KNEC Electromagnetic Fields and Communication Systems — June/July 2023 — Questions and Answers

KNEC Diploma Electrical and Electronic Engineering (Telecommunication Option) — Module III

Paper: Electromagnetic Fields Theory and Communication Systems (2602/304), June/July 2023. The paper has two sections: answer any two questions from Section A and any three from Section B. All questions carry equal marks.

Revision notice: The questions below are transcribed from the supplied KNEC paper, including the numerical data and question numbering. The worked responses are independently prepared revision guides for study; they are not the official KNEC marking scheme. Where a figure is printed in the source paper, the labels and operation are described in words so the lesson remains selectable and searchable.

SECTION A — ELECTROMAGNETIC FIELDS THEORY

Answer any TWO questions from this section.

Question 1

(a) State Maxwell’s equations for time varying fields in point form and explain the physical significance of each. (8 marks)

Answer:

1. Gauss’s law for electricity: ∇·D = ρ_v. Electric charge is the source or sink of electric flux.
2. Gauss’s law for magnetism: ∇·B = 0. Magnetic flux lines are continuous; isolated magnetic monopoles do not occur.
3. Faraday’s law: ∇×E = −∂B/∂t. A changing magnetic field produces a circulating electric field.
4. Ampère–Maxwell law: ∇×H = J + ∂D/∂t. Magnetic fields are produced by conduction current and by changing electric flux.

(b) The electric field intensity, E, in free space is given by E = 5 sin(10⁶πt − 4z) a_y. Determine the expressions for: (i) magnetic flux density, B; (ii) magnetic field intensity, H; (iii) electric flux density, D. (8 marks)

Answer: Let ψ = 10⁶πt − 4z. Using the stated free-space relations, η₀ = 120π Ω, μ₀ = 4π × 10⁻⁷ H/m and ε₀ = 8.854 × 10⁻¹² F/m:

• The wave travels in the −a_z direction. Therefore H = (1/η₀)(−a_z × E) = (5/120π) sinψ a_x A/m ≈ 0.01326 sinψ a_x A/m.
• B = μ₀H = 5/c sinψ a_x ≈ 1.67 × 10⁻⁸ sinψ a_x T.
• D = ε₀E = 4.427 × 10⁻¹¹ sinψ a_y C/m².

Note: The numerical phase constants are reproduced as printed. The field components above use the free-space constitutive relations supplied in the paper.

(c) A cuboid defined by 0 < x < 2, 0 < y < 2, 0 < z < 2 metres contains a volume charge density ρ_v = 15xyz μC/m³. Determine the total outward electric flux from the cube. (4 marks)

Answer: The total charge is

Q = ∫₀²∫₀²∫₀² 15xyz dx dy dz μC = 15(2)(2)(2) = 120 μC.

By Gauss’s law, Ψ_E = Q/ε₀ = 120 × 10⁻⁶/(8.854 × 10⁻¹²) ≈ 1.36 × 10⁷ N·m²/C, outward.

Question 2

(a) Distinguish between magnetic flux and magnetic field strength with respect to electrodynamics. (2 marks)

Answer: Magnetic flux Φ is the total magnetic field passing through a surface, measured in weber (Wb), Φ = ∫B·dA. Magnetic field strength H is the magnetising force per unit length, measured in ampere per metre (A/m); in a linear medium B = μH.

(b) With aid of a sketch, describe the Biot–Savart law of magnetism. (4 marks)

Answer: For a current element I dℓ, the small field at a point a distance r away is

dB = (μ I dℓ × ar)/(4πr2).

Thus dB is proportional to the current, the element length and sinθ, inversely proportional to r², and its direction is perpendicular to the plane containing dℓ and the observation point, given by the right-hand rule. A suitable sketch should show I dℓ, r, θ and dB.

(c) Figure 1 shows a rectangular coil, 0.5 m × 0.8 m. The coil rotates at ω = 25π radians/s in a magnetic field B = 0.7 a_z teslas. Determine the expression for: (i) magnetic flux; (ii) induced voltage. (5 marks)

Answer: Area A = 0.5 × 0.8 = 0.4 m². Taking the initial coil normal as +a_z and rotation about the y-axis,

Φ(t) = BA cosωt = (0.7)(0.4)cos(25πt) = 0.28 cos(25πt) Wb.

e(t) = −dΦ/dt = 0.28(25π)sin(25πt) = 7π sin(25πt) V ≈ 21.99 sin(25πt) V.

(d) Complete the table of electromagnetic-wave bands, frequency ranges and areas of application for the three bands printed in the paper: X-rays, ultra-violet and infra-red. (9 marks)

Answer:

Exact boundary values vary slightly between reference tables; the ordering and the physical applications are the key points.

Question 3

(a) Figure 2 shows a hysteresis loop of a magnetic material. (i) Identify the quantities labelled H_c, B_r and B₀. (ii) Define each quantity. (6 marks)

Answer: H_c is the coercive field: the reverse magnetising force required to reduce the flux density to zero after the material has been magnetised. B_r is the residual or remanent flux density: the value of B remaining when H returns to zero. B₀ is the saturation flux density shown at the end of the magnetisation curve, where further increases in H produce very little increase in B.

(b) The electric field of a plane wave in a dielectric is given by E = 3.77 × 10⁻² cos(2π × 10⁸t − (4π/3)z). The relative permeability of the dielectric μ_r = 1. Determine: (i) relative permittivity; (ii) dielectric impedance; (iii) wavelength in the dielectric; (iv) expression for H. (10 marks)

Answer: ω = 2π × 10⁸ rad/s and β = 4π/3 rad/m. In free space β₀ = ω/c = 2π/3 rad/m, hence β/β₀ = 2 and ε_r = 4. Therefore

• ε_r = 4;
• η = η₀√(μ_r/ε_r) = 120π/2 = 60π Ω ≈ 188.5 Ω;
• λ = 2π/β = 1.5 m;
• for an x-directed E field travelling in +z, H = (3.77 × 10⁻²/188.5) cos(2π × 10⁸t − 4πz/3) a_y ≈ 2.00 × 10⁻⁴ cos(…) a_y A/m.

(c) A vector potential A is given by A = 6(x² + y² + z²)a_x. The magnetic flux density B is defined by B = ∇ × A. Determine B. (4 marks)

Answer: With A_x = 6(x² + y² + z²) and A_y = A_z = 0,

B = (∂A_x/∂z) a_y − (∂A_x/∂y) a_z = 12z a_y − 12y a_z T.

SECTION B — COMMUNICATION SYSTEMS

Answer any THREE questions from this section.

Question 4

(a) Define each of the following with respect to TV systems: (i) hue; (ii) flyback. (2 marks)

Answer: Hue is the colour attribute that identifies a colour family, such as red, green or blue, by its dominant wavelength. Flyback is the rapid return of the scanning beam from the end of one line or field to the start of the next; picture information is blanked during the return.

(b) State any two of the following: (i) types of analogue colour TV systems; (ii) colour difference signals. (4 marks)

Answer: Analogue colour systems include NTSC, PAL and SECAM. Common colour-difference signals include R − Y and B − Y (other accepted quadrature forms include I and Q).

(c) Figure 3 shows a block diagram of a digital TV transmitter. Explain the functions of: (i) quantisation and PCM; (ii) data encryption; (iii) data compression; (iv) multiplexer. (8 marks)

Answer: Quantisation maps sampled video or audio amplitudes to a finite set of levels; PCM represents those levels as binary code words. Data encryption transforms the programme stream using a key so unauthorised receivers cannot use it. Data compression removes statistical redundancy and perceptually insignificant information to reduce the bit rate. The multiplexer combines the coded video, audio, service information and control data into one transport stream for transmission.

(d) Explain three challenges facing migration from analogue to digital TV transmission. (6 marks)

Answer: Challenges include the cost of new transmitters, studio equipment and set-top boxes; the need to educate viewers and maintain coverage during the transition; and spectrum/network-planning issues such as frequency reallocation, interference control and coordination with neighbouring services. A further practical challenge is the digital divide where some households cannot immediately afford compatible receivers.

Question 5

(a) Figure 4 shows a block diagram of a CW Doppler Radar Tx/Rx with a heterodyne receiver. (i) Identify the parts labelled x, y and z. (ii) State the function of each part. (6 marks)

Answer: X is the duplexer/circulator: it routes the transmitter signal to the antenna and the received echo to the receiver while providing isolation. Y is the mixer: it combines the received echo with the local oscillator to produce an intermediate-frequency difference signal containing the Doppler shift. Z is the IF filter/amplifier: it selects the required IF/Doppler band and provides gain before detection. The detector extracts the Doppler information and the display presents the target indication.

(b) A radar transmitter has a mean power of 500 kW, a pulse repetition frequency (PRF) of 2500 pps and a pulse width of 0.8 μs. Determine: (i) duty cycle; (ii) transmitted peak power; (iii) maximum unambiguous range. (6 marks)

Answer: Duty cycle D = τ × PRF = 0.8 × 10⁻⁶ × 2500 = 0.002 = 0.2%. Peak power P_p = P_mean/D = 500 × 10³/0.002 = 250 MW. Maximum unambiguous range R_u = c/(2PRF) = 3 × 10⁸/(2 × 2500) = 60 km.

(c) Figure 5 shows a diagram of a Doppler radar illumination for directing a missile to a target. (i) Describe its operation. (ii) State two limitations of the system. (8 marks)

Answer: The ground illuminator continuously transmits a CW signal towards the target. The target reflection contains a Doppler shift caused by target motion. The missile receives a rear/reference signal from the illuminator and a front/target-reflection signal. Its guidance receiver compares the two frequencies or phases to determine the target’s relative velocity and line-of-sight error, then commands the missile to intercept. Limitations include dependence on a high-power continuous illuminator, which can reveal the launching platform, and limited effectiveness against low-radar-cross-section, terrain-masked or heavily manoeuvring targets. The system is also vulnerable to electronic countermeasures.

Question 6

(a) Define each of the following with respect to satellite communication: (i) stabilization; (ii) footprint. (2 marks)

Answer: Stabilization is the control of a satellite’s attitude and/or orbit so its antennas and solar arrays maintain the required pointing direction. A footprint is the geographical area on Earth over which a satellite beam provides usable coverage.

(b) (i) Describe a VSAT. (ii) State three areas of VSAT applications. (5 marks)

Answer: A VSAT is a very small aperture terminal: a small-dish satellite earth station with an outdoor unit, indoor modem and antenna control equipment, normally linked through a hub or in a star/mesh network. Applications include banking and point-of-sale networks, rural/remote internet or telephone service, corporate branch networking, distance learning, and monitoring or disaster-response communications.

(c) Describe each of the following satellite types: (i) geosynchronous; (ii) polar; (iii) defence. (6 marks)

Answer: A geosynchronous satellite has an orbital period equal to Earth’s rotation; the special circular equatorial case is geostationary and appears fixed to an observer. A polar satellite travels approximately over the North and South Poles and, as Earth rotates beneath it, can provide near-global observation coverage. A defence satellite is designed for military services such as secure communications, surveillance, navigation, early warning or electronic intelligence.

(d) Table 2 gives: earth-station transmitter output at saturation = 200 W (33 dBW); earth-station transmit antenna gain = 64 dB; uplink atmospheric loss = 0.6 dB; free-space path loss = 206.5 dB; satellite antenna gain = 20 dB; uplink frequency = 4 GHz. Determine: (i) EIRP for the earth station, in dBW; (ii) carrier power density at the satellite; (iii) received power at the satellite, in dBW. (7 marks)

Answer: (i) EIRP = P_T(dBW) + G_T = 33 + 64 = 97 dBW.

At 4 GHz, λ = c/f = 3 × 10⁸/(4 × 10⁹) = 0.075 m. The carrier power density is

PFD = EIRP − L_atm − FSPL + 10 log₁₀(4π/λ²)

= 97 − 0.6 − 206.5 + 33.49 ≈ −76.6 dBW/m².

(iii) P_R = EIRP − L_atm − FSPL + G_R = 97 − 0.6 − 206.5 + 20 = −90.1 dBW.

Question 7

(a) Define each of the following with respect to waveguides: (i) characteristic impedance; (ii) cut-off frequency. (2 marks)

Answer: Characteristic or guide impedance is the ratio of the transverse electric field to the transverse magnetic field for a travelling wave in a specified guide mode. Cut-off frequency is the minimum frequency at which that mode propagates; below it the field is evanescent and does not carry power down the guide.

(b) Draw a labelled diagram of a travelling wave tube microwave amplifier and describe its operation. (8 marks)

Answer: A labelled diagram should show the electron gun, focusing magnets, slow-wave helix, RF input, RF output, attenuator, collector and vacuum envelope. The electron gun forms and accelerates the beam. The input RF wave travels along the helix at a reduced phase velocity and interacts continuously with the electrons. Electrons bunch as energy is exchanged with the RF wave; the bunched beam gives kinetic energy to the travelling wave, amplifying it. The attenuator prevents reflections and oscillation, while the collector absorbs the spent beam.

(c) A rectangular waveguide has dimensions W × h metres and propagates the TE_1,0 mode. The electric field across the guide is given by E_y = E₀ sin(πx/W). The guide impedance is Z_T. (i) Sketch the electric field across the guide. (ii) Show that the total power flow across the guide is W_T = E₀²hW/(4Z_T). (8 marks)

Answer: For TE_1,0, the field is zero at x = 0 and x = W and has one maximum of E₀ at x = W/2; it is uniform in y. Since H_x = E_y/Z_T, the average power is

W_T = ∫₀^h∫₀^W (E_y²/(2Z_T)) dx dy = (E₀²h/(2Z_T))∫₀^W sin²(πx/W)dx.

Because the integral is W/2, W_T = E₀²hW/(4Z_T).

(d) State the functions of each of the following microwave devices: (i) Hybrid-T; (ii) isolator. (2 marks)

Answer: A Hybrid-T is a four-port waveguide junction used to split or combine microwave power with useful phase relationships and isolation between ports. An isolator is a non-reciprocal device that passes power in one direction and absorbs reflected power, protecting the source from mismatch.

Question 8

(a) (i) Draw a labelled block diagram of an independent side-band AM (ISB-AM) transmitter. (ii) State with reasons one merit of ISB-AM. (iii) State two application areas of ISB-AM. (8 marks)

Answer: A suitable block diagram is: two independent message inputs → audio amplifiers/filters → two balanced modulators driven by a common carrier oscillator → upper/lower sideband filters → frequency converters or mixers → combiner/power amplifier → antenna. ISB transmits two independent channels in the two sidebands, so it doubles the information capacity for a given carrier arrangement and avoids wasting power in a large carrier. Applications include long-distance HF point-to-point telephone circuits, maritime/aeronautical communications and military or diplomatic radio links.

(b) State two merits of frequency modulation (FM) radio transmission. (2 marks)

Answer: FM has strong rejection of amplitude noise and interference because information is carried by frequency deviation; it also supports high-fidelity audio when an adequate channel bandwidth is provided. A constant-amplitude signal allows efficient limiter and power-amplifier stages.

(c) An FM reactance modulator uses an oscillator whose capacitance and inductance are 25 pF and 0.101 H respectively. When a modulating signal of 5 kHz is applied the effective capacitance becomes 24.92 pF. Determine: (i) carrier frequency; (ii) frequency deviation; (iii) modulation index; (iv) system bandwidth. (10 marks)

Answer: The unmodulated carrier is

f_c = 1/(2π√LC) = 1/[2π√(0.101 × 25 × 10⁻¹²)] = 100.16 kHz.

With C = 24.92 pF, f_max = 1/[2π√(0.101 × 24.92 × 10⁻¹²)] = 100.32 kHz. Hence Δf = f_max − f_c = 160.6 Hz (approximately).

Modulation index β = Δf/f_m = 160.6/5000 = 0.0321.

Using Carson’s rule, B_T ≈ 2(Δf + f_m) = 2(160.6 + 5000) = 10.32 kHz.