(a) Draw the sequence networks and calculate the load sequence impedances of a load circuit as shown in figure. The load circuit…

(a) Sequence networks and load sequence impedances The load has two balanced parts across lines R, Y, B: a star of z₁ = 4 + j6 Ω per phase whose star point N is earthed through z_n = j4 Ω, and a delta of z₂ = −j45 Ω per phase. (The printed delta leaves one corner unlabelled; it is taken as connected across all three lines.) Delta to star: z₂/3 = −j15 Ω per phase with an isolated star point. Positive and negative sequence. Balanced currents sum to zero at N, so no current flows in z_n. Both networks show z₁ in parallel with z₂/3, from the line terminal to the reference bus: Z₁ = Z₂ = (4 + j6)(−j15)/(4 + j6 − j15) = (90 − j60)/(4 − j9) = (900 + j570)/97 Z₁ = Z₂ = 9.28 + j5.88 Ω ≈ 10.98∠32.3° Ω (equal, as the load is static). Zero sequence. The three zero-sequence currents flow in phase through z₁ and return together through z_n, which therefore appears as 3z_n. The delta has no path to earth, so its branch is open for zero sequence: Z₀ = z₁ + 3z_n = 4 + j6 + j12 = 4 + j18 Ω ≈ 18.44∠77.5° Ω. The three sequence networks are independent: positive and negative are z₁ ∥ z₂/3; zero is z₁ + 3z_n in series.

(b) Block diagram and transfer function of the RC ladder Element equations (zero initial conditions): I₁ = (V_i − V₁)/R₁; V₁ = (I₁ − I₂)/(sC₁); I₂ = (V₁ − V₀)/R₂; V₀ = I₂/(sC₂). Each resistor becomes a gain block 1/R after a summing point (voltage difference → current) and each capacitor an integrator block 1/(sC) after a summing point (net current → voltage), with V₁ and V₀ fed back to the earlier summing points.

Reduction:

1. Inner loop (V₀ from I₂ through 1/(sC₂), fed back to the R₂ summing point): V₀/V₁ = 1/(1 + sR₂C₂).
2. Since I₂ = sC₂V₀, the feedback from V₁ to the I₁ − I₂ summing point is sC₂/(1 + sR₂C₂).
3. Middle loop: V₁/I₁ = (1 + sR₂C₂)/[sC₁(1 + sR₂C₂) + sC₂].
4. Outer loop with 1/R₁ and unity feedback: V₁/V_i = (1 + sR₂C₂)/[sR₁C₁(1 + sR₂C₂) + sR₁C₂ + 1 + sR₂C₂].
5. Cascade with step 1: V₀(s)/V_i(s) = 1/[R₁R₂C₁C₂s² + (R₁C₁ + R₁C₂ + R₂C₂)s + 1] The R₁C₂ term is the loading of stage 2 on stage 1; the dc gain is 1, as nodal analysis also confirms.

(c) Convolutional code Rate 1/3, K = 3: g₁ = 1 + D, g₂ = 1 + D², g₃ = 1 + D + D². With state (uₙ₋₁, uₙ₋₂) the outputs are v₁ = uₙ ⊕ uₙ₋₁, v₂ = uₙ ⊕ uₙ₋₂, v₃ = uₙ ⊕ uₙ₋₁ ⊕ uₙ₋₂. State diagram (a = 00, b = 10, c = 01, d = 11), branch input/output: a→b 1/111; b→c 0/101; b→d 1/010; c→a 0/011; c→b 1/100; d→c 0/110; d→d 1/001; a→a 0/000.

Splitting a into start and end and writing the node equations X_b = D³N + DN X_c, X_c = D²X_b + D²X_d, X_d = DN X_b + DN X_d, X_a′ = D²X_c gives T(D, N) = D⁷N/(1 − DN − D³N), so T(D) = D⁷/(1 − D − D³) = D⁷ + D⁸ + D⁹ + 2D¹⁰ + … d_free = 7 (input 100 gives 111, 101, 011), so up to three errors can be corrected. Catastrophic check: gcd(1 + D, 1 + D², 1 + D + D²) = 1, and the only zero-weight loop is the self-loop at a (the loop at d has weight 1). The code is not catastrophic.