
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 onthe 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 inthe 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,
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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 pi25,000^2] 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 = 2080watts. 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 is200MHz. 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 ofradius 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, tofind 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 forthe 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
noisepower.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 thatthe distance R is calculated from this condition, which
gives rise to the formula above equating the power received at
distance R with the powerrequired 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. Thetransmit antenna has low gain so can be made
omnidirectional. The power needed by the transmitter is of the
order of 1 kWatt when allowingfor 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 nullsmay 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 desireddirection. 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 theboresight 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 yaxis 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 theamplitudes 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 thesizes 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 drivingpoint impedances of the
elements will be different (they have differing local environments;
there are different environments for the end elementsthan for the
centre elements for example; so these types of elements will be
having different interelement couplings and mutual impedances),
sothe 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 they 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 travelto 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 aswe 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 angularposition 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 phasedifference 180 degrees (inversion) from each other) then
at 1/2 wavelength interelement spacing the contributions from
adjacent elements addup 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 takesthe signal to get from
an element to its immediate adjacent neighbour, and so together
with the 180 degrees of phase shift in the excitationcurrent 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, asabove....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 ofthe *array* pattern. Assuming that all the
excitation current amplitudes are equal, and in phase, then the
nulls will occur at angles theta wherethe 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 lieon 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 givean 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 toget 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 degreesphase 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 radiationpattern", 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 boresightdirection 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 elevationangle, 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, anddescribe 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 directivityis 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 onaxis 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 surface20,000 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) isabout 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 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 areceiver 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 ofwavelengths, 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 iskTB = 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 putsN 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 lineof 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
(DJJquestions)
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 asatellite 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 theEffective 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; thepointing 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 alarge 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 hasrotational 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 aneffective 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 thefootprint 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 whatconstraints 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 theboresight 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.
An "array antenna" has elements (say, N in number) which are
"similar": that is, they have the same radiation patterns and
polarisationproperties and are orientated in the same directions in
space. They don't have to have the same excitation currents
however. The excitationamplitude of the ith 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 1dimensional, or 2dimensional. The grid
does not have to be a straight line or a flat plane. The "array
pattern" or "arrayfactor" is the polar radiation plot of a
hypothetical collection of isotropic sources placed on the array
antenna grid or pattern, and each fed withthe 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 radiationpattern, by multiplying the gain of the
element by the gain of the array of isotropes for every particular
direction of propagation. Therequirements are that the element
patterns are all identical, and that the actual currents on the
elements (allowing for interelement couplings) areused to
determine the array factor.

A dipole element is omnidirectional in the Hplane, so the
radiation pattern is a circle centred on the element. The dipole
element has the classic"figure of eight" radiation pattern in the
Eplane. The nulls lie along the dipole rods. The array pattern for
four isotropes spaced by a halfwavelength 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 is6.68 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 theangle 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 nullaway 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 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 abovean 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 hornaperture 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 radiatedby 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 radiatedpower is maximum, this is
termed the boresight direction. The directivity of an antenna is a
dimensionless number representing the ratio of theradiated 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 areinput 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 thenear 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 transversepolarisation direction for radiated transverse
electromagnetic waves, one finds that this cannot be arranged for
radiation in every direction froma point. Therefore isotropic
radiators are impossible. Consider a radiator with E field
everywhere directed North on a reference sphere. Thedirection 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
acrossthis 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 fieldstrength 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
radiatedpower 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 hypotheticalantenna 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
theradiation. 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 areaof 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 is25.12*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 complexforward 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. Byredefining 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 theforward 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 unitlength = 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 is10 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
= 3.540watts.
A generator feeds the 75 ohm cable and antenna of the last part.
The generator has negligible internal impedance and is connected at
a voltagestandingwave 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 withZo = 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 sum38.73+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 somepower 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 itseffective 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 thisdirection 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 thatparticular 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 phaseand 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
contributingelements.
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 isachieved by incorporating variable phase
shifts in the feeds to the various elements. If along the new
boresight direction, the wave from thefurther 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 degreesand 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 inthe 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 themaximum 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 strengthPolarisation: direction of Electric field
vector projected on a plane normal to the propagation direction. It
can be linear, elliptical, orcircular (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, isotropicradiators 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
thepattern 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 afolded dipole which is a little shorter than a
halfwavelength. Behind this driven element is a single reflector,
of length about a half wavelengthand 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 ofelements) times (gain of dipole) which is
approximately (in dBi) 10 log[10](15) + 2dBi or 14 dBi. The antenna
construction can be varied quitewidely 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 30/1.8cms 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 elementpattern. [40%]
Explain the term very long baseline interferometry (VLBI) and
state its use. Estimate the resolution obtained with Earthbased
VLBI at 1 GHz usingthe 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 anarray of "subelements", so the array antenna can
be built up recursively. A "linear" array has elements spaced along
a single direction ordimension 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 befed 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 positionsof 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 ofelements.
"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 patterngain 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 alpharadians, 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 lobesinside 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
(DJJquestions)
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 inWatts/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 respectto 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 patternsurface is proportional to the power
radiated in that direction, taken in a plane containing the
direction of propagation and the electric fielddirection. 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 onehalf 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 somedirection 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 theradiation 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 isradiated 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 quadratureQuarter 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 adistance of 180 million kilometres. Assume a receiver
noise temperature of 70 K and a receiver bandwidth of 10 Hz.
Estimate the maximum receiversignaltonoise 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 thebeamwidth 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 whatconstraints 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 theantenna. 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 asare 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 structuresmay 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 otherscouple 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 theGreen's function of a little element of antenna current,
and pointwise matching the solutions to the given antenna feed
currents or voltages. In thecase 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 Hplanesection.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.
See antarray.html
[10%]
(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 sidelobes.
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 the1:2:1 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 becausethere 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
(DJJquestions)

Question 1.
(a)
Define the terms radiation impedance, feeder, antenna
efficiency, null, boresight, and VSWR2 bandwidth for a terrestrial
fixedlink antennainstallation.
Radiation impedance: the ratio of voltage to current at the
antenna terminals, at a specific frequency, expressed as a complex
number R+jXAntenna 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 boresightdirection.
VSWR=2 bandwidth: The reflection coefficient gamma from an
antenna having radiation impedance z (normalised to the feeder
impedance) isgiven 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
detectorof 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 befound
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. Thecontribution 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 profiletapers 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
roughlyindependent 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 noisetemperature 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
noisepower [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
theadjustment 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 offsetfrom the axis of the paraboloid so that the
reflected beam does not rehit the feed structure. The semiangle of
the horn feed is chosen byadjusting 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 istapered 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
acircular 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 abruptchanges or steps in the reflector illumination
profile, perhaps caused by spillover of the beam from the feed.
Ideally we would like a Gaussianillumination profile for the
reflector; alternatively the illumination may be tapered to zero at
the reflector edges. Obstacles should be avoided byusing 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
(DJJquestion)
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
directionNull: 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. Thenumerical 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. Thepractical halfwave dipole is
specified to have gain in decibels, after accounting for losses, of
2.06 dBi. What is the numerical gain of ANT withrespect 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 generateomnidirectional 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 radiationintensities 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. Thecontribution to the electric field in the farfield region
is proportional to (d/dt)[Integral of I(x) dx] and for a short
antenna the currentprofile tapers uniformly from feed to the end of
the rod. Thus the radiated power, which is proportional to E^2, is
proportional to (IL)^2which 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 thje maximum range at which two copolarised
lambda/2 dipoles can communicate in free space. Assume a system
noise temperature of100K. 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) wattsper 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 email7th April 2003.