(a) A 3-phase line is terminated by a star-connected resistive load of 800 ohm per phase. The line conductors are equilaterally…

(a) Surge impedance, reflected power and absorbed power (20 marks)

Surge impedance from the geometry. For conductors of radius r = 1 cm = 0.01 m spaced equilaterally at d = 2.0 m, the surge (natural) impedance per phase is

Z₀ = √(L/C) = 60 ln(d/r) = 138 log₁₀(d/r) = 138 log₁₀(200) = 138 × 2.301 = 317.5 ≈ 318 Ω.

(Including the internal flux linkage with r′ = 0.7788r would give about 325 Ω, a change of 2%.)

Coefficients at the termination. The load is a star-connected resistance of 800 Ω per phase, so Z_L = 800 Ω.

• Reflection coefficient: ρ = (Z_L − Z₀)/(Z_L + Z₀) = (800 − 318)/(800 + 318) = 482/1118 = 0.431.
• Refraction (transmission) coefficient: τ = 2Z_L/(Z_L + Z₀) = 1600/1118 = 1 + ρ = 1.431.

Waves. The incident voltage wave is e = 33 kV on each phase.

• Reflected voltage: ρe = 0.431 × 33 = 14.2 kV.
• Transmitted voltage across the load: τe = 1.431 × 33 = 47.2 kV.

Powers per phase.

• Incident power: P_i = e²/Z₀ = (33 × 10³)²/318 = 3.43 MW.
• Reflected power: P_r = (ρe)²/Z₀ = ρ²P_i = 0.186 × 3.43 = 0.637 MW.
• Power absorbed by the termination: P_t = (τe)²/Z_L = (47.2 × 10³)²/800 = 2.79 MW.

Check. P_i = P_r + P_t, that is, 3.43 = 0.637 + 2.79 MW, which confirms that the energy of the incident wave is either reflected or absorbed. The absorbed fraction is 1 − ρ² = 81.4%.

Three phases. If the same 33 kV wave travels on all three phases, the totals are: reflected 3 × 0.637 = 1.91 MW and absorbed 3 × 2.79 = 8.37 MW.

(b)(i) Entropy and information in messages (10 marks)

(I) H(X) = −Σ P(xᵢ) log₂P(xᵢ) = 0.4 log₂(1/0.4) + 0.3 log₂(1/0.3) + 0.2 log₂(1/0.2) + 0.1 log₂(1/0.1)

= 0.4 × 1.322 + 0.3 × 1.737 + 0.2 × 2.322 + 0.1 × 3.322

= 0.5288 + 0.5211 + 0.4644 + 0.3322 = 1.846 bits/symbol.

(II) The source is memoryless, so the probability of a message is the product of the symbol probabilities, and its information is I = −log₂P.

• Message x₁x₂x₁x₃: P = 0.4 × 0.3 × 0.4 × 0.2 = 0.0096, so I = −log₂0.0096 = 6.70 bits.
• Message x₄x₃x₃x₂: P = 0.1 × 0.2 × 0.2 × 0.3 = 0.0012, so I = −log₂0.0012 = 9.70 bits.

Comparison. A typical four-symbol message carries on average 4H(X) = 4 × 1.846 = 7.39 bits. The first message contains 6.70 bits, less than average, because it is made of the more probable symbols (x₁ twice, x₂ and x₃). The second contains 9.70 bits, more than average, because it includes the rarest symbol x₄ and repeats x₃. Both differ from 7.39 bits, since H(X) is only the mean information per symbol, and individual messages can carry more or less.

(b)(ii) (6,3) parity check code (10 marks)

H = [[1, 0, 1, 1, 0, 0], [1, 1, 0, 0, 1, 0], [0, 1, 1, 0, 0, 1]]. Here H = [Pᵀ | I₃] with Pᵀ = [[1, 0, 1], [1, 1, 0], [0, 1, 1]], so the code is systematic with the parity bits at the end.

(I) Generator matrix. P = [[1, 1, 0], [0, 1, 1], [1, 0, 1]], so

G = [I₃ | P] = [[1, 0, 0, 1, 1, 0], [0, 1, 0, 0, 1, 1], [0, 0, 1, 1, 0, 1]].

(Check: every row of G satisfies GHᵀ = 0 modulo 2.)

(II) Code word beginning with 101. With message m = (1, 0, 1), c = mG = row 1 + row 3 = (1, 0, 0, 1, 1, 0) + (0, 0, 1, 1, 0, 1) = 101011. The parity bits are c₄ = m₁ + m₃ = 0, c₅ = m₁ + m₂ = 1 and c₆ = m₂ + m₃ = 1.

(III) Decoding 110110. The syndrome is s = H rᵀ, with r = (1, 1, 0, 1, 1, 0):

• s₁ = r₁ + r₃ + r₄ = 1 + 0 + 1 = 0
• s₂ = r₁ + r₂ + r₅ = 1 + 1 + 1 = 1
• s₃ = r₂ + r₃ + r₆ = 1 + 0 + 0 = 1

So s = (0, 1, 1)ᵀ, which is not zero, so the received word contains an error. It equals the second column of H, so the error is in bit 2. Flipping bit 2 gives the corrected code word 100110, and the message is 100 (check: mG = 100 followed by the parity bits 1, 1, 0). The code has minimum distance 3, since all columns of H are distinct and non-zero, so it corrects any single error.

(c) Translay relay (10 marks)

The Translay relay is a balanced-voltage pilot-wire differential protection for feeders. A relay is installed at each end of the line and the two are joined by two pilot wires.

Construction. Each relay is an induction-disc relay with an upper and a lower electromagnet acting on a common disc. The upper magnet carries a primary winding P, fed from the line CT, and a secondary winding S. The secondary windings of the two ends are connected in opposition through the pilot wires, and in series with each lower-magnet coil L.

Operation. The primary current I produces flux Φ₁ in the upper magnet, which induces the emf E in the winding S, of magnitude proportional to I. The emfs of the two ends oppose each other in the pilot loop.

• Healthy line or external fault. The current entering the line at one end equals the current leaving it at the other end, so E₁ = E₂ and the net emf in the pilot loop is zero. No current flows in the pilot wires or in the lower coils L, no lower-magnet flux is produced, and the relay does not operate.
• Internal fault. The end currents no longer match. With infeed from both ends the current at one end reverses relative to the through direction, so the two emfs add in the loop and the net emf is E₁ + E₂; with infeed from one end only, the net emf is E₁ alone. Either way the net emf is not zero, and it drives a pilot current I_p through the coils L. The lower magnet produces a flux Φ_L, which is out of phase with Φ₁, and the two fluxes act on the disc with a torque proportional to Φ₁Φ_L sin α. The disc rotates and closes the trip contact of the circuit breaker.

Features. Only two pilot wires are needed, and, because the pilot current is small in normal conditions and external faults, the pilot-wire burden is low. The upper magnet is usually designed to saturate at high currents so that the emfs E₁ and E₂ stay nearly equal during heavy through-faults, despite CT errors, which improves stability. The scheme is sensitive to internal faults and is suitable for feeders of moderate length, up to the length allowed by the pilot-wire capacitance and insulation.