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The CORRECT CHOICE is (a) 15.18°

For explanation I would say: LENGTH of the H-plane horn \(ρ =\frac{h^2}{8δ}=\frac{(12λ)^2}{8×0.4λ}=45 λ\)

FLARE angle of the H-plane horn = \(2tan^{-1}⁡\frac{h}{2ρ}=2×tan^{-1}⁡\frac{12λ}{2×45λ}=15.18°\)

Correct answer is (B) 12.24 cm

The explanation is: DIAMETER \( = \SQRT{3λL}=\sqrt{\FRAC{3c}{F} L}=\sqrt{\frac{3×3×10^10}{900×10^6}×1.5}=12.24 cm \)

Right answer is (C) θ = 27.72

Explanation: Flare angle of horn antenna when axial length l and path DIFFERENCE d is given is

\(θ = 2 COS^{-1}⁡\frac{l}{l+δ}=2 cos^{-1}⁡\frac{10λ}{10.3λ}=27.72\)

Correct option is (B) \(\frac{67λ}{h_h} \)

EASY explanation: HPBW of H-plane horn is\(\frac{67λ}{h_h} \)where hh the APERTURE DIMENSION of H-plane horn.

Correct ANSWER is (a) Length \(ρ=\frac{h^2}{8δ}, flare\, ANGLE=2tan^{-1}⁡\frac{h}{2ρ}\)

Easiest EXPLANATION: Design equations of a horn antenna are:

Length \(ρ=\frac{h^2}{8δ},flare\, angle\, θ=2 TAN^{-1}⁡\frac{h}{2ρ}\)

The correct CHOICE is (b) 45 λ

Best explanation: LENGTH of the E-plane horn \(ρ =\FRAC{h^2}{8δ}=\frac{(12λ)^2}{8×0.4λ}=45 λ\)

Right choice is (d) Conical horn

Easy EXPLANATION: In conical horn, the WALLS of the circular waveguide are flared out. For pyramidal horn, the walls of the RECTANGULAR waveguide are flared out in both directions.

The CORRECT choice is (a) increased, decreased

Best EXPLANATION: Horn antenna is DESIGNED to IMPROVE the directivity and reduce the diffractions. It is used to properly match the waveguide to a large radiating aperture by shaping TRANSITION gradually. So, reflections are also decreased.

Right answer is (b) Horn antenna

For EXPLANATION I would SAY: Horn antenna is used for broadband signals due to its FLARED nature. It is used to properly MATCH the waveguide to a large radiating aperture by shaping transition gradually. It is used in 300MHz to 30GHZ frequency range.

The correct option is (c) Aperture Antenna

For explanation: HORN antenna BELONGS to Aperture antenna. Flaring DONE is done at the aperture of the rectangular WAVEGUIDE gives different types of horn antenna. DIPOLE belongs to Wire antenna. Yagi-Uda is an array antenna. Convex-plane is example for lens antenna.

Correct answer is (b) Exponential tapered pyramidal HORN

To EXPLAIN: In EXPONENTIALLY tapered pyramidal horn, the transition region is GRADUALLY tapered exponentially to minimize the reflections. These are mainly used when reflections in the waveguide are critical.

Right answer is (d) 6.7

Easiest explanation: HPBW of H-plane HORN is \(\FRAC{67λ}{h_h} = \frac{67λ}{10λ}=6.7\).

The correct option is (B) H-plane horn

To explain I would say: In Sectorial horn antenna, FLARING is done only in one DIRECTION.If the flaring is done in the MAGNETIC FIELD direction, then it is called H-plane horn antenna.

Right OPTION is (c) 7.59°

The explanation: OPTIMUM aperture ANGLE of the H-plane horn = \(tan^{-1}⁡\frac{h}{2ρ}=tan^{-1}⁡\frac{12λ}{2×45λ}=7.59°\)

Correct OPTION is (a) 30.15dB

The explanation: DIRECTIVITY \( = 10log_{10}\frac{7.5A_e}{λ^2} = 10log_{10}\frac{7.5×10λ×15λ}{λ^2} =30.51dB

\)

Right OPTION is (a) \(\frac{56λ}{h_e} \)

Best explanation: HPBW of E-plane HORN is \(\frac{56λ}{h_e} \)where he the APERTURE DIMENSION of E-plane horn.

The CORRECT option is (B) 22.36 cm

The explanation is: Aperture width \( = \sqrt{3λL}=\sqrt{\FRAC{3c}{f} L}=\sqrt{\frac{3×3×10^{10}}{900×10^6}×5}=22.36 cm\)

Right option is (b) Sectoral antenna

The best I can EXPLAIN: In Sectoral horn antenna, flaring is done only in one direction. Depending on the flaring direction with RESPECT to field PROPAGATION, it is DIVIDED into E-plane or H-plane horn antenna. Conical horn flared cross section is in shape of a cone.Pyramidal horn flared cross section is in shape of a four-sided pyramid. In exponential horn, the separation of sides INCREASES as a function of length.

The correct OPTION is (d) Improve Standing waves

The BEST EXPLANATION: Flaring is done to provide the impedance matching. So, the reflections are minimized so there will be no standing waves. Side lobes will be reduced so the FORWARD gain is ALSO improved.

Correct answer is (a) high, low

To explain I would say: The smaller FLARE angle means the beam is NARROWER. The BEAMWIDTH and directivity are inversely PROPORTIONAL to each other. If beamwidth increases then directivity will decrease. So, for a SMALL fare angle, the directivity is more and beamwidth is less.

Correct CHOICE is (b) Increases

Best explanation: The beam width of the horn antenna increases as the flare angle increases. The reflections are decreases as the flare angle increases. But at 900, the antenna approximates to an OPEN END waveguide and its gain decreases.

Correct answer is (c) UHF and SHF

For EXPLANATION I WOULD say: Horn antenna is used in the frequency range 300MHz to 30GHZ. So, it is used in UHF and SHF frequency bands.

HF- VHF: 3MHz to 300MHz

LF-UHF: 30KHZ to 3GHz

LF-VHF: 30kHz to 300MHz

Correct answer is (b) 19.36 cm

The explanation: APERTURE width \( = \sqrt{2λL}=\sqrt{\FRAC{2C}{f} L}=\sqrt{\frac{2×3×10^{10}}{800×10^6}×5}=19.36 cm\)

The CORRECT option is (a) 25π

For EXPLANATION: Physical area of CONICAL HORN A=πr^2=π×5^2=25π.

Right CHOICE is (a) True

To ELABORATE: The radiation PATTERN of H-plane Sectoral horn is narrower than the E-plane Sectoral horn. It is because of the dimension of the horn antenna in that direction.

Right answer is (a) \(θ= 2 cos^{-1}⁡\frac{L}{l+δ} \)

To EXPLAIN I would say: Flare angle of horn antenna when axial LENGTH l and path DIFFERENCE d is given is

\(θ= 2 cos^{-1}⁡\frac{l}{l+δ} \).