(a) Answer the following : (i) Write the recommended expression for time-current relationship of standard IDMT relay and draw…

(a)(i) The IDMT relay

The standard inverse definite minimum time characteristic recommended by IEC 60255 (and IS 3231) is

t = TMS × 0.14/[(PSM)^0.02 − 1] seconds,

where PSM, the plug-setting multiplier, is the fault current divided by the relay's current setting (I/Is), and TMS is the time-multiplier setting (0.1 to 1.0), which scales the whole curve. The related IEC curves have the same form with other constants: very inverse 13.5/(PSM − 1) and extremely inverse 80/(PSM² − 1).

The characteristic curve is drawn with PSM (1 to 20) on a logarithmic horizontal axis and operating time on the vertical axis. It starts at infinity just above PSM = 1, falls steeply, and flattens: with TMS = 1 the times are about 10 s at PSM = 2, 4.3 s at PSM = 5, 3.0 s at PSM = 10 and 2.3 s at PSM = 20. Above about PSM = 20 the curve becomes almost flat (the 'definite minimum time' part), because the electromagnet saturates in an electromechanical relay and the setting is deliberately limited in a numerical one. Lower TMS values give a family of curves of the same shape below it.

It suits electrical machines because a machine's own withstand curve is inverse: overload heating follows I²t, so a small overload can be tolerated for minutes while a heavy fault must be cleared in a fraction of a second. The IDMT curve tracks this thermal limit, rides through starting currents and short transient overloads without tripping, and its definite-minimum-time tail keeps the operating time predictable for heavy faults so that it grades with the up-stream and down-stream protection. The two settings (plug and time multiplier) make the relay easy to coordinate over a whole feeder.

(a)(ii) Earth-fault relay connections

Core-balance current transformer. All three phase conductors of the circuit pass together through the window of a single ring-type CT; the relay is connected across its secondary. Under normal load or a phase-to-phase fault the three currents sum to zero, the net flux in the core is zero and the relay carries no current. When an earth fault occurs on the protected circuit, the fault current returns through the earth and not through the window, so the vector sum of the three conductor currents is no longer zero; the residual current sets up flux in the core, a secondary current flows and the relay operates. Because the CT sees the residual directly and there is no summation of three separate CT errors, the setting can be very sensitive (a few per cent of full load); it is the usual arrangement for cables, motors and generators. The cable sheath earthing lead must be brought back through the window so that sheath current does not cause false operation.

Three-CT residual connection. A CT is fitted in each phase, the three secondaries are connected in star, and the earth-fault relay is placed in the common return (the neutral wire of the star). The current in the relay is the sum of the three secondary currents, i.e. the residual current 3I₀. Under balanced conditions and phase faults this sum is zero apart from the mismatch between the CTs, so the relay stays put; an earth fault produces a residual current and the relay operates. Phase-fault overcurrent relays may be connected in the same CT secondaries, one in each phase, with the earth-fault relay in the residual lead. The setting cannot be made as low as with a core-balance CT because the spill current from CT errors under heavy through-faults would cause maloperation; a stabilising resistor or a higher setting (10–40%) is used.

(b) The matched filter and the antipodal system

The signal s(t) is a pair of rectangular pulses of height A, one from t = 0 to 1 and one from t = 2 to 3, with a gap between them; its duration is T = 3 and its energy is E = ∫s²(t) dt = A²(1 + 1) = 2A².

(i) Impulse response. The filter matched to s(t) has h(t) = s(T − t) = s(3 − t). Reflecting the signal about t = 1.5 gives the same shape: h(t) = A for 0 ≤ t < 1 and 2 ≤ t < 3, and zero elsewhere. The sketch is s(t) itself.

(ii) Output for input s(t). The output is y(t) = s(t) * h(t), which equals the autocorrelation of s(t) shifted so that its peak falls at t = T = 3. The autocorrelation of two unit pulses separated by a gap is a large triangle of height 2A² (both pulses overlapping their own images) with two smaller triangles of height A² on either side (one pulse overlapping the other's image). The sketch therefore rises linearly from zero at t = 0 to A² at t = 1, falls to zero at t = 2, rises to the peak 2A² at t = 3, falls to zero at t = 4, rises to A² at t = 5 and returns to zero at t = 6. The sampling instant is t = 3, where y(3) = E = 2A².

(iii) Noise variance at t = 3. For white noise of spectral density N₀/2 through a filter with impulse response h(t), the output noise variance is

σ² = (N₀/2) ∫h²(t) dt = (N₀/2) × 2A² = N₀A².

(iv) Probability of error. The antipodal system sends s(t) for one symbol and −s(t) for the other, so the two sampled outputs are ±E = ±2A² with noise of variance σ² = N₀A². The optimum threshold is zero and

Pe = Q(E/σ) = Q(2A²/(A√N₀)) = Q(2A/√N₀) = Q(√(2E/N₀)) = ½ erfc(√(E/N₀)) = ½ erfc(A√(2/N₀)).

The result depends only on the energy per bit, not on the shape of s(t): the gap in the pulse costs nothing so long as the receiver is matched to it.

(c) Steady-state power limit

Let V = 1.0∠0° pu be the infinite-bus voltage, E the internal e.m.f. behind the generator reactance Xd = 0.5 pu, and X = 1.0 pu the series reactance, so that the total transfer reactance is 1.5 pu. The steady-state power limit is reached when the angle between E and V is 90°, so put E = |E|∠90° = j|E|.

The terminal voltage is the point between the two reactances. With no load tapped there, the current is the same in both, and Vt divides the voltage difference in the ratio of the reactances:

Vt = (E × X + V × Xd)/(Xd + X) = (j|E| × 1.0 + 1.0 × 0.5)/1.5 = (0.5 + j|E|)/1.5.

The terminal voltage is held at 1.20 pu, so |0.5 + j|E|| = 1.8, |E|² = 1.8² − 0.5² = 2.99, |E| = 1.729 pu.

The steady-state power limit is then

Pmax = |E||V|/(Xd + X) = 1.729 × 1.0/1.5 = 1.153 pu.

At this limit the terminal voltage leads the infinite bus by tan⁻¹(1.729/0.5) = 73.9°, and the power delivered through the line, |Vt||V| sin 73.9°/1.0 = 1.2 × 0.961 = 1.153 pu, checks. Because the limit is set by a 90° angle between E and V rather than across the line alone, it is less than the 1.2 × 1.0/1.0 = 1.2 pu that the line alone could carry at a line angle of 90°.