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

(a) Using the classical swing-equation model, neglect resistance and take both lines in service. The line equivalent reactance is ∦: XL = 0.4 ∥ 0.4 = (0.4×0.4)/(0.4+0.4) = 0.20 p.u. Total reactance from transient internal voltage E′ to infinite bus: X = Xd′ + Xt + XL = 0.20 + 0.10 + 0.20 = 0.50 p.u. Transformer-plus-line reactance from terminal to infinite bus: Xtl = Xt + XL = 0.10 + 0.20 = 0.30 p.u.

Let terminal voltage be Vt = |Vt|∠θ = 1.05∠θ p.u. Power delivered to infinite bus is P = (|Vt||V|/Xtl) sin θ. 0.8 = (1.05×1.0/0.30) sin θ sin θ = (0.8×0.30)/1.05 = 0.228571. θ = 13.21°, cos θ = 0.973527. Thus Vt = 1.05(0.973527 + j0.228571) = 1.02220 + j0.24000 p.u.

Current from terminal to infinite bus: I = (Vt − V)/(jXtl) = (0.02220 + j0.24000)/(j0.30) I = 0.8000 − j0.07401 p.u.

Now E′ behind Xd′: E′ = Vt + jXd′I = (1.02220 + j0.24000) + j0.20(0.8000 − j0.07401) = 1.03701 + j0.40000 p.u. |E′| = √(1.03701² + 0.40000²) = 1.1115 p.u. δ0 = tan⁻¹(0.40000/1.03701) = 21.09° = 0.368 rad.

Power-angle equation after the initial condition is Pe = (|E′||V|/X) sin δ = (1.1115×1.0/0.50) sin δ = 2.223 sin δ p.u.

Mechanical input is initially Pm = 0.8 p.u. Using swing equation (2H/ωs) d²δ/dt² = Pm − Pe. H = 5 MJ/MVA, f = 50 Hz, ωs = 2π×50 = 314.16 rad/s. M = 2H/ωs = 10/314.16 = 0.03183 s²/rad.

Therefore 0.03183 d²δ/dt² = 0.8 − 2.223 sin δ or d²δ/dt² = 25.13 − 69.83 sin δ rad/s², with initial conditions δ(0) = 0.368 rad, dδ/dt(0) = 0. Valid for constant Pm, constant E′, no damping, and both line sections in service.

(b) Nyquist stability criterion: For open-loop transfer function 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 open-loop poles in the right-half plane and N is the number of clockwise encirclements of the point −1 + j0 by the Nyquist plot of L(s).

Here L(s) = G(s)H(s) = [1/{s(s+1)}] × 2(s−2) = 2(s−2)/[s(s+1)].

Open-loop poles are s = 0 and s = −1. There are no right-half-plane poles, so P = 0 for Nyquist counting. However, because of the pole at s = 0, the open-loop system is not asymptotically or BIBO stable; it is only marginally stable.

Now L(jω) = 2(jω−2)/[jω(jω+1)] = 6/(ω²+1) + j 2(2−ω²)/[ω(ω²+1)].

For ω > 0:

• as ω → 0⁺, L → 6 + j∞,
• at ω = √2, L = 2 + j0,
• as ω → ∞, L → 0 − j0. The negative-frequency plot is its mirror image. The pole at the origin is indented into the right-half plane, producing a large arc centred at 6 + j0, passing through −∞ on the real axis. Hence the complete Nyquist plot encircles the critical point −1 + j0 once clockwise.

Now closed-loop characteristic equation: 1 + L(s) = 0 1 + 2(s−2)/[s(s+1)] = 0 s(s+1) + 2(s−2) = 0 s² + 3s − 4 = 0 (s − 1)(s + 4) = 0. The closed-loop poles are s = 1 and s = −4. Since one pole lies in the right-half plane, the closed-loop system is unstable.

Using Nyquist: P = 0, N = 1 clockwise encirclement, so Z = N + P = 1. Thus there is one closed-loop right-half-plane pole. The closed-loop system is unstable.

(c) Advantages of a spectrum analyzer:

• Displays amplitude versus frequency, so harmonics, sidebands, noise and spurious components are directly visible.
• Useful over a very wide frequency range, from audio to microwave.
• High dynamic range and calibrated amplitude/frequency readout.
• FFT types give fast real-time spectrum analysis for transient and modulated signals.

Disadvantages:

• Cost and complexity are high, especially for microwave and high-resolution models.
• Real-time bandwidth may be limited.
• Resolution bandwidth, sweep time and dynamic range involve trade-offs.
• Strong signals can overload the input, causing spurious responses.

Applications:

• RF/microwave transmitter and receiver testing.
• EMI/EMC measurements and spectrum monitoring.
• Harmonic distortion, intermodulation and phase-noise measurements.
• Audio, vibration and communication-signal analysis.