(a) Write the state space equation for the following circuit. [Circuit diagram: voltage source V(t), series resistance R and…

(a) State-space model of the series R-L-C circuit (10 marks)

The circuit has two independent energy stores, so it needs two state variables: the capacitor voltage v_C and the inductor current i (the same current flows through R, L and C). Take x₁ = v_C, x₂ = i and the input u = V(t).

Capacitor. i = C dv_C/dt, so ẋ₁ = (1/C) x₂.

KVL around the loop. V(t) = R i + L di/dt + v_C, so ẋ₂ = −(1/L) x₁ − (R/L) x₂ + (1/L) u.

In standard form ẋ = Ax + Bu:

[ẋ₁; ẋ₂] = [[0, 1/C], [−1/L, −R/L]] [x₁; x₂] + [0; 1/L] V(t).

Taking the capacitor voltage as output, y = v_C = [1 0] x, that is, C = [1 0] and D = 0.

Check. det(sI − A) = s² + (R/L)s + 1/(LC), the characteristic equation of the series RLC circuit, with ω_n = 1/√(LC) and ζ = (R/2)√(C/L).

(b) Oscilloscope terminal voltage and 3 dB frequency (10 marks)

Equivalent circuit. The source (1 V behind R_s = 600 Ω) drives the parallel combination of R_i = 1 MΩ and the total shunt capacitance C = C_i + C_cc = 30 + 100 = 130 pF.

Transfer function. With Z_in = R_i/(1 + jωR_iC),

V_i/V_s = Z_in/(R_s + Z_in) = [R_i/(R_i + R_s)] / [1 + jω(R_s ∥ R_i)C].

The low-frequency divider ratio is R_i/(R_i + R_s) = 10⁶/(10⁶ + 600) = 0.9994. The time constant uses R_s ∥ R_i = 599.64 Ω, so τ = 599.64 × 130 × 10⁻¹² = 7.80 × 10⁻⁸ s and the corner frequency is f_c = 1/(2πτ) = 2.04 MHz.

At 100 Hz. ωτ = 2π × 100 × 7.80 × 10⁻⁸ = 4.9 × 10⁻⁵, which is negligible, so

|V_i| = 0.9994/√(1 + (4.9 × 10⁻⁵)²) ≈ 0.9994 V ≈ 1 V, with a phase lag of only 0.003°. The capacitance has no effect at this frequency.

3 dB frequency. The output falls 3 dB below its low-frequency value at f = f_c = 1/(2πτ) ≈ 2.04 MHz. Measured strictly against 1 V, the 3 dB level is |V_i/V_s| = 0.708, which is reached at f = f_c√[(0.9994/0.708)² − 1] ≈ 2.03 MHz; the two values agree because the divider loss of 0.005 dB is negligible. The scope's bandwidth is limited by the source resistance and the cable plus input capacitance.

(c) Percentage differential relay: restraining-coil turns and the need for biasing (10 marks)

Turns for balance. The operating coil carries the difference of the two CT secondary currents, I₁ − I₂ = 3.4 − 3.0 = 0.4 A. The restraining coil, connected across the two secondaries with its mid-point tapped, carries the through (mean) current (I₁ + I₂)/2 = (3.4 + 3.0)/2 = 3.2 A. At the balance condition the operating and restraining ampere-turns are equal:

N_op (I₁ − I₂) = N_r (I₁ + I₂)/2

400 × 0.4 = N_r × 3.2, so N_r = 50 turns.

The relay therefore operates when the differential current exceeds N_r/N_op = 50/400 = 12.5% of the mean through current.

Why biasing is needed. An unbiased differential relay has a fixed pick-up. In the healthy state, or for a fault outside the protected zone (a through-fault), the CT secondary currents should be equal, but they are not: the CT ratios differ slightly, the magnetising currents differ, a tap changer alters the ratio, and the CTs may saturate unequally when the through current is heavy. The resulting spurious differential current is small at low current but grows with the through current. A fixed low setting would then trip wrongly on external faults, and raising the setting to avoid this would make the relay insensitive to internal faults of low current. Biasing (restraint) solves this: the restraining torque is proportional to the through current, so the pick-up rises automatically with the through current. The relay is stable for heavy through-faults and remains sensitive for internal faults, where the restraint is small and the differential current is large.

(d) HVDC modes of operation and sequence of events after a disturbance near the inverter (10 marks)

Modes of operation. An HVDC link can operate in:

• Monopolar mode: one conductor at one polarity with earth return (or a metallic return conductor).
• Bipolar mode: two poles of opposite polarity, each with its own converter; normally the current in the earth is nearly zero, and either pole can continue alone as monopolar during an outage.
• Homopolar mode: two (or more) conductors of the same polarity, with earth or a metallic return.

In normal operation, the rectifier keeps constant current (CC control) and the inverter holds a constant extinction angle (CEA control), and a voltage-dependent current-order limiter (VDCOL) reduces the current order when the dc voltage falls.

Sequence of events for a large disturbance near an inverter on CEA control, on a heavily loaded link:

1. The ac voltage at the inverter bus drops, so the commutating voltage falls.
2. The inverter dc voltage V_di = (3√2/π)E cos γ − (3/π)X_c I_d falls, while the rectifier voltage is unchanged, so the dc current I_d rises sharply.
3. The larger current and the lower voltage increase the overlap angle μ, so the extinction angle γ = β − μ falls below its set value.
4. The CEA controller advances the firing (increases β) to restore γ, but it acts only after a delay of a fraction of a cycle to a cycle.
5. If γ falls below the valve turn-off angle γ_min (typically about 15°), the outgoing valve is not turned off in time, and commutation failure occurs.
6. The bridge is short-circuited, so the inverter dc voltage collapses towards zero, the dc current surges, and the ac power delivered falls to zero; the rectifier CC control and VDCOL respond by reducing the current order.
7. When the disturbance clears and the ac voltage recovers, commutation succeeds and the link recovers within a few cycles, if the failure is not repeated; repeated commutation failures cause the bridge to be blocked and bypassed.

(e) Delta modulator: slope overload and output SQNR (10 marks)

The signal bandwidth is W = 3 kHz, so the Nyquist rate is 2W = 6 kHz and the sampling rate is f_s = 3 × 6 = 18 kHz. The step size is Δ = 0.25 V.

(e)(i) Largest amplitude without slope overload. The maximum slope the modulator can follow is Δf_s = 0.25 × 18000 = 4500 V/s. The maximum slope of A sin(2πf_m t) is 2πf_mA, so for no slope overload 2πf_mA ≤ Δf_s:

A_max = Δf_s/(2πf_m) = 4500/(2π × 1000) = 0.716 V.

(e)(ii) Output SQNR. The signal power is S = A²/2 = 0.716²/2 = 0.2565 W (normalised to 1 Ω). The quantizing noise power, uniformly spread over the range 0 to f_s, is Δ²/3, and the low-pass filter of bandwidth W passes only the fraction W/f_s of it:

N = (Δ²/3)(W/f_s) = (0.0625/3)(3000/18000) = 3.47 × 10⁻³ W.

SQNR = S/N = 0.2565/0.003472 = 73.9, which is 18.7 dB.

(The same result comes from SQNR = 3f_s³/(8π²f_m²W).)