Collected Antennas Problems
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Transmission line notes.
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*1023 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.
A linear antenna consists of 8 elements spaced a distance d metres
apart along the x axis. Describe the excitation amplitudes and phases
if the boresight direction of this antenna is to lie along the y axis.
How would the phasing change if it was desired to steer the beam
30 degrees from the y axis in the direction of the positive x axis?

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 offboresight.

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 interelement 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.
If this array has spacing d equal to half a wavelength, and the
adjacent elements are fed in antiphase, determine the boresight
direction and the angular position of the first null.

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 interelement
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 oppositelydirected 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.
If the array spacing is now reduced to 1/4 wavelength between adjacent
elements, describe how the currents on the
elements may be phased to give an array
gain of 8 in the positive x direction, and a null in the negative x
direction. Here we neglect interelement coupling effects.

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", "Eplane 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 farfield sphere.
 The Eplane 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 nonzero 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.
Explain why a vertical whip antenna may be expected to have
a roughly omnidirectional radiation pattern in the horizontal
(azimuth) plane, and describe its polarisation properties. How would
you generate the orthogonal polarisation for a similar omnidirectional
radiation pattern? Suggest a method of making an omnidirectional
antenna having right or left hand circular polarisation along every
direction in the azimuth plane.

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 Efield 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 Efield 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 angleindependent
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 semiangle
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 semiangle 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 semiangle of 0.0317 radians,
so the semiangle 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 = 1.36E23 * 300 * 200,000 = 8.3E16 watts. So at maximum range
the received power should be 2.6E14 watts which is a factor 2.6E8
times smaller than the transmitter power of 1 microwatt.
Thus the factor L can be as small as 2.6E8/(15.8E6) = 1.7E15
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
onemicrowatt transmitter.
Estimate, giving your reasons, the maximum lineofsight 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 YagiUda 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 semiangle 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 microwavetransparent 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 offsetfeed used
for beampointing.
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 halfwave 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 Hplane. 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.
Microwave Option paper 9798 (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 transmitterantenna 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 Eplane 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 freespace 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 9697 (DJJ's antennarelated 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 VV/(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, (ZLZo)/(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 106.460 = 3.540 watts.
A generator feeds the 75 ohm cable and antenna of the last part.
The generator has negligible internal impedance and is connected at a
voltage standingwave maximum. For a forward wave power of 10 watts,
calculate the rms voltage at the generator terminals. Give a qualitative
description of a method by which the antenna may be matched to the cable. [25%]

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 antennarelated question, from 9596
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 16A^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
Antennas notes.
Question 1.
Write definitions and notes on the terms boresight, polarisation,
null, isotropic radiator, efficiency, farfield radiation pattern,
and directivity in the context of antenna descriptions.
[35%]
Describe, giving quantitative detail, the construction of a typical
fifteen element linearly polarised YagiUda 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 YagiUda antenna has boresight gain 5.6 dBi. Calculate
the effective receive crosssectional 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 signaltonoise 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 timedependent.

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 halfwavelength 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 halfwavelength.
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 halfwavelength, 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 8.023E3 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 1.45E11) = 1.16E13 watts.
[25%]

The receiver noise power is kTB = (1.38E23)(450)(1E7)
or 6.21 E 14 watts so the signaltonoise 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 onedimensional and a twodimensional 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
Earthbased 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 3d space, with identical polarisation properties.
An "element" may consist of an array of "subelements", so the
array antenna can be built up recursively. A "linear" array has
elements spaced along a single direction or dimension in
3d space. An "area" array has elements spaced on a single
plane in 3d 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 3d 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 1E5 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,
Eplane radiation pattern, boresight direction, null, halfpower
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 Eplane radiation pattern is the far field contour of equal power density, such
that the distance from the origin of coordinates 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 halfpower 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 Efield 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 lowearthorbit 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 signaltonoise 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 = 4.076E23 w/m^2
Power received = 3079*4.076E23 watts = 1.25E19 watts
Boltzmann's constant k = 1.38E23 Joules/K, assume noise temperature 70K, bandwidth 10Hz
so kTB = 9.66E21 watts
so S/N ratio = 1.25E19/9.66E21 = 12.9 = 11.1 dB
[20%]
(iii)
Estimate the halfpower beamwidth of this 70 m Cassegrain
antenna at 8.45 GHz.

If the diameter of the beam at distance R metres is dR between halfpower
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 selfconsistent 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
halfwave dipoles.
(i)
Sketch the azimuth and elevation pattern for a halfwave
dipole. Explain which pattern is an Eplane section and which
is an Hplane 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 Hplane section.

The elevation pattern has nulls along the rods. It is an
Eplane 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%]
(ii)
Sketch the array factor for two isotropes spaced
(a) lambda/4, (b) lambda/2, and (c) lambda apart.
(iii)
Choose an element spacing and suitable drive amplitudes
for the elements so that the fourisotrope "array factor"
has only two main lobes, but no side lobes.

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 fixedlink
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 threedimensional space where there is
no farfield 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 = (z1)/(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)/(1modgamma)
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.
[25%]
(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)/(1modgamma) = 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.
[10%]
(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
farfield 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.
[15%]
(d)
Sketch the position of the nulls, in 3d 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 figureofeight
shape cos^2{theta}.
[10%]
(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.
[15%]
(f)
A freespace link is set up between two copolarised halfwave 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 [1.38E16]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
offsetfed reflector antenna.

An offset fed reflector antenna has a parabolic reflector of
appropriate crosssectional 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 semiangle 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 copolar to crosspolar conversion at
the reflection, so this kind of antenna is not so
easy to design for polarisation reuse applications.
It is desirable to have the profile of the reflector
dish accurate to +/ (1/20) wavelength.
[15%]
(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.
[15%]
(c)
The horn of part (b) illuminates a circular parabolic reflector
of area 80 square metres in a frontfed arrangement.
(i)
Determine the desirable feedreflector separation (in metres).

The beam semiangle 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 semiangle 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 semiangle.

The feed must be moved 31.5*1.7E3 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 threedimensional space where there is
no farfield 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 halfwave dipole. The practical halfwave
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 halfwave 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 coordinate system (theta, phi)
the directivity is independent of theta but not
independent of phi.

Isotropic: for all coordinate 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)
(e)
Calculate thje maximum range at which two copolarised 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%]
Copyright(c) D.Jefferies 1996, 1997, 1998, 1999, 2000, 2001, 2002, 2003.
D.Jefferies email
7th April 2003.