(a) A synchronous machine is connected to an infinite bus through a transformer and a double circuit line as shown in figure. The…

(a) Equation of motion of the rotor Use the classical model (constant E′ behind X_d′, resistances neglected, both lines in service). Two lines in parallel: X_L = 0.4/2 = 0.2 pu. Terminal to infinite bus: X_tl = 0.1 + 0.2 = 0.3 pu. Total X = 0.2 + 0.3 = 0.5 pu. Initial condition. P = (|V_t||V|/X_tl) sin θ, so 0.8 = (1.05/0.3) sin θ and θ = 13.21°, giving V_t = 1.05∠13.21° = 1.0222 + j0.24 pu. I = (V_t − V)/(jX_tl) = (0.0222 + j0.24)/(j0.3) = 0.8 − j0.074 pu. E′ = V_t + jX_d′I = 1.0222 + j0.24 + j0.2(0.8 − j0.074) = 1.0370 + j0.40 = 1.111∠21.09° pu, so δ₀ = 21.09° = 0.368 rad. Power–angle curve. P_e = (|E′||V|/X) sin δ = (1.111/0.5) sin δ = 2.223 sin δ pu, with P_m = 0.8 pu. Swing equation. (H/πf) d²δ/dt² = P_m − P_e, with H/πf = 5/(π × 50) = 0.0318 s²/rad: 0.0318 d²δ/dt² = 0.8 − 2.223 sin δ, i.e. d²δ/dt² = 25.13 − 69.84 sin δ rad/s², with δ(0) = 0.368 rad and dδ/dt(0) = 0. (δ in radians; no damping.)

(b) Nyquist criterion for the given loop Criterion: for L(s) = G(s)H(s), the number of closed-loop poles in the right-half plane is Z = N + P, where P is the number of right-half-plane open-loop poles and N is the number of clockwise encirclements of −1 + j0 by the Nyquist contour image (with indentation round poles on the imaginary axis). Here L(s) = [1/(s(s + 1))] × 2(s − 2) = 2(s − 2)/(s(s + 1)). The open-loop poles are s = 0 and s = −1, so P = 0: the open loop has no right-half-plane pole, though the pole at the origin makes it only marginally stable. L(jω) = 6/(ω² + 1) + j2(2 − ω²)/[ω(ω² + 1)]. For ω > 0: as ω → 0⁺, L → 6 + j∞ (a vertical asymptote at Re = 6); L crosses the real axis only at ω = √2, at +2; as ω → ∞, L → 0 from below the real axis. The ω < 0 branch is the mirror image. Indenting round s = 0 gives an infinite arc of clockwise sense through −∞ on the negative real axis, closing the contour on the left. This closed curve encircles −1 + j0 once clockwise: N = 1, so Z = 1 + 0 = 1.

The closed loop is unstable (one right-half-plane pole). Check: 1 + L = 0 gives s² + 3s − 4 = (s − 1)(s + 4) = 0, poles +1 and −4.

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