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

Antennas notes.

### 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 numerical 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).

[20%]

(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.

[15%]

(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(elevation), phi(azimuth)) the directivity is independent of phi but not independent of theta.

• 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.

[15%]

(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%]

(e)
Calculate the maximum range at which two co-polarised lambda/2 dipoles can communicate in free space. Assume a system noise temperature of 100K. Express your answer in terms of the transmitter power P watts, the system bandwidth B Hz, and the wavelength lambda metres.

• 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.

[30%]