(a) What are the limitations of (i) Proportional (P), (ii) Integral (I), (iii) Derivative (D), and (iv) PID Controllers? What is…

(a) Controllers and positive feedback. A proportional controller gives an output proportional to the instantaneous error. It improves speed and reduces rise time, but for a step input it generally leaves a finite steady-state error or offset, because a non-zero error is required to maintain the control output. An integral controller accumulates error with time; it can remove steady-state offset, but it makes the response slower, can produce overshoot, and if the error persists while the actuator is saturated it causes integral windup, which may lead to instability when the loop recovers. A derivative controller responds to the rate of change of error; it adds damping and anticipates future error, but it amplifies high-frequency noise, can drive actuators into saturation, and is sensitive to sudden disturbances or measurement spikes. A PID controller combines these actions, so it is flexible but requires careful tuning of Kp, Ki and Kd; poor tuning can combine offset, oscillation, noise amplification and windup. Positive feedback is used when the output is deliberately fed back in phase to reinforce the input. Its main application is oscillators, where the Barkhausen criterion, loop gain unity and zero phase shift, sustains sinusoidal oscillations. It is also used in regenerative amplifiers and Schmitt triggers, where positive feedback gives rapid switching and hysteresis.

(b) 8085 arithmetic instructions. After the opcode is fetched and decoded, the ALU performs the specified transfer. For ADD M, the byte in the memory location whose address is in the HL pair is added to the accumulator: (A) ← (A) + [(HL)]. The result is stored in A and the flags S, Z, AC, P and CY are set according to the result. For ADC M, the same memory byte is added to A together with the carry flag: (A) ← (A) + [(HL)] + CY. This is used for multi-byte addition; the result is in A and the flags are affected. For DAD rp, the 16-bit content of the specified register pair rp (BC, DE, HL or SP) is added to the 16-bit HL pair: (HL) ← (HL) + (rp). The result is stored in HL; only the carry flag CY is affected, indicating a carry out of bit 15. For SBI d8, the 8-bit immediate data d8 and the carry flag are subtracted from the accumulator: (A) ← (A) − d8 − CY. The result is in A and the flags are affected; it is used for multi-byte subtraction. For DCR reg, the selected 8-bit register is decremented by one: (reg) ← (reg) − 1. The result is stored in the same register; S, Z, AC and P are affected, but CY is not changed.

(c) Ohmmeter reading. The meter branch has Rm = 50 Ω and the shunt R2 = 50 Ω, so the two parallel branches have equal resistance and the current divides equally. At 0.5 FSD the meter current is Im = 25 μA. Therefore the shunt current is also 25 μA, and the total loop current is Ib = 50 μA. The total resistance required to draw this current from VB = 1.5 V is Rtotal = VB/Ib = 1.5 / 50×10⁻⁶ = 30,000 Ω. The internal resistance of the ohmmeter is R1 plus the parallel combination of Rm and R2: Rint = 15,000 + (50×50)/(50+50) = 15,000 + 25 = 15,025 Ω. Hence the unknown resistance connected across A and B is Rx = Rtotal − Rint = 30,000 − 15,025 = 14,975 Ω, approximately 15 kΩ. Thus the ohmmeter scale reading at half full-scale deflection is about 15 kΩ.

(d) Cable insulation loss. The insulation can be represented by a parallel equivalent circuit: an ideal capacitance C in parallel with a loss resistance R. The equivalent resistance is R = 1/(ωC tanδ), so the same loss can be written P = V²/R. The loss angle δ is the small angle between the total current I and the capacitive current IC. In the phasor diagram, take voltage V along the horizontal axis; the resistive current IR is in phase with V, the capacitive current IC leads V by 90°, and the total current I is the vector sum of IR and IC. Since δ is small, IR = IC tanδ. The dielectric power loss is the real power in R: P = V IR = V IC tanδ = V²ωC tanδ. Substituting the given values, P = (11×10³)² × 2π×50 × 9×10⁻⁶ × tan(0.05°) ≈ 16.7 W. The small δ means the total current is almost purely capacitive, and only the in-phase component IR consumes real power; this loss appears as heat in the insulation.

(e) Constellation diagram. A constellation diagram plots the possible transmitted symbols in the in-phase (I) and quadrature (Q) plane. Each point represents a phase and amplitude of the modulated carrier; the distance from the origin is proportional to the signal amplitude and the angle gives the phase. For PSK with the same transmitted symbol energy Es, all points lie on a circle of radius √Es. For M = 2, BPSK has two points opposite each other, usually at 0° and 180°, i.e. +√Es and −√Es on the I axis. For M = 4, QPSK has four points separated by 90°; commonly they are placed at 45°, 135°, 225° and 315°, or equivalently at 0°, 90°, 180° and 270°, all on the same circle. For M = 8, 8-PSK has eight points separated by 45°, for example at 0°, 45°, 90°, 135°, 180°, 225°, 270° and 315°. A sketch would show a circle of radius √Es with the corresponding points marked on it: two diametrically opposite points for BPSK, four points at the corners of a square for QPSK, and eight equally spaced points for 8-PSK. In all three cases the radius remains √Es, so increasing M increases the number of phase states without changing the symbol energy, but the angular separation becomes smaller and the noise margin decreases. Thus, the answer follows from the governing mechanism in each case: error action in control, register transfer in the 8085, current division in the ohmmeter, dielectric loss in the insulation, and phase-state geometry in the constellation.