DIPLOMA IN ELECTRICAL AND ELECTRONIC ENGINEERING
MODULE III — ELECTROMAGNETIC FIELDS AND COMMUNICATION SYSTEMS
Time: 3 hours | Maximum attempted marks: 100 | Paper: Original revision mock
Revision notice: This is an original practice paper prepared from the structure, command words, mark balance and difficulty of the supplied KNEC Module III paper. It is not an official KNEC examination paper. The worked answers are independently prepared revision guides and are not an official KNEC marking scheme.
INSTRUCTIONS TO CANDIDATES
- This paper has two sections: Section A and Section B.
- Answer any TWO questions from Section A and any THREE questions from Section B.
- All questions carry 20 marks. Answer all parts of each selected question.
- Show all working for calculations, use appropriate units and draw clear labelled diagrams.
- Use a non-programmable scientific calculator where necessary. Take
c = 3 × 10⁸ m/s,η₀ = 120π Ωandμ₀ = 4π × 10⁻⁷ H/m.
SECTION A: ELECTROMAGNETIC FIELDS THEORY
Answer any TWO questions from this section.
QUESTION 1 (20 marks)
(a) State Coulomb’s law of electrostatics. (2 marks)
(b) Two equal positive point charges of 20 μC each are located in free space at (−3, 0, 0) and (3, 0, 0). Determine the electric field at (0, 4, 0) and the force on a −10 μC charge placed at that point. (10 marks)
(c) State three co-ordinate systems used in electromagnetic field analysis. (3 marks)
(d) A magnetic material has relative permeability 6 and is in a magnetic field strength of 15 A/m. Determine its magnetic flux density and magnetization. (5 marks)
Answer
(a) The force between two stationary point charges is proportional to the product of their charges and inversely proportional to the square of their separation, acting along the line joining them: F = (1/(4πε))(q₁q₂/R²)aR. Like charges repel and unlike charges attract.
(b) For either charge, the distance to P is R = √(3² + 4²) = 5 m. The x-components cancel and the y-components add:
E = 2[kq(4ay)/5³] = 2[(8.99 × 10⁹)(20 × 10⁻⁶)(4)/125]ay.
Therefore E ≈ 1.15 × 10⁴ay N/C. For q₀ = −10 μC, F = q₀E, so F ≈ −0.115ay N.
(c) Rectangular/cartesian, cylindrical and spherical co-ordinate systems.
(d) B = μ₀μrH = 4π × 10⁻⁷ × 6 × 15 ≈ 1.13 × 10⁻⁴ T. For a linear material, M = (μr − 1)H = 5 × 15 = 75 A/m.
QUESTION 2 (20 marks)
(a) State Biot–Savart law. Points P₁ = (1, 2, 0) and P₂ = (4, 6, 0) are associated with a current element of 5πaz μA·m at P₁. Determine the unit vector from P₁ to P₂ and the incremental field strength at P₂. (11 marks)
(b) State Maxwell’s equations in integral form for time-varying fields. (4 marks)
(c) State five desirable properties of dielectric materials. (5 marks)
Answer
(a) dH = (I dℓ × aR)/(4πR²). Here R12 = 3ax + 4ay, |R12| = 5 m, and aR = 0.6ax + 0.8ay.
dH = [5π × 10⁻⁶az × (0.6ax + 0.8ay)]/(4π × 25).
Since az × ax = ay and az × ay = −ax:
aR = 0.6ax + 0.8ay and dH = (−4.0 × 10⁻⁸ax + 3.0 × 10⁻⁸ay) A/m.
(b)
∮ E · dℓ = −d/dt ∫S B · dS.∮ H · dℓ = ∫S J · dS + d/dt ∫S D · dS.∮S D · dS = ∫V ρvdv.∮S B · dS = 0.
(c) High insulation resistance, high dielectric strength, suitable relative permittivity, low dielectric loss/tangent and good thermal/mechanical stability with low moisture absorption.
QUESTION 3 (20 marks)
(a) A uniform plane wave in free space has electric field E = 60 cos(ωt + 12z)ax V/m. Determine the angular frequency, wavelength, magnetic field intensity and average power density. (8 marks)
(b) A wave crosses an interface at z = 0. In medium 1, μr1 = 2 and B₁ = (2ax + 1.5ay + 0.8az) T. In medium 2, μr2 = 6. With no free surface current, determine the angle made by B₁ with the normal and the expression for B₂. (8 marks)
(c) State two sources of electromagnetic radiation and two properties of electromagnetic waves in free space. (4 marks)
Answer
(a) β = 12 rad/m; the plus sign means propagation in the −z direction. Therefore ω = βc = 12 × 3 × 10⁸ = 3.60 × 10⁹ rad/s, and λ = 2π/β = 0.5236 m.
Using η₀ = 120π Ω, H = (−az × 60ax)/η₀, hence H = −0.1592 cos(ωt + 12z)ay A/m.
Sav = E₀²/(2η₀) = 60²/(2 × 120π) ≈ 4.77 W/m², in the −z direction.
(b) |B₁| = √(2² + 1.5² + 0.8²) = 2.6249 T. Thus the angle with the normal is θ = cos⁻¹(0.8/2.6249) ≈ 72.26°; the complementary angle from the interface is about 17.74°.
Normal B is continuous and tangential H is continuous. Therefore B₂n = 0.8az, while B₂t = (μ₂/μ₁)B₁t = (6/2)(2ax + 1.5ay). Hence B₂ = (6ax + 4.5ay + 0.8az) T.
(c) Sources include accelerating charges and time-varying currents. In free space, E and H are mutually perpendicular and transverse to propagation, and the waves travel at 3 × 10⁸ m/s with E/H = η₀.
SECTION B: COMMUNICATIONS SYSTEMS
Answer any THREE questions from this section.
QUESTION 4 (20 marks)
(a) State three merits of digital radar systems. (3 marks)
(b) Draw a labelled block diagram of a continuous-wave Doppler radar and describe its operation. (8 marks)
(c) A 1.2 GHz pulsed radar produces a minimum received power of 500 pW at 20 km. The antenna capture area is 6 m² and the target cross-sectional area is 12 m². Determine the peak pulsed power radiated and the minimum received power at 35 km. (9 marks)
Answer
(a) Digital processing improves clutter rejection and measurement accuracy; digital data can be stored, filtered and tracked automatically; and the system is programmable and has good repeatability/noise immunity.
(b)
[RF oscillator] ──┬──> [Power amplifier] ─> [Tx antenna] ─> Moving target
└──> [Reference] [Rx antenna]
│
v
[RF amplifier]
│
[Reference] ─────────> [Mixer]
│
v
[IF/filter] ─> [Doppler detector]
│
v
[Display]
The carrier is transmitted continuously. A moving target returns an echo shifted by fd = 2v/λ. Mixing the echo with the reference produces the Doppler beat frequency, which is filtered, detected and displayed as radial velocity information.
(c) λ = c/f = 0.25 m. For a monostatic radar, Pt = Pr4πR⁴λ²/(Ae²σ).
Pt = (500 × 10⁻¹²)(4π)(20 × 10³)⁴(0.25)²/(6² × 12), so Pt ≈ 1.45 × 10⁵ W = 145.4 kW.
Since received power varies as 1/R⁴, Pr35 = 500(20/35)⁴ ≈ 53.3 pW.
QUESTION 5 (20 marks)
(a) Define amplitude gain and bandwidth as used in an AM transmitter. (2 marks)
(b) Draw a labelled block diagram of a high-level AM transmitter. (6 marks)
(c) An FM modulator has a fixed tuning capacitance of 150 pF in parallel with a 75 pF varactor at a carrier frequency of 80 MHz. Determine the tuned-circuit inductance and the change in varactor capacitance when the frequency is increased to 88 MHz. (6 marks)
(d) A 120 kHz band carries a baseband signal of bandwidth 6 kHz. Determine the number of AM and SSB channels it can accommodate, and state two advantages of VSB over DSB-AM. (6 marks)
Answer
(a) Amplitude gain is Vout/Vin. Bandwidth is the frequency range occupied; conventional AM has BW = 2fm(max).
(b)
[Microphone] -> [AF voltage amplifier] -> [AF power amplifier] ─┐
v
[RF oscillator] -> [RF buffer/driver] ----------------> [Final RF power amplifier]
│
v
[Output filter] -> [Antenna]
The AF power amplifier provides the high-level modulation power at the final RF power amplifier, where carrier amplitude is varied by the audio signal.
(c) At 80 MHz, C = 150 + 75 = 225 pF. Therefore L = 1/[(2π × 80 × 10⁶)²(225 × 10⁻¹²)] ≈ 17.59 nH.
At 88 MHz, Ctotal ≈ 185.95 pF; hence Cvaractor = 185.95 − 150 = 35.95 pF. The capacitance decreases by 39.05 pF.
(d) AM channel width = 2 × 6 = 12 kHz, so AM channels = 120/12 = 10. SSB channel width = 6 kHz, so SSB channels = 120/6 = 20. VSB uses less bandwidth/power than DSB-AM and is easier to filter while preserving low-frequency information needed for wideband signals.
QUESTION 6 (20 marks)
(a) Define azimuth angle, apogee and perigee as used in satellite communication. (3 marks)
(b) Draw a labelled block diagram of an earth-station subsystem. (5 marks)
(c) State the approximate frequency ranges of Ku-band and C-band. (4 marks)
(d) An earth station at 12 GHz radiates 5 kW toward a satellite 40,000 km away. The transmitting antenna gain is 50 dB and the received power is 1.2 μW. Determine the wavelength, receiving antenna gain and free-space path loss. (8 marks)
Answer
(a) Azimuth is the horizontal bearing measured from a reference direction, normally true north. Apogee is the point farthest from earth in an elliptical orbit. Perigee is the nearest point.
(b)
TRANSMIT: [Baseband] -> [MUX/codec] -> [Modulator] -> [IF amp]
-> [Up-converter] -> [HPA] -> [Feed/antenna] -> Satellite
RECEIVE: Satellite -> [Antenna/feed] -> [LNA/LNB] -> [Down-converter]
-> [IF amp] -> [Demodulator] -> [Decoder] -> [Baseband]
(c) Ku-band is approximately 12–18 GHz. C-band is approximately 4–8 GHz.
(d) λ = 3 × 10⁸/(12 × 10⁹) = 0.025 m.
Lfs = 20log₁₀(4πR/λ) = 20log₁₀[4π(4 × 10⁷)/0.025] ≈ 206.07 dB.
From Pr = Pt + Gt + Gr − Lfs, Gr = 10log₁₀(1.2 × 10⁻⁶/5000) + 206.07 − 50, so Gr ≈ 59.87 dB.
QUESTION 7 (20 marks)
(a) Define monochromaticity, brightness and hue as applied to TV systems. (3 marks)
(b) State one possible cause for each: sound with no video; picture flicker; vertical lines on the screen; and power ON with neither picture nor sound. (4 marks)
(c) A CCTV system has three cameras, a DVR, router and wireless laptop. Draw a labelled block diagram. (7 marks)
(d) A TV standard has 625 lines per frame and 25 frames per second. The screen is 32 inches wide and 18 inches high. Determine line frequency, line period and aspect ratio. (6 marks)
Answer
(a) Monochromaticity is colour purity or dominance of one wavelength. Brightness is perceived light intensity. Hue is the basic colour determined by dominant wavelength.
(b) Sound but no video may indicate a video IF/detector or video amplifier fault. Flicker may indicate unstable vertical scanning. Vertical lines may indicate a horizontal deflection fault. Power ON with no picture or sound may indicate a common power supply, tuner/IF or system-control fault.
(c)
[Camera 1] ─┐ [Camera 2] ─┼──> [DVR] ──> [Router] ))) [Wireless laptop] [Camera 3] ─┘ └──> [Local monitor/storage]
(d) fH = 625 × 25 = 15,625 Hz = 15.625 kHz. TH = 1/15,625 = 64 μs. Aspect ratio = 32:18 = 16:9.
QUESTION 8 (20 marks)
(a) Describe the HLR, VLR and AuC in a GSM cellular network. (6 marks)
(b) With the aid of a labelled diagram, describe the operation of a travelling-wave tube. (8 marks)
(c) A circular waveguide has an internal diameter of 5 cm and operates at 10 GHz in TE1,1 mode. Using Kc = 1.84, determine cut-off wavelength, guide/group wavelength and TE characteristic impedance. (6 marks)
Answer
(a) HLR is the permanent subscriber and service database. VLR is the temporary database for subscribers currently in a location area. AuC stores security data and produces authentication/ciphering parameters used to verify subscribers and protect calls.
(b)
[Electron gun/cathode] ===== electron beam =====> [Collector]
│ │
v v
[Focusing magnets] [Helix slow-wave structure]
↑ RF input
│
RF output/coupler
The gun forms a focused beam. The helix slows the RF wave so that its axial phase velocity is close to the beam velocity. The RF field bunches electrons and extracts their kinetic energy, amplifying the travelling wave. The output coupler takes the amplified RF signal and the collector receives spent electrons.
(c) Radius a = 0.025 m. λ = 3 × 10⁸/10 × 10⁹ = 0.030 m.
λc = 2πa/Kc = 2π(0.025)/1.84 ≈ 0.08537 m, so fc ≈ 3.51 GHz.
λg = λ/√[1 − (fc/f)²] ≈ 0.03204 m.
ZTE = η₀/√[1 − (fc/f)²] ≈ 402.7 Ω.
End of answered mock examination. Attempt the paper under timed conditions before reading the worked answers.