(a) Determine the inductance of a single-phase transmission line consisting of three conductors of 2 cm radii in the 'go'…

(a) Inductance of the composite-conductor line

Composite conductor A (the 'go') is the three strands a, b, c, radius 2 cm, one above the other 4 m apart; composite conductor B (the 'return') is the two strands a', b', radius 4 cm, 4 m apart, in a second column 6 m away, with a' level with a and b' level with b. The inductance of each composite conductor is L = 2 × 10⁻⁷ ln(Dm/Ds) H/m, with Dm the mutual GMD between the two groups and Ds the self GMD of the group.

Mutual GMD. The six distances from a strand of A to a strand of B are: a–a' = 6, a–b' = √(6² + 4²) = 7.211, b–a' = 7.211, b–b' = 6, c–a' = √(6² + 8²) = 10, c–b' = 7.211 m.

Dm = (6 × 7.211 × 7.211 × 6 × 10 × 7.211)^(1/6) = (360 × 7.211³)^(1/6) = (1.350 × 10⁵)^(1/6) = 7.16 m.

Self GMD of A. With r' = 0.7788 × 0.02 = 0.01558 m, the nine distances are r', 4, 8 (from a), 4, r', 4 (from b) and 8, 4, r' (from c):

DsA = (r'³ × 4 × 8 × 4 × 4 × 8 × 4)^(1/9) = (0.01558³ × 16384)^(1/9) = (0.0619)^(1/9) = 0.734 m.

Self GMD of B. With r' = 0.7788 × 0.04 = 0.03115 m: DsB = (r' × 4 × 4 × r')^(1/4) = √(4 × 0.03115) = 0.353 m.

Inductances: LA = 2 × 10⁻⁷ ln(7.16/0.734) = 2 × 10⁻⁷ × 2.278 = 4.56 × 10⁻⁷ H/m = 0.456 mH/km, and LB = 2 × 10⁻⁷ ln(7.16/0.353) = 2 × 10⁻⁷ × 3.010 = 6.02 × 10⁻⁷ H/m = 0.602 mH/km.

The loop inductance of the single-phase line is L = LA + LB = 1.06 mH/km (1.06 × 10⁻³ H per km of line). The return side has the higher inductance despite its larger strands because it has fewer of them and hence a smaller self GMD.

(b) Terminal voltage of the generator

Work in per unit on a 100 MVA base with voltage bases 11 kV, 220 kV and 66 kV in the three sections. Both transformers are rated at the base, so XT1 = 0.12 pu and XT2 = 0.10 pu. The line reactance on the 220 kV base is Xline = 100/(220²/100) = 100/484 = 0.2066 pu. Total series reactance X = 0.12 + 0.2066 + 0.10 = 0.4266 pu.

At the load bus V₂ = 60/66 = 0.9091 pu, taken as reference. The load is P = 0.6 pu at 0.9 pf lagging, so Q = 0.6 tan(cos⁻¹ 0.9) = 0.6 × 0.4843 = 0.2906 pu and the current is

I = (P − jQ)/V₂* = (0.6 − j0.2906)/0.9091 = 0.660 − j0.320 pu.

The generator terminal voltage on the 11 kV base is

V₁ = V₂ + jXI = 0.9091 + j0.4266(0.660 − j0.320) = 0.9091 + 0.1364 + j0.2816 = 1.0455 + j0.2816 pu,

|V₁| = √(1.0455² + 0.2816²) = 1.083 pu, i.e. 1.083 × 11 = 11.9 kV line-to-line, at an angle of 15.1° ahead of the load-bus voltage. The generator must be held at about 11.9 kV for the substation bus to stay at 60 kV.

(c) Arc in a circuit breaker and the restriking voltage

As the contacts begin to part, the area of contact shrinks to a few points, the current density and the resistance at these points rise sharply, and the metal there is heated to incandescence. Electrons are emitted from the hot cathode spot by thermionic emission, and, because the gap is at first only a few microns wide, the voltage gradient across it is very high and field emission adds to them. The emitted electrons, accelerated by the field, ionise the gas in the gap by collision; the resulting column of ionised gas is a good conductor and the current continues to flow through it as an arc. The arc is sustained because its own energy input (arc voltage times current) keeps the column hot enough for thermal ionisation to continue, the cathode spot stays hot enough for emission, and the surrounding medium is continuously ionised faster than it recombines. The arc is extinguished only when, at a natural current zero, the dielectric strength of the gap recovers faster than the voltage across it rises: this is the race between recovery voltage and dielectric recovery that decides breaker performance.

Restriking voltage. Consider a single phase of a solidly grounded system, with the source inductance L per phase (including the generator and transformer) and the stray capacitance C across the breaker contacts and to ground. Let the fault be cleared at a current zero; in an inductive circuit this occurs when the source voltage is at its peak, v = Vm, and for the few microseconds of the transient Vm can be taken as constant. After current zero the current through L flows into C, so, with the arc extinguished,

L di/dt + vc = Vm, i = C dvc/dt, giving LC d²vc/dt² + vc = Vm.

With vc(0) = 0 (the arc had short-circuited the capacitance) and i(0) = 0, the solution is

vc(t) = Vm(1 − cos ωn t), ωn = 1/√(LC).

This is the restriking voltage across the contacts: it oscillates at the natural frequency fn = 1/(2π√(LC)), reaches a peak of 2Vm at t = π√(LC), and its maximum rate of rise is Vm ωn (the average up to the first peak being 2Vm/(π√(LC))). Here Vm = √2 VL/√3 is the peak phase voltage, the neutral being solidly grounded. Damping resistance in the circuit reduces the peak below 2Vm.

(d) Mean square quantization error of a non-uniform quantizer

Divide the input range into intervals of widths Δᵢ, the ith interval having representation level yᵢ at its centre. Let f(x) be the probability density of the input amplitude. The mean square quantization error is

σq² = Σᵢ ∫(interval i) (x − yᵢ)² f(x) dx.

Because each Δᵢ is small compared with the excursion of the signal, f(x) is nearly constant over an interval and can be taken as its value fᵢ there. Writing e = x − yᵢ, which runs from −Δᵢ/2 to Δᵢ/2,

σq² ≈ Σᵢ fᵢ ∫(−Δᵢ/2 to Δᵢ/2) e² de = Σᵢ fᵢ Δᵢ³/12.

But fᵢΔᵢ is the probability pᵢ that the input lies in the ith interval, so

σq² ≈ (1/12) Σᵢ Δᵢ² pᵢ,

which is the required result. For a uniform quantizer every Δᵢ = Δ and Σ pᵢ = 1, and the expression reduces to the familiar Δ²/12; the non-uniform quantizer (a compander, for instance) makes Δᵢ small where pᵢ is large, which is how it lowers the error for signals such as speech that spend most of their time at small amplitudes.

(e) Syndromes and cosets in the standard array

Let C be an (n, k) linear block code with parity-check matrix H, so that a vector v is a code word if and only if its syndrome s = vHᵀ is zero. The standard array lists the 2^(n−k) cosets of C: each row is e + C = {e + c : c ∈ C} for a coset leader e, and the rows partition the space of all 2ⁿ vectors.

Same syndrome within a coset. Any vector in the coset e + C is of the form v = e + c with c ∈ C. Its syndrome is

s(v) = (e + c)Hᵀ = eHᵀ + cHᵀ = eHᵀ,

because cHᵀ = 0. Every vector in the row therefore has the same syndrome, that of its coset leader.

Uniqueness to the coset. Suppose two cosets e₁ + C and e₂ + C had the same syndrome, e₁Hᵀ = e₂Hᵀ. Then (e₁ − e₂)Hᵀ = 0, so e₁ − e₂ is a code word c, e₁ = e₂ + c, and e₁ lies in e₂ + C. Since cosets are either identical or disjoint, the two cosets are the same one. Hence distinct cosets have distinct syndromes.

There are 2^(n−k) cosets and 2^(n−k) possible syndromes, so the correspondence is one-to-one; this is what allows syndrome decoding: compute s = rHᵀ, look up the coset leader with that syndrome, and subtract it from r.