Question 1
A certain coaxial cable has characteristic impedance 50 ohms and velocity factor 0.63. Calculate the wave
velocity on this cable in cm/nanosecond; also the inductance and capacitance per unit length of the cable.
 The wave velocity in vacuum times the velocity factor is equal to the wave velocity on the cable.
The wave velocity in vacuum is 30 cm per nanosecond, or 3 times 10^8 metres per second. Thus
the wave velocity is 30*0.63 = 18.90 cm per nanosecond or in SI units, 1.89 times 10^8 metres
per second.
 The inductance per unit length L and capacitance per unit length C, in H/m or F/m respectively,
are related to the velocity u in SI units and the characteristic impedance Zo in ohms by the
relationships Zo = sqrt(L/C) and u = 1/sqrt(LC). Thus we see that L = Zo/u and C = 1/(Zo*u).
Using these we find L = 50/(1.89*10^8) = 264.55 nH/m and C = 1/(50*1.89*10^8) = 105.8 pF/m.
Let us suppose that this cable feeds a load impedance 75 + j150 ohms. Calculate the complex reflection
coefficient gamma, and express it both as real and imaginary parts, and also as modulus and phase angle.
 gamma = (ZL-Zo)/(ZL+Zo) = (75+j150-50)/(75+j150+50) = (1+j6)/(5+j6) = (honestly, by
calculator) 0.6721+j0.3934 = 0.7788*exp(j30.34deg)
Determine the normalised load impedance in this case and plot it on the Smith chart. Measure gamma
from the Smith chart plot and show that it agrees with the value you calculated.
 Normalised load impedance = ZL/Zo = 1.5+j3
Calculate the return loss in dB.
 The return loss = -20 log10(|gamma|) = -20 log10(.7788) = 2.17dB
At a frequency of 1.2 GHz calculate the wavelength. How long a cable is needed to make a 1/4
wavelength section?
 at 1.2 GHz the wavelength in free space or vacuum is 30/1.2 cm or 25 cm. On the cable the
wavelength is smaller than this by the velocity factor 0.63. On the cable, the wavelength is
25*0.63 = 15.75 cm.
 For a 1/4 wavelength section we need 15.75/4 = 3.9375cm = 39.375mm of cable.
An 80 cm length of this cable is connected to the load. Calculate the input impedance (to the combination
of load and cable) from the formula for the transformation of impedance along the line. Repeat
graphically using the Smith chart. Compare results.
 How many wavelengths of 15.75 cm are there in the 80 cm of cable? well, there are 80/15.75 =
5.0794 wavelengths. The phase shift is 360 degrees per wavelength, or 1828.5714 degrees. But
because exp(j theta) is cyclic in theta every 360 degrees, this is entirely equivalent to a phase shift
of .0794*360 = 28.58 degrees.
 The reflection coefficient at the input differs in phase from the reflection coefficient at the load by
-2*28.58 degrees, or -57.17 degrees. The modulus of the reflection coefficient is unchanged along
the line in a lossless line. At the input therefore, gamma(in) = gamma*exp(-j57.17) = 0.7788
exp(j[30.34-57.17]) = 0.6950-j0.3515
 Now Zin/Zo = (1+gamma(in))/(1-gamma(in)) = (1.695-j0.3515)/(0.3050+j0.3515) = 1.8165j3.246
 So Zin = 90.83-j162.3 ohms.
For this example, determine the two values of entirely real impedance which can be found along the line.
What is the VSWR on this line for this load impedance?
 The maximum value of impedance occurs at a VSWR maximum where the voltages add and the
currents subtract, but all quantities are in phase. At this point the voltage is (1+|gamma|)*(V+)
and the current is (1-|gamma|)*(I+). The impedance is real and is equal to Zo*(1+|gamma|)/(1|gamma|) = 50*(1.7788/0.2212) = 402.1 ohms = 8.04*Zo = VSWR*Zo
 By a similar argument, the minimum value of impedance = Zo/VSWR = 50/8.04 = 6.218 ohms.
 Clearly, the max voltage is (1+|gamma|) and the min voltage (1-|gamma|) times V+ in each case,
so the VSWR = (1+|gamma|)/(1-|gamma|) = 8.04 as above.
Question 2.
A certain 75 ohm coax cable feeds an antenna of (assumed) radiation impedance 80 + j15 ohms. Calculate
the return loss. Calculate the proportion of forward wave power which is radiated.
 As in Qn 1, gamma = (80+j15-75)/(80+j15+75) = (5+j15)/(155+j15) = 0.0412+j0.0928 =
0.1015*exp(j66.04deg) so the return loss is -20*log10(0.1015) = 19.87dB.
 The reflected power proportion is |gamma|*|gamma| = 0.1015^2 = 0.0103 or 1.03 percent. Thus
98.97 percent is radiated.
 Comment. This isn't bad. However, suppose your antenna is presented with transmitter power of
100 kWatt. The 1.03 percent of reflected power represents 1030 watts which is quite a lot of heat
to absorb somewhere in the transmitter.
The antenna is to be matched using a single series shorted stub. Calculate the position of attachment and
length of this stub in terms of numbers of wavelengths. Calculate the physical stub dimensions for signals
at 400MHz on cable of velocity factor 0.70.
 This is where the SMITH chart really comes in useful. The normalised antenna load impedance is
80/75+j15/75 = 1.067+j0.2000
 We plot this on the SMITH chart, place a ruler from the chart centre via the plotted point to
intersect the "wavelengths towards generator" scale and read off an angle 0.156 wavelengths.
 We now transform the impedance towards the generator at constant radius (|gamma|) until we hit
the R=Zo circle at X = -j0.20 and plot this second point. The angle for this point is 0.366
wavelengths towards the generator.
 We have moved through an angle 0.366-0.156 wavelengths or 0.21 wavelengths towards the
generator, to transform the impedance from 1.067+j0.200 to 1.000-j0.20 (within the accuracy of
our plot).
 We cut the transmission line here and insert a pure reactive shorted stub having normalised
reactance 0.20j. On the generator side of the series stub we see 1.000-j0.20+j0.20 = 1.000+j0
normalised impedance, and we have a match.
 The stub length is, from the SMITH chart plots of short point and +j0.20 point, 0.0315
wavelengths.
 In free space the wavelength at 400 MHz is 30/0.4 = 75 cm. The velocity factor is 0.7 so on the
transmission line the wavelength at 400 MHz is 0.7*75= 52.5 cm. The short series stub is placed
52.5*0.21 cm from the load, or 11.03 cm from the load. The stub length is 52.5*0.0315 = 1.65
cm.
Calculate the return loss for this arrangement at 390MHz, assuming the antenna impedance has not
changed.
 At 390 MHz the wavelength is 52.5*(400/390) = 53.85 cm and the stub attachment point is
11.03/53.85 wavelengths from the load. This is 0.2048 wavelengths from the load. The stub
length of 1.65 cm is 0.0306 wavelengths so its (normalised) reactance (from the SMITH chart) is
j0.195
 Transforming the antenna load impedance 0.2048 wavelengths towards the generator we find the
point 1.02-j0.21 normalised impedance, so adding the j0.195 of the stub we find the point 1.02j0.015 for the impedance.
 As before, the return loss is found from |gamma| for this impedance which is modulus[(1.02j0.015-1)/(1.02-j0.015+1)] which my calculator makes 0.0124, so the return loss is 20*log10(0.0124) = about 38dB.
 Comment. This matching network isn't really necessary in this particular scenario; we would be
quite happy with a 20 dB return loss from the bare antenna. But if we put a simple matching
network in we can improve the matching which gives 20 dB reduction in reflected power over a
range 390-410 MHz. This is a bandwidth of 20 MHz, good enough for a TV channel and more,
and is a "fractional bandwidth" of 20/400 = 5%. Over this range of frequencies the matched
antenna radiates at least 99.99% of the incident power.
Question 3.
A coaxial cable feeds a load (2+j1)Zo. The line is lossless and has characteristic impedance Zo. Define,
and find values for, the following:The Voltage Standing Wave Ratio (VSWR) on the line.
 The VSWR is a property of the standing wave pattern on the line. It is a dimensionless number,
equal to the ratio of voltage at a standing wave pattern maximum to the voltage at a standing
wave pattern minimum, 1/4 of a transmission line wavelength away along the line.
 Numerically, the VSWR = (1+|s|)/(1-|s|) where |s| is the modulus of the complex reflection
coefficient or scattering parameter somewhere along the line. Strictly speaking, the VSWR is
only defined precisely in terms of the wave pattern on lossless lines. However, the definition in
terms of |s| extends the concept of the VSWR to lines having attenuation or loss, and in this case
the value of the VSWR depends on where it is measured along the line.
 In this case, s = (ZL-Zo)/(ZL+Zo) = (1+j)/(3+j) and |s| = sqrt(2/10) = 0.4472 so the VSWR =
2.618
The return loss, in deciBels.
 The return loss is defined as the number of dB by which the reflected power from a load is less
than the incident power. Again, it is a function of position along the line only on lossy lines.
 In this case, the return loss = -20 log10(|s|) = 6.99dB
The position (in numbers of guide wavelengths from the load) of the first standing wave voltage
minimum.
 The position of the first standing wave minimum is measured from the position of the load, and it
is where a probe measuring the total voltage on the line would pick up the smallest signal, as the
probe was run back along the line from the load. It is also the place where the scattering
parameter s has phase angle 180 degrees modulo 360 degrees.
 In this case, the phase of the scattering parameter s at the load is arctan(1/1)-arctan(1/3) = 45.0018.43 degrees = 26.57 degrees. At the first VSWR minimum the phase of s is -180 degrees. A
change in the s parameter phase angle of -360 degrees represents a translation of 0.5 of a line
wavelength away from the load and towards the generator. In this case the phase shift is -206.57
degrees or 0.2869 wavelengths from the load to the first SW minimum.
The value of impedance (ratio of total line voltage to total line current) at the first standing wave voltage
minimum.
 The definition is inherent in this part of the question. Here, the voltage and currents at the
standing wave minimum are those that would be measured if we could include a current meter in
one of the wires of the transmission line, and measure the voltage between the wires of the
transmission line, both measurements being taken at the standing wave minimum.
 At the first standing wave minimum, the total line voltage = (V+)[1-|s|] and the total line current
= (I+)[1+|s|]. The ratio of these is therefore Zo([1-|s|]/[1+|s|]), since V+ = ZoI+
 In our case using the results above, the impedance = Zo/(VSWR) = Zo/2.618 = 0.3820*Zo
The value of impedance at the first standing wave voltage maximum.
 By direct analogy to the above part of the question, the value of impedance at the first standing
wave maximum is Zo*VSWR = 2.618*Zo
The scattering parameter (complex reflection coefficient) at the load.
 The scattering parameter at the load is the complex ratio (backward wave voltage)/(forward wave
voltage), at the load, where the voltages are regarded as complex phasors describing the
oscillating voltages.
 In this case, as before, s = (ZL-Zo)/(ZL+Zo) = (2+j1-1)/(2+j1+1) = (1+j)/(3+j) =
0.4472*exp(j26.57) = 0.4 + j0.2
The scattering parameter at the first standing wave voltage minimum.
 The scattering parameter varies only in phase along a lossless line. Therefore its magnitude is
0.4472 and the phase angle is -180 degrees at the first standing wave minimum. In real and
imaginary parts it is -.4472+j0
The input impedance if the line length is 1.73 wavelengths. and Zo is 75 ohms.

The input impedance is the ratio of line voltage to current at the input, or generator end, of the
line.
 In our case Zin/Zo = (1+s)/(1-s) where s is taken at the line input. Again, |s| doesn't vary along the
line so it is still 0.4472 at the line input. The line length is 1.73 wavelengths = 3*0.5+.23
wavelengths. Each 0.5 wavelength represents a change in phase of the s parameter of -360
degrees and can be ignored. Thus we are left with a phase shift input -> load -> input of 360*.23*2 degrees = -165.60 degrees. At the input therefore the scattering parameter has phase
26.57-165.60 = -139.03 degrees.
 Converting to real and imaginary parts, at the input the scattering parameter s = -0.3377-j0.2932
so Zin = Zo(0.6623-j0.2932)/(1.3377+j0.2932) with Zo=75 ohms. The trusty calculator finds Zin
= 31.99 -j23.45 ohms
The scattering parameter at the input to the transmission line.
 See the previous section
Question 4.
Define the term "scattering matrix" for a 2-port microwave network. Explain carefully the output and
input variables related by the scattering parameters. Explain why the scattering parameters have
dimensionless complex character. Explain clearly the meaning of the term "reference plane" with respect
to which the scattering parameters are defined and measured. Explain what difference in s parameters
occurs if the reference plane is moved further from the 2-port. Explain why scattering parameters are
usually given in the form "modulus and phase angle".
 Most of the section above is dealt with in the notes on scattering parameters .
 If the reference plane is moved the s parameters change phase but not modulus. Thus it is best to
express them in this format.
Write down the 2 port scattering matrix for a lossless transmission line of electrical length 1/5 lambda,
giving the scattering parameters in both "modulus and phase" and then "real and imaginary"
representations.
 The round trip distance is 2/5 wavelength, or 360*2/5 degrees which is 144 degrees. The phase
retards as we move with the wave. One pass of the line is therefore -72 degrees, and the line is
lossless. Thus s12=s21 = 1 angle -72 degrees. s11=s22 = 0 because there are no inherent
reflections within the length of line.
 In terms of real and imaginary parts, s12=s21= 0.309-j0.951
This transmission line is connected to an antenna which has scattering parameter 0.1 angle 0 degrees.
Determine the fraction of the forward wave power which is radiated. Design a matching network to be
placed at the input to the transmission line so that the generator sees a perfect match at its terminals.
 The reflected wave power is 0.1*0.1 = 1% of the incident power. Thus 0.99 of the incident power
is radiated. (99%).
One might think this is a good enough match. However, at the input of the transmission line the
reflection coefficient (1-port s parameter) is 0.1 angle -144 degrees using the results of the last
section. This is 0.2 wavelengths towards the generator from the antenna normalised impedance of
(1+0.1)/(1-0.1) = 1.2222 + j0
which gives us an input impedance of (read graphically)
0.84 - j0.1
from the SMITH plot 0.2 wavelengths towards the generator
or we can CALCULATE the reflection coefficient gamma
(0.1 angle -144 degrees) = -0.0809 - j0.0588
which gives us an input impedance of (calculated)
(1 - 0.0809 - j0.0588)/(1 + 0.0809 + j0.0588)
which evaluates to be
0.8449 - j0.1004
which is satisfactorily in agreement with the SMITH plot.



We therefore need a single stub match to the impedance
0.8449 - j0.1004
A stub of series impedance -j0.2 placed 0.186 wavelengths further towards the generator does
this. The stub length for an impedance -j0.2 is 0.468 wavelengths from a short.
Question 5
Calculate the range of frequencies over which only the lowest order waveguide mode will propagate in
rectangular waveguides having the following cross sectional dimensions.
20cm by 15cm 2cm by 1.5cm 2mm by 1.5mm
2cm by 1cm
2cm by 0.8cm
2cm by 2cm
2cm by 4cm

The cutoff wavelength for the TE10 mode is twice the longest cross section dimension.
Remembering 30 = freq(GHz) times wavelength we have cutoff for the 2cm guides as 30/4 =
7.5GHz. For the 20 cm guide it is 750MHz and for the 2mm guide it is 75 GHz.
 The next mode is either the TE20 mode at twice the cutoff frequency of the TE10 mode, or the
TE01 mode if this is lower.
o Thus, 20cm by 15cm has 750MHz to 1GHz
2cm by 1.5cm, 7.5GHz to 10GHz, and 2mm by 1.5mm, 75GHz to 100GHz.
2cm by 1cm and 2cm by 0.8cm both have 7.5GHz to 15GHz.
2cm by 2cm has two degenerate modes, TE10 and TE01 both with cutoff 7.5GHz. There
is thus no band of frequency over which only one mode will propagate.
2cm by 4cm has long dimension 4cm so the range is 3.75GHz to 7.5GHz.
For the 2cm by 1cm waveguide, determine the lowest frequency of the lowest mode which will propagate.
Calculate the guide group and phase velocities, and the propagation angle alpha, for frequencies 100MHz
300MHz and 1GHz higher than the cutoff frequency.
 From the geometry of TE10 mode propagation, refer to the book of your choice, if the free space
wavelength is lambda, the guide width a, the propagation angle alpha, and the guide wavelength
lambdaG we can write
o sin(alpha) = lambda/(2a)
o cos(alpha) = lambda/lambdaG
o tan(alpha) = lambdaG/(2a)
 The cutoff of the 2cm by 1cm guide is 7500MHz so we want the values above at 7600, 7800, and
8500MHz.
 The values of lambda respectively are 3.9474cm, 3.8462cm, and 3.5294cm.
 The values of alpha are 80.70degrees, 74.06degrees, and 61.93degrees.
 The values of lambdaG are 24.41cm, 14.0cm, and 7.50cm.

The group velocities are c cos(alpha) = 0.4848, 0.8239, and 1.412, all in units of 10^8 metres per
second.
 The phase velocities are c/cos(alpha) = 18.56, 10.92, and 6.376, all in units of 10^8 metres per
second.
Question 6.
Calculate the scattering matrix of a 25 cm length of waveguide of internal rectangular cross section 5cm
by 2.2cm at a frequency of 4.0 GHz. Hint, you will need to find out the guide wavelength.
 The cutoff frequency is 3GHz. The guide wavelength is 11.3389cm. There are 2.2048 guide
wavelengths in the 25cm length of guide.
 The residual phase shift after allowing whole cycles is -73.73 degrees.
 The s parameters are s11=s22=0, s12=s21 = 1 angle -73.73 degrees.
The wave impedance of a waveguide mode is defined as the ratio of the transverse component of electric
field strength to the transverse component of magnetic field strength. Show that the wave impedance
depends on the angle alpha and the guide wavelength, and derive expressions for the dependencies.
 For a TE mode, the electric field is wholly transverse and the same as it is in a free space wave.
The magnetic field is turned by cos(alpha) from being transverse, so is smaller than the free space
wave value by cos(alpha). Thus the wave impedance is 120 pi/cos(alpha) or 120 pi
lambdaG/lambda.
 For a TM mode the electric field is rotates through alpha but the magnetic field is entirely
transverse. The wave impedance is thus 120 pi cos(alpha) or 120 pi lambda/lambdaG.
Calculate the wave impedance for the conditions of the first part of this question. Hint, you need to know
that in a free space wave the wave impedance is 120 pi ohms.
 Applying the results above, the impedance at 4GHz is 570 ohms to the nearest ohm.
Question 7
Show that a condition for stability of a 2 port is that there is no reflection gain for connections looking in
to either port; and that such a circuit is unconditionally stable for any passive load impedance and passive
generator source impedance.
 Here, the important point is that there must be no reflection gain at either port for any conditions
of passive load attached to the other port. If this condition is not met, the return power is more
than the incident power; we have in effect made a "reflection amplifier" which can cause
oscillations in a reactive load formed by a length of transmission line attached to an open or short
circuit.
 The condition for a length of line to oscillate is that the round trip gain, accounting for the losses
of the line and the reflection coefficients at each end, be greater than one.
 For a passive load the magnitude of the reflection coefficient is always less than one. Therefore,
for there to be round trip gain we require the other reflection to have magnitude greater than one.
If it doesn't, no oscillations will occur.
A certain 2-port has s parameters s11 = 0.1 angle -30 degrees, s12 = 0.05 angle -60 degrees, s21 = 30
angle -60 degrees, and s22 = 0.15 angles -20 degrees. All these s parameters are taken at a spot frequency.

let us write these s parameters formally as



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


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


{0.1 -30}
{30 -60}
{0.05 -60}
{0.15 -20}
the defining equations are
b1 = s11a1 + s12a2
b2 = s21a1 + s22a2
Discuss which values of load impedance on port 2 just give rise to reflection gain on port 1.

let us call the reflection coefficient at the load

on port 2 gamma.
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
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
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Then a2 = gamma b2 because the output b2 of the 2-port
is reflected from the load to form a2.
Using the equation b2 = s21a1 + s22a2
then we find a2(1/gamma - s22) = s21a1
Using the equation b1 = s11a1 + s12a2
then we find b1 = a1[s11 + s12s21/(1/gamma - s22)]
We want b1 to have magnitude less than a1
for there to be no reflection gain at port 1.
Thus the complex number s11 + s12s21/(1/gamma -s22)
has to lie inside the unit circle.
One can substitute the values above; I estimate
that if the reflection coefficient is less than
about |gamma| = 0.55 the circuit will be stable.
That puts the output transmission line VSWR as
being less than about 3.5
Question 8.
Write down the S matrices of the following microwave components.... You may assume they have ideal
characteristics.
A 30 dB amplifier
An isolator with 20dB front to back transmission ratio
A perfect 3 port circulator
A perfect 4 port dual directional coupler, with coupling strength -30 dB
A 2 metre length of 50 ohm coaxial cable at 400MHz, assuming a velocity factor of 2/3
 answers.....
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The 30dB amplifier
0
0
{31.6 -theta}
0
The 20 dB isolator
0
{0.1 - phi}
{1 -theta)
0
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The circulator
0
0
{1 -phi}
{1 -phi} 0
0
0
{1 -phi} 0
Here the circulation is 123123123....
The dual directional coupler
0
{0.9995 -theta}
{0.9995 -theta} 0
{0.0316 -rho} 0
0
{0.0316 -phi}
{0.0316 -rho}
0
0
{0.0316 -phi}
0
{0.9995 -psi}
{0.9995 -psi}
0
Here we see the S matrix is reciprocal.
Adding the moduli squared in the columns gives 1
Also, theta-rho = phi-psi - 180 degrees
and, theta-phi = rho-psi - 180 degrees
The 2 metre length of cable
Well, the velocity is 20 cm/nsec so a wavelength
at 0.4GHz is 50cm and the line is 4 wavelengths long.
0
1
1
0
A generator is connected via the isolator described above to one of the ports of the three port circulator.
The circulator has a mismatch on its two other ports giving a reflection coefficient of 0.2 angle 0 on each
load connected to the ports. The phase shift between successive ports of the circulator is -45 degrees.
Calculate the magnitude of the reflection coefficient first at the driven port of the circulator, and second,
at the generator terminals.
 The reflection coefficient at port 2 has size 0.2 so the wave incident on port 3 is 0.2 of the size of
the incident wave. It is reflected by the load on port 3 with another factor 0.2 so the size of the
reflection coefficient at port 1 is 0.04. Now the phase shift is -90 degrees if you need to calculate
the effective input impedance.
 The isolator reduces the reflection coefficient by another 20 dB or factor of 10. Thus at the
generator terminals the reflection coefficient is 0.004 in size.
Question 9
Define the terms "characteristic impedance" and "velocity factor" for a lossless coaxial cable. Derive
formulas for the relationships between these quantities and the inductance and capacitance of a 1 metre
length of cable. [30%]
 Characteristic impedance is the ratio of voltage to current on a coaxial cable carrying waves
travelling in a single direction only. It is that impedance, which, when used as a load results in
zero reflected energy, or a VSWR of 1.
 Velocity factor:- the velocity of waves on a coaxial cable, in SI units, divided by the velocity of
light in a vacuum, also in SI units. The velocity factor is therefore a dimensionless number.
Typically it is about 2/3 for most coaxial cables.
 Consider a cable with inductance L Henries per metre, and capacitance C Farads per metre, as the
quantities per unit length under consideration. Draw a diagram with series inductance Ldx and
shunt capacitance Cdx. This represents a little length of cable dx metres long. We solve Kirchoff's
laws to write down the voltage drop along the cable dV = - Ldx di/dt and the current increment
due to the shunt current through the capacitance di = - Cdx dV/dt. This gives us the Telegraphers'
equations dV/dx = -Ldi/dt and di/dx = -CdV/dt which when combined give us the wave equation
in V or i with wave velocity 1/sqrt(LC) and characteristic impedance sqrt(L/C).
 The velocity factor eta is therefore [1/(3E8)][1/sqrt(LC)] and the characteristic impedance Zo =
sqrt(L/C)
Explain why a coaxial cable connected to a load impedance Z has, in general, an input impedance which
depends on the length of the cable. Derive a formula for this input impedance, for a cable of length d
metres, having velocity factor eta and characteristic impedance Zo ohms at a frequency of f Hz. Under
what conditions is the input impedance independent of the length of this cable? [35%]
 The wavelength on the cable is lambda, and f lambda is the velocity of waves. The propagation
constant beta is defined as 2 pi/lambda and forward waves propagate as exp[-j beta distance]
whereas backward waves travel as exp [+j beta distance]. If there is any reflected power at all,
these phase angles vary differently along the line, and the phase angle between the forward and
backward wave voltages and currents depends on position. Since forward waves have V+ = +Zo
I+ and backward waves have V- = -Zo I-, the impedance V/I = [V+ + V-]/[I+ + I-] depends on
distance unless V- and I- are zero, that is, there is no backward wave. In general there will be
waves travelling in each direction, so in general the input impedance depends on the length of
cable connected to the load Z.
 If the load is placed at distance x = 0, (with the input to the line at x = -d) then at the load we have
Z/Zo = [V+ + V-]/[V+ - V-] = [1+s]/[1-s] where s is the reflection coefficient V-/V+ at x=0. At x
= -d, the input to the line, the reflection coefficient s = [Z-Zo]/[Z+Zo] must be multiplied by the
phase factor exp [-j 2 beta d] to account for the delay in the wave propagation. Let us assume this
new reflection coefficient is called s'. Then the input impedance Zin is given by Zin/Zo =
[1+s']/[1-s']. Pull all this together and express in terms of the quantities given in the question. If
V- = 0 or if s = 0 then Z=Zo and the line is matched; the input impedance does not then depend
on the length of the cable.
A certain coaxial feeder cable has wave velocity 0.20 metres per nanosecond and a nominal characteristic
impedance of 50 ohms. It also has loss of 1dB/m at a frequency of 1 gigaHertz. It is connected to a load
impedance of 75 ohms. Estimate the input impedance of this cable at a distance 5.73 metres from the
load, for signals at 1GHz. [35%]
 The wave velocity is 20 cm per nanosecond so at 1 GHz the wavelength is 20 cm, and 5.73
metres represents 28.65 wavelengths of cable between the input and the load. The one way loss at
1GHz is 5.73 dB. At the load, the reflection coefficient s = [75-50]/[75+50] = 1/5. The round trip
attenuation generator-load-generator is 2 times 5.73 dB which is an amplitude reduction of 3.74
so at the input the size of the reflection coefficient is 1/5 times 1/3.74 = 1/18.71
 The round trip phase delay is a whole number of wavelengths plus 0.3 wavelengths so the
reflection coefficient at the input is s(in) = 1/18.71 exp[-j 108 degrees] and using the formula
above we find at the input Zin = 48.13 - j4.91 ohms, that is, not far from a good match.
Question 10.
Describe the principles behind the use of a network analyser to measure the scattering parameters of a
two-port microwave network. [30%]
 A network analyser consists of a synthesised frequency source and two bridges which measure
port currents and voltages at the two ports. The ratios of outward to inward wave amplitudes and
phases are obtained directly from the bridge measurements. The network analyser sets a
frequency of the source, and applies an input signal first to port 1 and then to port 2. For each of
these two excitation conditions, the bridge derives the proportional complex scattered amplitude
from each port, making four complex dimensionless measurements in all.
 Calibration of the instrument is carried out by presenting it with known scattering circuits; usually
a short, a termination, and a "through" length of transmission line of known electrical length. The
instrument modifies the measurements which it has made on the unknown scatterer and allows for
its own imprecisions which are obtained from the calibration procedure. The synthesised source
scans at spot frequencies across the range of frequencies specified externally by the software, and
presents the results as a graph of s parameters against frequency. Presentations can be either real
and imaginary parts of the s parameter, or modulus and phase, or the s parameter can be plotted as
points on the Smith chart.
Derive a matrix formula linking the two-port y-parameters with the scattering parameters. [30%]
 In matrix notation, the matrix of n by n y parameters for an n-port network is written here as Y.
Then the n vector of port currents I is related to the n vector of port voltages V by I=YV.
 Suppose Y represents normalised admittances; that is we have divided all the y parameters by the
characteristic admittance of the transmission line in which the n-port network is embedded. Y is
then a dimensionless matrix.
 If we consider the system in terms of normalised matrix elements, for which we may consider Zo
= Yo = 1, then the incoming wave amplitudes are proportional to V+I and the outgoing wave
amplitudes are proportional to V-I. (See the development of the solution to the next part of this
question.) Thus we can write [V-I] = S[V+I] where S is the scattering matrix. (by definition). A
simple substitution shows that S = [1-Y][1+Y]^(-1), where we have post-multiplied the matrix [1Y] by the inverse of the matrix [1+Y].
 Looking at this in the inverse way, S+SY = [1-Y] so Y[S+1] = [S-1] and we obtain Y by postmultiplying the matrix [S-1] by the inverse of the matrix [S+1]
Explain how measurements of terminal currents and voltages at the port terminals may be used to
determine the incoming and outgoing wave amplitudes for waves on the connecting transmission lines of
known characteristic impedance Zo. A totally absorbing two-port network has no outgoing waves at
either port under any excitation conditions. List its scattering parameters.[40%].
 We start from the incoming wave voltage V+ and current I+ and the outgoing wave voltage Vand current I-. The definition of Zo and the direction of power flow gives us V+=ZoI+ and V-=ZoI-. The total port voltage V is V+ + V- and the total port current I is I+ + I-. Multiplying I by
Zo we see ZoI = V+ - V- by substitution.
 Adding and subtracting V and ZoI gives us V+ = (V + ZoI)/2 and V- = (V - ZoI)/2. We have
expressed the incoming and outgoing wave voltage amplitudes in terms of the port voltage V,
port current I, and characteristic impedance Zo.
 If there are no outgoing waves at all for any combination of input waves, all the scattering
parameters are identically zero. Listing them, s11=s12=s21=s22=0.
 Funnily enough, in the actual exam in 1995 no one got this bit right. That demonstrated to me that
they hadn't understood what the S matrix actually represents.
Question 11.
Derive formulas for the phase velocity and group velocity of waves on a lossless rectangular waveguide
supporting a TE10 mode. Show that the product of these velocities is c^2. [35%]
 There were two acceptable ways of answering this part of the question, one involving a graphical
construction showing the waveguide field patterns and angle of propagation, and the other using
the definitions (1) that the phase velocity = [angular frequency] /[propagation constant] and (2)
that the group velocity = d[angular frequency] /d[propagation constant] (which is the derivative
of angular frequency with respect to the propagation constant).
 As this is a standard piece of bookwork I refer you to J D Kraus Electromagnetics or other
standard text of your choice.
 Geometrically, the guide wavelength is given by lambda/cos(alpha) where alpha is the angle of
propagation with respect to the guide axis, so the guide phase velocity is c/cos(alpha) since f
lambda = c. The guide group velocity is the component of velocity along the guide axis and is c
cos(alpha). Clearly the product of velocities is c^2.
 Algebraically, one obtains the same results from the waveguide formula using the relationships
given above.
A certain cavity wavemeter has an accurate micrometer which gives a reading of the distance between the
moving short and the fixed short. The waveguide (a standard X band WG90 type) has dimensions of
0.900 inches by 0.400 inches. Determine the micrometer readings when the wavemeter absorbs at the
following frequencies: (i) 8.58 GHz, (ii) 10.04 GHz, (iii) 11.38 GHz. [35%]
by the frequency in GHz. (That gives it in cm, and the waveguide dimensions are in inches so you can
convert one to the other, as long as you state which units you are using for the end result). The free space
wavelengths in cm are (i) 3.4965 (ii) 2.9880 (iii) 2.6362. We keep a few more significant figures at this
stage of the calculation as rounding occurs last.
derive it). The critical guide dimension is a=0.900 inches = 2.286 cm so 2a = 4.572 cm and (2a)^2 =
20.9032. The waveguide formula gives the guide wavelength as sqrt[1/{(1/lambda^2)- (1/2a)^2}]. The
respective guide wavelengths are (i) 5.4267 cm (ii) 3.9477 cm (iii) 3.2266 cm.
horts the cavity wavemeter is resonant and
absorbs power from the waveguide to which it is coupled. That gives us a micrometer reading of half the
guide wavelength in each case, always assuming that we can neglect perturbations of the field
distributions due to the coupling hole.
1.61 cm or 0.635 inches. It is good practice to give the result to the same number of significant figures as
the original data in the question.
Describe how such a wavemeter can be used in conjunction with a 30dB dual directional coupler to
measure the frequency of a microwave test bench. Stating your assumptions, estimate the achievable
precision of the measurement. [30%].
 The directional coupler samples 1/1000 of the forward wave power and feeds it up a side arm to a
detector. The side arm is coupled by means of a small hole in the common wall to the cavity
wavemeter. The cavity wavemeter has a loaded Q of about 1000 to 10,000 depending on how the
inside surfaces are coated (silver plating is common), on the surface roughness, and on the
coupling strength to the side arm waveguide. This gives an achievable resolution in frequency of
between 0.01% and 0.1%, or 1 to 10 MHz at X band.
Question 12.
Give a brief description of the physical principles governing the behaviour of non-reciprocal ferrite
microwave components. Discuss the practical limitations imposed by power handling and frequencyselective effects. [40%]
 In ferrite biased by a DC magnetic field, the spins of the unpaired electrons precess around the
direction of the bias field. The strength of the bias field governs the rate of precession, and it may
be tuned so that the precession frequency lies near the frequency of microwaves inside the ferrite
loaded device. By placing the ferrite in a region where the local RF magnetic field produced by
the passing microwaves rotates in local direction once a cycle, a resonance condition may be set
up between the precessing spins and the microwave magnetic field. The spins then absorb energy
from the microwave radiation; the spins pass this energy to phonons in the ferrite by a process
known as "spin-lattice relaxation", and the consequence is the ferrite gets warm, heated by the
microwave energy.
 If the microwave radiation is travelling in the opposite direction along the transmission line, the
spins and microwave magnetic field are rotating in opposite directions and there is no resonant
absorption. Thus the ferrite device has low insertion loss for waves travelling in this direction.
 In general, even if the resonance condition is not satisfied exactly, there will be different
magnetic properties of the ferrite for forward and backward microwave propagation. This gives
rise to a difference in wave velocity which may be used to construct interference devices such as
the three port circulator, where the transmission is cyclic from ports 1->2->3->1-> etc. Such
devices may be substantially lossless to the microwaves.
 In a field displacement isolator, the differing field patterns for forward and backward wave
propagation give maximum electric field in different places across the waveguide. A resistive
card is placed at a region where the forward wave electric field is small, but the backward wave
electric field is large. This therefore gives non-reciprocal attenuation properties. The advantage of
this arrangement is that the ferrite does not get hot; as its magnetic properties are temperature
dependent, this is an advantage.
 Thus power handling in a field displacement device is higher than in a resonance absorption
device. Typically, bandwidths of 2:1 frequency ratio between the top and bottom frequency of
operation are commonplace. This is because the resonant peak of the spin system is broad, and
many devices are constructed to lie on the shoulder of the resonance and make use of the different
permeability tensors for forward and backward wave propagation.
An imperfect isolator has forward loss of 0.2 dB and unwanted reverse transmission of -11 dB. It is used
to connect a transmitter amplifier (HPA) to an irregular passive load which can have any (positive)
resistive and/or reactive impedance. Using a SMITH chart, draw an estimated contour which represents
the limits of the input impedance of the loaded isolator. [35%]
 The round trip attenuation is 11.2 dB which is an amplitude attenuation factor of 3.63 (the return
wave is 3.63 times smaller than the incident wave if all the power is reflected beyond the
isolator). Thus the maximum modulus of the s parameter or reflection coefficient is 1/3.63 =
0.275.
 The input impedance must therefore lie within a circle centred on the SMITH chart centre, of
radius 0.275 of the radius of the periphery of the SMITH chart.
The outputs of three HPAs in differing frequency bands are to be combined so that no HPA output is
driven by the others. With the aid of a diagram, explain how this may be achieved by means of a
circulator tree arrangement. [25%]
 The diagram you need must show each HPA passing its output through a bandpass filter which
has a stop band for the outputs of the other HPAs. The filter outputs are fed into the circulator tree
arrangement; each circulator has three ports 1,2,3 and port 1 is fed by the filter, port 2 of the
lower circulator is fed to port 3 of the one above it. At the top, port 2 feeds the output
transmission line, and at the bottom port 3 is terminated. Power travels cyclically 123123123 etc.
Return power is never presented to a filter output, passing from port 2 to 3 (top) to 2 to 3 (middle)
to 2 to 3 (bottom) where it is absorbed in the termination.
 Since the filters are perfect reflectors of power in their stop bands, it is clear that no HPA output
can drive any other HPA's output terminals.
Question 1 3
Define the terms "characteristic impedance", "velocity factor", "forward travelling wave", "reflection
coefficient", and "return loss", in the context of propagation on a transmission line. [25%]
 The characteristic impedance is the ratio of voltage to current in a forward travelling wave, when
there is no reverse travelling wave. A transmission line presents the characteristic impedance to a
generator for times until the first reflection arrives at the generator.
 The velocity factor eta is the ratio of the "wave velocity or wave phase velocity" to the "speed of
light in vacuum".


The forward travelling wave is any disturbance travelling from generator to load.
The reflection coefficient gamma is the complex ratio of backward wave amplitude to the forward
wave amplitude, considering monochromatic waves (waves of a single frequency).
 The return loss is the amount in dB by which the reflected wave power is less than the forward
wave power. Numerically, the return loss is -20 log[10](|gamma|).
A building is wired with Ethernet cable having velocity factor 0.6. There is a break in this cable 22 metres
from a time domain reflectometer (TDR) which emits rectangular pulses of rise and fall times 0.5
nanoseconds and duration 5 nanoseconds. What is the time delay until the return of the first reflection
from this break? [35%]
 The pulse velocity is 0.6 times 30 cm per nanosecond, or 18 cm per nanosecond.
 The round trip distance to the break is 2 times 22 times 100 cm, or 4400 cm.
 The round trip time is 4400/18 = 244 nanoseconds to the nearest nanosecond. This is 0.244
microseconds.
A shunt open circuit spur of length 12 metres is added at a distance 10 metres from the TDR. The rest of
the cable is undamaged and is correctly terminated. Describe quantitatively the pulse sequence seen on
the TDR. [40%]
 Draw a diagram with G the generator, which has internal impedance Zo matched to the line so
that it doesn't produce reflections, placed 10 metres from P the junction which feeds a shunt
matched load Zo and a side arm 12 metres long to Q which is an open circuit.
 Measuring the impedance at G in the steady state we of course see just a matched load Zo. One
might be fooled by this into thinking that there would be no reflections from the system.... This is
the point of the question.
 The reflection coefficient at P is formed by the Zo of the matched arm in parallel with Zo for the
open spur. Thus gamma = (0.5-1)/(0.5+1) which is -1/3.
 The reflection coefficient at Q is similarly +1.
 The round trip time GPG is 2000/18 = 111 nanoseconds to the nearest nanosecond.
 The round trip time PQP is 2400/18 = 133 nanoseconds.
 The transmission coefficient at the junction P in any direction is calculated as follows: power
reflected = (-1/3)(-1/3) = 1/9 of the incident power. Power transmitted = 1-1/9 = 8/9 of the
incident power. This splits so that 4/9 of the incident power is transmitted down each of the
connected arms,
 The transmitted amplitude on each arm is therefore 2/3 of the incident amplitude.
 The sequence of pulses at the generator end is therefore amplitude 1 at time zero (initial pulse)
followed by -1/3 at 111 nanoseconds and then a pulse which has travelled once up and down the
spur, having amplitude 2/3 times 1 times 2/3 = 4/9 at 111+133 = 244 nanoseconds. The next pulse
has travelled twice up and down the spur, and has amplitude 2/3 times 1 times -1/3 times 1 times
2/3 or -4/27 at time 111 + 2 times 133 = 377 nanoseconds. If you can get this far you can work
out the rest for yourselves. The next pulse has travelled three times up and down the spur...
Question 14
Describe the construction and principles of operation of a waveguide dual directional coupler having
coupling strength -30 dB and fractional bandwidth about 10% for use at X band frequencies. [30%]
 For a 30dB dual directional coupler, narrow band, a two hole waveguide coupler is sufficient.
Draw two waveguides with the broad face common, with two small x shaped holes on the centre
line of the common face and spaced a quarter of a guide wavelength apart.
 The coupling strength depends on the hole size. The forward coupled waves add in phase, giving
a combined coupling strength with coupled power 1/1000 of the incident power in the other
waveguide. The waveguide dimensions are about 0.900 inches by 0.400 inches (WG90).
A -30dB coupler is used to sample the forward and backward wave amplitudes on a waveguide connected
to an antenna. The detectors on the sampling arms consist of totally absorbing matched diode detectors
which have accurate square law characteristics. Determine the ratio of output voltages on the detectors if
the reflection causes a VSWR of 1.8 on the main arm of the coupler. [25%]

The VSWR is 1.8 so the modulus of the reflection coefficient, |gamma|, is (1.8-1)/(1.8+1) =
0.8/2.8 = 0.2857 and the square of this is 0.0816 so 8.16% of the power is reflected. The detector
voltages are proportional to the coupled wave powers, so the backward wave detector voltage is
about 8% of the forward wave detector voltage. The ratio of voltages is therefore 12.25.
A mismatch is introduced at each of the detector diodes, giving a reflection coefficient of magnitude 0.1.
For the same conditions as part (b), estimate the ratio of output voltages at the detectors. [25%]
 We are not told of the phase delays in the system. The first assumption is that the phase is
arbitrary and we add powers. The power at the backward wave detector is therefore 1/12.25 of the
forward wave power plus 0.1 times 0.1 of the power incident on the forward wave detector. Now
1/12.25 + 1/100 = 1/10.9. Of course there will be some reflected power on the forward wave
detector which comes from the backward wave detector, but this is less than 1/1000 of the
original forward wave detector power and we neglect it.
 If you like you can do a phasor addition wherein the amplitudes rather than the powers are added.
This is not necessary in this particular example.
 We learn from this example that it is most important to have a good match on the detectors,
particularly the one sampling the forward wave power.
Explain how the fractional bandwidth of the dual directional coupler may be increased to 50% or more.
[20%]
coupling strength as the two original holes. They are spaced so that the resultant reverse wave
amplitude phasor is small; the contributions from each hole add in a straight line for forward
wave coupling, but for reverse coupling they add in a polygonal manner until the resultant is
minimised. The bandwidth enhancement depends on the number of holes. For this application a
collection of 20 holes spread out over several guide wavelengths would work nicely.
Question 15
Describe the principal uses of components in microwave transmission systems which contain ferrites.
Explain what is the most important property of the components which is due to the presence of the ferrite.
[30%]
 Ferrites have the property that their wave propagation behaviour can depend on the direction of
power flow along a transmission line into which they are incorporated. This property is referred
to as being "non-reciprocal". This is contrasted with most other non-magnetic passive
components where transmission does not depend on the direction of power flow between two of
the ports of a network. Ferrite devices share the property of non-reciprocity with active devices
such as transistor amplifiers.
 They are used to "isolate" a generator from reflections caused by load mismatch. They can be
used in power combiners and in switching applications. They make excellent microwave absorber
for lining anechoic chambers; the magnetic field due to microwaves is large near a metal
reflecting surface, and magnetic loss placed here is more effective at absorbing power than is
resistive dielectric loss; electric transverse fields are small close to a metal surface.
 Circulators are used to separate forward waves from backward waves in many applications. They
can be used in combination with a passive matched load to construct an equivalent 2-port to an
isolator.
A ferrite isolator is imperfect, having forward loss of 1 dB and a forward/backward transmission ratio of
18 dB. It is connected between a generator and a load. Calculate the VSWR on the generator side of the
isolator if the load consists of
(i) A short circuit
(ii) A mismatch produced by a load 2Zo where Zo is the characteristic impedance of the transmission
system.
[35%]
 The forward loss is 1 dB, and the backward insertion loss is 18dB more than this, that is, it is
19dB in total. The round trip attenuation for passes both ways through the isolator is therefore
20dB, and the reflected wave amplitude is only 1/10 of what it would be in the absence of the
isolator.
 If the load is a short, all the power is reflected. Therefore on the generator side of the isolator the
return wave amplitude is 0.1 times the incident wave amplitude, and the VSWR is [1 + 0.1]/[1 0.1] which is 1.22
 If the load is 2Zo, the reflection coefficient from the load is [2-1]/[2+1] = 1/3. The return wave
amplitude on the generator side of the isolator is therefore 1/30 of the forward wave amplitude,
and the VSWR is [1+1/30]/[1-1/30] which is 1.069
 In the absence of the isolator, the VSWR values for these two cases would have been infinity and
2.0 respectively. Thus we see the isolator, which does not have a tremendously large front to back
ratio, is nevertheless very effective at reducing the possible values of VSWR seen at the generator
side of the isolator.
Show, with a diagram, how a circulator tree may be used to combine the outputs of a number of
amplifiers in a frequency division multiplex system, without any of the amplifiers driving power into the
other amplifiers. Explain the use of bandpass filters in this arrangement. [35%]
Question 16
Write comprehensive notes on TWO of the following topics. [50% for each topic].
(i) Scattering parameter descriptions of microwave n-port networks.
 See the notes on scattering parameters
(ii) Gunn diode microwave reflection amplifiers.
 For the purposes of this answer we can consider a Gunn diode load on a transmission line to be a
shunt combination of negative resistance and capacitance. The capacitance C has normalised
susceptance s =[omega C/Yo] where Yo is the characteristic admittance of the transmission line.
The shunt negative resistance has a normalised conductance value -g where g is a positive
number.
 Using a normalised load admittance yL = -g + js together with the formula for reflection
coefficient or scattering parameter gamma = [1-yL]/[1+yL] we see that gamma = [(1+g) -js]/[(1g) + js] from which we find that the magnitude of the reflection coefficient is [(1+g)^2 + s^2]/[(1g)^2 + s^2].
 For small and very large values of g the reflection gain is about unity. There is a maximum
reflection gain when g=1, that is the real part of the shunt negative admittance of the Gunn diode
equals the transmission line characteristic admittance. The gain is then [1+ 4/s^2].
 One can also do a bandwidth calculation similarly to find the range of frequencies between which
the gain is more than half its peak value.
 For large gain-bandwidth products the shunt capacitance and susceptance needs to be as small as
possible.
 The reflected power is separated from the incident power using a circulator. The circulator
forward transmission is 123123 etc, and the signal source is connected to port 1. The Gunn diode
is connected to port 2 and a matched load (or isolator driven load) is connected to port 3. It is
important that the source does not see reflection gain on its own transmission line.
(iii) Microwave bandpass filter design and construction.
 Filters are usually designed to work between known generator and load impedances. Microwave
filters are no exception and it is usual to take the reference impedances of source and load to be
the transmission line characteristic impedance.
 For the purposes of this answer it is assumed that it is known how to design lumped component
bandpass filters working between known specified impedance levels.
 Reactive components can be generated, at least at a single frequency, by shorted lengths of
transmission line of known length.
 A bandpass filter is usually constructed from reactive components only, as this keeps its in-band
insertion loss low and its noise performance is good.

Admittance to impedance transformation can be accomplished by means of quarter wave sections
of transmission line. Thus we see that we can used shunt reactive stubs placed at intervals along a
transmission line to make a filter.
 The shunt elements can be physical lengths of line, or in waveguide they and be irises, or
apertures between coupled cavities.
 Microwave filters have a frequency response which, in principle, repeats indefinitely as the
frequency is raised and the line lengths become progressively longer by added amounts of half a
wavelength. Thus they are by no means direct equivalents of their lumped constant templates.
 They can be fabricated from coupled microstrip lines, dielectric resonator sections, waveguide
with irises, or acoustic delay lines. They are reciprocal and it is often assumed that they are
lossless.
 For a lossless bandpass filter the sum of the reflected and transmitted powers must equal the
incident power. Thus filters can be constructed in reflection mode as well as in transmission
mode; in reflection mode the pass band becomes a stop band and vice versa.
 Group delay as well as amplitude response is usually an important design parameter of the
microwave filter.
Question 17
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 16|A|^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.
Question 18
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 |V-||V-|/(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, (ZL-Zo)/(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 10-6.460 = 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 standing-wave 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.
Question 19
Describe the s-parameter representation of the properties of a linear microwave 2-port circuit, giving the
defining equations and also stating precisely which variable are related by the s-parameters. [25%]
 The s-parameters relate inward and outward wave complex amplitudes. They are measured at a
"reference plane" with respect to each port. The reference planes are positions along the
transmission lines feeding the ports; if moved, the phases of the s-parameters will alter. It is usual
to define the wave amplitudes so that the square moduli of the amplitudes represent the wave
power flows.
 The amplitudes are complex phasors and so the s-parameters are complex quantities. If we call
the input amplitudes a1 and a2 at ports 1 and 2 respectively, and the outward wave amplitudes b1
and b2 similarly, the defining equations are....







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b1 = s11 a1 + s12 a2
b2 = s21 a1 + s22 a2
where the four complex s-parameters are s11 s12 s21 s22.
In terms of the port 1 rms voltage V and rms current I, and the port 1 transmission line
characteristic impedance Zo, we have for the normalised wave amplitudes a1 and b1
a1 = (V+ZoI)/(2Zo)
b1 = (V-ZoI)/(2Zo)
and similar relationships hold for the amplitudes a2 and b2
State, giving reasons for your choice, which microwave components are described by the following
collections of s-parameters. [15% for each of the three parts below]
s11=0
s12=1exp(-j720)
s21=1exp(-j720)
s22=0

This is a section of lossless transmission line, two wavelengths long (720 degrees of phase shift =
2 times 360). All the incident power is transferred to the output; there is no reflection at all if the
output port is terminated.
s11=0
s12=0.07 exp(-j40)
s21=1exp(-j60)
s22=0

This is an isolator, having no insertion loss in the forward direction, so that all the power in port 1
comes out of port 2 and there is no intrinsic reflection from the isolator. In the reverse direction,
the isolator produces no intrinsic reflection, but attenuates the wave passed back to the port 1 by 20 log10(0.07) dB which is 23.1 dB. It is a good isolator.
s11=0.1exp(-j30)
s21=9.7exp(-j80)
s12=0.3exp(-j80)
s22=0.15exp(-j60)

This is an amplifier. It has forward gain of 20log10(9.7) or 19.7dB, and reverse transmission loss
of -20log10(0.3) or 10.45dB. If the output is matched, the input return loss is 20dB, and if the
input is matched and the output driven, the return loss from the output port is -20log10(0.15) or
16.5dB. If placed between gross miss-matches it would oscillate because of the round trip gain
being greater than unity.
A one-port circuit is created by taking a two-port device D, with generalised scattering matrix S, and
adding a load to port 2 of D. The load has complex reflection coefficient gamma. Derive an algebraic
formula for the complex reflection coefficient seen looking into port 1 of D. If a generator is attached to
port 1 of D, what is the condition for this arrangement to be stable for all values of generator source
impedance? [30%]
 The wave leaving port 2 of D is reflected by the load and the reflected wave forms the input wave
to port 2. Thus we have a2 = gamma b2. Substituting into the defining s-parameter equations we
find the following formulae....

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








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b1 = s11 a1 + s12 gamma b2
b2 = s21 a1 + s22 gamma b2
From the second of these equations we find, rearranging, that
b2(1-s22 gamma) = s21 a1
and we substitute for b2 in the first equation so that
b1 = s11 a1 + (s12 s21 gamma a1)/(1 - s22 gamma)
The complex reflection coefficient at port 1 is just b1/a1
which is clearly
(b1/a1) = s11 + (s12 s21 gamma)/(1 - s22 gamma)
If this quantity has modulus less than unity, there can be no
reflection gain, and the circuit arrangement will be
unconditionally stable.
Question 20
Describe the behaviours of an ideal isolator and an ideal 3-port circulator. Give examples of the
applications of these components. State which kind of material, used in their construction, gives rise to
their unique properties. [40%]
 An ideal isolator transmits waves in one direction without any attenuation, but possibly some
phase shift; it absorbs completely waves in the reverse direction. Reverse waves see a matched
impedance at the output port of the isolator so at that port s22 is zero and so there is no reflection
of waves impinging on the output port (2) of the isolator. Ideally the isolator works over a
significant bandwidth and is essentially a non-resonant device.
 An ideal 3-port circulator transmits power from port 1 to port 2, from port 2 to port 3, and from
port 3 to port 1 without attenuation. Intrinsically it has no loss. It is matched on each port; thus
there is no reflected wave from port 1 if ports 2 and 3 are terminated and not driven. The same
remarks apply to cyclic rotation of the port numbers.
 Isolators are used to prevent the pulling of low Q oscillators such as Gunn oscillators by resonant
loads whose coupling may be time-dependent. They are used to protect high power generators
and amplifiers from the effects of gross miss-match at the loads. They are used to reduce the
VSWR on transmission line and the reverse transmission through amplifiers which might
otherwise go unstable.
 Circulators are used in power combiners, and to separate forward from backward waves so that
the return loss of a load may be measured directly. They are used with antennas to separate the
received signal from the transmitted signal.
 Both isolators and circulators contain magnetised ferrite; ferrite is insulating magnetic material
and when it is placed in a steady magnetic field the spins in the unpaired electrons in the ferrite
precess around the DC magnetic field and couple with differing strengths to forward and
backward waves, which have rotating RF magnetic fields in opposite senses of rotation. This
gives the devices the intrinsic and unique property, for passive circuits, of non-reciprocal
behaviour.
An ideal 3-port circulator is embedded in a transmission line of characteristic impedance 50 ohms. An
engineer attempts to measure the complex reflection coefficient at port 1 when ports 2 and 3 are each
loaded with 75 ohms resistance. Assuming that the phase shift between successive ports is -60 degrees,
calculate the complex reflection coefficient that the engineer would expect to measure. [35%]
 The reflection coefficient from the 75 ohm load at the port reference planes on ports 2 and 3 is
given by gamma = (75-50)/(75+50) = 25/125 = 0.2 = 1/5.
 The signal travels without loss from port 1 to port 2, with a phase shift of -60 degrees en route. It
is reflected at the load and reduced in amplitude by a factor of 5. It has another -60 degrees of
phase shift to get to port 3, where it is reduced in amplitude by another factor of 5. Finally, it has
another -60 degrees or phase shift to get to port 1 again. The total reduction in amplitude is 1/5
times 1/5 and the total phase shift is 3 times -60 degrees. Thus the output wave at port 1 is 0.04
angle -180 degrees with respect to the input wave; and this is the complex reflection coefficient.
Explain how a circulator may be used in place of an isolator, stating any advantages of this arrangement.
[25%]
 If the circulator has a matched load placed on port 3, and the load of interest is attached on port 2
and the generator on port 1, any reflected power from the load on port 2 is all absorbed in the
matched termination on port 3. It is therefore a perfect isolator and has better isolation properties
than a 2-port isolator. The power is absorbed by a resistive load, which can be cooled by a fan or
water circulation, and the ferrite does not get hot. This is a technological advantage as its
magnetic properties are temperature dependent.
Question 21
Give a justification of the formula, stated below, for the fundamental TE10 mode in a rectangular
waveguide of cross-sectional dimensions a (cm) by b (cm).
1
1
1
--- = --- --(lambdag)^2 (lambda)^2
(2a)^2
Sketch the waveguide cross-section, indicating clearly which dimension is a. State the equivalent formula
for the guide wavelength of the TE11 mode. [40%]
 In a waveguide, the electrical conductivity of the metal of the guide walls results in the parallel
components of electric field, and the perpendicular components of time-varying magnetic field,
vanishing at the walls.
 In a wave propagating in free space in a single direction only (a "plane wave") the magnetic field,
electric field, and direction of travel, are all mutually at right angles to each other. Thus to satisfy
the boundary conditions in our rectangular guide it is necessary for there to be more than one
propagating plane wave giving rise to the field patterns.
 If we regard propagation as being at an angle to the guide walls, there are three components of
propagation constant; namely, two transverse and one along the guide. The repetition distance or
wavelength associated with the propagation along the guide is called the "guide wavelength"
lambdag. The wavelength associated with the vector resultant of these three orthogonal
propagation constants is just the free space wavelength lambda.
 In order for the fields to vanish at the guide walls there must be standing wave patterns in the
transverse propagation. The transverse propagation constants must have wavelengths or repetition
distances such that an integral number of half-wavelengths fit inside the lateral guide dimensions.
If a is the longest lateral dimension, then 2a must be the longest transverse guide repetition
distance or wavelength, and the reciprocal of this times 2 pi is the associated transverse
propagation constant.
 Adding the components by Pythagoras theorem, and cancelling the factors of 2 pi, we arrive at
the formula stated. It is not necessary, for the fundamental mode, to have standing waves patterns
in both lateral directions at once, for we are allowed electric fields normal to guide walls, and
changing magnetic fields parallel to guide walls.
 The equivalent formula for the TE11 mode is




1/(lambdag)^2 = 1/(lambda)^2 - 1/(2a)^2 - 1/(2b)^2
If a = 3 cm and b = 1.2 cm, calculate the cutoff frequencies of the TE10, TE01, and TE11 modes,
assuming that the waveguide is filled with air. Calculate the guide wavelength, the phase velocity, and the
group velocity, for propagating signals having a frequency 100MHz above the TE10 mode cutoff
frequency. [35%]
 At the cutoff frequency lambdag increases without limit and the waves bounce laterally across the
guide without making progress along it. Using the relation [1/(lambdag)] = 0 we find the
following free space guide wavelength cutoffs:

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TE10 = 6 cm
TE01 = 2.4 cm
TE11 = sqrt(1/(1/36 + 1/5.76)) = 2.228 cm
using (frequency times wavelength) = 3 times 10^10 cm per sec,
the velocity of microwave radiation in air, we find cutoff
frequencies of 5GHz, 12.5GHz, and 13.46GHz respectively.
The signal frequency for this problem is therefore 5.1GHz and the corresponding free space
wavelength is 3/0.51 = 5.8824 cm. Using the waveguide formula we find that lambdag = 29.85
cm. Multiplying this by the signal frequency 5.1 GHz we find the guide phase velocity 1.522
times 10^11 cm per second, and using the result given in the following part of the question the
guide group velocity is 5.912 times 10^9 cm per second.
Give a derivation of the relationship: (guide phase velocity)x(guide group velocity) = (velocity of light in
waveguide medium)^2. [25%]
 Given the angular frequency omega, and the guide propagation constant beta, the phase velocity
which is frequency times guide wavelength is easily seen to be omega/beta. The group velocity is
d(omega)/d(beta), the derivation of which is not needed here, and expressing the waveguide
formula given in the earlier part of this question in terms of omega and beta, having translated the
free space wavelength lambda into frequency by using c = frequency times free space
wavelength, the result follows transparently on differentiation.
 It is also possible to use a geometrical construction showing the propagation directions in the
waveguide as being angle alpha to the guide wall. Inspection shows that the phase velocity is
given by c/cos(alpha) and the group velocity by c times cos(alpha) from which the result also
follows transparently.
There was a wide spread of quality in the written answers. Most people managed to write a significant
amount on the questions they attempted. Question 4 was popular, being largely bookwork. There were not
many good answers to the s-parameter question 2 particularly on the descriptions of components, and the
problem, although the bookwork part of the question was quite well handled. Question 1 caused
difficulties for some people who didn't realise that what isn't reflected from the antenna has to be
transmitted. Many people transposed the reflected and transmitted powers here. The precision of some
people in giving definitions needed improvement.
Most people seemed a lot happier with descriptive answers than with answers requiring a little thought
and calculation. There were some over-long answers to question 3 which must have taken time away from
attempts at other questions. Incidentally, there are three sections of phase shift to get round a three port
circulator, not two.
Finally, well done everyone in what is generally accepted to be a difficult subject. I promise microwaves
will become clearer to you all in the future as the ideas sink in deeper. Remember you only had 13 weeks
from your first introduction to the subject until you were asked to write an exam. Remember also the
objective of an exam is not primarily to "grade" people but to fix the ideas more firmly in memory by
generating a little stress.
Question 22
Define the terms "Velocity factor", "Characteristic Impedance", "Return Loss", "Complex reflection
coefficient" and "Phase delay" for waves travelling on a lossless transmission line. [25%]
A certain coaxial line is lossless, and has inductance L Henries/metre and capacitance C Farads/metre.
Give expressions for the characteristic impedance in ohms and the velocity factor (dimensionless) for this
line. Calculate the inductance and capacitance of a 1 metre length of this line if it has impedance 50 ohms
and velocity factor 0.6. [25%]
Calculate the complex reflection coefficient and return loss (dB) for reflections at a load impedance
75+j13 ohms connected to this 50 ohm transmission line. State the complex reflection coefficient in real
and imaginary parts, and also as modulus and phase angle. How does the complex reflection coefficient
transform as the reference plane at which it is measured moves towards the generator? [25%]
For the load impedance above, and by using the SMITH chart, estimate the position (in wavelengths from
the load) of the reference plane at point P at which the real part of the transformed admittance is equal to
the line characteristic admittance of 0.02 Seimens.. Estimate the length of a shunt short circuit stub that
would provide a matching network if attached to the transmission line at point P. [25%]
 Outline solution 1.
 The velocity factor is the ratio of the speed of waves on the transmission line to the speed c of
light in vacuum, where c = 3E8 metres/sec. The velocity factor is a dimensionless number less
than one. The characteristic impedance is the impedance of an infinitely long length of
transmission line, or of a finite length of line at times before any reflection has arrived back at the
measuring source. The return loss is the number of dB by which the return wave power is less
than the forward wave power. The complex reflection coefficient is the complex ratio of return
wave amplitude to forward wave amplitude. The phase delay is the amount of phase shift suffered
by the wave in travelling along a certain length of transmission line. [25%]
 The characteristic impedance Zo is given by sqrt(L/C), and the wave velocity by the quantity
1/(sqrt(LC)). By the definition of velocity factor above, we have to divide the wave velocity
1/(sqrt(LC)) by the velocity of light c in vacuum. Thus the velocity factor eta =
[1/(sqrt(LC))]*[1/3E8] and is dimensionless. If Zo = 50 and eta = 0.6 we readily calculate L/C =
2500 and LC = 3.09E-17 so that L = 278 nH and C = 111 pF for a 1 metre length of line. [25%]
 The load impedance is 75 + j13 ohms, so the normalised load impedance for a 50 ohm line is
75/50 + j 13/50 ohms, or 1.5 + j0.26. Call this normalised impedance zL. Then the complex
reflection coefficient gamma is given by the formula gamma = (zL-1)/(zL+1) =
(0.5+j0.26)/(2.5+i0.26) = 0.21 + j0.08 or in polar form 0.22 angle 21.54 degrees. The return loss
is given by -20 log[10](0.22) or 13dB to the nearest dB. As the reference plane is moved towards
the generator the modulus of the complex reflection coefficient stays the same (assuming the line
is lossless), but the phase angle is reduced by 360 degrees for each half-wavelength of
displacement of the reference plane from the load. [25%]
 The normalised load admittance yL = 1/zL gives us the same complex reflection coefficient
gamma. yL calculates to be 1/(1.5+j0.26) = 0.65-j0.11. The expression for the transformed
admittance y'L in terms of the transformed complex reflection coefficient gamma' at P is y'L =
(1+gamma')/(1-gamma') A SMITH chart plot gives the point P at 0.17(5) wavelengths towards
the load, where the real part (conductance) of the transformed normalised admittance is 1 and the
residual normalised susceptance as +0.48. A further SMITH chart plot gives the length of a
shorted stub needed to make a normalised susceptance of -0.48 as about 0.18 wavelengths [25%]
Question 23.
State the electromagnetic boundary conditions that constrain electromagnetic wave propagation at the
interface between air and a perfect conductor. Define what is meant by the term "transverse
electromagnetic wave", and give a diagram showing the directions of propagation, electric field, and
magnetic field in such a wave. Give another diagram showing the orientation of a stack of metal plates
spaced an arbitrary distance apart which may be introduced into this transverse electromagnetic wave
without disturbing it. [30%]
Give a qualitative description of how the boundary conditions are satisfied for propagation in a
rectangular hollow metal waveguide of cross sectional dimensions A metres by B metres, with A > B. If
the large dimension A is 1 cm calculate the lowest mode cutoff frequency of this guide, if it is air-filled.
[30%]
A certain researcher proposes a scheme to make a waveguide for use over the frequency range 110GHz to
140GHz, using rectangular metal pipe filled with a substance of relative dielectric constant 9 at these
frequencies. Suggest suitable waveguide dimensions, and calculate the guide wavelength at a frequency
of 125GHz. Comment on the probable sources of attenuation in such a waveguide, and estimate the
surface roughness (in microns) of the internal guide walls that it would be possible to tolerate. [40%]
 Outline solution 2.
 The electric fields meet the perfect conductor walls at right angles. There can be no component of
electric field parallel to the metal wall, close to the wall. The time-varying magnetic fields are
parallel to the metal walls but have no perpendicular-to-the-wall component. The direction of
magnetic field is at right angles to the local surface current on the metal walls. In a transverse
electromagnetic wave, the electric field vector and magnetic field vector are both at right angles
to each other and to the direction of travel of the wave. A stack of metal sheets may be placed so
that the electric field is at right angles to the surfaces of the plates, and the magnetic field parallel
to the surfaces of the plates, without violating the boundary conditions. [30%]
 In a rectangular pipe waveguide the electric fields can run normal to the A faces and the magnetic
fields can lie parallel to the A faces thereby satisfying the boundary conditions on the A faces.
Standing waves can exist with a transverse component of propagation between the B faces, which
can be regarded as short circuit ends of transmission line, without requiring the fields to be zero
everywhere within the guide. The lowest mode cutoff frequency happens when a half wavelength
of free space radiation just fits into the longest transverse dimension of the guide. In the case here
that gives a free space wavelength of 2 cms which occurs at a frequency of 15GHz, which is
therefore the cutoff frequency of a 1cm guide. [30%]
 If the relative dielectric constant is 9, the wave velocity in the material filling the guide is slower
than the speed of light in vacuum by a factor of sqrt(9) = 3. If we assume the cutoff frequency of
the required guide is 100GHz, in free space the wavelength would be 3 mm and in the dielectric
the wavelength would be 1 mm. The guide cross section dimensions could therefore be 1/2 mm
by 1/4 mm, or 500 by 250 microns. Attenuation would be provided by dielectric loss, and by the
resistance of the guide walls since the skin depth would be small. At 125 GHz the free space
wavelength in the dielectric would be 1/1.25 = 0.8 mm, and using the waveguide formula we
discover the guide wavelength to be 1/(sqrt[(1/(0.8)^2 - 1] = 1.33 mm. It would be appropriate to
keep the surface roughness below about 1/100 of a wavelength, so we would like a surface polish
to better than 10 microns. [40%]
Question 24.
Define the term "scattering matrix" and state carefully the quantities related by this matrix. For a 2-port
network state how many elements there are in this matrix. Define the term "reciprocity" and state the
maximum number of independent elements there can be for a reciprocal scattering matrix used to describe
a 3-port junction. [30%]
Give a brief description of how a ferrite displacement isolator works. A certain isolator has forward
insertion loss of 0.5dB and reverse insertion loss of 11.5dB. It is placed between a generator and a short
circuit load impedance. Calculate the minimum value of normalised resistive impedance which can be
seen by the generator in this arrangement. [40%]
With the aid of a diagram, show how four 3-port ferrite circulators may be made to combine the output
from four high power amplifiers to feed a common antenna. Assume the amplifiers operate on different
channels with spacing of channels equal to the channel widths. Explain how it may be arranged that no
amplifier feeds power into any of the others, and explain what happens to reflected power from the
antenna if it is mismatched by placing an obstacle in its near field region. [30%]
 Outline solution 3.
 The scattering matrix for an N port network consists of N*N complex numbers which are
dimensionless. Let us call these numbers sij where the suffices i and j run over 1-N each. Suppose
there is an input wave amplitude aj on the jth port, and we measure the output on the ith port with
all the other ports terminated. Reference planes are defined for each port at which the input and
output wave amplitudes are defined. Then the element sij is numerically equal to the complex
ratio of the output complex amplitude bi to the input complex amplitude aj. For a 2-port network
there are 4 elements, s11, s12, s21, s22. The term reciprocity applies to most networks for which
the scattering parameters are the same under interchange of the ports. Thus for a reciprocal 2-port
network, s12=s21. In practice this means that the insertion characteristics between ports i and j
are independent of the direction of travel of the wave between i and j. For a reciprocal 3-port
junction we can have 6 independent s parameters, s11, s22, s33, s12, s23, s13. [30%]
 A ferrite isolator contains magnetically biased insulating magnetic material called ferrite. The
electron spins in the ferrite precess around the direction of the applied magnetic bias field in a
certain direction. The local rf magnetic field in the forward and backward waves rotate either in
the same direction as the precessing electrons, or in the opposite direction. This gives the ferrite
an effective permeability that differs for forward and backward propagating waves. The field
distributions in the waveguide have different patterns for the forward and backward waves ("field
displacement") and a resistive card placed in the guide will attenuate forward waves differently
from backward waves. In the problem, the round trip loss is 12 dB and so the return wave
amplitude is 1/3.98 times the amplitude of the incident wave, since the magnitude of the
reflection at the short is unity. Thus the minimum value of normalised resistive impedance is
given by [1-1/3.98]/[1+1/3.98] = 0.60. [40%]
 Each high power amplifier (HPA) directs its output through an associated bandpass filter which
has a stop band for the frequencies of all the other amplifiers, into a circulator tree of 3-port
circulators. The bottom circulator passes its HPA signal to the next one up, and so on, and any
return power to the bottom circulator is absorbed in a matched termination on its spare port. The
top 3-port circulator has its output port connected to the antenna. The bandpass filters are perfect
reflectors for signals in their stop bands. Thus the signal from the bottom HPA proceeds via each
circulator, rflected from the other filters, to the antenna. Clearly no power from the lower HPAs
enters any of the higher HPAs because it is prevented by the stopbands of the associated filters.
Return power from the antenna is passed directly to the termination at the bottom of the tree.
[30%]
Question 25.
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 transmitter-antenna 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 E-plane 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 free-space 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%]
Question 25.
a)
Define the terms "characteristic impedance", "wave velocity", "velocity factor", "reflection coefficient"
and "voltage standing wave ratio VSWR" for waves on a coaxial transmission line.
[25%]
b)
Derive a formula for the VSWR in terms of the magnitude of the reflection coefficient. Explain, with a
formula, why and how the impedance measured at the generator end depends both on the normalised load
impedance ZL and the frequency f.
[30%]
c)
A certain coaxial cable has loss 0.4 dB per wavelength at a frequency of 100 MHz. The cable has velocity
factor 0.6 and characteristic impedance 50 ohms. It feeds an antenna load of 75 + j 25 ohms. If the cable
is 10m long, estimate the VSWR at the mid point of the cable, and also at the generator end.
[30%]
d)
Estimate, using the SMITH chart, the impedance seen by the generator when connected to this system.
Calculate the return loss at the generator.
[15%]
Outline solution (1).
 Definitions.
o Characteristic impedance: equal to sqrt(L/C) for waves on a lossless transmission line
having distributed inductance L Henries/metre and distributed capacitance C
Farads/metre. It is the ratio of voltage to current in a forward travelling wave on the line;
it is equal to the input impedance of a very long length of line for times shorter than the
time it takes the reflections to return from the far end. It is that impedance which, when
used as a load, gives rise to no reflected wave, or a perfect match.
o Wave velocity. On a coaxial transmission line the waves propagate as TEM modes. The
wave velocity is therefore either the phase velocity or the group velocity; it is the speed
(and direction) with which the wave crests travel along the coaxial cable, and it is also the
speed with which the energy and the modulations travel.
o Velocity factor: This is a dimensionless number, less than unity, which is the ratio of the
speed of waves on the coaxial line to the speed of light in vacuum, 3E8 metres per
second. Typically the velocity factor for coax lies between 0.55 and 0.8; it is closely
given by 1/sqrt(epsilon) where epsilon is the "effective relative dielectric constant" of the
insulator forming the line spacer.
o Reflection coefficient: A complex dimensionless number which describes, at a given
point along the transmission line called the "reference plane", the ratio of return wave
amplitude to forward wave amplitude. The phase of the reflection coefficient changes as
the reference plane is moved along the transmission line. On a lossless line, the
magnitude of the reflection coefficient is the same everywhere along the line. There are
important special cases of the reflection coefficient when the reference plane coincides
with the terminals of the load impedance, or the generator terminals.
o


VSWR: The Voltage Standing Wave Ratio is a dimensionless number representing the
ratio of the maximum voltage of the standing waves on the transmission line, at a
reference plane where the forward and backward waves are in phase, to the minimum
voltage at a reference plane (lambda/4 away from the maximum) where the forward and
backward waves are in anti-phase. The VSWR is a sensitive measure of the mismatch on
the line, and is closely related to the magnitude of the reflection coefficient.
[25%]
Calling the forward wave complex amplitude V+ and the backward wave complex amplitude V-,
the magnitude |gamma| of the reflection coefficient gamma is given by |V+| / |V-| so the VSWR is
given by
VSWR = (|V+| + |V-|)/(|V+| - |V-|) = (1 + |gamma|)/(1 - |gamma|)
At the load, the reflection coefficient is gammaL, say, and at the generator it is gammaG =
gammaL (exp{2j 2 pi d/lambda}) for a line of length d and waves of wavelength lambda. Using
the relationships
gammaL = (ZL-Zo)/(ZL+Zo) and gammaG = (Zg-Zo)/(Zg+Zo)
we can determine Zg in terms of the quantities ZL, Zo, lambda, and d by algebra.
[30%]
The wave velocity is 0.6*30 = 18 cm/nanosec. So at 100MHz = 0.1 GHz, the wavelength lambda
= 180 cm =1.8 metres. So the 10 metre cable is 10/1.8 wavelengths long. Total cable loss, one
pass, = 10*0.4/1.8 = 2.2dB. Round trip loss midpoint-load-midpoint = 2.2 dB. Round trip loss
generator-load-generator = 4.4 dB. Magnitude of reflection coefficient at load = 1/(sqrt[13]). (see
last part to this question).
2.2 dB loss is a factor 1.66 in power or 1.29 in amplitude.
4.4 dB loss is a factor 2.75 in power or 1.66 in amplitude.
At the mid point, |gamma| = (1/sqrt[13])/1.29 so VSWR = 1.55
At the generator, |gamma| = (1/sqrt[13])/1.66 so VSWR = 1.40
[30%]

At the generator, the line length to the load is 5.5555 lambda or integer number of half
wavelengths plus 0.0555 wavelengths. Plot 1.5 + j 0.5, transform 0.0555 lambda towards the
generator, read off |gamma| = 0.28 so that the generator end reflection coefficient including the
loss is 0.17, read off Zg/Zo = 1.4 - j 0.05.
[15%]
Question 26.
a)
State the boundary conditions for microwave electromagnetic field components adjacent to a perfect
conductor. Explain the difference between Transverse Electric [TE] modes and Transverse Magnetic
[TM] modes, and show how these modes satisfy the boundary conditions in rectangular waveguide.
[30%]
b)
Determine the cutoff frequencies of the TE10, TE01, TM11, and TE11 modes in a waveguide filled with
a lossless dielectric of relative permittivity epsilon(r) = 6.5, for guide dimensions 1.8 cm by 0.8 cm. State
which of these modes are degenerate .
[30%]
c)
List any advantages and disadvantages of using dielectric filled waveguide, compared to air-filled.
[20%]
d)
Estimate the power handling capacity of this guide for the TE10 mode if the dielectric breakdown
strength is 10^7 V/m
[20%]
Outline solution (2).
 The parallel component of E and the normal component of dH/dt must be zero adjacent to a
perfect conductor. So the E field always meets a conductor at right angles, and a changing H field
always lies parallel to a conductor surface adjacent to the surface. In a TE mode there is a
longitudinal H field, and in a TM mode there is a longitudinal E field. Here, longitudinal is taken
to mean in the direction of propagation along the axis of the waveguide. A sketch of mode
patterns is needed here, see the waveguide notes .
 The relative dielectric constant is 6.5, so the wave velocity interior to the guide is c/(sqrt(6.5) =
1.18E8 metres per second. The cutoff wavelength for TE10 is twice the broad guide dimension,
or 3.6 cm so the cutoff frequency for the TE10 mode is 0.118/(0.036) GHz = 3.27 GHz. The
cutoff wavelength for the TE01 mode is twice the narrow guide dimension, so the cutoff
frequency works out to be 7.36GHz by a similar argument. The TE11 and TM11 modes are
degenerate and share a cutoff frequency of 8.05 GHz.
[30%]
 Advantages of dielectric-loaded waveguide are; increased breakdown field strength and power
handling capacity, and reduced size for a given frequency of use. Disadvantages include
fabrication problems, dielectric loss leading to attenuation, material dispersion as well as
geometric dispersion, and possible arcing at the wall-dielectric interfaces.
[20%]
 Within the TE10-only propagation band, we assume the group velocity is (1.18E8)/2 metres per
second. The energy density per unit length of the guide is of the order (epsilon0)(6.5)(E^2)A
where A is the guide cross-section in square metres, E is the maximum r.m.s. electric field
strength in the wave, and epsilon0 = 8.85E(-12) Farads/metre is the permittivity of free space.
This energy density is of the order of 0.4 Joules per metre length of guide, so the maximum
power flow is 0.4*(1.18E8)/2 = 24 Megawatts.
[20%]
Question 27.
a)
Define the term scattering matrix and give examples of the scattering matrix values (s-parameters) for
each of the following devices:a lambda/3 length of lossless coaxial cable
a perfect 3-port circulator
a perfect 4-port magic tee in waveguide.
[40%]
b)
At its terminals, a certain antenna has s11 = - j0.2. Using the SMITH chart, or otherwise, determine the
driving point impedance at the input to a 50 ohm feeder of length 21.67 lambda which is connected to this
antenna. Assume negligible loss in the feeder.
[40%]
c)
Explain the construction of a single short-circuit stub match to an antenna. Indicate how the bandwidth of
the stub match might be maximised.
[20%]
Outline solution (3).
 An n-port microwave circuit has n input waves and n output waves. We relate the incoming wave
amplitudes ai to the outgoing wave amplitudes bi by the matrix equation bi = sij aj where the sij
are the (complex dimensionless) scattering parameters, which can be assembled into a matrix
called the scattering matrix . In matrix notation we can write this relationship as B = SA.
o
o
o


For a lambda/3 transmission line we have s11=s22 = 0. It is a 2-port network. The other s
parameters are given by s12=s21 = exp{-j 2 pi/3}; they have unity magnitude and
negative phase parts.
For a perfect 3-port circulator with the sequence 1->2->3->1.... we have s11=s22=s33 = 0
and forward transmission s21=s32=s13 = 1 exp {j -theta} with theta some phase angle
depending on the construction. The other s parameters s12,s23, s31 are all zero as there is
no reverse propagation around the circulator.
For a perfect 4-port magic tee, the ports may be numbered (1 and 4) straight through, 2
H-plane arm, and 3 E-plane arm. Then all the scattering parameters have magnitude
1/sqrt(2), and the matrix may be written as 1/sqrt(2) times {
s11=s22=s33=s44=s14=s41=s23=s32 = 0, s21=s31=s13=s12 =s43=s34=1, s24=s42 = -1}
[40%]
The magnitude of s11 is 1/5 and the angle -90 degrees. The length of the line is (round trip
distance) 2*(-0.17)*360 degrees (we have subtracted 43 whole wavelengths from the round trip
phase shift) so the input reflection coefficient gamma has magnitude 1/5 and phase angle (-90122.4) degrees, so Zin/Zo = [1 + gamma]/[1 - gamma] = 0.6967 + j 0.1556 so Zin = 34.8 + j7.78
ohms.
[40%]
For this part, see the stub matching notes
[20%]
Question 28.
a)
Explain the term isotropic radiator. How do practical antennas differ from the isotropic ideal? Define the
terms directivity and gain of an antenna and give an expression for the efficiency of a front feed
Cassegrain reflector antenna.
[30%]
b)
A formula for the gain gi (dBi) of a reflector antenna having effective area Ae square metres is
gi = 10 log[10](4 pi Ae) - 20 log[10](lambda)
for radiation of wavelength lambda metres. Explain why the actual reflector physical area is larger than
Ae and estimate the actual area needed for gi = 48 dBi at 13 GHz. State any assumptions.
[20%]
c)
If the radiated power is assumed to be distributed uniformly within a cone of semi-angle theta, determine
theta if gi = 48 dBi.
[20%]
d)
Explain why sidelobes can arise in practical dish antennas, and state how the sidelobes may be reduced.
Identify any compromises to the antenna performance involved in reducing sidelobes.
[30%]
Outline solution (4).
 Definitions.
o The term isotropic refers to a hypothetical antenna which radiates uniformly in every
direction in space. Since electromagnetic waves are transverse, an isotropic antenna can
only be approximated. Practical antennas have at least two directions along which they
cannot radiate; these directions are along a weighted average direction of current flow in
the antenna source region.
o
o
o
o
Directivity: radiated power density in a certain direction at a certain distance, divided by
the radiated power density at the same distance, from an isotropic source having the same
total integrated radiated power as the antenna in question.
Gain: the same as directivity but allowing for antenna loss in the near field and the
radiating structure. The isotropic source now has the same total radiated power as that
accepted by the antenna in question from its feeder.
The efficiency of an antenna is the proportion of total accepted input power that is
ultimately radiated. Numerically, the efficiency = (the gain) / (the directivity) and it
doesn't matter which directions are taken (providing they are the same) for establishing
the gain and the directivity.
Sources of loss in an antenna include absorbing objects in the near field, resistive loss in
the current-bearing structures, and spillover from the feed to a reflector antenna.
[30%]
The (effective area)/(the actual area) = (the aperture efficiency) and the aperture efficiency may
be about 70-85% for a typical Cassegrain reflector antenna. Assuming an aperture efficiency of
80%, and using the formula G = (4 pi Ae)/(lambda^2) for the gain of the antenna in terms of the
effective area Ae, for this case we find G = 10^(48/10) = 63096, and the wavelength at 13 GHz is
0.03/1.3 metres = 0.0231 metres so that Ae = 2.68 square metres and the actual area is Ae/(0.8) =
3.35 square metres.
[20%]
 The gain is approximately equal to the solid angle of the whole sphere (4 pi steradians) divided
by the solid angle of the cone of radiation (pi theta^2) whence theta = 0.0080 radians = 0.456
degrees is the cone semi-angle.
[20%]
 Sidelobes arise mathematically from the Fourier transform of a uniformly illuminated aperture.
They may be regarded as the rings of a diffraction pattern formed by a uniformly illuminated
aperture having the shape of the reflector. They may be reduced by using aperture illumination
taper, where the fields are reduced at the edges of the reflector from their values at the centre. For
minimum sidelobes, the taper should approximate to a Gaussian. The sidelobes are only reduced
by the trade-off of a wider main beam, so the antenna needs to be larger for a given beamwidth.
[30%]
Question 29
(a)
Define the terms "characteristic impedance", "wave velocity", "velocity factor", "dispersion", "complex
reflection coefficient".
[10%]
(b)
Give a formula that relates the velocity factor and characteristic impedance of a transmission line to the
inductance and capacitance of a 1 metre length of line.
[10%]
(c)
Calculate the inductance and capacitance of a 10 metre length of 300 ohm parallel wire ribbon cable
which has velocity factor 0.9
[20%]
(d)
A transmission line junction is formed from four microstrip lines, all having the same characteristic
impedance, and connected together in the form of a cross. Power is input to port 1 and ports 2,3 and 4 are
all terminated in matched loads. Evaluate the size of the reflection coefficient at port 1. Determine how
many dB of signal loss occurs between the in-port and one of the out-ports.
[40%]

(e)
Describe qualitatively, giving reasons, what happens to a rectangular pulse of current propagating along a
uniform but lossy dispersive microstrip transmission line.
[20%]
Outline solution (1).
Definitions.
 Characteristic impedance: equal to sqrt(L/C) for waves on a lossless transmission line having
distributed inductance L Henries/metre and distributed capacitance C Farads/metre. It is the ratio
of voltage to current in a forward travelling wave on the line; it is equal to the input impedance of
a very long length of line for times shorter than the time it takes the reflections to return from the
far end. It is that impedance which, when used as a load, gives rise to no reflected wave, or a
perfect match.
 Wave velocity. On a coaxial transmission line the waves propagate as TEM modes. The wave
velocity is therefore either the phase velocity or the group velocity; it is the speed (and direction)
with which the wave crests travel along the coaxial cable, and it is also the speed with which the
energy and the modulations travel.
 Velocity factor: This is a dimensionless number, less than unity, which is the ratio of the speed of
waves on the coaxial line to the speed of light in vacuum, 3E8 metres per second. Typically the
velocity factor for coax lies between 0.55 and 0.8; it is closely given by 1/sqrt(epsilon) where
epsilon is the "effective relative dielectric constant" of the insulator forming the line spacer.
 Dispersion: this is a technical term for the case where the phase velocity of waves depends on
frequency or wavelength, and a superposition (Fourier) of waves "disperses" in distance as it
travels along the line.
 Complex reflection coefficient: A complex dimensionless number which describes, at a given
point along the transmission line called the "reference plane", the ratio of return wave complexamplitude to forward wave complex-amplitude. The phase of the reflection coefficient changes as
the reference plane is moved along the transmission line. On a lossless line, the magnitude of the
reflection coefficient is the same everywhere along the line. There are important special cases of
the reflection coefficient when the reference plane coincides with the terminals of the load
impedance, or the generator terminals.
[10%]
 Calling the velocity factor eta and the velocity of light c ; with the capacitance of the cable C
Farads per metre and the inductance of the cable L Henries per metre we have
Zo = characteristic impedance = sqrt (L/C)
velocity of waves on the line = (eta c) = 1/sqrt(LC)
so L = Zo/(eta c) and C = 1/(Zo eta c) by simple algebra.
[10%]
 If Zo = 300 ohms, eta = 0.9 and c = 3E8 then for 10 metres of cable the inductance L is 11.1
micro-Henries and the capacitance C is 123 pF.
[20%]
 At the microstrip junction, the three outgoing terminated transmission lines each present a shunt
impedance Zo. There are thus three Zo impedances in parallel at the end of the input line, and
they form a load impedance ZL = (1/3)Zo. The reflection coefficient at the junction is therefore
[1/3 - 1]/[1/3 + 1] or -1/2. The amount of power reflected is therefore [-1/2][-1/2] or 1/4, and so
3/4 of the incident power is transmitted. By symmetry, it has to divide equally between the three
outgoing transmission lines and so the transmitted power into each load is 1/4 of the incident
power, and the loss is 6dB.
[40%]
 Wave shape on lossy dispersive microstrip. The rectangular pulse will spread out and diminish
with distance away from the generator. The wave shape may become oscillatory for sufficiently
large distances and strong dispersion. The integrated energy in the wave will become less as the
wave propagates.
[20%]
Question 30
(a)
Give a description of the formalism of a scattering matrix representation of a two-port microwave circuit,
stating clearly what each of the scattering parameters represents physically.
[30%]
(b)
A certain microwave component has the following s-parameters
s11 = 0.1 angle -30 degrees
s22 = 0.3 angle -60 degrees
s21 = 8.4 angle -120 degrees
s12 = 0.05 angle -90 degrees
State what kind of component is represented by these s-parameters.
[10%]
(c)
The component is supplied on port 1 with 10mW power level while port 2 is connected to a matched load.
Determine the power output on port 2.
[10%]
(d)
This component is embedded in a 50 ohm transmission system. Determine the input impedance at port 1
if port 2 is
 (i) terminated by an impedance of 50 ohms
 (ii) short circuited
[30%]
(e)
Write down the s-matrix of a perfect microwave 3-port circulator 1>2>3>1 with 120 degrees phase delay
between successive ports.
[20%]
Outline solution (2).
 The S-matrix links incoming complex wave amplitudes a to outgoing complex wave amplitudes b
. The incoming power on port i is (ai)(ai)* = the squared modulus of ai, and similarly for the b
amplitudes. In matrix notation
















b1 | s11 s12 | a1
=|
|
b2 | s21 s22 | a2
or
b =
S
a
or fully
b1 = s11 a1 + s12 a2
b2 = s21 a1 + s22 a2
Now, if port 1 is driven and port 2 is terminated then a2 = 0 and b1 = s11 a1, and b2 = s21 a1.
so |s11|^2 is the proportion of input power reflected, and |s21|^2 is the proportion of input power
transmitted. The phase angles of s11 and s21 represent the phase shifts on reflection and
transmission, respectively.
Similarly for port 1 terminated and port 2 driven.
[30%]
 The component is an amplifier as the magnitude of s21 is larger than unity and the reverse
transmission s12 is small.
[10%]
 For 1 milliwatt of input power, the output power delivered to the matched load on port 2 is
8.4*8.4 milliwatts, from the property above. This is 70.56 milliwatts. Thus for 10mW of input
power, the power delivered to the load will be 0.7 watts approximately.
[10%]
 Now Zo is 50 ohms. If port 2 is terminated then b1 = s11 a1 and the reflection coefficient at port
1 is s11. Therefore Zin/Zo = (1+s11)/(1-s11) so Zin = 59.7 - j 6.0 ohms.
[15%]
 If there is a short circuit on port 2, the reflection coefficient from the short is -1 and so a2 = - b2
is the wave amplitude returned to port 2 from the reflection at the short.
Thus b1 = s11 a1 - s12 b2
and b2 = s21 a1 - s22 b2
where we have substituted (-b2) for a2 in the equations listed in the first part of the answer.
Solving these equations we obtain b1 in terms of a1 to get the reflection coefficient gamma at the
input port, which is given by
gamma = [s11 - (s12 s21)/{1+s22}]
and so, as before,
Zin = Zo[1+gamma]/[1-gamma] = 113 - j46 ohms on putting in the numbers.
[15%]
 For the three port circulator we have s11 = s22 = s33 = 0 as it it intrinsically matched on each
port, and we also have s12 = s23 = s31 = 0 as there is no reverse cyclic transmission, and we are
left with s21 = s32 = s13 = magnitude 1 angle -120 degrees.
[20%]
Question 31.
(a)
State the boundary conditions for electromagnetic fields at the interface between air and a perfect
conductor.
[10%]
(b)
A rectangular waveguide has cross section 4.8 cm (wide face) by 2.4 cm (narrow face). Determine the
cutoff frequencies for each of the following modes: TE10 TE01 TE20 TM11
[20%]
(c)
Which two of the modes above are degenerate? Give reasons.
[10%]
(d)
Determine the attenuation coefficient in dB/metre for the TM11 mode, at a frequency 100 MHz below its
cutoff.
[30%]
(e)
Explain, giving reasons, why waveguide is usually used at frequencies well away from mode cutoffs.
[15%]
(f) For the waveguide described above, state the range of frequencies over which it might best be used.
[15%]
Outline solution 3
 The parallel component of E, and the normal component of dH/dt must be zero adjacent to a
perfect conductor. So the E field always meets a conductor at right angles, and a changing H field
always lies parallel to a conductor surface adjacent to the surface.
[10%]
 For the "m,n" mode (whether TE or TM) we recall the waveguide formula
(1/lambdacutoff)^2 = (m/2a)^2 + (n/2b)^2
and we have frequency(cutoff) times lambdacutoff = 3E8 metres/sec
So we obtain
cutoff frequencies TE10 = 3.125GHz, TE01=TE20 = 6.25GHz, TM11 = 6.988GHz.
[20%]
 The TE01 and TE20 modes are degenerate; they have the same cutoff frequencies and the guide
wavelengths for each mode will be the same at any given frequency.
[10%]
 We now use the (remembered) formula
(1/lambdaguide)^2 = (1/lambda)^2 - (1/lambdacutoff)^2
and lambda * frequency = 3E8 metres per second
to find (1/lambdaguide)^2 = (f^2 - fc^2)/(3E8)^2.
as fc is greater than f, lambdaguide is imaginary and the attenuation coefficient alpha is given by
alpha = (2 pi)/(modulus of lambdaguide)
Again, putting in the numbers we find the attenuation alpha = 24.8 Nepers per metre = 215
dB/metre.
[30%]
 Close to cutoff, the group velocity of the propagating mode becomes very low and the waveguide
mode becomes highly dispersive. The evanescent (non-propagating but attenuating) modes
extend further physically from imperfections in the guide walls, or scattering objects, and
therefore can couple unwanted energy and signal along the guide. For this guide, we might
suggest 4 to 6 GHz as a sensible single-mode (TE10) range.
[30%]
Question 32.
(a)
Distinguish between the terms gain and directivity of an antenna.
[10%]
(b)
Explain why a dipole antenna necessarily has a maximum directivity which is greater than unity.
[10%]
(c)
An array antenna consists of two vertically orientated half-wave dipoles placed with feeds a distance one
wavelength apart and orientated along the line joining their feed points. In both azimuth and elevation
planes, sketch radiation patterns of (i) the dipoles considered as isolated elements, (ii) the array pattern of
two isotropes placed on the dipole centres, and (iii) the total radiation pattern of the antenna.
[50%]
(d)
Explain what is meant by the term "pattern multiplication".
[10%]
(e)
Determine the boresight directivity (in dBd) of the array antenna shown above, and determine the
directions of nulls in the radiation pattern.
[20%]
Outline solution (4).
 The gain of an antenna is defined as the ratio of the power radiated per unit solid angle (in a
certain direction, possibly along boresight if no direction is specified) to that power which would
be radiated by an isotropic antenna, having the same total accepted input power. The directivity
on the other hand refers the ratio of the power radiated per unit solid angle to the integrated power
radiated over all directions (but in the far field; that is, at long distances from the antenna
structure) by the antenna. It is a measure of the intrinsic "concentration of energy" properties of
the antenna, and does not take account of loss due to absorption in the near field, spillage,
blockage, or due to resistive ohmic dissipation in the antenna structure.
[10%]
A dipole antenna has nulls along the rod axes, and so to compensate must have directivity greater
than one in some other directions.
[10%]
 The dipole axes are vertical, and the azimuth plane is in the horizontal directions. Since the dipole
has rotational symmetry in the horizontal plane, there can be no preferred direction to the
radiation, and so the azimuth plots of both the element and the array patterns are omnidirectional
(that is, circles). In the elevation plane a single dipole has nulls along the rod axes, and a
maximum at right angles to the rod axes. The pattern looks like a "figure of eight" on its side
(rotated through 90 degrees from normal) and when combined with the azimuth pattern, the total
3-d pattern resembles a toroid (doughnut with hole).
For the two isotropes spaced lambda apart, the path difference is lambda/2 for a direction at 30
degrees to the azimuth plane. (arcsine 30 = 1/2). So the nulls in the elevation pattern are at 60
degrees to the axis along which the isotropes are spaced. There are lobe maxima at zero degrees,
90, 180 and 270 degrees.
When we combine the azimuth patterns by pattern multiplication, the total radiation pattern in the
azimuth plane remains circular (omnidirectional). However, the elevation pattern multiplication
puts an additional null along the rod axes into the array pattern, so there are now a total of 6 lobes
and 6 nulls as we rotate around in elevation.
Sketches would help and get marks here.
[50%]
 Pattern multiplication may be summarised algorithmically as
o choose a direction (azimuth, elevation)
o determine element directivity in this direction
o determine array factor directivity in this direction
o plot the product of these directivities (in this direction) on a 3d plot
o repeat for all (azimuth elevation) angles at reasonable intervals to derive the total antenna
radiation pattern.
[10%]
 The array factor gain is 2 (there are two elements) and the element factor gain is just that of a
single dipole. Therefore the total antenna gain is 3dBd with respect to a dipole, or 5.2 dBi with
respect to an isotropic source.
As derived above, the null directions are at 0 degrees 60 degrees 120 degrees 180 degrees 240
degrees 300 degrees with respect to the rod axes, in the elevation plane.
[20%]