Microwave Option paper 99-00 (DJJ questions)

navigation page

Question 1.

(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 complex-amplitude 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 2.

(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

(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

(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 3.

(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 4.

(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
  • choose a direction (azimuth, elevation)
    
  • determine element directivity in this direction
    
  • determine array factor directivity in this direction
    
  • plot the product of these directivities (in this direction) on a 3d plot
    
  • 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%]