Microwave Option paper 98-99 (DJJ questions)

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Question 1.

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.
  • 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.
  • 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.
  • 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 2.

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 3.

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.
  • 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 (-90-122.4) 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 4.

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

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.
  • 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.
  • 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%]