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Antenna Fundamentals and Parameters Questions and Answers

21.

Assertion (A): A thick dipole has a wider impedance bandwidth than a thin one.

Reason (R): Increasing the conductor diameter lowers the Q of the resonant structure.

Select the correct answer from the options below.

Answer: A

Both are true and the second is the reason. A fatter element has a lower ratio of stored to dissipated energy, so the impedance varies more slowly with frequency and the useful band widens. The same idea explains cage dipoles, biconical antennas and the whole family of broadband structures, whose common feature is a large effective diameter. The limiting case is the biconical antenna, whose impedance is frequency independent because its geometry is defined entirely by angles.

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Answer: B

Polarisation is the figure traced by the tip of the electric field vector as time passes, observed in the direction of propagation, and it may be linear, circular or elliptical. A horizontal dipole radiates a horizontally polarised wave. The direction of maximum radiation is the beam direction, current sense is a circuit matter, and the ratio of the fields is the intrinsic impedance of the medium. Matching polarisation between transmitter and receiver is as important as matching impedance.

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Answer: C

Orthogonal linear polarisations are in principle completely isolated, and in practice twenty to thirty decibels of rejection is achieved, which is why polarisation is used to reuse frequencies. A three decibel loss is what occurs between a linear and a circular antenna, where half the power is in the wrong sense. The claim that polarisation affects only the pattern is simply wrong, and six decibels corresponds to nothing standard. Cross polarisation discrimination is a specified quantity in any serious link.

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Answer: D

Two orthogonal elements fed with equal amplitude and ninety degrees of phase difference produce a field whose vector rotates once per cycle, which is circular polarisation. Any departure from equal amplitude or from quadrature gives an ellipse instead. Feeding in phase gives a linear polarisation at forty-five degrees, which is what the last option describes as well. Helical antennas achieve the same end geometrically, by winding the conductor rather than by splitting the feed.

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Answer: A

Axial ratio is the ratio of the major to the minor axis of the polarisation ellipse, so zero decibels is perfectly circular and a small value such as one decibel is very nearly so. Linear polarisation corresponds to an infinite ratio. Satellite and GPS specifications commonly call for three decibels or better over the coverage angle. It has nothing to do with feed loss or with match, and confusing it with those is easy because all three are quoted in decibels.

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Answer: B

A rotating vehicle or a wave whose plane has been rotated by the ionosphere would fade badly on a linear system, whereas circular polarisation is insensitive to rotation about the propagation axis. Faraday rotation at lower microwave frequencies makes this decisive. It gives no gain advantage and in fact costs three decibels against a matched linear pair, rain affects it at least as much and depolarises it, and generating it needs extra hardware rather than less.

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Answer: C

Effective aperture is the received power divided by the power density of the incident wave, so it is an area with the dimensions of square metres whether or not the antenna has any obvious physical opening. A half wave dipole has an effective aperture of about 0.13 square wavelengths despite being a thin wire. For a reflector it is the physical area multiplied by the aperture efficiency, typically fifty to seventy per cent.

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28.

Effective aperture and directivity are related by \( A_{e} = D\lambda^{2}/4\pi \). What aperture does a directivity of 100 give at 1 m wavelength?

Answer: D

Substituting directly:

\[ A_{e} = \tfrac{100 \times 1}{4\pi} \approx 7.96 \]

giving about 8 square metres. The relation is universal, holding for wire antennas and reflectors alike, and it shows that for a fixed directivity the collecting area shrinks as the square of the frequency. That is one reason why link budgets at high frequency need physically large dishes to keep the same performance.

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Answer: A

The Friis equation gives the received power in a free space link from the transmitted power, both antenna gains, the wavelength and the separation, and it is the foundation of every link budget. Terminal current relates to radiated power through the radiation resistance, field strength and antenna height belong to ground wave propagation formulas, and noise temperature is a separate calculation. The equation assumes free space, matched polarisation and far field conditions, and every departure is added as a further term.

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Answer: B

Power density falls as the inverse square of range, so path loss rises as the square of distance, twenty decibels per decade or six decibels for each doubling. The fourth power law belongs to a two ray ground reflection model at long range over a plane earth, which is a different situation with a different physical cause. Path loss also rises as the square of frequency at constant gain, which is a consequence of the shrinking effective aperture rather than of anything happening in the space between.

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