Subtopic: Fundamentals of Biomolecules, enzyme kinetics
Question Statement:
In the open-loop process shown in the figure, the input U(s), the transfer function
Gp(s) and the output Y(s) are given in the Laplace domain in terms of the Laplace
variable s. For this process, which of the following is true?
(where M, šp, KP, are the magnitude of the input, the characteristic time and the gain
for the process, respectively)
Options:
(A) y(t) = Kp(1 - e-t špā )
(B) y(t) = MKp(1 - e-t špā )
(C) y(t) = Kp(M - e-t špā )
(D) y(t) = Kp(1 - Me-t špā )
Official Answer Key:
B
Step-by-Step Solution & Explanation
### **Official Solution & Concept Analysis (GATE BT 2026 Q.36)**
**1. Governing Concept**:
- **Topic**: Process Control & First-Order System Dynamic Response
- For a standard first-order system with transfer function:
$G(s) = \frac{Y(s)}{U(s)} = \frac{K_p}{\tau s + 1}$
- Subjected to a step change of magnitude $M$ in input ($U(s) = M/s$):
$Y(s) = \frac{K_p M}{s(\tau s + 1)} \implies y(t) = K_p M \left(1 - e^{-t/\tau}\right)$
- At $t = \tau$ (one time constant): $y(\tau) = K_p M (1 - e^{-1}) \approx 0.632 K_p M$ (63.2% of total response).
- The steady-state ultimate value is $y(\infty) = K_p M$.
**2. Option Analysis**:
- Evaluating the response curve parameters against step input magnitude and time constant confirms Option (B) matches the canonical first-order step response.
- **Option (B) is CORRECT**.
- **Options (A), (C), and (D) are INCORRECT**: Mismatch the steady-state gain $K_p$ or the 63.2% response time constant $\tau$.
**3. Final Answer**:
- **Correct Choice**: **B**
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