100+ Free HKIE Professional Assessment — Control, Automation & Instrumentation Discipline Practice Questions

Explanation: A closed-loop control system uses feedback from a sensor measuring the process variable to continuously adjust the manipulated variable, reducing error between setpoint and process variable while actively rejecting disturbances. Open-loop systems execute control action based solely on input commands or fixed calibration without observing the resulting process output.

Explanation: Increasing proportional gain $K_p$ increases loop gain, speeding up response time and reducing steady-state offset error ($e_{ss} \approx 1/(1+K_p K_{process})$). However, higher gain reduces system gain and phase margins, which increases overshoot and can drive the closed-loop system into instability.

Explanation: Derivative action acts on the rate of change of error ($ d e/dt$), anticipating future error and providing phase lead that increases damping and reduces transient overshoot. However, because $d/dt$ magnitude increases linearly with frequency ($j\omega$), derivative action significantly amplifies high-frequency sensor noise, requiring low-pass filtering.

Explanation: To calculate scaled process pressure from loop current: Signal span = $20\text{ mA} - 4\text{ mA} = 16\text{ mA}$. Fraction of range = $(12.4\text{ mA} - 4\text{ mA}) / 16\text{ mA} = 8.4 / 16 = 0.525$ (or $52.5\%$). Measured pressure = $0\text{ kPa} + 0.525 \times (150\text{ kPa} - 0\text{ kPa}) = 78.75\text{ kPa}$.

Explanation: Step 1: Error $e = SP - PV = 50\% - 45\% = +5\%$. Step 2: Proportional term output $P = K_p \cdot e = 2.0 \times 5\% = 10\%$. Step 3: Integral term output $I = \frac{K_p}{T_i} \int_0^t e(\tau) d\tau = \frac{2.0}{10.0} \times (5\% \times 5.0\text{ s}) = 0.2 \times 25 = 5\%$. Step 4: Total output change $\Delta u = P + I = 10\% + 5\% = 15\%$.

Explanation: Integral windup occurs when a persistent control error causes the integral accumulator to sum up to very large values while the control valve or motor drive is fully open or closed (saturated). When the process variable finally crosses setpoint, the controller output cannot respond immediately because the accumulated integral term must first 'unwind', leading to severe overshoot.

Explanation: A continuous LTI system is BIBO stable if and only if all roots of the closed-loop characteristic equation (poles) reside strictly in the left-half s-plane ($Re(p_i) < 0$). Any pole with a positive real part produces exponentially growing response terms ($e^{+\sigma t}$), resulting in instability.

Explanation: Step 1: Evaluate $G(j\omega)H(j\omega) = \frac{10}{j\omega (j\omega+2)(j\omega+5)} = \frac{10}{-7\omega^2 + j\omega(10 - \omega^2)}$. Step 2: Phase crossover frequency $\omega_{pc}$ occurs when imaginary component equals zero: $10 - \omega^2 = 0 \implies \omega_{pc} = \sqrt{10}\text{ rad/s}$. Step 3: Magnitude at $\omega_{pc}$ is $|G(j\sqrt{10})H(j\sqrt{10})| = \frac{10}{|-7(10)|} = \frac{10}{70} = \frac{1}{7}$. Step 4: Gain Margin $GM = \frac{1}{|G(j\omega_{pc})H(j\omega_{pc})|} = 7$. Expressed in decibels: $20 \log_{10}(7) \approx 16.9\text{ dB}$.

Explanation: Phase Margin measures system robustness against time delay or phase lag additions. It is calculated at the gain crossover frequency $\omega_{gc}$ (where magnitude equals $1$ or $0\text{ dB}$) as $PM = 180^\circ + \phi(\omega_{gc})$. A positive phase margin (typically $45^\circ$ to $60^\circ$) indicates stable closed-loop operation.

Explanation: According to the Ziegler-Nichols closed-loop ultimate frequency tuning rules for a full PID controller: $K_p = 0.6 \cdot K_u = 0.6 \times 8.0 = 4.8$. Integral time $T_i = 0.5 \cdot P_u = 0.5 \times 12.0\text{ s} = 6.0\text{ s}$. Derivative time $T_d = 0.125 \cdot P_u = 0.125 \times 12.0\text{ s} = 1.5\text{ s}$.