GATE 2024 BT Q.12: solution and answer

Official answer: ASource: Official answer key, GATE 2024 BT (IISc Bengaluru)

1. The order of a process is the highest power of $s$ in the denominator of its transfer function. That is also the order of the differential equation the transfer function comes from. Here the denominator ${\tau }_{p}s+1$ has degree 1.

2. Take the inverse Laplace transform, with $y$ and $u$ as deviation variables and zero initial conditions: ${\tau }_p\frac{dy}{dt}+y=K_{p}u(t)$. This is a first-order linear ODE. $K_p$ is the steady-state gain (output units per input unit) and ${\tau }_p$ is the time constant (in $\text{s}$ or $\text{min}$).

3. The single pole is at $s=-1/{\tau }_p$. A time constant is positive, so this is one real negative pole and the process is self-regulating. A step of size $M$ gives $y(t)=K_{p}M\left(1-e^{-t/{\tau }_p}\right)$, which reaches 63.2% of its final value $K_{p}M$ at $t={\tau }_p$.

4. Why not (C): a purely capacitive (integrating) process has $G(s)=\frac{K}{s}$. Its pole is at $s=0$ and there is no constant term in the denominator, so after a step its output ramps without bound. The $+1$ in $\frac{K_p}{{\tau }_{p}s+1}$ makes the output settle instead. Physically this is a single capacity with a resistance, such as a tank draining through a valve.

5. Why not (B) or (D): a multi-capacity process is two or more first-order lags in series, $\frac{K_p}{({\tau }_{1}s+1)({\tau }_{2}s+1)\cdots }$, so its denominator has degree 2 or more. A second-order process has an $s^2$ term, $\frac{K_p}{{\tau }^2{s}^2+2\zeta \tau s+1}$, where $\zeta$ is the damping ratio. The given denominator is only of degree 1, so it is neither.

6. So the process is first order: option (A).