Antenna Fundamentals and Parameters Questions and Answers

Answer: D

Path loss contains the wavelength squared, so at fixed gains it rises by six decibels for each doubling of frequency. The underlying reason is that a fixed gain antenna has a smaller effective aperture at higher frequency. If instead the physical aperture is held constant, gain rises and the effect reverses, which is why millimetre wave links are feasible at all. Transmitter power, bandwidth and efficiency are set by other considerations entirely.

Answer: A

Both are true and the reason is correct. Path loss rises as frequency squared but each antenna's gain rises as frequency squared as well, so with fixed apertures at both ends the net effect is an improvement proportional to frequency squared. This is why satellite and point to point links have migrated steadily upward in frequency. Atmospheric absorption and rain attenuation eventually reverse the advantage, which is what sets the practical ceiling for each kind of link.

Answer: B

Above about ten gigahertz rain begins to attenuate seriously, and specific absorption bands of oxygen and water vapour add further loss, so link availability rather than equipment capability sets the limit. Antenna gain and available bandwidth both improve with frequency, and receiver noise temperature is manageable. The design response is shorter hops, higher fade margins and site diversity, which is why millimetre wave links are measured in hundreds of metres rather than tens of kilometres.

Answer: C

A line of sight path needs not merely optical clearance but clearance of a substantial part of the first Fresnel zone, or diffraction loss will be significant. Raising the antennas is the usual way to achieve it. Gain is a property of the antenna and does not change with height, receiver noise is unaffected, and polarisation is set by orientation. Path profiling with Fresnel zone clearance is the first calculation done for any microwave hop.

Answer: D

The zone is an ellipsoid with the two antennas as foci, so its radius is largest halfway along and tapers to nothing at each end. That is where obstacle clearance is hardest to achieve and where a hilltop or building matters most. Since the radius also grows with wavelength, lower frequency links need more clearance for the same path, which is an unexpected consequence of the diffraction physics and a common source of surprise in planning.

Answer: A

The ratio of the electric to the magnetic field in a plane wave in vacuum is about 377 ohms, or 120 pi, a property of the medium rather than of any antenna. It appears in the expressions for radiation resistance and for power density and is worth knowing for that reason. Fifty ohms is a cable convention, 73 the dipole feed impedance and 300 that of a folded dipole and twin feeder, and none of them is a property of space itself.

Answer: B

Power density is the field squared divided by the intrinsic impedance:

\[ S = \tfrac{1^{2}}{377} \approx 0.00265 \]

watts per square metre, that is about 2.65 mW per square metre. The relation is the bridge between a field strength measurement, which is what an EMC receiver reports through its antenna factor, and a power density limit, which is how exposure standards are written. Being able to move between the two quickly is essential in compliance work.

Answer: C

The isotropic radiator is a mathematical fiction radiating uniformly over the whole sphere, used as the reference against which gain is quoted. It cannot be built, because no physical antenna can radiate uniformly over a sphere with a single polarisation. Equal patterns in the two planes describes a symmetric beam rather than an isotropic one, reciprocity gives equal transmit and receive gain to every passive antenna, and polarisation is a separate matter.

Answer: D

A half wave dipole in the clear has a gain of 2.15 dBi and an impedance near 73 ohms, both known accurately enough to serve as a comparison standard, and it is cheap and easy to reproduce. Its gain is modest rather than high, its pattern is omnidirectional only in the plane perpendicular to its axis, and like every antenna it is affected by nearby objects, which is why a reference measurement is made on a proper range.

Answer: A

The dipole radiates most strongly broadside and not at all along its own axis, giving a toroidal or doughnut shaped pattern with the wire through the hole. In the plane perpendicular to the wire the pattern is a circle, which is why a vertical dipole gives omnidirectional coverage in azimuth. A uniform sphere would be the impossible isotropic case, and no simple dipole produces a narrow beam in any direction.