(a) The figure shows a unity feedback system with G(s) = 2/[s(s+1)(s+2)(s+3)]. (i) Sketch the approximate polar plot of…

(a) Polar plot, stability and gain margin

G(jω) = 2/[jω(1 + jω)(2 + jω)(3 + jω)], a type-1 system of order 4. Magnitude |G| = 2/[ω√(1 + ω²)√(4 + ω²)√(9 + ω²)] and phase φ = −90° − tan⁻¹ω − tan⁻¹(ω/2) − tan⁻¹(ω/3).

(i) Polar plot. As ω → 0⁺, |G| → ∞ at −90°, and the real part tends to the finite value −(2/6)(1 + 1/2 + 1/3) = −0.611, so the plot starts from an asymptote at Re = −0.611 parallel to the negative imaginary axis. As ω increases the phase decreases through −180° in the third quadrant, then passes through the second and first quadrants and approaches the origin along the positive real axis at −360° (|G| → 0). The plot crosses the negative real axis once.

(ii) Stability. The phase crossover frequency satisfies tan⁻¹ω + tan⁻¹(ω/2) + tan⁻¹(ω/3) = 90°. ω = 1 rad/s gives 45° + 26.57° + 18.43° = 90°, so ω_pc = 1 rad/s. Here |G(j1)| = 2/(1 × √2 × √5 × √10) = 0.2, so the plot crosses the negative real axis at −0.2. This point is to the right of −1 + j0, so the critical point is not encircled. There is no open-loop pole in the right half plane, hence the closed-loop system is stable.

(iii) Gain margin. GM = 1/|G(jω_pc)| = 1/0.2 = 5, that is 20 log₁₀ 5 = 13.98 dB. A Routh check on s⁴ + 6s³ + 11s² + 6s + 2K = 0 gives stability for K < 5, where K = 1 is the given gain, so the gain can be raised five times.

(b) Ammeter range extension

Shunt R_sh = I_m R_m/(I − I_m), with V_m = I_m R_m.

Ammeter 1: V_m = 0.2 × 1 = 0.2 V, so R_sh1 = 0.2/(20 − 0.2) = 0.0101 Ω. Ammeter 2: V_m = 0.25 × 2 = 0.5 V, so R_sh2 = 0.5/(20 − 0.25) = 0.0253 Ω.

The effective resistance of each extended ammeter is V_m/20: R₁′ = 0.2/20 = 0.01 Ω and R₂′ = 0.5/20 = 0.025 Ω. In parallel, the 15 A divides inversely to these resistances: I₁ = 15 × 0.025/(0.01 + 0.025) = 10.71 A, and I₂ = 15 × 0.01/0.035 = 4.29 A. These are the readings, since each meter is scaled to 20 A full scale.

(c) Short-circuit capacity after interconnection

Reactance of each station at 11 kV: X₁ = 11²/1500 = 0.0807 Ω and X₂ = 11²/1000 = 0.121 Ω. The cable reactance is X_c = 0.7 Ω.

For a fault at station 1, the other station reaches it through the cable, so X_eq1 = X₁ ∥ (X_c + X₂) = 0.0807 ∥ 0.821 = 0.0734 Ω. The short-circuit capacity is S₁ = 11²/0.0734 = 1647 MVA.

For a fault at station 2, X_eq2 = X₂ ∥ (X_c + X₁) = 0.121 ∥ 0.7807 = 0.1048 Ω, so S₂ = 11²/0.1048 = 1155 MVA.