Microwave Option paper 97-98 (DJJ questions)

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Question 1.

Define the terms "Velocity factor", "Characteristic Impedance", "Return Loss", "Complex reflection coefficient" and "Phase delay" for waves travelling on a lossless transmission line. [25%]

A certain coaxial line is lossless, and has inductance L Henries/metre and capacitance C Farads/metre. Give expressions for the characteristic impedance in ohms and the velocity factor (dimensionless) for this line. Calculate the inductance and capacitance of a 1 metre length of this line if it has impedance 50 ohms and velocity factor 0.6. [25%]

Calculate the complex reflection coefficient and return loss (dB) for reflections at a load impedance 75+j13 ohms connected to this 50 ohm transmission line. State the complex reflection coefficient in real and imaginary parts, and also as modulus and phase angle. How does the complex reflection coefficient transform as the reference plane at which it is measured moves towards the generator? [25%]

For the load impedance above, and by using the SMITH chart, estimate the position (in wavelengths from the load) of the reference plane at point P at which the real part of the transformed admittance is equal to the line characteristic admittance of 0.02 Seimens.. Estimate the length of a shunt short circuit stub that would provide a matching network if attached to the transmission line at point P. [25%]

• Outline solution 1.

• The velocity factor is the ratio of the speed of waves on the transmission line to the speed c of light in vacuum, where c = 3E8 metres/sec. The velocity factor is a dimensionless number less than one. The characteristic impedance is the impedance of an infinitely long length of transmission line, or of a finite length of line at times before any reflection has arrived back at the measuring source. The return loss is the number of dB by which the return wave power is less than the forward wave power. The complex reflection coefficient is the complex ratio of return wave amplitude to forward wave amplitude. The phase delay is the amount of phase shift suffered by the wave in travelling along a certain length of transmission line. [25%]

• The characteristic impedance Zo is given by sqrt(L/C), and the wave velocity by the quantity 1/(sqrt(LC)). By the definition of velocity factor above, we have to divide the wave velocity 1/(sqrt(LC)) by the velocity of light c in vacuum. Thus the velocity factor eta = [1/(sqrt(LC))]*[1/3E8] and is dimensionless. If Zo = 50 and eta = 0.6 we readily calculate L/C = 2500 and LC = 3.09E-17 so that L = 278 nH and C = 111 pF for a 1 metre length of line. [25%]

• The load impedance is 75 + j13 ohms, so the normalised load impedance for a 50 ohm line is 75/50 + j 13/50 ohms, or 1.5 + j0.26. Call this normalised impedance zL. Then the complex reflection coefficient gamma is given by the formula gamma = (zL-1)/(zL+1) = (0.5+j0.26)/(2.5+i0.26) = 0.21 + j0.08 or in polar form 0.22 angle 21.54 degrees. The return loss is given by -20 log[10](0.22) or 13dB to the nearest dB. As the reference plane is moved towards the generator the modulus of the complex reflection coefficient stays the same (assuming the line is lossless), but the phase angle is reduced by 360 degrees for each half-wavelength of displacement of the reference plane from the load. [25%]

• The normalised load admittance yL = 1/zL gives us the same complex reflection coefficient gamma. yL calculates to be 1/(1.5+j0.26) = 0.65-j0.11. The expression for the transformed admittance y'L in terms of the transformed complex reflection coefficient gamma' at P is y'L = (1+gamma')/(1-gamma') A SMITH chart plot gives the point P at 0.17(5) wavelengths towards the load, where the real part (conductance) of the transformed normalised admittance is 1 and the residual normalised susceptance as +0.48. A further SMITH chart plot gives the length of a shorted stub needed to make a normalised susceptance of -0.48 as about 0.18 wavelengths [25%]

Question 2.

State the electromagnetic boundary conditions that constrain electromagnetic wave propagation at the interface between air and a perfect conductor. Define what is meant by the term "transverse electromagnetic wave", and give a diagram showing the directions of propagation, electric field, and magnetic field in such a wave. Give another diagram showing the orientation of a stack of metal plates spaced an arbitrary distance apart which may be introduced into this transverse electromagnetic wave without disturbing it. [30%]

Give a qualitative description of how the boundary conditions are satisfied for propagation in a rectangular hollow metal waveguide of cross sectional dimensions A metres by B metres, with A > B. If the large dimension A is 1 cm calculate the lowest mode cutoff frequency of this guide, if it is air-filled. [30%]

A certain researcher proposes a scheme to make a waveguide for use over the frequency range 110GHz to 140GHz, using rectangular metal pipe filled with a substance of relative dielectric constant 9 at these frequencies. Suggest suitable waveguide dimensions, and calculate the guide wavelength at a frequency of 125GHz. Comment on the probable sources of attenuation in such a waveguide, and estimate the surface roughness (in microns) of the internal guide walls that it would be possible to tolerate. [40%]

• Outline solution 2.

• The electric fields meet the perfect conductor walls at right angles. There can be no component of electric field parallel to the metal wall, close to the wall. The time-varying magnetic fields are parallel to the metal walls but have no perpendicular-to-the-wall component. The direction of magnetic field is at right angles to the local surface current on the metal walls. In a transverse electromagnetic wave, the electric field vector and magnetic field vector are both at right angles to each other and to the direction of travel of the wave. A stack of metal sheets may be placed so that the electric field is at right angles to the surfaces of the plates, and the magnetic field parallel to the surfaces of the plates, without violating the boundary conditions. [30%]

• In a rectangular pipe waveguide the electric fields can run normal to the A faces and the magnetic fields can lie parallel to the A faces thereby satisfying the boundary conditions on the A faces. Standing waves can exist with a transverse component of propagation between the B faces, which can be regarded as short circuit ends of transmission line, without requiring the fields to be zero everywhere within the guide. The lowest mode cutoff frequency happens when a half wavelength of free space radiation just fits into the longest transverse dimension of the guide. In the case here that gives a free space wavelength of 2 cms which occurs at a frequency of 15GHz, which is therefore the cutoff frequency of a 1cm guide. [30%]

• If the relative dielectric constant is 9, the wave velocity in the material filling the guide is slower than the speed of light in vacuum by a factor of sqrt(9) = 3. If we assume the cutoff frequency of the required guide is 100GHz, in free space the wavelength would be 3 mm and in the dielectric the wavelength would be 1 mm. The guide cross section dimensions could therefore be 1/2 mm by 1/4 mm, or 500 by 250 microns. Attenuation would be provided by dielectric loss, and by the resistance of the guide walls since the skin depth would be small. At 125 GHz the free space wavelength in the dielectric would be 1/1.25 = 0.8 mm, and using the waveguide formula we discover the guide wavelength to be 1/(sqrt[(1/(0.8)^2 - 1] = 1.33 mm. It would be appropriate to keep the surface roughness below about 1/100 of a wavelength, so we would like a surface polish to better than 10 microns. [40%]

Question 3.

Define the term "scattering matrix" and state carefully the quantities related by this matrix. For a 2-port network state how many elements there are in this matrix. Define the term "reciprocity" and state the maximum number of independent elements there can be for a reciprocal scattering matrix used to describe a 3-port junction. [30%]

Give a brief description of how a ferrite displacement isolator works. A certain isolator has forward insertion loss of 0.5dB and reverse insertion loss of 11.5dB. It is placed between a generator and a short circuit load impedance. Calculate the minimum value of normalised resistive impedance which can be seen by the generator in this arrangement. [40%]

With the aid of a diagram, show how four 3-port ferrite circulators may be made to combine the output from four high power amplifiers to feed a common antenna. Assume the amplifiers operate on different channels with spacing of channels equal to the channel widths. Explain how it may be arranged that no amplifier feeds power into any of the others, and explain what happens to reflected power from the antenna if it is mismatched by placing an obstacle in its near field region. [30%]

• Outline solution 3.

• The scattering matrix for an N port network consists of N*N complex numbers which are dimensionless. Let us call these numbers sij where the suffices i and j run over 1-N each. Suppose there is an input wave amplitude aj on the jth port, and we measure the output on the ith port with all the other ports terminated. Reference planes are defined for each port at which the input and output wave amplitudes are defined. Then the element sij is numerically equal to the complex ratio of the output complex amplitude bi to the input complex amplitude aj. For a 2-port network there are 4 elements, s11, s12, s21, s22. The term reciprocity applies to most networks for which the scattering parameters are the same under interchange of the ports. Thus for a reciprocal 2-port network, s12=s21. In practice this means that the insertion characteristics between ports i and j are independent of the direction of travel of the wave between i and j. For a reciprocal 3-port junction we can have 6 independent s parameters, s11, s22, s33, s12, s23, s13. [30%]

• A ferrite isolator contains magnetically biased insulating magnetic material called ferrite. The electron spins in the ferrite precess around the direction of the applied magnetic bias field in a certain direction. The local rf magnetic field in the forward and backward waves rotate either in the same direction as the precessing electrons, or in the opposite direction. This gives the ferrite an effective permeability that differs for forward and backward propagating waves. The field distributions in the waveguide have different patterns for the forward and backward waves ("field displacement") and a resistive card placed in the guide will attenuate forward waves differently from backward waves. In the problem, the round trip loss is 12 dB and so the return wave amplitude is 1/3.98 times the amplitude of the incident wave, since the magnitude of the reflection at the short is unity. Thus the minimum value of normalised resistive impedance is given by [1-1/3.98]/[1+1/3.98] = 0.60. [40%]

• Each high power amplifier (HPA) directs its output through an associated bandpass filter which has a stop band for the frequencies of all the other amplifiers, into a circulator tree of 3-port circulators. The bottom circulator passes its HPA signal to the next one up, and so on, and any return power to the bottom circulator is absorbed in a matched termination on its spare port. The top 3-port circulator has its output port connected to the antenna. The bandpass filters are perfect reflectors for signals in their stop bands. Thus the signal from the bottom HPA proceeds via each circulator, rflected from the other filters, to the antenna. Clearly no power from the lower HPAs enters any of the higher HPAs because it is prevented by the stopbands of the associated filters. Return power from the antenna is passed directly to the termination at the bottom of the tree. [30%]

Question 4.

Define the terms "isotropic radiator", "boresight direction", "directivity", "gain", and "E- plane radiation pattern" in the context of antenna design. Explain why it is impossible to construct an isotropic radiator in practice. [30%]

A certain transmitter has a final amplifier that delivers 10 kW to an antenna with 85% efficiency. The antenna has boresight directivity of 14dBi above an isotropic source. Calculate the r.m.s. electric field strength at a distance of 50 km from an ISOTROPIC source radiating 10 kW with 100% efficiency, and compare it with the field strength at this distance from the hypothetical transmitter-antenna combination described above. [40%]

Give a formula relating the effective area of an antenna to its boresight gain and to the wavelength of the radiated signal. Estimate the area of a horn aperture antenna required to give a gain of 14dBi at 13 GHz. If such a dish were used to transmit the 10kW signal described above, estimate the power flow in watts per square metre at a distance of 10m from this horn along boresight, and comment on the safety implications. [30%]

• Outline solution 4.

• An isotropic radiator is one that radiates uniformly in all directions, both in azimuth and elevation. The power density and field strengths radiated by an isotrope do not depend at all on the direction of radiation. For a directional antenna, if there is a single direction in which the radiated power is maximum, this is termed the boresight direction. The directivity of an antenna is a dimensionless number representing the ratio of the radiated power density on boresight to the radiated power density at the same place in space which would exist if radiated by an ideal isotrope having the same total radiated power. Often the directivity is given in dB. The gain is similarly defined, except that the powers referred to are input powers to the antenna. The gain is less than the directivity if the antenna is less than 100% efficient, due to resistive loss, absorption in the near field, spillover and blockage. The E-plane radiation pattern is a polar plot of either the power radiated or the field amplitude radiated as a function of direction in a plane containing the electric field vector of a linearly polarised antenna. Since there has to be a unique transverse polarisation direction for radiated transverse electromagnetic waves, one finds that this cannot be arranged for radiation in every direction from a point. Therefore isotropic radiators are impossible. Consider a radiator with E field everywhere directed North on a reference sphere. The direction is undefined on the polar axis. [30%]

• The surface area of a sphere of radius 50Km is 4*pi*50*50*10^6 square metres, and the 10 kW isotropic power is spread uniformly across this area at a power density of 0.32 microwatts per square metre. The power density p in a free-space wave is related to the rms electric field strength e by p = e*e/Zo where Zo is 377 ohms (120 pi ohms), the impedance of free space. From this we calculate e = 11 mV/metre approximately. The radiated power is 0.85*10,000 or 8500 watts. The antenna directivity of 14dBi increases the effective isotropic radiated power on boresight by a factor 10^1.4 or 25.12 so the e.i.r.p is 213.5 kW. The field strength on boresight at 50Km from our hypothetical antenna is therefore 11 * sqrt(21.35) = 50.75 mV/metre. [40%]

• The gain G and effective area A of an antenna are related by the formula G = 4*pi*A/(lambda)^2, where lambda is the wavelength of the radiation. At 13 GHz lambda = 3/1.3 cms = 2.31 cms and we found above that the 14dBi directivity is a factor of 25.12. From these, the area of the dish is A = 25.12*2.31*2.31/(4 pi) = 10.65 square cms. At 10m from the horn, in the far field region, the boresight power density is 25.12*10000/(4 pi 10*10) = nearly 200 watts per square metre, or 20 mW per square cm. This is comparable with the US allowed safety limits of 10mW/ square cm averaged over 6 mins. [30%]