GATE 2026 EE Q.52: solution and answer

GATE 2026 · EEQ.52 · 2 marks · MCQ

A system is characterized by the following state equation and output equation ( $u$ : input, $x$ : state vector, $y$ : output)

\dot{x} = [a - a b 0] x + [10] u

y = [12] x

What are the values of $a$ and $b$ for which the poles of the transfer function are at $- 2 + j 3$ and $- 2 - j 3$ ?

(A)
$a = 4, b = 3.25$
(B)
$a = - 4, b = 3.25$
(C)
$a = 4, b = - 3.25$
(D)
$a = - 4, b = - 3.25$

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Answer and solution

ANSWER

$a = - 4$ , $b = - 3.25$ . Option (D).

Official answer: DSource: Official answer key, GATE 2026 EE (IIT Guwahati)

The poles of $G (s) = C (s I - A)^{- 1} B$ are the roots of the characteristic equation $∣ s I - A ∣ = 0$ , provided no pole is cancelled by a zero. We check for cancellation at the end.
Characteristic polynomial: $∣ s I - A ∣ = s - a a - b s = s (s - a) - (- b) (a) = s^{2} - a s + ab$
Desired polynomial from the poles $- 2 \pm j 3$ : $(s + 2)^{2} + 3^{2} = s^{2} + 4 s + 13$
Compare the coefficients of $s$ : $- a = 4 \Rightarrow a = - 4$ .
Compare the constant terms: $ab = 13 \Rightarrow b = \frac{13}{- 4} = - 3.25$ .
Check for cancellation. $(s I - A)^{- 1} = \frac{1}{s ^{2} - a s + ab} [s - a b s - a]$ , so $(s I - A)^{- 1} B = \frac{1}{s ^{2} - a s + ab} [s - a]$ and $G (s) = [12] \frac{1}{s ^{2} - a s + ab} [s - a] = \frac{s - 2 a}{s ^{2} - a s + ab} = \frac{s + 8}{s ^{2} + 4 s + 13}$ The zero at $s = - 8$ cancels neither pole, so the transfer function does have poles at $- 2 \pm j 3$ .
So $a = - 4$ and $b = - 3.25$ , which is option (D).

KEY IDEA

The denominator of the transfer function is $∣ s I - A ∣$ . Set it equal to the polynomial built from the required poles, $(s + 2)^{2} + 9 = s^{2} + 4 s + 13$ , and compare coefficients. Then check that the numerator does not cancel a pole.

WATCH OUT FOR

The usual slip is a sign error in the determinant. Writing $s (s - a) - ab$ instead of $s (s - a) + ab$ (the off-diagonal product is $(- b) (a) = - ab$ , and it is subtracted) gives $ab = - 13$ and points to (A) or (B). A second slip is matching $- a$ to $+ 4$ as $a = 4$ , which leads to (A) or (C). Remember that a pole at $- 2$ needs a positive coefficient of $s$ in the characteristic polynomial.

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GATE 2026 EE Q.52: solution and answer

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