KNEC Electromagnetic Fields Theory and Communication Systems — March/April 2023
Paper code: 2602/304. Programme: Diploma in Electrical and Electronic Engineering, Telecommunication Option, Module III. The paper contains eight questions in two sections. Answer any two from Section A and any three from Section B.
Revision notice: The questions are transcribed from the supplied KNEC paper. The worked responses are independently prepared revision guides and are not the official KNEC marking scheme. Where the source paper uses a diagram, the required labels and operation are described in text.
SECTION A — ELECTROMAGNETIC FIELDS THEORY
Answer any TWO questions from this section.
Question 1
(a) Differentiate between electric field intensity and electric flux density with respect to electrostatics. (2 marks)
Answer: Electric field intensity E is the force on a unit positive test charge, measured in N/C or V/m. Electric flux density D is electric flux per unit area normal to the field, measured in C/m2; in a linear medium D = εE.
(b) (i) State Coulomb’s law of electrostatics. (ii) Two point charges of 20 μC each are located in free space at (−4, 1, 0) and (4, 1, 0). Determine the electric field at (0, 0, 5) and the force on a 10 μC charge located at (0, 0, 5). All distances are in metres. (12 marks)
Answer: Coulomb’s law states that the force between two point charges is along the line joining them and has magnitude F = (1/4πε) |q1q2|/r2; like charges repel and unlike charges attract.
For either source charge, R = √42 m. The two x-components cancel. The field at P is
E = kq/R3[(4,−1,5) + (−4,−1,5)] = k(20 × 10−6)/423/2(0,−2,10).
Hence E(P) ≈ (0 ax − 62.9 ay + 314.5 az) kN/C. For q = 10 μC, F = qE ≈ (−0.629 ay + 3.145 az) N.
(c) A parallel-plate capacitor has a plate area of 20 cm2 and is separated by 15 mm. A voltage v = 15 sin ωt is applied across the plates. Obtain an expression for the displacement current within the capacitor. (6 marks)
Answer: A = 20 × 10−4 m2, d = 0.015 m and C = ε0A/d = 1.1805 × 10−12 F. Therefore id = C dv/dt = 1.1805 × 10−12(15ω cosωt) A = 17.71 × 10−12ω cosωt A.
Question 2
(a) Differentiate between a perfect conductor and a perfect dielectric. (2 marks)
Answer: A perfect conductor has infinite conductivity and zero electric field inside under electrostatic conditions. A perfect dielectric has zero conductivity, stores electric energy, and ideally has no conduction current.
(b) Define each of the following with respect to magnetic materials: (i) reluctance; (ii) remanence. (2 marks)
Answer: Reluctance is the opposition offered by a magnetic circuit to flux, ℜ = l/(μA), measured in AT/Wb. Remanence is the flux density left in a magnetic material when the magnetising field is returned to zero.
(c) With the aid of a circuit diagram describe the procedure for determining the B–H curve of a magnetic material by reversal method. (8 marks)
Answer: Place the test ring/core in a magnetising-coil circuit with a controllable DC current and a reversal switch. Measure current to obtain H = NI/l. Reverse the current in small steps and use a ballistic galvanometer or fluxmeter connected to a search coil to measure the change of flux; calculate B = Φ/A. Repeat through positive saturation, zero field, negative saturation and back, plotting B against H. The resulting loop gives saturation, remanence and coercive field.
(d) (i) Explain the Poynting theorem. (ii) The magnetic field intensity is H(z,t) = 50 sin(ωt − βz) A/m. Determine the total power passing through a cylindrical area of radius r = 70 mm on the x = 0 plane. (8 marks)
Answer: Poynting’s theorem is the electromagnetic statement of conservation of energy: the decrease of stored field energy plus the outward power flow equals the work done on charges. The Poynting vector is S = E × H W/m2. For a uniform free-space wave, E = η0H and the time-average power through area A = πr2 is Pav = η0H02A/2. With r = 0.07 m, H0 = 50 A/m and η0 ≈ 120π Ω, Pav ≈ 72.6 kW, assuming the plane is normal to propagation.
Question 3
(a) State three properties of electromagnetic waves in free space. (3 marks)
Answer: They are transverse waves; E, H and the direction of propagation are mutually perpendicular; E/H = η0; and they travel in free space at c ≈ 3 × 108 m/s.
(b) Define each of the following with respect to a conducting medium: (i) permittivity constant; (ii) permeability; (iii) impedance. (3 marks)
Answer: Permittivity ε measures the electric flux density produced by electric field, D = εE. Permeability μ measures the magnetic flux density produced by magnetic field, B = μH. Wave impedance η is the ratio E/H for the travelling wave.
(c) A 6 GHz plane wave propagates through a lossless medium with εr = 8, μr = 2 and E = 0.8 sin(ωt − βz)ay V/m. Determine: (i) wave velocity; (ii) wave impedance; (iii) wavelength; (iv) phase constant; (v) expression for H. (14 marks)
Answer: For a lossless medium, v = c/√(μrεr) = 7.50 × 107 m/s. η = η0√(μr/εr) = 120π/2 = 60π Ω ≈ 188.5 Ω. Thus λ = v/f = 0.0125 m, β = 2π/λ = 160π rad/m ≈ 502.7 rad/m, and H = E/η in the corresponding transverse direction: H = 0.00424 sin(ωt − βz)ax A/m for propagation in +z.
SECTION B — COMMUNICATION SYSTEMS
Answer any THREE questions from this section.
Question 4
(a) Draw a labelled block diagram of a high-level modulation AM transmitter and describe its operation. (8 marks)
Answer: The diagram should show the audio/message source, AF amplifier, carrier oscillator, modulator, RF power amplifier and antenna. In a high-level transmitter, the carrier oscillator and message are applied to the modulator at low level; the AM signal is then amplified by the RF power amplifier to the required output power before radiation. The carrier and two sidebands carry the information.
(b) A DSB-AM transmitter radiates 200 kW without modulation and 250 kW when a sinusoidal modulating signal is applied. Determine: (i) modulation index; (ii) power of each sideband; (iii) transmitter efficiency. (6 marks)
Answer: PT = Pc(1 + m2/2). Hence 250/200 = 1 + m2/2, giving m = 0.707. Total sideband power = 250 − 200 = 50 kW, so each sideband = 25 kW. Efficiency η = PSB/PT = 50/250 = 20%.
(c) With the aid of a circuit diagram describe the operation of a varactor diode-based FM modulator. (6 marks)
Answer: A reverse-biased varactor is connected in the tuned circuit of an oscillator. The modulating voltage changes the diode capacitance; since f = 1/(2π√LC), the oscillator frequency changes in proportion to the instantaneous modulating voltage while amplitude remains approximately constant. A bias network sets the centre frequency and RF coupling extracts the FM signal.
Question 5
(a) Define each of the following with respect to TV systems: (i) chroma; (ii) interlaced scanning. (2 marks)
Answer: Chroma is the colour information of a television signal, represented by hue and saturation. Interlaced scanning divides each frame into two fields, scanning alternate lines in successive fields to reduce flicker without doubling the channel bandwidth.
(b) Draw a labelled block diagram of a monochrome TV transmitter. (6 marks)
Answer: Show the camera, video pre-amplifier, sync separator/generator, blanking and mixing circuits, vision modulator, RF amplifier and antenna. The camera converts brightness into video; synchronising and blanking pulses are added; the composite signal modulates the vision carrier and is amplified for transmission.
(c) An HDTV standard has 1250 interlaced lines per frame and 30 frames per second. Determine: (i) line scan frequency; (ii) field scan frequency; (iii) time to scan one line. (6 marks)
Answer: Line frequency = 1250 × 30 = 37.5 kHz. With two interlaced fields per frame, field frequency = 2 × 30 = 60 Hz. Line period = 1/37,500 = 26.67 μs.
(d) A CCTV system consists of digital cameras, multiplexer, monitor screen, server, router and modem. State the function of each device. (6 marks)
Answer: Cameras capture and digitise scenes. The multiplexer combines several camera streams. The monitor displays live or recorded images. The server stores footage and provides management or viewing services. The router directs packets between the CCTV network and other networks. The modem converts signals for the access link, such as a broadband or cellular connection.
Question 6
(a) State the two types of signal losses in rectangular waveguides. (2 marks)
Answer: The principal losses are conductor or ohmic loss in the guide walls and dielectric loss in the material filling the guide.
(b) With the aid of a labelled diagram, describe the operation of a two-cavity klystron amplifier. (8 marks)
Answer: A diagram should show electron gun, buncher cavity, drift space, catcher cavity, focusing magnets and collector. The buncher cavity velocity-modulates the electron beam. Faster electrons catch slower electrons in the drift space, forming bunches. The bunches induce a larger RF voltage in the catcher cavity and transfer kinetic energy to the output signal; the collector receives the spent beam.
(c) A rectangular waveguide has internal dimensions of 2 cm × 4 cm and operates at 8 GHz. For the stated TE mode, determine the cut-off wavelength, waveguide impedance and group velocity. (8 marks)
Answer: Use the mode printed in the source figure and the rectangular-guide relations fc,mn = (c/2)√[(m/a)2 + (n/b)2], λc = c/fc, β = k√[1 − (fc/f)2], ηTE = η0/√[1 − (fc/f)2] and vg = c√[1 − (fc/f)2]. Substitute a = 0.04 m, b = 0.02 m and the exact TE indices shown on the supplied paper. This preserves the required calculation even where the scan’s mode subscript is not legible in OCR.
(d) State two merits of semiconductor microwave devices. (2 marks)
Answer: They are small, light and mechanically rugged, and they require low operating voltage and power. They also support high-frequency operation and can be integrated with control circuits.
Question 7
(a) Define each of the following with respect to satellite communication: (i) Effective Isotropic Radiated Power (EIRP); (ii) uplink frequency. (2 marks)
Answer: EIRP is the equivalent isotropic radiated power in the direction of maximum antenna radiation, EIRP = transmitter power × antenna gain, or PT(dBW) + GT(dB). Uplink frequency is the frequency used for transmission from an earth station to a satellite.
(b) With the aid of diagrams, describe satellite access modes: (i) FDMA; (ii) TDMA. (8 marks)
Answer: In FDMA, each earth station or carrier is assigned a separate frequency slot and stations transmit simultaneously. In TDMA, stations share the same carrier frequency but transmit in allocated, synchronised time slots. A correct diagram should show separate frequency channels for FDMA and sequential bursts in a repeating frame for TDMA.
(c) State the frequency ranges for the C and Ku satellite frequency bands. (4 marks)
Answer: Common engineering ranges are C-band approximately 4–8 GHz and Ku-band approximately 12–18 GHz. State the precise convention used by the course tables when writing the final examination response.
(d) A geosynchronous satellite operates at 4 GHz and radiates 2 W from an antenna of gain 12 dB. The receiving earth antenna has gain 40 dB. Determine: (i) free-space earth loss; (ii) power received at the earth’s surface. The orbit altitude is 36,000 km. (6 marks)
Answer: λ = c/f = 0.075 m and d = 36,000 km = 3.6 × 107 m. Free-space loss L = 20 log10(4πd/λ) ≈ 255.6 dB. PT = 10 log102 = 3.01 dBW, so PR = PT + GT + GR − L ≈ 3.01 + 12 + 40 − 255.6 = −200.6 dBW.
Question 8
(a) Define each of the following with respect to RADAR: (i) duty cycle; (ii) range. (2 marks)
Answer: Duty cycle is pulse width divided by pulse repetition period, or τ/T = τ × PRF. Range is the distance from the radar to the target, determined from echo delay by R = ct/2.
(b) Figure 1 shows a block diagram of a digital RADAR receiver. (i) State three advantages of digital RADAR over analogue RADAR. (ii) Explain the functions of DSP, PLL and LNA. (9 marks)
Answer: Digital radar offers flexible software processing, better repeatability and calibration, easier storage and integration with tracking algorithms, and improved clutter/interference rejection. The DSP digitises and processes the received samples to detect and estimate targets. The PLL locks an oscillator to a reference and provides coherent frequency/phase control. The LNA amplifies the weak antenna signal while adding minimal noise.
(c) (i) Define tracking with respect to RADAR. (ii) Describe phased-array tracking and angle tracking. (5 marks)
Answer: Tracking is continuously measuring target position and motion and maintaining the radar beam on the target. A phased array changes the relative phase of many antenna elements electronically to steer the beam without mechanically moving the antenna. Angle tracking compares signals from offset beams or antenna quadrants and uses the error to steer the beam towards the target.
(d) A RADAR system transmits pulses of 2 μs duration at a repetition rate of 1 kHz. Determine: (i) maximum range; (ii) minimum range. (4 marks)
Answer: Maximum unambiguous range Rmax = c/(2PRF) = 3 × 108/(2 × 1000) = 150 km. Minimum range Rmin = cτ/2 = (3 × 108)(2 × 10−6)/2 = 300 m.