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

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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.


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(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.

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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.


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(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.

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

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

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(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.

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(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.
  

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