(a) An LTI system with the following state-space representation is given : ẋ = [0 1] x + [0] u [0 -0.5] [k] y = [1 0]…

(a)

The plant transfer function is G(s) = C(sI − A)⁻¹B = k/[s(s + 0.5)].

For a 2% settling time, Ts = 4/(ζωₙ) = 2 s, so ζωₙ = 2 rad/s. Given ω_d = 2 rad/s, we have ωₙ = √((ζωₙ)² + ω_d²) = √(4 + 4) = 2√2 rad/s, and ζ = (ζωₙ)/ωₙ = 1/√2. Desired characteristic equation: s² + 2ζωₙ s + ωₙ² = s² + 4s + 8 = 0. Desired closed-loop poles: s = −2 ± j2.

Let the phase-lead compensator be G_c(s) = K_c (s + z)/(s + p), with p > z. Choose z = 0.5 to cancel the plant pole at −0.5. Then the loop transfer function is L(s) = G_c(s) G(s) = K_c k/[s(s + p)]. The closed-loop characteristic equation is s² + p s + K_c k = 0. Comparing with s² + 4s + 8 = 0 gives p = 4 and K_c k = 8, so K_c = 8/k. Thus the designed compensator is G_c(s) = (8/k)(s + 0.5)/(s + 4).

For passive realization, the lead network (s + 0.5)/(s + 4) is realized by an RC circuit: input to output through R1 || C, and output to ground through R2. Its transfer function is H(s) = α(s + 1/(R1 C))/(s + 1/((R1||R2) C)), where α = R2/(R1 + R2). Set 1/(R1 C) = 0.5 and 1/((R1||R2) C) = 4. Then R1 C = 2 s and (R1||R2) C = 0.25 s. This gives R2/(R1 + R2) = 0.125, so R1 = 7 R2. Choose C = 1 μF, then R1 = 2 MΩ and R2 = 2/7 MΩ = 285.7 kΩ. Including the network's own dc attenuation α = 1/8, the extra gain needed after the passive network is 64/k (a resistive attenuator can supply this only if 64/k ≤ 1, i.e. k ≥ 64; otherwise an amplifier is needed). For example, if the plant gain is k = 8, then K_c = 1 and an amplifier of gain 8 makes the realization of (s + 0.5)/(s + 4) exact.

(b)(i)

Use register pair BC as a down counter and set the zero flag when it reaches zero: DCX B MOV A, B ORA C JNZ LOOP At BC = 0000H, ORA C sets Z = 1.

(b)(ii)

LDA 2050H. Since [2050H] = F8H, after execution A = F8H.

(b)(iii)

H = 20H, L = 50H, so HL = 2050H. MVI M, 3AH. This stores 3AH at memory location 2050H.

(b)(iv)

SBI 25H. A = 37H, CY = 1 (borrow set). A ← A − 25H − CY = 37H − 25H − 1 = 11H. Thus A = 11H.

(b)(v)

CMA. A = 89H = 10001001₂. Complement = 01110110₂ = 76H. Thus A = 76H.

(c)(i)

Direct shunt. R_m = 10 Ω, I_m = 1 mA = 0.001 A, I = 1 A. n = I/I_m = 1000. R_sh = R_m/(n − 1) = 10/999 Ω. At ΔT = −5 °C: R_m′ = 10[1 + 0.004(−5)] = 9.8 Ω. R_sh′ = (10/999)[1 + 0.00015(−5)] = (10/999)(0.99925) = 9.9925/999 Ω. Current through meter at full-scale I = 1 A: I_m′ = 1 × R_sh′/(R_m′ + R_sh′) = (9.9925/999)/(9.8 + 9.9925/999) = 9.9925/9800.1925 = 0.00101962 A. Error = (I_m′ − I_m)/I_m × 100 = (0.00101962 − 0.001)/0.001 × 100 = 1.962%. The reading is high by 1.962%.

(c)(ii)

90 Ω manganin in series with moving coil. Branch resistance R_br = 10 + 90 = 100 Ω. R_sh = R_br/(n − 1) = 100/999 Ω. At ΔT = −5 °C: R_m′ = 9.8 Ω. R_s′ = 90[1 + 0.00015(−5)] = 89.9325 Ω. R_br′ = 9.8 + 89.9325 = 99.7325 Ω. R_sh′ = (100/999)(0.99925) = 99.925/999 Ω. I_m′ = 1 × R_sh′/(R_br′ + R_sh′) = (99.925/999)/(99.7325 + 99.925/999) = 99.925/99732.6925 = 0.00100193 A. Error = (I_m′ − I_m)/I_m × 100 = (0.00100193 − 0.001)/0.001 × 100 = 0.193%. The reading is high by 0.193%.