94-95 microwave option solutions, DJJ questions.

Paper date Monday 24 April 1995, 2pm-5pm.

Outline solutions only, variations acceptable...

Question 1

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 2.

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 [1-Y] 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 post-multiplying 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 V- and 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 3.

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%]

We start our solution by calculating the three free space wavelengths for the problem, by dividing 30 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.

We now find the guide wavelengths from the waveguide formula (you can quote this from memory or 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.

Now when half a guide wavelength fits between the shorts 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.

The micrometer readings are therefore (i) 2.71 cm or 1.07 inches, (ii) 1.97 cm or 0.777 inches, and (iii) 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 4.

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 frequency-selective 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.

D.Jefferies

18 November 1996