(a) A discrete memoryless source generates three independent message symbols m_1, m_2 and m_3 with probabilities 0·9, 0·08 and…

(a) Huffman code and its extension

Given P(m₁) = 0.9, P(m₂) = 0.08, P(m₃) = 0.02. The source entropy is H = −Σ p log₂ p = 0.9 × 0.152 + 0.08 × 3.644 + 0.02 × 5.644 = 0.541 bit/symbol.

(i) Huffman code. Combine m₂ and m₃ (0.10), which then pairs with m₁. With '0' for the most probable message: m₁ = 0, m₂ = 10, m₃ = 11. Average length L = 0.9 × 1 + 0.08 × 2 + 0.02 × 2 = 1.1 bits/symbol. A fixed-length code needs 2 bits, so the compression ratio is 2/1.1 = 1.82. Efficiency η = H/L = 0.541/1.1 = 49.2%.

(ii) Extension code (pairs of symbols). The nine pair probabilities are m₁m₁ 0.81, m₁m₂ and m₂m₁ 0.072 each, m₁m₃ and m₃m₁ 0.018 each, m₂m₂ 0.0064, m₂m₃ and m₃m₂ 0.0016 each, m₃m₃ 0.0004. Huffman coding gives one valid code: m₁m₁ = 0, m₂m₁ = 11, m₁m₂ = 100, m₃m₁ = 1011, m₁m₃ = 10100, m₂m₂ = 101010, m₃m₂ = 1010111, m₂m₃ = 10101100, m₃m₃ = 10101101.

The average length per pair is 0.81 × 1 + 0.072 × 2 + 0.072 × 3 + 0.018 × 4 + 0.018 × 5 + 0.0064 × 6 + 0.0016 × 7 + 0.0016 × 8 + 0.0004 × 8 = 1.3976 bits, so L = 1.3976/2 = 0.6988 bit/symbol. Compression ratio = 2/0.6988 = 2.86. Efficiency = 0.541/0.6988 = 77.4%, a clear gain over 49.2%.

(b) Universal relay torque equation

The universal torque equation of an electromagnetic relay is T = K₁I² + K₂V² + K₃VI cos(θ − τ) + K₄, where V and I are the relay voltage and current, θ is the angle of V/I, τ is the maximum-torque angle, and K₄ is the spring torque (taken negative). The relay operates when T > 0, and the boundary is T = 0.

(i) Impedance relay. Put K₃ = 0 and K₄ = 0, with K₂ negative: K₁I² − K₂′V² = 0, so Z = V/I = √(K₁/K₂′), a constant. The characteristic is a circle on the R-X plane centred at the origin. The relay operates for impedance inside the circle, and does not operate outside it.

(ii) Reactance relay. Put K₂ = K₄ = 0 and τ = 90°: K₁I² − K₃VI sin θ = 0, so Z sin θ = X = K₁/K₃, a constant. The characteristic is a straight line parallel to the R-axis, and the relay operates when X is below the setting (toward the R-axis). It does not operate above the line.

(iii) Mho relay. Put K₁ = K₄ = 0: −K₂′V² + K₃VI cos(θ − τ) = 0, so Z = (K₃/K₂′) cos(θ − τ). This is a circle of diameter K₃/K₂′ passing through the origin, at angle τ to the R-axis. The relay operates inside the circle and does not operate outside it. It is inherently directional.

(c) Single-phase distributor

Take the voltage at C as the reference: V_C = 240∠0° V, Z = 0.15 + j0.25 Ω per section.

I_C = 150∠−36.87° = 120 − j90 A and I_B = 150∠−53.13° = 90 − j120 A. The current in AB is the sum, I_AB = 210 − j210 A.

V_B = V_C + I_C Z = 240 + (120 − j90)(0.15 + j0.25) = 240 + (18 + j30 − j13.5 + 22.5) = 280.5 + j16.5 V.

V_A = V_B + I_AB Z = 280.5 + j16.5 + (210 − j210)(0.15 + j0.25) = 280.5 + j16.5 + (31.5 + j52.5 − j31.5 + 52.5) = 364.5 + j37.5 V.

Supply voltage |V_A| = 366.4 V, at a phase angle of 5.87° leading the voltage at the far end.