBE occurs when θ =0.

Right option is (a) -13.26

The explanation: The aperture of antenna at the end determines its SHAPE. If the FIELD is uniform in amplitude and phase ALONG the RECTANGULAR aperture then it is called a uniform rectangular aperture antenna. The FIRST side-lobe occurs at -13.26dB.

Right option is (a) UHF and EHF

The best I can EXPLAIN: ANTENNA with an aperture at the END is known as aperture antenna. Example is waveguide. These antennas usually operated at UHF and EHF frequencies (300MHz to 300GHz).

LF – 30 KHZ – 300 kHz

MF – 300k-3MHZ

HF – 3MHz -30MHz