(a) Explain the ratio error and phase angle error of current transformer. 10 marks (b) The two top rows of a Routh table of a…

(a) Current-transformer error A current transformer is not an ideal current source; its excitation current I0 distorts both the magnitude and phase of the secondary current. Let Ip be the primary current, Is the secondary current, and Kn the nominal ratio, so that ideally Kn Is = Ip. The ratio error is (Kn Is - Ip)/Ip × 100%. It is usually negative because a part of the primary ampere-turns is consumed by I0, making Is smaller than the ideal value. The phase angle error δ is the angle between Ip and the reversed secondary current -Is. It arises because I0 has a magnetizing component lagging the flux and a core-loss component in phase with the flux. In the phasor diagram, Ip is the vector sum of the ideal secondary component (Ns/Np)Is and I0; the small angle between Ip and -Is is δ. These errors are important in Indian power-grid metering and protection, where CT class affects revenue metering and relay operation. They are reduced by using a high-permeability core to lower I0, stranded conductors to reduce leakage and circulating effects, and minimizing the secondary burden to keep core loss and secondary voltage low.

(b) Routh table and left-half-plane roots Using the given top rows, the characteristic polynomial is s^4 + 5s^3 + 10s^2 + 20s + 24. The Routh recurrence gives the s^2 row as (5×10 - 1×20)/5 = 6 and (5×24 - 1×0)/5 = 24. The next row initially gives (6×20 - 5×24)/6 = 0, so the auxiliary polynomial is formed from the s^2 row: 6s^2 + 24 = 0. Its derivative, 12s, replaces the zero row. The completed table is: s^4 row 1, 10, 24; s^3 row 5, 20, 0; s^2 row 6, 24, 0; s^1 row 12, 0, 0; s^0 row 24, 0, 0. The first column has no sign change, so there are no right-half-plane roots. The auxiliary equation gives s = ±j2, two roots on the imaginary axis. Since the polynomial is fourth degree, the remaining two roots lie in the left half s-plane. Factoring confirms (s^2 + 4)(s^2 + 5s + 6) = 0, so the left-half-plane roots are s = -2 and s = -3.

(c) Piezoelectric transducer undershoot For a piezoelectric transducer, the generated charge is q = dF, and the equivalent parallel RC circuit gives C dv/dt + v/τ = dq/dt, where τ = RC. For a step force F0, the output is v(t) = dF0 e^(-t/τ). A rectangular pulse of duration T is a step applied at t = 0 and removed at t = T, so for t > T the response is v(t) = dF0(e^(-t/τ) - e^(-(t-T)/τ)). Immediately after the pulse ends, at t = T+, the response is v(T+) = dF0(e^(-T/τ) - 1) = -dF0(1 - e^(-T/τ)), which is the undershoot relative to the initial positive step dF0. To keep this undershoot within 5%, 1 - e^(-T/τ) ≤ 0.05. Hence e^(-T/τ) ≥ 0.95, so T/τ ≤ -ln 0.95 = 0.0513. Therefore τ ≥ T/0.0513 ≈ 19.5T, which is normally taken as τ ≈ 20T.

(d)(i) Shannon-Fano code Sort the probabilities as 0.40, 0.19, 0.16, 0.15, 0.10. The first split is after x2 because 0.40 + 0.19 = 0.59 is closer to 0.5 than 0.40 is; assign 0 to x1 and x2, and 1 to x3, x4 and x5. Subdividing gives x1 = 00, x2 = 01. In the 1-group, split after x3, giving x3 = 10 and x4, x5 = 11; then x4 = 110 and x5 = 111. The code lengths are 2, 2, 2, 3 and 3, so L = 0.40×2 + 0.19×2 + 0.16×2 + 0.15×3 + 0.10×3 = 2.25 bits/symbol. The entropy is H = -Σ p log2 p = 0.40 log2(1/0.40) + 0.19 log2(1/0.19) + 0.16 log2(1/0.16) + 0.15 log2(1/0.15) + 0.10 log2(1/0.10) ≈ 0.5288 + 0.4552 + 0.4230 + 0.4105 + 0.3322 = 2.1498 bits/symbol. The efficiency is H/L × 100 ≈ 2.1498/2.25 × 100 = 95.54%.

(d)(ii) Huffman code and comparison For Huffman coding, combine the two smallest probabilities: 0.10 + 0.15 = 0.25; then 0.16 + 0.19 = 0.35; then 0.25 + 0.35 = 0.60; finally 0.40 + 0.60 = 1.00. One valid prefix code is x1 = 0, x4 = 100, x5 = 101, x2 = 110 and x3 = 111. The lengths are 1, 3, 3, 3 and 3, so L = 0.40×1 + 0.19×3 + 0.16×3 + 0.15×3 + 0.10×3 = 2.20 bits/symbol. The efficiency is 2.1498/2.20 × 100 ≈ 97.72%. Huffman coding is therefore shorter by 0.05 bit/symbol and more efficient than Shannon-Fano coding. The reason is that Huffman coding is an optimal prefix code, while Shannon-Fano coding only approximates equal-probability bisection; in modern communication systems this small gain can reduce required bandwidth.

(e) 8085 instruction classes The 8085 instruction set is functionally classified as data transfer, arithmetic, logical, branch, stack, I/O and machine control. Examples are MOV A,B for data transfer; ADD B for arithmetic; ANA B for logical; JMP 2000H for branch; PUSH PSW for stack; IN 01H for I/O; and HLT for machine control. These classes cover the operations used in legacy industrial controllers and embedded instrumentation, where 8085-based designs still appear in Indian process-control panels.

Thus, CT error is caused by excitation current, the Routh table shows two left-half-plane roots, the piezoelectric undershoot condition follows from exponential decay, Huffman coding outperforms Shannon-Fano coding, and 8085 instructions are grouped by the operation they perform.