Collected Antennas Problems

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Antennas

Question A1

Define the term "isotropic radiator". A certain transmit antenna has boresight gain which is a factor 2.6 over isotropic. Express this gain as dBi.

• An isotropic radiator is a hypothetical source radiating power equally in all directions. The power density incident on a large sphere centred on the source does not depend on the position on the surface of the sphere.
• A numerical power gain of 2.6 is 4.15 dBi since 10 log[10](2.6) = 4.15.

This transmit antenna is fed with a signal of a certain power level, 800 Watts of which is accepted. Assuming that there are no scattering obstacles in the beam or the near field, and that there is no attenuation along the path, calculate the power density in watts/square metre, and the rms electric field, at a point at a range of 25km from the antenna along its boresight.

• At a radius of 25km = 25,000 metres, an isotropic radiator having 800 Watts total radiated power would give a power density of 800/[4 pi 25,000^2] Watts/square metre = 102 nanoWatts/square metre.
• The actual antenna has boresight gain of 2.6 so the effective isotropic radiated power for the actual antenna, on boresight, is 2.6*800 = 2080 watts. This allows for any loss of power in the resistive and other loss in the antenna ("efficiency factor") since the gain rather than the directivity is quoted, and this includes any loss.
• Therefore the received power density at 25km is 102*2.6 = 265.2 nanoWatts/sq metre.
• This power is the same as (E^2)/Zo where E is the r.m.s. electric field in Volts/metre and Zo is the impedance of free space, 120 pi = 377 Ohms.
• Thus E^2 = 120 pi 2.6 102 10^(-9) and the electric field is 10 mV/metre.

Question A2.

A receiver is fed by an array antenna. The array consists of a broadside arrangement of 8 identical elements connected with equal weights and the same phases to the receiver. Each element has boresight gain of 6 dBi perpendicular to the plane of the broadside array. The frequency of the link is 200MHz. Calculate the array pattern gain, the total gain, and the effective area of the receive antenna array.

• The element gain of 6dBi is a factor element gain of 3.98 = 10^(0.6). The array gain for equally weighted and phased elements is equal to the number of elements, and in this case is 8.
• The combined gain is 8 * 3.98 = 31.85.
• The effective area A is obtained from the formula gain = 4 pi A/[lambda^2]. The wavelength lambda at 200MHz is 1.5 metres. The gain is 31.85 so the antenna effective area is 5.7 square metres.

If the transmitting system of Question A1 is pointing at this array from a distance of 100km, calculate the total received signal power.

• From the answer to question A1, the power density at 25 km was 102*2.6 nanoWatts/square metre. 100km is four times further, so the power has reduced by a factor 1/[4^2] = 1/16 because of the inverse square law. The area of a sphere of radius 100km is 16 times that of a sphere of radius 25km.
• The received power density at 100km is therefore 2.6*102/16 = 16.58 nanoWatts/sq metre.
• So the total received power is 5.7 * 16.58 = 94.5 nanoWatts. We multiply the power density by the effective area of the receive antenna, to find the total power received.

If the receiver noise power is due to thermal noise in 10MHz bandwidth at a source temperature of 300K, calculate the possible range of the link for the receiver signal to noise ratio to be greater than 10dB. Comment on your result.

• The noise power in a bandwidth B of 10 MHz at a temperature T of 300K is given by kTB = 1.38*10^(-23)*300*10^7 = 4.14*10^(-14) Watts. Here, k is Boltzmann's constant, 1.38*10-23 Joules/degree K. A signal of 4.14*10^(-13) Watts is 10 times larger than this noise power.
• The received signal power to give 10dB S/N ratio needs to be 10 times larger than this noise power(a factor 10 = 10dB), and so the received signal power at a distance R of 47,777 or about 48,000km will be at this level since [100/R]^2 * 94.5 * 10-(9) = 4.14 * 10^(-13). Note that the distance R is calculated from this condition, which gives rise to the formula above equating the power received at distance R with the power required by the receiver.
• Thus with these antenna arrangements we can get a 10MHz bandwidth TV signal a distance 1.2 times the circumference of the earth. The transmit antenna has low gain so can be made omnidirectional. The power needed by the transmitter is of the order of 1 kWatt when allowing for notional antenna losses. The fractional bandwidth for this example is 10/200 = 0.05 or 5%, not unreasonable for this kind of antenna.

Question A3

Define the term "uniform array" as applied to a linear array antenna. Explain the term "null placement" and also indicate with an example how the nulls may be placed in specified directions for the radiation into the "array factor" from a uniform linear array antenna.

• A uniform array has elements spaced at equal intervals in the plane. For a linear uniform array antenna all the elements lie along a straight line at equal spacings.
• For a given element spacing, the phase shift between any two element currents can be chosen to give perfect cancellation in any desired direction. This is called "null placement".
• See this figure for an example.

  

• Provided all the elements are fed in phase, the boresight direction will lie along the y axis whatever the amplitude distribution. Choosing the amplitudes of the elements controls the radiation pattern in directions off-boresight.
• The simplest answer to this part of the question assumes that all the elements are fed with equal amplitudes. The amplitudes in question are the sizes of the currents on the elements, not the voltages fed to the elements. This is because it is the currents which radiate, and since the driving point impedances of the elements will be different (they have differing local environments; there are different environments for the end elements than for the centre elements for example; so these types of elements will be having different inter-element couplings and mutual impedances), so the currents will differ between elements if they are all fed with the same voltages.
• The extra phase shift between the radiation from adjacent elements spaced a distance d, for a plane wavefront propagating at 30 degrees to the y axis in the direction of the positive x axis, is (d/lambda)sin(30)(2 pi) radians. The wave starting from element on the right has less far to travel to get to the far field measurement point, so we have to retard its phase by this amount with respect to the adjacent element on the left. Thus as we move along the array from negative x to positive x, adjacent elements have phase shifts progressively less by (d/lambda)(1/2)(2*pi) radians.

• Suppose the phasing is +-+-+-+-, (this is my way of indicating a horizontal line of isotropic sources with adjacent elements fed with phase difference 180 degrees (inversion) from each other) then at 1/2 wavelength inter-element spacing the contributions from adjacent elements add up in phase in the horizontal direction, or along the x axis. This is because there is an additional 180 degrees of phase delay in the time it takes the signal to get from an element to its immediate adjacent neighbour, and so together with the 180 degrees of phase shift in the excitation current of the neighbour, the contributions add in phase. Therefore the two boresight directions are along the positive and negative x axes.
• The nulls may be determined from pattern multiplication. The array may be constructed by multiplying the element pattern (spacing lambda/2)

                         +-
  

  by the array pattern (spacing lambda)

                        o o o o 
  

  Here, we have brought two isotropes together, a half wavelength apart, and fed in antiphase, to make an "element" denoted by +-
  The radiation pattern from this "element" looks like the figure below, taken from the array antenna pages,

  

  and we have to multiply this "element pattern" by the "array pattern" formed from the array of four isotropes fed in phase and spaced lambda, as above....I haven't plotted this one for you but it is a simple extension of the xmaple program given on the array pages to do so.

  Now the only nulls of the *element* pattern lie along +/- y (which is the vertical axis, up and down) so all we have to do is to find the nulls of the *array* pattern. Assuming that all the excitation current amplitudes are equal, and in phase, then the nulls will occur at angles theta where the four phasors in the far field have resultant zero when added vectorially. They either add up to form the four sides of a square, or else they lie on top of each other forming two groups of two oppositely-directed phasors. Thus nulls occur for (d/lambda)sin(theta)(2 pi) = (pi/2) or (pi) or (3pi/2) and since (d/lambda) = 1 we find sin(theta) = (1/4) or (1/2) or (3/4) and so theta (with respect to the y axis) is either 14.5 degrees, 30 degrees, or 48.6 degrees.

• Travelling in the positive x direction along the elements of the array, the phase delay for (lambda/4) spacing is -(pi/2) radians for the radiation to get from one element to the next. The contributions must all add in phase along positive x, so there should be -(pi/2) radians or -90 degrees phase shift between successive elements, as we move to the right (positive x direction).
• By a similar argument, adjacent element contributions cancel in the negative x direction. The array factor power gain is 8.

Question A4.

With the aid of a sketch, explain the terms "boresight direction", "main beam", "azimuth angle", "elevation angle", "sidelobes", "nulls", "E-plane radiation pattern", and "vertical polarisation".

• See the following figures:- (these are very rough sketches only)

  

  

• The boresight direction is at zero azimuth and elevation angles and is the direction of strongest radiation in the polar pattern,
• The main beam consists of the radiation between boresight direction and the first null.
• The azimuth angle is measured between the boresight direction and the radiation direction in the horizontal plane.
• The elevation angle is measured between the boresight direction and the radiation direction in the vertical plane.
• The sidelobes consist of continuous regions of radiation between nulls, discounting the main beam.
• Nulls are directions in which there is no radiation. Nulls consist of lines or points on a far-field sphere.
• The E-plane radiation pattern is a plot of the power radiated as a function of angle away from boresight, in the plane defined by the boresight direction and the electric field vector. It is defined for linear polarisation cases only.
• Vertical polarisation occurs when the electric field vector has components in the vertical direction and, if the radiation is at a non-zero elevation angle, also along the projection of the radiation direction onto the azimuth plane. The electric field vector is always at right angles to the direction of propagation.

• The source currents lie along the antenna conductors, and for a vertical whip antenna the source current direction is vertical. There is nothing to define an azimuth direction (in the absence of scattering objects in the near field) so by usual symmetry arguments the radiation must not depend on the azimuth direction. Thus the antenna is omnidirectional.
• The E-field in the far field region is parallel to the projection of the source currents onto the plane a which lies at right angles to the selected radiation direction. In this case there will be a vertical component of E-field so the polarisation is vertical.
• To generate omnidirectional horizontal polarisation the source currents must lie in the azimuth plane. The circularly symmetric way of doing this is to use a loop in the azimuth plane.
• To generate circular polarisation we can use a combination of a loop antenna and a rod antenna, with the rod vertical on the axis of the horizontal loop. The currents are arranged to be in phase quadrature. The handedness of the polarisation depends on which current leads the other.

Explain why you would expect an omnidirectional antenna to have boresight directivity greater than unity.

• An antenna which has boresight directivity equal to one is necessarily isotropic. This is because if there exists a direction in which the directivity is less than one, there must be a corresponding direction (defining the boresight) in which the directivity is greater than one to maintain the average, and therefore the assumption of unity gain is contradicted.
• An omnidirectional antenna is only omnidirectional in the azimuth plane, and has zero directivity in directions at right angles to the azimuth plane. This is physically because the angle-independent source currents may lie along the direction at right angles to the azimuth plane; longitudinal currents do not radiate the necessarily transverse waves. Alternatively the currents form loops in the azimuth plane and the magnetic fields on axis of the loop are longitudinal. Therefore it has to have boresight directivity greater than unity, by the argument above. Of course, the gain (after allowing for antenna loss) can be less than unity or equal to unity even.

Calculate the beam solid angle for an antenna of gain 36 dBi. For a circular antenna beam from an antenna of gain 36 dBi pointing at a plane surface 20,000 km distant, orientated at right angles to boresight, estimate the circular footprint radius at the -1dB contour, assuming illumination of 0dB on boresight at this distance.

• 36 dBi represents a numerical power gain of 10^3.6 = 3981 and so the beam solid angle is (4 pi)/3981 = 3.157 millisteradians. We can assume the power is uniformly distributed within the beam solid angle and zero outside; setting the beam semi-angle at 0.0317 radians or 1.816 degrees. The gain at the -1dB contour is 3162 and at the -3dB contour is 1995. If we assume that the power gain falls off as [(sin(theta))/theta]^2 with the -3dB contour set at an angle given by the beam semi-angle as calculated above, then the value of theta for which [(sin(theta))/theta]^2 is -3dB (about 1/2) is about 1.39 radians, and the value of theta for which [(sin(theta)/theta]^2 is -1dB (about 0.8) is about 0.82 radians. Thus the -1dB contour is roughly at a factor 0.82/1.39 = 0.59 times the beam semi-angle of 0.0317 radians, so the semi-angle at the -1dB contour is about 0.0187 radians. At a range of 20,000 km this defines a circular footprint of radius 20,000*0.0187 = 374 km.

Estimate the maximum range (in number of wavelengths) for free space propagation between two antennas of gain 36 dBi pointing at each other, for a transmitter amplifier power of 1 microwatt and a system noise temperature of 300K in a bandwidth of 200kHz. You can assume the range is set at a receiver S/N ratio of 15dB.

• The free space divergence factor ("loss") is L = [lambda/(4 pi R)]^2 at range R and wavelength lambda, so the ratio of received power to transmitter power is L*(3981)^2 = 15.8E6 times L since 36dBi is a gain of 3981. If we call the range R = N*lambda where N is the number of wavelengths, then L = (1/(4 pi N))^2
• For a Signal/Noise ratio of 15dB we need a signal about 10^1.5 = 32 times larger than the noise, which for 200kHz bandwidth at 300 Kelvin is kTB = 1.36E-23 * 300 * 200,000 = 8.3E-16 watts. So at maximum range the received power should be 2.6E-14 watts which is a factor 2.6E-8 times smaller than the transmitter power of 1 microwatt. Thus the factor L can be as small as 2.6E-8/(15.8E6) = 1.7E-15 and this puts N at about 1.9 million wavelengths.
• An example. For a frequency of 13 GHz the wavelength is 3/1.3 cm so the physical range is about 45 km. This isn't bad for a one-microwatt transmitter.

Estimate, giving your reasons, the maximum line-of-sight range of a terrestrial microwave link using two 30cm square cross section pyramidal horns at 12GHz, for the transmission of PAL TV signals using some appropriate analogue modulation of a 15mW Gunn source.

• This part is now left as an exercise for the student.

Question MJU

Here is a question and answer given to me by Professor Underhill, transcribed exactly as I have it in front of me.

For a three element Yagi-Uda antenna explain why the element lengths are not the same. (hint: phasing of element currents?)

Given that an exact half wave dipole has an input impedance of 73+j42.5 ohms and for a particular thickness the dipole behaves as a transmission line of 500 ohms, calculate in units of wavelengths the lengths of:

(i) a director with a reactance of j10 ohms

(ii) a reflector giving a current phase lag of approximately 30 degrees.

• (i) = 0.50 lambda for reflector length
• (ii): 500 cot(kl) = 10 + 42.5 gives 2l = 0.467 lambda for director length.

I'm sorry I can't comment on this any further.
If you have any questions please email Professor Underhill at
m.underhill@ee.surrey.ac.uk

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

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.

Microwave Option paper 97-98 (DJJ antenna question)

Question 4.

Define the terms "isotropic radiator", "boresight direction", "directivity", "gain", and "E- plane radiation pattern" in the context of antenna design. Explain why it is impossible to construct an isotropic radiator in practice. [30%]

A certain transmitter has a final amplifier that delivers 10 kW to an antenna with 85% efficiency. The antenna has boresight directivity of 14dBi above an isotropic source. Calculate the r.m.s. electric field strength at a distance of 50 km from an ISOTROPIC source radiating 10 kW with 100% efficiency, and compare it with the field strength at this distance from the hypothetical transmitter-antenna combination described above. [40%]

Give a formula relating the effective area of an antenna to its boresight gain and to the wavelength of the radiated signal. Estimate the area of a horn aperture antenna required to give a gain of 14dBi at 13 GHz. If such a dish were used to transmit the 10kW signal described above, estimate the power flow in watts per square metre at a distance of 10m from this horn along boresight, and comment on the safety implications. [30%]

• Outline solution 4.

• An isotropic radiator is one that radiates uniformly in all directions, both in azimuth and elevation. The power density and field strengths radiated by an isotrope do not depend at all on the direction of radiation. For a directional antenna, if there is a single direction in which the radiated power is maximum, this is termed the boresight direction. The directivity of an antenna is a dimensionless number representing the ratio of the radiated power density on boresight to the radiated power density at the same place in space which would exist if radiated by an ideal isotrope having the same total radiated power. Often the directivity is given in dB. The gain is similarly defined, except that the powers referred to are input powers to the antenna. The gain is less than the directivity if the antenna is less than 100% efficient, due to resistive loss, absorption in the near field, spillover and blockage. The E-plane radiation pattern is a polar plot of either the power radiated or the field amplitude radiated as a function of direction in a plane containing the electric field vector of a linearly polarised antenna. Since there has to be a unique transverse polarisation direction for radiated transverse electromagnetic waves, one finds that this cannot be arranged for radiation in every direction from a point. Therefore isotropic radiators are impossible. Consider a radiator with E field everywhere directed North on a reference sphere. The direction is undefined on the polar axis. [30%]

• The surface area of a sphere of radius 50Km is 4*pi*50*50*10^6 square metres, and the 10 kW isotropic power is spread uniformly across this area at a power density of 0.32 microwatts per square metre. The power density p in a free-space wave is related to the rms electric field strength e by p = e*e/Zo where Zo is 377 ohms (120 pi ohms), the impedance of free space. From this we calculate e = 11 mV/metre approximately. The radiated power is 0.85*10,000 or 8500 watts. The antenna directivity of 14dBi increases the effective isotropic radiated power on boresight by a factor 10^1.4 or 25.12 so the e.i.r.p is 213.5 kW. The field strength on boresight at 50Km from our hypothetical antenna is therefore 11 * sqrt(21.35) = 50.75 mV/metre. [40%]

• The gain G and effective area A of an antenna are related by the formula G = 4*pi*A/(lambda)^2, where lambda is the wavelength of the radiation. At 13 GHz lambda = 3/1.3 cms = 2.31 cms and we found above that the 14dBi directivity is a factor of 25.12. From these, the area of the dish is A = 25.12*2.31*2.31/(4 pi) = 10.65 square cms. At 10m from the horn, in the far field region, the boresight power density is 25.12*10000/(4 pi 10*10) = nearly 200 watts per square metre, or 20 mW per square cm. This is comparable with the US allowed safety limits of 10mW/ square cm averaged over 6 mins. [30%]

Microwave Option paper 96-97 (DJJ's antenna-related questions)

Question 1

For waves travelling on a real coaxial cable connected to a linear antenna, define the following terms: (i) characteristic impedance (ii) complex reflection coefficient (iii) return loss (iv) complex voltage amplitude (v) propagation constant. [25%]

• The characteristic impedance is the ratio of voltage to current in a wave travelling in a single direction on transmission line, where the current sense is taken in the direction of travel of the wave.

• The complex reflection coefficient, measured at a point along the transmission line, is the ratio of complex backward wave voltage to complex forward wave voltage at that point.

• The return loss is the amount in dB by which the reflected POWER is less than the incident POWER.

• The "complex voltage amplitude" is the size of the voltage phasor at a point along the line, with phase angle determined by the origin of time. By redefining the zero of time the voltage amplitude can always be made real.

• The propagation constant is the amount by which the phase of the forward wave decreases (in radians) per unit distance travelled along the line, in the forwards direction. Numerically, it is equal to 2 pi divided by the wavelength.

Assuming that the line is nearly lossless, express the forward and backward wave power flows, in watts, in terms of the quantities defined above. [10%]

Derive an expression for the stored energy per unit length, on the cable, for waves travelling in a single direction only. [15%]

• If the forward wave "complex voltage amplitude" is written V+, and the backward wave "complex voltage amplitude" is written V-, then the forward wave power flow is |V+||V+|/(2Zo) where |V+| represents the modulus of the forward wave "complex voltage amplitude". The 2 is necessary to convert from peak to rms value.

• Similarly the backward wave power flow is |V-||V-|/(2Zo).

• The (stored energy per unit length) times the (propagation velocity) equals the (forward wave power flow). Thus the stored energy per unit length = |V+||V+|/(2Zo times wave velocity). Looking at the units, (energy)/(length) times (length/time) gives us (energy/time) = power [because Joules/sec=watts]

A 75 ohm cable, assumed lossless, feeds an antenna having radiation resistance (30+j120) ohms at the signal frequency. If the forward wave power is 10 watts, calculate the return loss and the radiated power. [25%]

• The characteristic impedance Zo = 75 ohms, and the load impedance ZL = 30+j120 ohms. The reflection coefficient is, at the load terminals, (ZL-Zo)/(ZL+Zo) = -45+j120 divided by 105+j120 which gives us 0.3805+j0.7080 whose modulus squared is 0.6460.

• The return loss in dB is therefore -10 log10 (0.6460) which is 1.898 dB, which is the amount by which the return power is smaller than the incident power. The return power is therefore 10 times 0.6460 = 6.46 watts, and the radiated power is what is left, namely 10-6.460 = 3.540 watts.

• Now the forward wave voltage modulus is given by |V+| and we know from earlier parts of the question that |V+||V+|/(2Zo) = 10 watts with Zo = 75 ohms. Thus |V+| = sqrt[10 times 2 times 75] = sqrt[1500] so the forward wave voltage size is sqrt[1500] = 38.73 volts. The backward wave power flow is similarly 6.46 watts so the backward wave voltage size is sqrt[6.46 times 2 times 75] = sqrt[969] = 31.13 volts.

• At a voltage standing wave maximum, the forward and backward wave voltage phasors add in phase, so the peak generator voltage is the sum 38.73+31.13 = 69.86 volts; to find the rms voltage we divide by sqrt(2) to find 49.90 volts.

• The generator may be matched by using a single shorted stub which may be either series or shunt; by a double or triple stub tuner, or with some power loss by an isolator, although in this case the radiated power will be less than that supplied by the generator. In the case of a stub match, the reflections from the stub(s) just cancel the reflection from the load.

DJJ's antenna-related question, from 95-96

Question 5

Define the terms "directivity", "gain", "efficiency", and "effective area" for an antenna array. Explain how the boresight gain of an antenna is related to its effective area and efficiency. [25%]

• The directivity of an antenna is the ratio of the power directed along boresight to the power which would be radiated equally in all directions if the antenna was isotropic. Usually the directivity is assumed to be equal to the boresight gain assuming no losses in the antenna.
• The gain of an antenna is usually specified in the direction of maximum radiation, the boresight, and is equal to the total power radiated in this direction assuming it was radiated isotropically (uniformly in all directions) divided by the input power to the antenna structure. The gain is less than the directivity because of antenna losses, spillage, blockage, and so on.
• The efficiency of an antenna is the gain divided by the directivity. It is a measure of what proportion of the input power is usefully radiated.
• The effective area of an antenna is numerically equal to the gain times 4 pi/(lambda^2), and is the area of a perfect antenna which had that particular value of gain. The effective area is usually (in the case of an aperture antenna) somewhat smaller than the physical area.

A square array of 4 by 4 elements consisting of lambda/2 dipoles each having a maximum gain 3dB is fed by 16 signals of adjustable relative phase and equal amplitudes. Assuming 90% efficiency, calculate the boresight gain when the signals are all in phase. [50%]

• Let us call the amplitude of the individual signals driving each element A. Then the input power to an element is |A|^2 (modulus of A squared, or AA*). The efficiency is 90%, or 0.9, and the square root of this is 0.95 which represents how much of the individual amplitude A is radiated.
• The gain of an individual element, in terms of power, is 2, or in terms of amplitude, is sqrt(2). This is 3dB.
• When all the signals add up in phase, the boresight amplitude is A times sqrt(2) times 0.95 times 16 since there are 16 equally contributing elements.
• However the total input power is 16|A|^2, that is, 16 times the input power to each element.
• The boresight power is (16 times 0.95 times sqrt(2))^2 in units of |A|^2 so the power gain is 16 times 0.95^2 times 2 which is 28.8 or 14.6 dB

Explain with a diagram how the direction of the beam from this array may be altered without moving the positions of the elements. [25%]

• The principle here is to delay the arrival in signals in time at successive elements across the array. The principle is called "phased array" and is achieved by incorporating variable phase shifts in the feeds to the various elements. If along the new boresight direction, the wave from the further element has to travel a distance d further than the wave from its neighbour, the phase shift we need to incorporate is theta = 360 degrees times d/lambda.
• The direction of the beam boresight is at an angle phi from the broadside direction of the array. Here, a simple diagram will show that sin(phi) = d/(element spacing) or phi = arcsin [(lambda theta)/360 degrees times the element spacing] where the phase shift theta is expressed in degrees and the element spacing and lambda have the same units.
• Clearly, for our square array we can steer the beam in both azimuth and elevation by applying this principle to both directions of the array plane.

MSc Map Antennas and Propagation module exam 1999 (DJJ questions)

navigation page

Question 1.

Write definitions and notes on the terms boresight, polarisation, null, isotropic radiator, efficiency, far-field radiation pattern, and directivity in the context of antenna descriptions.
[35%]

Describe, giving quantitative detail, the construction of a typical fifteen element linearly polarised Yagi-Uda 800MHz television receiving antenna. State the number of elements which are directly driven from the feeder, and explain why only a single reflector element is needed. Estimate, giving reasons, an upper limit to the maximum boresight gain (dBi) of this antenna.
[25%]

A receive Yagi-Uda antenna has boresight gain 5.6 dBi. Calculate the effective receive cross-sectional area of this antenna at 1.8 GHz. Estimate the maximum power which can be received by this antenna from a transmitting source directly overhead (along boresight) at a distance of 100 km, assuming the transmitter power is 1 watt and the transmit antenna gain is 2.6 dBi.
[25%]

If the receive channel noise temperature is 450 K, estimate the receiver signal-to-noise ratio (dB0 for this link, for 10 MHz bandwidth.
[15%]

Outline solution 1.

• Definitions
  • Boresight: direction(s) of maximum gain, or directivity, or radiated field strength
  • Polarisation: direction of Electric field vector projected on a plane normal to the propagation direction. It can be linear, elliptical, or circular (RH or LH). It can be time-dependent.
  • Null: A direction at which there is zero radiation. It is a point or line from a continuous set; one cannot have zero radiation over a range of angles between two lobes.
  • Isotropic radiator: A hypotehtical source radiating equally in all directions in three dimensions. Sinve EM waves are transverse, isotropic radiators can only be approximated in practice, but cannot be accurately realised.
  • Efficiency: The amount of power accepted by the antenna from the feed that is actually radiated. Gain/directivity.
  • Far field radiation pattern: A plot of the gain or directivity as a function of azimuth and elevation directions, at a sufficient distance that the pattern shape does not depend on the distance from the source.
  • Directivity: Ratio of field strength in a certain direction to the field strength at that distance, from an isotropic radiator having the same total integrated radiated power as the antenna under investigation.
  
  [35%]

• At 800 MHz, a half-wavelength is 18.75 cms. The antenna has a single element driven from the feed (75 ohm coax usually) which consists of a folded dipole which is a little shorter than a half-wavelength. Behind this driven element is a single reflector, of length about a half wavelength and spacing about lambda/5 from the driven element In front of the driven element are 13 directors, each about 15% shorter than a half-wavelength, and spaced by about 1/3 lambda. The diameter of the rods is typically 5 mm. As the reflector returns most of the power to the forward direction, there are only small fields behind it so extra reflector elements have little effect. An estimate of the gain is (number of elements) times (gain of dipole) which is approximately (in dBi) 10 log[10](15) + 2dBi or 14 dBi. The antenna construction can be varied quite widely without greatly affecting the forward gain, and we have not considered broadbanding techniques.
  [25%]

• A gain of 5.6 dBi is a numerical factor of 10^(5.6/10) = 3.63. This factor is equated to (4 pi Ae)/(lambda^2) and lambda at 1.8 GHz is 30/1.8 cms or 0.1667 metres. So the value of the effective area Ae is 80.23 sq cms or 8.023E-3 sq metres The transmitter power e.i.r.p is 1 watt times 10^(2.6/10) or 1.82 watts. The received signal strength is 1.82/(4 pi R^2) watts/square metre at a distance R metres, so at 100 km = 1E5 metres, the received power is (Ae 1.45E-11) = 1.16E-13 watts.
  [25%]

• The receiver noise power is kTB = (1.38E-23)(450)(1E7) or 6.21 E -14 watts so the signal-to-noise ratio is 11.6/6.21 = 1.87 or 2.72dB.
  [15%]

Question 2.

Explain, illustrating with examples and sketches, the terms array antenna, element, array pattern, element pattern, and pattern multiplication . Distinguish between the element placings in a one-dimensional and a two-dimensional array.
[30%]

An array antenna is formed from two elements consisting of 50 dBi gain aperture antennas, which are separated in space by 100,000 wavelengths. Calculate the boresight gain of the array. Estimate how many interference fringes of the array pattern lie within the -3dB contours of the element pattern.
[40%]

Explain the term very long baseline interferometry (VLBI) and state its use. Estimate the resolution obtained with Earth-based VLBI at 1 GHz using the maximum possible practical separation of the elements.
[30%]

Outline solution 2.

• An array antenna comprises a collection of spaced "similar" elements. "Similar" means that they all have the same radiation patterns, considered individually, and they are all orientated in the same direction in 3-d space, with identical polarisation properties. An "element" may consist of an array of "sub-elements", so the array antenna can be built up recursively. A "linear" array has elements spaced along a single direction or dimension in 3-d space. An "area" array has elements spaced on a single plane in 3-d space, in 2 dimensions. The elements do not have to be fed with the same amplitudes and phases of signal, but they do have to be fed with the same signals in terms of the frequency spectrum or Fourier decomposition.

  The "array pattern" consists of the far field interference pattern set up by a collection of hypothetical isotropic radiators, located at the positions of the actual elements, and fed with the same signals as the actual elements.

  The "element" pattern is the far field radiation pattern of a single element, on a single site, having the same orientation as the actual array of elements.

  "Pattern multiplication":- pointwise multiplication of the "element pattern" by the "array pattern" to obtain the total radiation pattern for the antenna array. Pointwise multiplication means that we choose a direction in 3-d space, determine the array pattern gain and the element pattern gain in this direction, multiply them to find the total gain, then repeat for all possible radiation directions.
  [30%]

• For two elements the gain is twice that of one element, so the array pattern gain is 10 log[10](2) = 3 dBi and the total gain is 50 + 3 = 53 dBi.

  A single element has numerical gain 10^(50/10) = 100,000 numerical gain. If all the radiation is contained within a cone of semiangle alpha radians, then we have that 100,000 = 4 pi steradians divided by the solid angle of the cone, so 100,000 = (4 pi)/(pi (alpha^2)/4) whence 1E5 = 16/(alpha^2) and alpha = 12.6 milliradians.

  For an array spacing of 1E5 wavelengths, the angular spacing of the lobes is very nearly 1E-5 radians. So there are about 1260 array lobes inside the main beam of a single element.
  [40%]

• At 1GHz, the wavelength lambda is 30 cms. The maximum separation of antennas on the Earth's surface, such that they can both see in the same direction in space, is about 11,000 km. This is 3.67E6 wavelengths, so the angular resolution between adjacent nulls is 1/[3.67E6] radians, or 2.7 E -7 radians, or 0.056 arc seconds.
  [30%]

MSc Map Antennas and Propagation module exam 2000 (DJJ questions)

Question 1.

(a)
Define the terms effective aperture, gain, efficiency, E-plane radiation pattern, boresight direction, null, half-power beamwidth and polarisation for a large Cassegrain reflector antenna. Use diagrams to illustrate your answers, where appropriate.

• Effective aperture of a receive antenna is the {power delivered to receiver in Watts}/{average radiant power flux density across the antenna in Watts/square metre.}
• Gain of a receive antenna is the {power delivered (Watts) by the antenna to the receiver}/ {Power which would be delivered to the receiver by a hypothetical perfect isotropic receive antenna} The gain in general depends on the directional orientation of the receive antenna with respect to the source.
• Efficiency is {Power delivered to receiver}/{Total power incident on antenna}.
• The E-plane radiation pattern is the far field contour of equal power density, such that the distance from the origin of co-ordinates to the pattern surface is proportional to the power radiated in that direction, taken in a plane containing the direction of propagation and the electric field direction. It is only really defined for linearly polarised antennas.
• The boresight is the direction, or the directions if there are more than one, of maximum radiated intensity.
• A null in the radiation pattern lies along a direction of minimum radiation intensity, ideally zero.
• The half-power beamwidth is the angle between directions, on a plane radiation polar plot section, where the radiated power has fallen to one half its value on boresight.
• Polarisation is the projection of the E-field vector, in the far field region, onto a plane normal to the propagation direction. Polarisation may be undefined, linear, circular (LH or RH), or elliptical.

  [30%]

(b)
Explain why all practical antennas necessarily have maximum directivity greater than unity.

• The polarisation direction is, in general, transverse to the propagation direction. Consider the complete sphere surrounding the origin: at some direction of propagation, the polarisation vector cannot be uniquely defined, so there can be no radiation in such a direction. Alternatively, there is no radiation along a direction where the average current in the antenna structure is directed. Therefore there must be at least one null. If the radiation falls to zero in a certain direction, it must exceed unity in some other direction, for the average directivity value is unity (all the power is radiated uniformly from an isotrope with unity in every direction).

  [10%]

(c)
Give three methods which might be used to generate circular polarisation for a low-earth-orbit satellite antenna communication system.

• Helical antenna structure
• Crossed dipoles or yagis fed in phase quadrature
• Quarter wave plate for microwave aperture antennas

  [15%]

(d)
A deep space communication system uses a Cassegrain antenna of diameter 70m at a frequency of 8.45 GHz.

(i)
Determine the gain of this dish (in dBi) assuming an aperture efficiency of 80%

• Numerical gain = {efficiency factor}*4 pi A/lambda^2
  with {efficiency factor} = 0.8
  A = {pi 70^2}/4 (area of a circle of diameter 70 metres)
  lambda = 3E8/8.45E9 = 3.55cm = 0.0355m
  Numerical gain = 30.7E6
  dBi gain = 10log[10]{30.7E6} = 74.9 dBi

  [10%]

(ii)
Determine the power received by this dish from a transmission from a satellite having antenna gain 2.2 dBi and transmitter power of 10 W at a distance of 180 million kilometres. Assume a receiver noise temperature of 70 K and a receiver bandwidth of 10 Hz. Estimate the maximum receiver signal-to-noise ratio.

• The effective area of the dish = 0.8*{pi 70^2}/4 = 3079 square metres
  2.2 dBi is equivalent to a numerical factor 10^0.22 = 1.66
  10 watts times 1.66 = 16.6 watts e.i.r.p.
  180E6 km = 1.8E11 metres
  Power density at antenna = 16.6/{4 pi 1.8^2 10^22} watts per square metre = 4.076E-23 w/m^2
  Power received = 3079*4.076E-23 watts = 1.25E-19 watts
  Boltzmann's constant k = 1.38E-23 Joules/K, assume noise temperature 70K, bandwidth 10Hz
  so kTB = 9.66E-21 watts
  so S/N ratio = 1.25E-19/9.66E-21 = 12.9 = 11.1 dB

  [20%]

(iii)
Estimate the half-power beamwidth of this 70 m Cassegrain antenna at 8.45 GHz.

• If the diameter of the beam at distance R metres is dR between half-power points then the beam solid angle is {pi d^2}/4 steradians and the beamwidth is d radians.
• The gain is {4 pi}/{beam solid angle} = 30.7E6 numerical, so d = 4/(sqrt{30.7}) milliradians = 0.72 milliradians = 0.04 degrees about.

  [15%]

Question 2.

(a)
Define the terms element, element factor, array factor, pattern multiplication, and total radiation pattern for an array antenna. State what constraints have to be applied to the individual elements for pattern multiplication to be possible.

• Element: one of a number n of identical radiating structures orientated in the same direction in space contributing to the total radiation from the antenna. They do not have to be fed with identical amplitudes and phases, but the signal to each element has to be the same.
• Element factor: The radiation pattern (gain or directivity as a function of direction) of a single isolated element of the array.
• Array factor: The radiation pattern of a collection of isotropes placed on the element centres and fed with the same amplitudes and phases as are applied to the actual elements of the array.
• Pattern multiplication: Pointwise multiplication of the element pattern by the array factor to obtain the total radiation pattern for the array.

  [25%]

(b)
Distinguish between active arrays and passive arrays and discuss to what extent the method of moments calculation process for antenna structures may be applied to array antennas.

• An active array consists of elements each of which is driven by a physical feed. Passive arrays have one element actively driven, and the others couple to it electromagnetically through the near field.
• The method of moments derives the radiated field pattern and also the antenna currents from a self-consistent matrix calculation using the Green's function of a little element of antenna current, and pointwise matching the solutions to the given antenna feed currents or voltages. In the case of multiply fed antennas, power may be transferred between the feeds. In the case of passive array antennas, any power delivered to the driven element must eventually be radiated or absorbed in resistive loss.

  [35%]

(c)
An active array antenna is to be constructed from four half-wave dipoles.

(i)
Sketch the azimuth and elevation pattern for a half-wave dipole. Explain which pattern is an E-plane section and which is an H-plane section.

• Looking along the rod direction of the dipole, there is no structure to determine a preferred direction in azimuth, so the radiation pattern is a circle centred on the rod. Since the H field is at right angles to the E field, and the E field lies along the rod, the azimuth pattern is an H-plane section.
• The elevation pattern has nulls along the rods. It is an E-plane section and has very approximately a cos^2{theta} distribution where theta is the elevation angle, being zero at right angles to the rod.

  [10%]

• See antarray.html

  [10%]

• Space the four elements by lambda/2 and feed them with the excitation pattern 1:3:3:1 which is a combination of 1:2:1 spaced lambda/2 and the 1:2:1 is a combination of 1:1 spaced lambda/2.

  [10%]

(iv)
Choose an orientation and spacing for the dipole elements so that the entire array antenna has maximum directivity of about 8 dBi.

• For a single dipole the element gain is about 2dBi so we need another 6dBi from the array factor. This is a numerical factor of 4, and because there are 4 elements in the array, the boresight gain will be close to 4 for the array factor if we feed them in phase and with equal amplitudes. The spacing does not affect the boresight gain.

  [10%]

MSc Map Antennas and Propagation module exam 2001 (DJJ questions)

Question 1.

(a)
Define the terms radiation impedance, feeder, antenna efficiency, null, boresight, and VSWR2 bandwidth for a terrestrial fixed-link antenna installation.

• Radiation impedance: the ratio of voltage to current at the antenna terminals, at a specific frequency, expressed as a complex number R+jX
• Antenna efficiency: Total radiated power in the far field divided by the power accepted by the antenna from the feed.
• Null: A direction in three-dimensional space where there is no far-field radiation.
• Boresight: A direction along which the radiation from an antenna is maximum in the far field. There may be more than one unique boresight direction.
• VSWR=2 bandwidth: The reflection coefficient gamma from an antenna having radiation impedance z (normalised to the feeder impedance) is given by gamma = (z-1)/(z+1) which is a complex number depending on frequency. The size of gamma is the modulus of this number gamma. Call the modulus of gamma modgamma. Then the VSWR is defined by VSWR = (1+modgamma)/(1-modgamma) and it is a sensitive detector of how much power is reflected by the antenna impedance mismatch. Clearly, if z=1 then VSWR = 1. The VSWR2 bandwidth is the difference in frequencies at which VSWR is less than or equal to 2.

(b)
Calculate the percentage of power reflected from an antenna if the VSWR on the feed is 2:1.

• If the VSWR = 2 then (1+modgamma)/(1-modgamma) = 2 and modgamma = (1/3) therefore. The percentage of power reflected may be found from the square modulus of gamma, so in this case (modgamma)^2 = 1/9 and 11% of incident power is reflected.

(c)
Explain why the radiation resistance of a short rod antenna is approximately proportional to the square of its length.

• Radiation is from accelerated charge, which is equivalent to the rate of change of current I in a little length dL of the antenna rod. The contribution to the electric field in the far-field region is proportional to (d/dt)[Integral of I(x) dx] and for a short antenna the current profile tapers uniformly from feed to the end of the rod. Thus the radiated power, which is proportional to E^2, is proportional to (IL)^2 which equals RradI^2 so the radiation resistance Rrad is proportional to L^2.

(d)
Sketch the position of the nulls, in 3-d space, for a short dipole antenna.

• The nulls lie along the rod directions. The azimuth plane radiation pattern is a circle, and the elevation pattern is a figure-of-eight shape cos^2{theta}.

(e)
Explain how the radiation impedance and bandwidth of a short dipole antenna depend on the diameter/length ratio of its rods.

• For a short dipole, the radiation reactance Xrad is appropriate to a small capacitance. Zrad = Rrad + jXrad.

  This capacitance gets larger as the rods get fatter. Thus Xrad decreases as D/lambda increases. For a given length L, Rrad is roughly independent of L, so as Xrad gets smaller, the bandwidth of the antenna increases.

(f)
A free-space link is set up between two co-polarised half-wave dipole antennas at a frequency of 800MHz and a bandwidth of 10MHz.

(i)
Derive a formula for the free space maximum range (in kilometres) as a function of the transmitter power in watts and the receiver noise temperature in Kelvin.
• The frequency is 800MHz = 0.8GHz so the wavelength = 30/0.8 cm = 37.5 cm The gain (numerical) of a lossless dipole = 1.65 (== 2.22 dBi) so the effective area of the receive dipole = (lambda)^2*[ Gain/{4 pi}] = 0.0185 square metres.

  For a transmitter power of P watts, the effective isotropic radiated power on boresight is 1.65P from the transmitting dipole.

  The power density at distance R kilometres is [1.65P]/[4 pi 10^6 R^2] watts per square metre and this has to be equal to the noise power [1.38E-16]T watts in bandwidth 10MHz = 10E7 Hz.

  Equating these terms we find R = 4191 sqrt(P/T) km.


[15%]

(ii)
Evaluate the range for a 10 watt transmitter with a 20dB link S/N ratio if the receiver noise temperature is 290K.

• For 20dB S/N ratio we need 100 times larger power at the receiver, or 10 times larger sqrt(power) so the range = 4191*{1/10}*sqrt(10/290) = 77.8 km. The factor 10 (in the sqrt) is the transmitter power in watts, the factor 1/10 in the range is the adjustment for 20dB S/N ratio.

[10%]

Question 2.

(a)
Describe the principal components, construction, and properties of an offset-fed reflector antenna.

• An offset fed reflector antenna has a parabolic reflector of appropriate cross-sectional shape fed from the front by a horn feed which is offset from the axis of the paraboloid so that the reflected beam does not rehit the feed structure. The semi-angle of the horn feed is chosen by adjusting the gain of the feed. The feed beam completely illuminates the reflector antenna without too much spillover.

  There is no blockage in this kind of antenna, so it has minimal sidelobe production from diffraction around the feed. The feed illumination is tapered so that the intrinsic sidelobes of the reflector are small.

  There is co-polar to cross-polar conversion at the reflection, so this kind of antenna is not so easy to design for polarisation re-use applications.

  It is desirable to have the profile of the reflector dish accurate to +/- (1/20) wavelength.

(b)
A pyramidal square horn antenna is excited by the TE10 mode in rectangular waveguide at 12GHz. It has boresight gain 22.0 dBi. Calculate the dimensions of the horn mouth, assuming constant phase across the horn aperture.

• Assume the square has side a metres. At 12 GHz the wavelength is 30/12 = 2.50 cms = 0.025 metres.

  The E plane profile of field across the horn mouth is a rectangular pulse, the H plane profile is a cosine distribution.

  The aperture efficiency is (1/pi)Integral from -pi/2 to +pi/2 of {cos(theta) d(theta)} = 2/(pi). The effective area = 2(a^2)/(pi).

  The gain is 22 dBi = 158.5 numerical, = (4 pi) (effective area)/(lambda^2) so a = 0.1113 metres = 11.13 cm.

(c)
The horn of part (b) illuminates a circular parabolic reflector of area 80 square metres in a front-fed arrangement.

(i)
Determine the desirable feed-reflector separation (in metres).

• The beam semi-angle of the horn feed is 2/sqrt(G) radians where G is the numerical gain, so the beam semi angle = 0.16 radians. For a circular reflector dish of 80 square metres area, the diameter is 10 metres. For a beam semi angle of 0.16 radians the feed distance needs to be 31.5 metres.


[10%]

(ii)
Assuming 2% blockage and 15% spillover, estimate the boresight gain.

• The gain is 0.98*0.85*(4 pi)*80/(0.025^2) = 1.33E6 = 61 dBi.

[10%]

(iii)
Estimate the beam semi-angle in milliradians.

• The beam semi angle is 2/sqrt(1.33) milliradians = 1.7 milliradians.

[10%]

(iv)
Estimate how far the feed should be moved across boresight to steer the main beam through the beam semi-angle.

• The feed must be moved 31.5*1.7E-3 metres = 5.5cm approx.

[10%]

(d)
Write notes on the causes of sidelobe production in reflector antennas. Explain what steps may be taken to minimise the sidelobes.

• Sidelobes are produced by two principal causes: Diffraction around obstacles in the beam, particularly in the near field, and also by abrupt changes or steps in the reflector illumination profile, perhaps caused by spillover of the beam from the feed. Ideally we would like a Gaussian illumination profile for the reflector; alternatively the illumination may be tapered to zero at the reflector edges. Obstacles should be avoided by using an offset fed arrangement, see above, or else kept to a minimum cross section so that less power is scattered into the sidelobes.
  

  [30%]

MSc Map Antennas and Propagation module exam 2002 (DJJ question)

Question 1.

(a)
Define the terms numerical gain, boresight, polarisation, null, and isotropic radiator. Illustrate your answers with diagrams and write notes where needed.

• The numerical gain of an antenna is the radiation intensity (power density in watts per steradian) produced by an antenna in a direction (theta, phi) divided by the radiation intensity of an isotropic 100% efficient antenna which has the same accepted input power.
• Boresight: There may be more than one unique boresight direction. It is the direction (theta, phi) for which the umerical gain is a maximum.
• Polarisation: The trajectory of the projection of the E field vector on a plane normal to the propagation direction
• Null: A direction in three-dimensional space where there is no far-field radiation.
• Isotropic radiator: A radiator that emits equally in all directions around a sphere centred on the source. Impossible to realise in practice. The numerical gain is independent of (theta, phi).
  

(b)
A certain antenna, ANT, consists of a number of parallel straight rods. It has numerical gain 8.4 with respect to a practical half-wave dipole. The practical half-wave dipole is specified to have gain in decibels, after accounting for losses, of 2.06 dBi. What is the numerical gain of ANT with respect to an isotrope? Express this gain also in dBi and indicate two directions in space where nulls must lie. You may ignore any residual radiation from the feed.

• A half-wave dipole of gain 2.06 dBi has numerical gain 10^(0.206) = 1.607 with respect to an isotrope. So the gain of ANT = 8.4*1.607 = 13.498 numerical. This is 11.303 dBi.
• The nulls lie along the rod directions.

(c)
Distinguish between the terms omnidirectional and isotropic. Explain how a loop antenna in combination with a dipole, may be used to generate omnidirectional circular polarisation in a horizontal plane. Sketch the radiation pattern in an elevation plane.

• Omnidirectional: for some co-ordinate system (theta, phi) the directivity is independent of theta but not independent of phi.

• Isotropic: for all co-ordinate systems (theta, phi), the directivity is independent of both theta and phi.

• A combination of a loop antenna and a dipole placed along its axis of rotation, with the elements fed in phase quadrature to give equal radiation intensities in the plane of the loop in the far field, will generate omnidirectional circular polarisation in the azimuth plane.

(d)

(i)
Justify the statement that "the radiation resistance of a short dipole is proportional to the square of its length".

• Radiation is from accelerated charge, which is equivalent to the rate of change of current I in a little length dL of the antenna rod. The contribution to the electric field in the far-field region is proportional to (d/dt)[Integral of I(x) dx] and for a short antenna the current profile tapers uniformly from feed to the end of the rod. Thus the radiated power, which is proportional to E^2, is proportional to (IL)^2 which equals RradI^2 so the radiation resistance Rrad is proportional to L^2.

[10%]

(ii)
Indicate how the radiation resistance of a small loop antenna scales with diameter.


• The radiation resistance of a small loop decreases as (diameter/wavelength)^4 as the loop gets smaller. The approximation breaks down for loop perimeter approaching a wavelength.

[10%]

• The numerical gain of a half wave dipole is 1.66 so the effective aperture, which is gain(lambda^2)/(4 pi) is 1.66 (lambda^2)/(4 pi)

  Also, for transmitter power P watts and transmit antenna gain 1.66, the generated power density at distance R metres is 1.66P/(4 pi R^2) watts per square metre.

  The received power at distance R metres is (1.66)^2 P (lambda^2)/(4 pi R)^2 watts which has to be equal or larger than the noise power at temperature T Kelvin and bandwidth B Hz, which is kTB watts.

  Equating these terms and rearranging, one obtains R = sqrt[P/kTB]*1.66 lambda/(4 pi) which on putting in the numbers is sqrt[P/B]*3.566E9 metres.