(a) A closed-loop system is shown in the figure below. The maximum overshoot for the system is 40% and the peak time is 2…

(a) Values of A and B

Reduce the block diagram from the inside out. The inner loop has the forward path A/(Js) with feedback B, so its transfer function is (A/Js)/(1 + AB/Js) = A/(Js + AB). This is followed by the integrator 1/s, and the outer loop is unity feedback, so the open-loop transfer function is G(s) = A/[s(Js + AB)] and the closed-loop transfer function is

C(s)/R(s) = A/(Js² + ABs + A).

With J = 1 kg-m² this is the standard second-order form ωn²/(s² + 2ζωn s + ωn²) with ωn² = A and 2ζωn = AB.

The damping ratio follows from the overshoot: Mp = exp(−πζ/√(1−ζ²)) = 0.40, so πζ/√(1−ζ²) = −ln 0.4 = 0.9163, ζ/√(1−ζ²) = 0.2917, ζ² = 0.0851(1 − ζ²), ζ = 0.28.

The natural frequency follows from the peak time: tp = π/(ωn√(1−ζ²)) = 2 s, so ωn = π/(2 × 0.960) = 1.636 rad/s.

Hence A = ωn² = 2.68 (N-m per rad, the gain of the torque loop) and AB = 2ζωn = 2 × 0.28 × 1.636 = 0.916, giving B = 0.916/2.68 = 0.342 (N-m per rad/s, the velocity-feedback constant).

(b) Nyquist plot and range of K

The open-loop transfer function is L(s) = K/[(s² + 2s + 2)(s + 1)]. Its poles are at −1 ± j1 and −1, all in the left half-plane, so P = 0 and the closed loop is stable when the Nyquist plot of L(jω) does not encircle the point −1 + j0.

Multiplying out the denominator, (s² + 2s + 2)(s + 1) = s³ + 3s² + 4s + 2, so L(jω) = K/[(2 − 3ω²) + jω(4 − ω²)].

The plot for K > 0 starts at ω = 0 on the positive real axis at K/2, phase 0°. As ω rises the phase falls through −90° (each pole contributes lag) and the magnitude falls monotonically. The plot crosses the negative real axis where the imaginary part of the denominator vanishes, ω(4 − ω²) = 0, i.e. at the phase-crossover frequency ω = 2 rad/s, where L(j2) = K/(2 − 12) = −K/10. For ω → ∞ the magnitude goes to zero with the phase approaching −270°, so the curve enters the origin from the direction of the positive imaginary axis. The ω < 0 half is the mirror image about the real axis. The complete locus is therefore a closed curve that leaves K/2 on the real axis, sweeps clockwise through the fourth and third quadrants, crosses the negative real axis at −K/10 and returns through the second and first quadrants to K/2.

For K > 0 the point −1 is not encircled provided the real-axis crossing lies to the right of it: −K/10 > −1, i.e. K < 10. For K < 0 the plot is the same curve rotated by 180°: it starts at K/2 on the negative real axis, and −1 is encircled once the start point passes −1, i.e. for K < −2. The Routh array of the characteristic polynomial s³ + 3s² + 4s + (2 + K) confirms this: the s¹ row is (10 − K)/3 and the s⁰ row is 2 + K, both positive only for −2 < K < 10.

The system is stable for −2 < K < 10.

(c) Instruction cycle of the 8085

An instruction cycle is the time taken to fetch and execute one instruction. It is made up of one to five machine cycles, and each machine cycle of three to six T-states, a T-state being one period of the processor clock. The machine cycles the 8085 uses are: opcode fetch (4 T-states, or 6 for instructions such as CALL and conditional returns), memory read (3), memory write (3), I/O read (3), I/O write (3), interrupt acknowledge (6), and bus idle (for DAD and the internal cycles of RST/TRAP). The opcode fetch is always the first machine cycle; ALE goes high in T1 to latch the low-order address, the opcode is read on AD7–AD0 in T2–T3 and decoded in T4. As an example, STA 2050H needs four machine cycles: opcode fetch (4 T), memory read of the low address byte (3 T), memory read of the high address byte (3 T) and memory write of the accumulator (3 T), 13 T-states in all.

The shortest instruction is a single opcode fetch of 4 T-states (MOV r, r; NOP; ADD r); the longest is 18 T-states (CALL: 6 + 3 + 3 + 3 + 3). At 3 MHz one T-state is 1/(3 × 10⁶) = 0.333 μs, so the execution time of an instruction lies between 4 × 0.333 = 1.33 μs and 18 × 0.333 = 6.0 μs.

(d) Distributed capacitance and inductance of the coil

Let Cd be the self-capacitance of the coil and L its inductance. At each resonance the total capacitance is the external capacitor in parallel with Cd:

f₁ = 1/[2π√(L(C₁ + Cd))] with C₁ = 100 pF at f₁ = 10 MHz, f₂ = 1/[2π√(L(C₂ + Cd))] with C₂ = 16 pF at f₂ = 20 MHz.

Since f₂ = 2f₁, (C₁ + Cd) = 4(C₂ + Cd): 100 + Cd = 64 + 4Cd, so Cd = 36/3 = 12 pF.

Then L = 1/[(2πf₁)²(C₁ + Cd)] = 1/[(6.283 × 10⁷)² × 112 × 10⁻¹²] = 1/(3.948 × 10¹⁵ × 1.12 × 10⁻¹⁰) = 2.26 × 10⁻⁶ H, i.e. L ≈ 2.26 μH.

(e) Wattmeter reading with the pressure coil across the other two lines

With the current coil in the red line and the pressure coil across red and neutral the wattmeter reads the per-phase power: W₁ = Vph I cos φ = 5000 W. The phase voltage is 400/√3 = 230.9 V, so cos φ = 5000/(230.9 × 28) = 0.773 and φ = 39.4°, taken as lagging.

Now the pressure coil is connected across the other two lines. Take VRN as reference; for the sequence RYB, VYN lags by 120° and VBN by 240°. The line voltage between the yellow and blue lines, VYB = VYN − VBN, has magnitude 400 V and lags VRN by 90°, i.e. it is in quadrature with the phase voltage of the current-coil phase. The current IR lags VRN by φ. The angle between VYB and IR is 90° − φ, so the reading is

W₂ = VL I cos(90° − φ) = VL I sin φ = 400 × 28 × sin 39.4° = 400 × 28 × 0.634 = 7.10 kW.

If the coil's marked terminal is put on the blue line instead, the applied voltage is VBY = −VYB and the pointer deflects backwards by the same amount; the pressure-coil leads are then interchanged to read +7.10 kW. This is the reactive-power method of measurement: the reading is VL I sin φ, so the total reactive power of the balanced load is √3 × 7.10 = 12.3 kVAR, which checks against √3 × 400 × 28 × 0.634 = 12.3 kVAR.