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 2

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....

·

· 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) s12=0.3exp(-j80)

s21=9.7exp(-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....

·

· 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 3.

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 4.

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:

· 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.

## Comments on the answers received.

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.