(a) The figure below shows the single-line diagram of a generator connected through parallel transmission lines to an infinite…

(a)

Method: per-unit power-angle equation and swing equation. From the diagram, the generator terminal connects through a 0.10 pu transformer reactance to two parallel lines of 0.40 pu each. The parallel lines give X_line = 0.40/2 = 0.20 pu, so the reactance from generator terminal to infinite bus is X_rest = 0.10 + 0.20 = 0.30 pu. Hence the total reactance from internal emf E′ to infinite bus is X = 0.20 + 0.30 = 0.50 pu.

Let V_t = 1∠θ and V∞ = 1∠0. Since P = V_t V∞ sin θ / X_rest = 1 pu, sin θ = 0.30 = 3/10. For stable operation, cos θ = √(1 − 9/100) = √91/10. Thus V_t = (√91/10) + j(3/10) pu.

Generator current is I = (V_t − V∞)/(j X_rest). Therefore E′ = V_t + j X′_d I = V_t + (X′_d/X_rest)(V_t − V∞) = V_t + (2/3)(V_t − 1). So E′ = (5/3)V_t − 2/3 = ((√91 − 4)/6) + j(1/2) pu. Thus |E′| = √[((√91 − 4)/6)² + (1/2)²] = (1/3)√(29 − 2√91) ≈ 1.05 pu.

Power-angle equation: P_e = (|E′| V∞ / X) sin δ = [ (1/3)√(29 − 2√91) / 0.50 ] sin δ = (2/3)√(29 − 2√91) sin δ pu ≈ 2.10 sin δ pu, where δ is the rotor angle of E′ with respect to the infinite bus. At the given operating point, δ₀ = arcsin(1/2.10) ≈ 28.44°.

For H = 4 MJ/MVA and f = 50 Hz, M = H/(π f) = 4/(50π) = 2/(25π) ≈ 0.02546 s². The swing equation is M d²δ/dt² = P_m − P_e. Hence (2/(25π)) d²δ/dt² = P_m − (2/3)√(29 − 2√91) sin δ. Equivalently, d²δ/dt² = (25π/2)(P_m − P_e) rad/s² ≈ 39.27(P_m − 2.10 sin δ) rad/s². In degrees, d²δ/dt² = 2250(P_m − 2.10 sin δ) deg/s². Initially P_m = 1 pu.

(b)

A rate-1/3 encoder is not fixed by rate alone; the standard constraint-length-3 choice is used: g1 = 1 + D + D², g2 = 1 + D², g3 = 1 + D. Input u enters shift register D1 then D2. Modulo-2 adders give y1 = u ⊕ D1 ⊕ D2, y2 = u ⊕ D2, y3 = u ⊕ D1. Diagram: u → D1 → D2; y1 is XOR of u, D1, D2; y2 is XOR of u, D2; y3 is XOR of u, D1.

Code tree: start state 00 (state = D1D2); upper branch = input 0, lower = input 1. Each branch is labelled output triplet / next state, and branches are listed by the input path that reaches them. Level 1 (from 00): input 0 → 000/00; input 1 → 111/10. Level 2: path 00 → 000/00; path 01 → 111/10; path 10 → 101/01; path 11 → 010/11. Level 3: path 000 → 000/00; 001 → 111/10; 010 → 101/01; 011 → 010/11; 100 → 110/00; 101 → 001/10; 110 → 011/01; 111 → 100/11. From level 3 onward the tree repeats, because nodes with the same state generate identical subtrees (this is what the trellis exploits).

(c)

Let C_s be capacitance of each core to sheath and C_c the capacitance between any two cores. (i) Three cores bunched to sheath: C_b = 3 C_s = 8 µF ⇒ C_s = 8/3 µF. (ii) One core to other two and sheath: C_1 = C_s + 2 C_c = 5 µF. Hence 2 C_c = 5 − 8/3 = 7/3 µF ⇒ C_c = 7/6 µF. Capacitance to neutral per phase: C_n = C_s + 3 C_c = 8/3 + 3(7/6) = 8/3 + 7/2 = 37/6 µF ≈ 6.17 µF.

Total charging kVA for 3-phase at V_L = 11 kV, f = 50 Hz: ω = 2π f = 100π rad/s. Q = ω C_n V_L² = 100π × (37/6 × 10⁻⁶) × (11000)² = (223850π/3) var = (22385π/300) kVA ≈ 234.4 kVA.

Final answers: (a) P_e = (2/3)√(29 − 2√91) sin δ pu ≈ 2.10 sin δ pu; (2/(25π)) d²δ/dt² = P_m − P_e. (b) Encoder and code tree as above. (c) C_n = 37/6 µF ≈ 6.17 µF; total charging kVA ≈ 234.4 kVA.