GATE 2023 EC Q.43: solution and answer

GATE 2023 · ECQ.43 · 2 marks · MCQ

The state equation of a second order system is

\dot{x} (t) = A x (t), x (0) is the initial condition.

Suppose $λ_{1}$ and $λ_{2}$ are two distinct eigenvalues of $A$ and $v_{1}$ and $v_{2}$ are the corresponding eigenvectors. For constants $α_{1}$ and $α_{2}$ , the solution, $x (t)$ , of the state equation is

(A)
$i = 1 \sum 2 α_{i} e^{λ_{i} t} v_{i}$
(B)
$i = 1 \sum 2 α_{i} e^{2 λ_{i} t} v_{i}$
(C)
$i = 1 \sum 2 α_{i} e^{3 λ_{i} t} v_{i}$
(D)
$i = 1 \sum 2 α_{i} e^{4 λ_{i} t} v_{i}$

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Answer and solution

ANSWER

The solution is option (A): $x (t) = i = 1 \sum 2 α_{i} e^{λ_{i} t} v_{i}$ . The constants $α_{i}$ come from expanding the initial condition in the eigenvectors, $x (0) = α_{1} v_{1} + α_{2} v_{2}$ .

Official answer: ASource: Official answer key, GATE 2023 EC (IIT Kanpur)

With no input, the solution of $\dot{x} (t) = A x (t)$ is $x (t) = e^{A t} x (0), e^{A t} = k = 0 \sum \infty \frac{A ^{k} t ^{k}}{k !}$
The eigenvectors are linearly independent because $λ_{1} \neq = λ_{2}$ . Suppose $c_{1} v_{1} + c_{2} v_{2} = 0$ . Multiply on the left by $(A - λ_{2} I)$ to get $c_{1} (λ_{1} - λ_{2}) v_{1} = 0$ , so $c_{1} = 0$ , and then $c_{2} = 0$ . Two independent vectors in a two-dimensional state space form a basis. (If $A$ is real, its eigenvalues can be a complex-conjugate pair. The eigenvectors and the $α_{i}$ are then complex, so the basis is of $C^{2}$ , not only $R^{2}$ .) So the initial condition has a unique expansion $x (0) = α_{1} v_{1} + α_{2} v_{2}$
Use the eigenvector property. From $A v_{i} = λ_{i} v_{i}$ we get $A^{k} v_{i} = λ_{i}^{k} v_{i}$ for every $k$ . Put this into the series: $e^{A t} v_{i} = k = 0 \sum \infty \frac{λ _{i}^{k} t ^{k}}{k !} v_{i} = e^{λ_{i} t} v_{i}$
$e^{A t}$ is linear, so $x (t) = e^{A t} (α_{1} v_{1} + α_{2} v_{2}) = α_{1} e^{λ_{1} t} v_{1} + α_{2} e^{λ_{2} t} v_{2} = i = 1 \sum 2 α_{i} e^{λ_{i} t} v_{i}$ This is option (A).
Check by substitution: $\dot{x} = i = 1 \sum 2 α_{i} λ_{i} e^{λ_{i} t} v_{i} = i = 1 \sum 2 α_{i} e^{λ_{i} t} A v_{i} = A x$ At $t = 0$ it gives $x (0) = α_{1} v_{1} + α_{2} v_{2}$ . So both the equation and the initial condition are satisfied.
Rule out (B), (C) and (D). Take $y (t) = \sum_{i = 1}^{2} α_{i} e^{k λ_{i} t} v_{i}$ with $k = 2, 3, 4$ . Then $\dot{y} - A y = i = 1 \sum 2 (k - 1) α_{i} λ_{i} e^{k λ_{i} t} v_{i}$ Here $v_{1}, v_{2}$ are independent, $e^{k λ_{i} t} \neq = 0$ and $k - 1 \neq = 0$ . So this is zero only if $α_{i} λ_{i} = 0$ for both $i$ . That is a special case (for example, $x (0)$ along an eigenvector whose eigenvalue is zero), not the solution for a general initial condition. In fact $e^{k λ_{i} t} v_{i}$ are the modes of $\dot{x} = k A x$ , a different system. Options (B), (C) and (D) are wrong.
Answer: (A) $i = 1 \sum 2 α_{i} e^{λ_{i} t} v_{i}$

KEY IDEA

This is modal decomposition. Along an eigenvector, the state transition matrix acts as a scalar: $e^{A t} v_{i} = e^{λ_{i} t} v_{i}$ . Distinct eigenvalues give independent eigenvectors, so any $x (0)$ splits as $\sum α_{i} v_{i}$ . Each part then evolves on its own as the mode $e^{λ_{i} t} v_{i}$ , and the free response is the weighted sum of these modes.

WATCH OUT FOR

The usual trap is thinking "second order" puts a 2 into the exponent and choosing (B). The order of the system is the number of modes (two terms in the sum), not a factor on $λ_{i}$ . The term $e^{2 λ_{i} t} v_{i}$ is a mode of $\dot{x} = 2 A x$ , not of $\dot{x} = A x$ . Substituting it back leaves an extra factor of 2. A second slip is to take the $α_{i}$ as the components of $x (0)$ . They are its coordinates in the eigenvector basis.

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GATE 2023 EC Q.43: solution and answer

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