MSc Map Antennas and Propagation module exam 1998 (DJJ questions)

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Question 1.

Define the terms "directivity", "gain", "efficiency", "polarisation", and "effective aperture" in the context of antenna design. [20%]

Using a diagram, illustrate the terms "boresight direction", "azimuth angle", "elevation angle", "main beam", and "sidelobes". Explain why an isotropic source cannot be constructed in practice, and distinguish carefully between the terms "isotropic" and "omnidirectional". Explain why an omnidirectional antenna necessarily has a maximum directivity greater than unity. [20%]

A hypothetical isotropic radiator has unity gain and effective aperture (lambda^2)/(4 pi) for radiation of wavelength lambda. Calculate the gain of a satellite dish antenna of effective aperture 3.75 square metres at a frequency of 14 GHz, and estimate the diameter of the parabolic reflector needed to implement such a dish, given an aperture efficiency factor of 0.65. [30%]

Estimate the pointing accuracy needed for an antenna dish of boresight gain 45dBi and explain with examples what technological steps can be taken to alleviate the effects of wind loading. [30%]

Outline solution 1.

• Directivity :- (Power per steradian in a specified direction, as radiated by the antenna)/(Power per steradian radiated by isotropic source radiating the same total power).

  Gain:- as directivity but normalised to input power of the antenna; that is, (Power per steradian in a specified direction) /(the power per steradian radiated by an isotrope having the same total radiated power as the input power of the antenna under consideration.)

  Efficiency:- Gain/Directivity

  Polarisation:- The projection of the tip of the E phasor onto a plane surface normal to the direction of propagation; the curve traced out (ellipse, circle, line) has characteristics which define the polarisation.

  Effective Aperture:- If the incoming power density is S watts per square metre, the power delivered to the receiver, in watts, = S times the Effective Aperture in square metres. The gain = (4 pi Effective Aperture)/(lambda^2).

• A diagram showing a polar radiation plot for an antenna is presented below. I have drawn the half power azimuth directions as straight lines; the pointing accuracy required for the antenna may be taken as lying within these lines. The azimuth angle is the direction from boresight in the plane; the elevation angle is the direction from boresight out of the plane.

  

• An isotropic source radiates equally in all directions in both azimuth and elevation angles. The power density is uniformly distributed around a large sphere centred on the antenna. An omnidirectional antenna radiates uniformly in all azimuth directions, but has a deep null in the orthogonal elevation direction. This is because the radiated E fields are transverse, in the direction of the currents in the antenna. When the current is directed straight at the field point, there can be no transverse component of electric field generated. For a loop antenna, which also has rotational symmetry, there can be no transverse H field for propagation along the "axis of the loop" direction. If an omnidirectional antenna has less radiation than an isotrope (in a certain direction), it must compensate by having more radiation than an isotrope along boresight directions. Thus the maximum directivity is necessarily greater than unity.

• At a frequency of 14 GHz, the wavelength lambda = 0.03/1.4 metres = 2.143 cm. The gain = (4 pi effective area)/(lambda^2) = (4 3.142 3.75)/(0.02143^2) = 102,625 or about 50 dBi. The aperture efficiency is 0.65 so the area of the parabolic dish we need to produce an effective aperture of 3.75 square metres is 3.75/0.65 = 5.77 square metres, which for a circular dish gives a diameter of 2.71 metres. Allowing for unforeseen losses due to weathering, we might make the dish diameter 2.8 metres.

• Assuming a uniformly illuminated circular beam footprint, the (area of the footprint)/(the area of the sphere at radius R) = 1/(power gain of the antenna), from simple proportional considerations. The power gain is 45 dBi or 31623. The area of the footprint is (pi a^2) where "a" is the footprint radius, and a/R = the beam semi-angle theta in radians. Thus theta = sqrt(4/31623) = 0.0112 radians = 0.64 degrees so we must point our antenna to within about 38 minutes of arc of the target. In an antenna of radius about a metre, the dish should be sited away from turbulent gusts of wind around the edges of buildings. Bracing can be applied. Alternatively the antenna can be encased in a microwave-transparent plastic "radome". It is also possible to reduce the wind loading by using FLAPS technology, where the parabolic dish profile is synthesised by electrical phase shifts in a plane array of dipole elements, each element consisting of a conducting coating on thin dielectric wires separated by distances of the order of a wavelength. The flat surface can be mounted close to the building and parallel to a wall, and an offset-feed used for beam-pointing.

Question 2.

Describe what is meant by the term "array antenna". Define the terms "array factor", "element pattern", and "pattern multiplication". Explain what constraints are imposed on the individual elements in order for the properties of the array antenna to be calculated by pattern multiplication. [25%]

An array consists of four half-wave dipoles, spaced in a straight line by a distance of one half wavelength along a line at right angles to their rods. Sketch the element pattern in the H plane and in the E plane. Sketch the array pattern and identify the direction(s) of the main beam(s). Calculate the boresight gain in the case that all elements are fed with equal amplitudes which are in phase. [35%]

It is desired to steer the beam by 30 degrees from boresight in the H-plane. Calculate the required phase shift between currents on adjacent elements. [15%]

Explain the term "Very Long Baseline Interferometry" (VLBI) and estimate the resolution of a VLBI system consisting of two dishes at a frequency of 8GHz, each of diameter 20m, spaced by a distance of 1000km. [25%]

Outline solution 2.

• An "array antenna" has elements (say, N in number) which are "similar":- that is, they have the same radiation patterns and polarisation properties and are orientated in the same directions in space. They don't have to have the same excitation currents however. The excitation amplitude of the i-th element may be written A exp(-j phi) where A is the amplitude and phi the phase. The elements are spaced on a grid, or pattern which can be 1-dimensional, or 2-dimensional. The grid does not have to be a straight line or a flat plane. The "array pattern" or "array factor" is the polar radiation plot of a hypothetical collection of isotropic sources placed on the array antenna grid or pattern, and each fed with the same amplitude A and phase phi as the actual antenna elements. The "element pattern" is the pattern of any of the similar elements and does not depend on the location of the element within the grid. "Pattern multiplication" can then be used to find the total array antenna polar radiation pattern, by multiplying the gain of the element by the gain of the array of isotropes for every particular direction of propagation. The requirements are that the element patterns are all identical, and that the actual currents on the elements (allowing for inter-element couplings) are used to determine the array factor.

• A dipole element is omnidirectional in the H-plane, so the radiation pattern is a circle centred on the element. The dipole element has the classic "figure of eight" radiation pattern in the E-plane. The nulls lie along the dipole rods. The array pattern for four isotropes spaced by a half-wavelength and fed with equal excitation amplitudes and phases is given in the diagram below. Note there is rotational symmetry of this pattern around the line joining the elements.

  

  The element boresight gain for a dipole is about 1.67, and the array pattern gain for the four elements is 4. Thus the total array antenna gain is 6.68 or 8.24dBi and this occurs in the two directions at right angles to the antenna rods and to the line joining the elements.

• To steer the beam through an angle psi from boresight, the extra distance the wavefront has to go from the adjacent element is (element spacing)[sin(psi)] to catch up with the original radiation. In our case psi is 30 degrees so sin(30) = 1/2 and the element spacing is lambda/2 so the extra distance to be travelled is lambda/4 or 90 degrees of phase shift.

• Many narrow lobes ("interference fringes") appear in the array pattern for large element spacings; this sets the resolution of the system to the angle between adjacent nulls. For pointing in azimuth and elevation we might use three elements spaced at the vertices of a triangle. For the problem, at 8 GHz the wavelength lambda is 0.03/0.8 = 0.0375 metres and the VLBI element spacing is 1000km = 10^6 metres. The first null away from boresight occurs when the path difference between the elements is lambda/2, or an angle from boresight of psi = lambda/(2*10^6) radians = 0.0039 seconds of arc. Thus the angle between nulls is 0.0078 seconds. This represents the angle subtended by my thumbnail at a range of about 13,000 km, or at a distance from the UK to Australia.