DJJ's solns 95-96 Option paper

Question 1

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 2

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%]

To increase the fractional bandwidth we add more coupling holes, having the same total 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 3

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%]

See the answer to question 4 on 94-95 paper

Question 4

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]/[(1-g) + js] from which we find that the magnitude of the reflection coefficient is [(1+g)^2 + s^2]/[(1-g)^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 5

Define the terms "directivity", "gain", "efficiency", and "effective area" for an antenna array. Explain how the boresight gain of an antenna is related to its effective area and efficiency. [25%]

The directivity of an antenna is the ratio of the power directed along boresight to the power which would be radiated equally in all directions if the antenna was isotropic. Usually the directivity is assumed to be equal to the boresight gain assuming no losses in the antenna.
The gain of an antenna is usually specified in the direction of maximum radiation, the boresight, and is equal to the total power radiated in this direction assuming it was radiated isotropically (uniformly in all directions) divided by the input power to the antenna structure. The gain is less than the directivity because of antenna losses, spillage, blockage, and so on.
The efficiency of an antenna is the gain divided by the directivity. It is a measure of what proportion of the input power is usefully radiated.
The effective area of an antenna is numerically equal to the gain times 4 pi/(lambda^2), and is the area of a perfect antenna which had that particular value of gain. The effective area is usually (in the case of an aperture antenna) somewhat smaller than the physical area.

A square array of 4 by 4 elements consisting of lambda/2 dipoles each having a maximum gain 3dB is fed by 16 signals of adjustable relative phase and equal amplitudes. Assuming 90% efficiency, calculate the boresight gain when the signals are all in phase. [50%]

Let us call the amplitude of the individual signals driving each element A. Then the input power to an element is |A|^2 (modulus of A squared, or AA*). The efficiency is 90%, or 0.9, and the square root of this is 0.95 which represents how much of the individual amplitude A is radiated.
The gain of an individual element, in terms of power, is 2, or in terms of amplitude, is sqrt(2). This is 3dB.
When all the signals add up in phase, the boresight amplitude is A times sqrt(2) times 0.95 times 16 since there are 16 equally contributing elements.
However the total input power is 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.

D.Jefferies email

12th December 1997