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