GATE 2025 EC Q.53: solution and answer

GATE 2025 · ECQ.53 · 2 marks · MSQ

Consider a system where $x_{1} (t), x_{2} (t)$ , and $x_{3} (t)$ are three internal state signals and $u (t)$ is the input signal. The differential equations governing the system are given by

\frac{d}{d t} x_{1} (t) x_{2} (t) x_{3} (t) = 200 0 - 2 0 000 x_{1} (t) x_{2} (t) x_{3} (t) + 111 u (t) .

Which of the following statements is/are TRUE?

(A)
The signals $x_{1} (t), x_{2} (t)$ , and $x_{3} (t)$ are bounded for all bounded inputs
(B)
There exists a bounded input such that at least one of the signals $x_{1} (t), x_{2} (t)$ , and $x_{3} (t)$ is unbounded
(C)
There exists a bounded input such that the signals $x_{1} (t), x_{2} (t)$ , and $x_{3} (t)$ are unbounded
(D)
The signals $x_{1} (t), x_{2} (t)$ , and $x_{3} (t)$ are unbounded for all bounded inputs

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Answer and solution

ANSWER

Only (B) is true. The bounded input $u (t) = 1$ drives $x_{3} = x_{3} (0) + t$ (and $x_{1}$ ) to infinity. But $x_{2}$ , with its pole at $s = - 2$ , stays bounded for every bounded input, so (C) and (D) are false and (A) fails.

Official answer: BSource: Official answer key, GATE 2025 EC (IIT Roorkee)

$A$ is diagonal, so the three states are decoupled: $\overset{x}{˙}_{1} = 2 x_{1} + u$ , $\overset{x}{˙}_{2} = - 2 x_{2} + u$ , $\overset{x}{˙}_{3} = u$ . The modes are $λ = 2, - 2, 0$ .
Each state is a first-order system driven by $u$ . $x_{1}$ sees $\frac{1}{s - 2}$ (pole in the right half-plane), $x_{2}$ sees $\frac{1}{s + 2}$ (pole in the left half-plane) and $x_{3}$ sees $\frac{1}{s}$ (pole at the origin).
Take the bounded input $u (t) = 1$ . Then $x_{3} (t) = x_{3} (0) + t \to \infty$ whatever the initial state is. With zero initial state, $x_{1} (t) = \frac{e ^{2 t} - 1}{2}$ also goes to $\infty$ . So the states are not bounded for every bounded input: (A) is false and (B) is true.
$x_{2}$ is BIBO stable. If $∣ u ∣ \leq M$ , then $∣ x_{2} (t) ∣ \leq ∣ x_{2} (0) ∣ e^{- 2 t} + \int_{0}^{t} e^{- 2 (t - τ)} ∣ u (τ) ∣ d τ \leq ∣ x_{2} (0) ∣ + \frac{M}{2}$ . So $x_{2}$ stays bounded for every bounded input and every initial state.
Since $x_{2}$ can never be unbounded, no bounded input makes all three states unbounded. (C) is false.
For the same reason it is not true that the states are unbounded for all bounded inputs, because $x_{2}$ is always bounded. (A concrete case: $u (t) = 0$ with zero initial state gives $x_{1} = x_{2} = x_{3} = 0$ .) (D) is false.
The only true statement is (B).

KEY IDEA

With a diagonal $A$ , the states decouple into first-order systems $\frac{1}{s - λ _{i}}$ . Every $b_{i} = 1$ , so every mode is driven by the input, and each state's boundedness is decided by its own eigenvalue alone. $λ = 2$ (right half-plane) and $λ = 0$ (a pole at the origin, whose step response is a ramp) can be made unbounded by a bounded input. $λ = - 2$ cannot. So the option you need is "at least one state is unbounded", not "all states are unbounded".

WATCH OUT FOR

Treating the whole system as one unit and choosing (C) because it is unstable, without noticing that $x_{2}$ ( $λ = - 2$ ) stays bounded for every bounded input. Another mistake is to count the $λ = 0$ state as marginally stable and so bounded: under a constant input, a pole at the origin integrates the input into a ramp.

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GATE 2025 EC Q.53: solution and answer

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