JEMSHAH REVISION MOCK EXAMINATION

DIPLOMA IN ELECTRICAL AND ELECTRONIC ENGINEERING

MODULE III — ELECTROMAGNETIC FIELDS AND COMMUNICATION SYSTEMS

Time: 3 hours   |   Maximum attempted marks: 100   |   Paper: Original revision mock — Paper A

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

  1. This paper has two sections: Section A and Section B.
  2. Answer any TWO questions from Section A and any THREE questions from Section B.
  3. All questions carry 20 marks. Answer all parts of each selected question.
  4. Show working, use SI units and draw clear labelled diagrams where required.
  5. Take ε₀ = 8.85 × 10⁻¹² F/m, μ₀ = 4π × 10⁻⁷ H/m, c = 3 × 10⁸ m/s and η₀ = 120π Ω.

SECTION A: ELECTROMAGNETIC FIELDS THEORY

Answer any TWO questions from this section.

QUESTION 1 (20 marks)

(a) Distinguish between electric field intensity and electric flux density. (4 marks)

(b) An 8 μC point charge is in free space at the origin. Determine the electric field intensity and electric flux density at P(0, 0, 2) m. (6 marks)

(c) A parallel-plate capacitor has plate area 0.01 m², separation 2 mm and dielectric relative permittivity 4. Determine its capacitance and energy stored when charged to 100 V. (6 marks)

(d) State four properties of electrostatic field lines. (4 marks)

Answer

(a) Electric field intensity E is force per unit positive test charge, measured in N/C or V/m. Electric flux density D is electric flux per unit area, measured in C/m², and in a linear medium D = εE.

(b) R = 2 m along az. Thus E = kq/R² = (8.99 × 10⁹)(8 × 10⁻⁶)/4, giving E = 1.80 × 10⁴az N/C. Also D = ε₀E ≈ 1.59 × 10⁻⁷az C/m².

(c) C = ε₀εrA/d = (8.85 × 10⁻¹²)(4)(0.01)/(0.002) = 1.77 × 10⁻¹⁰ F, or 177 pF. W = ½CV² = ½(1.77 × 10⁻¹⁰)(100)² = 8.85 × 10⁻⁷ J.

(d) Field lines start on positive charge and end on negative charge or infinity; they never cross; the tangent gives field direction; and line density indicates field magnitude. In electrostatic equilibrium, field lines meet a conductor normally.

QUESTION 2 (20 marks)

(a) State Maxwell’s equations in integral form for time-varying fields. (4 marks)

(b) A long straight conductor carries 5 A. Determine the magnetic field intensity and flux density at a radial distance of 0.2 m in free space. (6 marks)

(c) State the electromagnetic boundary conditions at an interface. (4 marks)

(d) State six desirable properties of dielectric materials. (6 marks)

Answer

(a) ∮E·dℓ = −d/dt∫B·dS; ∮H·dℓ = ∫J·dS + d/dt∫D·dS; ∮D·dS = ∫ρvdv; and ∮B·dS = 0.

(b) For a long conductor, H = I/(2πρ) = 5/(2π × 0.2) = 3.98 A/m, in the circumferential direction. Hence B = μ₀H = 4π × 10⁻⁷ × 3.98 ≈ 5.00 × 10⁻⁶ T.

(c) Tangential E is continuous; tangential H changes by the free surface current density; normal B is continuous; and normal D changes by the surface charge density.

(d) High resistivity, high dielectric strength, suitable permittivity, low dielectric loss, low moisture absorption and good thermal/mechanical stability.

QUESTION 3 (20 marks)

(a) A uniform plane wave in free space has E = 80 cos(ωt − 15x)ay V/m. Determine its direction of propagation, angular frequency, wavelength, magnetic field intensity and average power density. (8 marks)

(b) State the Poynting theorem and the expressions for electric and magnetic energy density. (6 marks)

(c) State two sources of electromagnetic radiation and two properties of electromagnetic waves in free space. (6 marks)

Answer

(a) The phase term is ωt − βx, so propagation is in the +x direction and β = 15 rad/m. ω = βc = 4.50 × 10⁹ rad/s, and λ = 2π/15 = 0.4189 m.

H = (ax × 80ay)/η₀ = 0.2122 cos(ωt − 15x)az A/m. Average power density is Sav = 80²/(2 × 120π) ≈ 8.49 W/m² in the +x direction.

(b) Poynting theorem expresses conservation of electromagnetic energy: the power flowing out of a volume plus the rate of increase of stored energy equals the power supplied to loss. In differential form, ∇·(E × H) = −J·E − ∂u/∂t. Electric energy density is ue = ½E·D; magnetic energy density is um = ½B·H.

(c) Sources include accelerating charges and time-varying currents. In free space, E and H are mutually perpendicular and transverse to propagation; waves travel at 3 × 10⁸ m/s and have 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 GHz pulsed radar gives a minimum received power of 250 pW at 18 km. The antenna capture area is 5 m² and the target cross-sectional area is 10 m². Determine the peak pulsed radiated power and the minimum received power at 30 km. (9 marks)

Answer

(a) Digital processing provides good clutter rejection and measurement accuracy; data can be stored and automatically tracked; and the system is programmable, repeatable and relatively noise immune.

(b)

[RF oscillator] ──┬──> [RF amplifier] ─> [Tx antenna] ─> Moving target
                   └──> [Reference]                         [Rx antenna]
                                                               │
                                                               v
                                                      [RF amplifier] -> [Mixer]
                                                                            │
                                                                            v
                                                               [IF/filter] -> [Doppler detector]
                                                                                  │
                                                                                  v
                                                                                [Display]

The transmitted continuous wave is reflected by a moving target with a Doppler shift. The echo is mixed with the reference, and the beat frequency fd = 2v/λ is filtered and displayed as radial velocity.

(c) λ = 0.3 m. Using Pt = Pr4πR⁴λ²/(Ae²σ) gives Pt ≈ 1.19 × 10⁵ W = 118.7 kW. At 30 km, Pr = 250(18/30)⁴ ≈ 32.4 pW.

QUESTION 5 (20 marks)

(a) Define amplitude gain and bandwidth in an AM transmitter. (2 marks)

(b) Draw a labelled high-level AM transmitter. (6 marks)

(c) An FM tuning circuit has a fixed 180 pF capacitor in parallel with a 60 pF varactor at 75 MHz. Determine the inductance and the change in varactor capacitance at 84 MHz. (6 marks)

(d) A 150 kHz band carries a baseband signal of bandwidth 7.5 kHz. Determine the number of AM and SSB channels and state two advantages of VSB over DSB-AM. (6 marks)

Answer

(a) Amplitude gain is the ratio of output to input amplitude. Bandwidth is the frequency range occupied; for conventional AM it is twice the highest modulating frequency.

(b)

[Audio source] -> [AF voltage amplifier] -> [AF power amplifier] ─┐
                                                                      v
[RF oscillator] -> [RF driver] ----------------------> [Final RF amplifier]
                                                                      │
                                                                      v
                                                              [Filter] -> [Antenna]

The AF power amplifier modulates the final RF power amplifier at high level.

(c) Total capacitance at 75 MHz is 240 pF. L = 1/[(2π × 75 × 10⁶)²(240 × 10⁻¹²)] ≈ 18.76 nH. At 84 MHz total capacitance is 191.33 pF, so the new varactor value is 11.33 pF. The decrease is 48.67 pF.

(d) AM width is 15 kHz, giving 150/15 = 10 channels. SSB width is 7.5 kHz, giving 150/7.5 = 20 channels. VSB uses less bandwidth/power than DSB-AM and is easier to filter while preserving low-frequency information.

QUESTION 6 (20 marks)

(a) Define azimuth angle, apogee and perigee in satellite communication. (3 marks)

(b) Draw a labelled block diagram of an earth-station subsystem. (5 marks)

(c) State the approximate ranges of Ku-band and C-band. (4 marks)

(d) An earth station operating at 10 GHz radiates 3 kW towards a satellite 38,000 km away. The transmitting antenna gain is 46 dB and the received power is 2 μW. Determine wavelength, receiving antenna gain and free-space path loss. (8 marks)

Answer

(a) Azimuth is the horizontal bearing from a reference direction, normally true north. Apogee is the farthest orbital point from earth and perigee is the nearest.

(b)

TRANSMIT: [Baseband] -> [Modulator] -> [IF amplifier] -> [Up-converter]
          -> [HPA] -> [Feed/antenna] -> Satellite
RECEIVE:  Satellite -> [Antenna/feed] -> [LNA] -> [Down-converter]
          -> [IF amplifier] -> [Demodulator] -> [Baseband]

(c) Ku-band is approximately 12–18 GHz; C-band is approximately 4–8 GHz.

(d) λ = 3 × 10⁸/(10 × 10⁹) = 0.030 m. Lfs = 20log₁₀[4π(38 × 10⁶)/0.03] ≈ 204.04 dB. From Friis’ equation, Gr = 10log₁₀(2 × 10⁻⁶/3000) + 204.04 − 46 ≈ 66.28 dB.

QUESTION 7 (20 marks)

(a) Define monochromaticity, brightness and hue as applied to TV systems. (3 marks)

(b) Give one possible cause of sound without video, flicker, vertical lines and no picture/sound with power ON. (4 marks)

(c) A CCTV system has three cameras, DVR, router and wireless laptop. Draw a labelled system block diagram. (7 marks)

(d) A standard has 625 lines per frame and 25 frames per second. A 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; brightness is perceived light intensity; hue is the basic colour or dominant wavelength.

(b) Video IF/detector or video amplifier fault; unstable vertical scan; horizontal deflection fault; and a common power-supply/tuner/IF fault respectively.

(c)

[Camera 1] ─┐
[Camera 2] ─┼──> [DVR] ──> [Router] ))) [Wireless laptop]
[Camera 3] ─┘       └──> [Local monitor/storage]

(d) fH = 625 × 25 = 15.625 kHz; TH = 64 μs; aspect ratio 32:18 = 16:9.

QUESTION 8 (20 marks)

(a) Describe the HLR, VLR and AuC in GSM. (6 marks)

(b) With the aid of a labelled diagram, describe a travelling-wave tube. (8 marks)

(c) A circular waveguide has internal diameter 6 cm and operates at 8 GHz in TE1,1 mode. With Kc = 1.84, determine cut-off wavelength, guide/group wavelength and characteristic impedance. (6 marks)

Answer

(a) HLR permanently stores subscriber identity and service information. VLR temporarily stores details of subscribers in its location area. AuC generates authentication and ciphering parameters used to validate subscribers and protect communication.

(b)

[Electron gun] ===== focused electron beam =====> [Collector]
       │                          │
       v                          v
 [Focusing magnets]       [Helix slow-wave structure]
                                  ↑ RF input   -> RF output

The gun forms the beam, the helix slows the RF wave and its field bunches electrons. The bunched electrons transfer kinetic energy to the travelling RF wave, producing amplification; the collector receives the spent beam.

(c) a = 0.03 m, λ = 0.0375 m. λc = 2πa/1.84 ≈ 0.10244 m, so fc ≈ 2.93 GHz. λg = λ/√[1 − (fc/f)²] ≈ 0.04030 m. ZTE = 120π/√[1 − (fc/f)²] ≈ 405 Ω.

End of answered mock examination. Attempt the paper under timed conditions before reading the worked answers.

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