GATE 2024 EC Q.53: solution and answer

GATE 2024 · ECQ.53 · 2 marks · MSQ

Q.53 Consider a system $S$ represented in state space as
$\frac{d x}{d t} = [01 - 2 - 3] x + [10] r, y = [2 - 5] x .$
Which of the state space representations given below has/have the same transfer function as that of $S$ ?

(A)
$\frac{d x}{d t} = [0 - 2 1 - 3] x + [01] r, y = [12] x$
(B)
$\frac{d x}{d t} = [0 - 2 1 - 3] x + [10] r, y = [02] x$
(C)
$\frac{d x}{d t} = [- 1 0 0 - 2] x + [- 1 3] r, y = [11] x$
(D)
$\frac{d x}{d t} = [- 1 0 0 - 2] x + [11] r, y = [12] x$

See the question in the paper

Tap to open full size

Answer and solution

ANSWER

Options (A) and (C). Both have the same transfer function as $S$ : $G (s) = \frac{2 s + 1}{s ^{2} + 3 s + 2} = \frac{- 1}{s + 1} + \frac{3}{s + 2}$ .

Official answer: A, CSource: Official answer key, GATE 2024 EC (IISc Bengaluru)

Transfer function of $S$ , from $G (s) = C (s I - A)^{- 1} B$ . Here $s I - A = [s - 1 2 s + 3]$ , so $det (s I - A) = s (s + 3) + 2 = s^{2} + 3 s + 2 = (s + 1) (s + 2)$ and $adj (s I - A) = [s + 3 1 - 2 s]$ .
$adj (s I - A) B = [s + 3 1]$ , so $C adj (s I - A) B = 2 (s + 3) - 5 = 2 s + 1$ . Therefore $G_{S} (s) = \frac{2 s + 1}{( s + 1 ) ( s + 2 )} = \frac{2 s + 1}{s ^{2} + 3 s + 2}$
(A) Controllable canonical form. $A = [0 - a_{0} 1 - a_{1}]$ with $a_{0} = 2$ , $a_{1} = 3$ is the companion matrix of $s^{2} + 3 s + 2$ , and $B = [01]^{T}$ . For this form $C = [b_{0} b_{1}]$ gives the numerator $b_{1} s + b_{0}$ . With $C = [12]$ the numerator is $2 s + 1$ , so $G_{A} = \frac{2 s + 1}{s ^{2} + 3 s + 2}$ . This matches.
(B) Same $A$ , but $B = [10]^{T}$ . Now $s I - A = [s 2 - 1 s + 3]$ , so $adj (s I - A) = [s + 3 - 2 1 s]$ and $adj (s I - A) B = [s + 3 - 2]$ . Then $C (\cdot) = 0 \cdot (s + 3) + 2 \cdot (- 2) = - 4$ , so $G_{B} = \frac{- 4}{s ^{2} + 3 s + 2}$ . This does not match.
Diagonal (modal) form. When $A = diag (λ_{1}, λ_{2})$ , $G (s) = \sum_{i} \frac{c _{i} b _{i}}{s - λ _{i}}$ . Expand $G_{S}$ in partial fractions. The residue at $s = - 1$ is $\frac{2 ( - 1 ) + 1}{- 1 + 2} = - 1$ , and the residue at $s = - 2$ is $\frac{2 ( - 2 ) + 1}{- 2 + 1} = 3$ . So $G_{S} = \frac{- 1}{s + 1} + \frac{3}{s + 2}$ Check: $- (s + 2) + 3 (s + 1) = 2 s + 1$ .
(C) $λ_{1} = - 1$ , $λ_{2} = - 2$ , with $c_{1} b_{1} = 1 \cdot (- 1) = - 1$ and $c_{2} b_{2} = 1 \cdot 3 = 3$ . So $G_{C} = \frac{- 1}{s + 1} + \frac{3}{s + 2}$ . This matches.
(D) $c_{1} b_{1} = 1 \cdot 1 = 1$ and $c_{2} b_{2} = 2 \cdot 1 = 2$ , so $G_{D} = \frac{1}{s + 1} + \frac{2}{s + 2} = \frac{3 s + 4}{( s + 1 ) ( s + 2 )}$ . This does not match.
All four options share the characteristic polynomial $s^{2} + 3 s + 2$ , so they all have poles at $- 1$ and $- 2$ . Only the numerators tell them apart, and only (A) and (C) give $2 s + 1$ . Answer: (A) and (C).

KEY IDEA

A state-space realisation is not unique. Every option here has the same characteristic polynomial $s^{2} + 3 s + 2$ , so the poles cannot tell them apart and only the numerator can. For a controllable canonical form, read the numerator straight from $C$ (when $B = [01]^{T}$ ). For a diagonal form, compare the products $c_{i} b_{i}$ with the partial-fraction residues of $G_{S}$ .

WATCH OUT FOR

Deciding on eigenvalues alone. All four options have poles at $- 1$ and $- 2$ , so that test passes every one of them. A second trap is option (B): it has the companion $A$ matrix, but $B = [10]^{T}$ is not the controllable-canonical input vector, so you cannot read the numerator off $C$ . Its actual transfer function is $- 4/ (s^{2} + 3 s + 2)$ . In the diagonal forms, only the products $c_{i} b_{i}$ matter, not $b_{i}$ or $c_{i}$ on their own. That is why (C) matches even though its $C = [11]$ looks nothing like the original.

From the GATE 2024 EC paper. Every question of this paper · Download the full paper · All 16 State space questions, with notes

GATE 2024 EC Q.53: solution and answer

Answer and solution

More State space questions