GATE EC Control Systems Previous Year Questions with Solutions

Full question on page 37 of the official paper (PDF)

GATE 2026, Q.40

Control Systems2 marksMCQ

Unity negative feedback is applied around $C(s)G(s)$ , with $G(s) = \dfrac{1}{(s+1)(s+2)}$ and $C(s) = K(s + \alpha)$ . Which $(K, \alpha)$ puts the closed-loop poles at $-3 \pm j\sqrt{5}$ ?

Block diagram: R(s) enters a summing junction, then C(s) and G(s) in cascade give Y(s), with unity negative feedback from Y(s) to the junction. — Figure from the official GATE 2026 EC paper, IIT Guwahati

Option A: (2, 3)
Option B: (3, 4)
Option C: (2, 4)
Option D: (3, 3)

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Key idea: Match the characteristic polynomial to the one the target poles give.

Official answer: B

The closed-loop characteristic equation is $1 + C(s)G(s) = 0$ :

(s+1)(s+2) + K(s+\alpha) = s^2 + (3 + K)s + (2 + K\alpha)

\begin{aligned} &(s+1)(s+2) \\ &\qquad + K(s+\alpha) \\ &\quad = s^2 \\ &\qquad + (3 + K)s \\ &\qquad + (2 + K\alpha) \end{aligned}

The required poles give

(s + 3)^2 + 5 = s^2 + 6s + 14

Matching coefficients: $3 + K = 6$ , so $K = 3$ ; and $2 + 3\alpha = 14$ , so $\alpha = 4$ . Option (B).

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Full question on page 38 of the official paper (PDF)

GATE 2026, Q.41

Control Systems2 marksMCQ

Find $Y(s)/U(s)$ for $\dot{x} = \begin{bmatrix} -4 & -1.5 \\ 4 & 0 \end{bmatrix}x + \begin{bmatrix} 2 \\ 0 \end{bmatrix}u$ and $y = \begin{bmatrix} 1.5 & 0.625 \end{bmatrix}x$ .

Option A: $\dfrac{3s}{s^2+4s-6}$
Option B: $\dfrac{3s+5}{s^2+4s-6}$
Option C: $\dfrac{3s+5}{s^2+4s+6}$
Option D: $\dfrac{3s}{s^2+4s+6}$

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Key idea: The transfer function is $C(sI - A)^{-1}B$ .

Official answer: C

sI - A = \begin{bmatrix} s + 4 & 1.5 \\ -4 & s \end{bmatrix}, \qquad \det(sI - A) = s^2 + 4s + 6

\begin{aligned} sI - A &= \begin{bmatrix} s + 4 & 1.5 \\ -4 & s \end{bmatrix}, \\ \det(sI - A) &= s^2 + 4s + 6 \end{aligned}

(sI - A)^{-1} = \frac{1}{s^2 + 4s + 6}\begin{bmatrix} s & -1.5 \\ 4 & s + 4 \end{bmatrix}

(sI - A)^{-1}B = \frac{1}{s^2 + 4s + 6}\begin{bmatrix} 2s \\ 8 \end{bmatrix}

\frac{Y(s)}{U(s)} = \frac{1.5(2s) + 0.625(8)}{s^2 + 4s + 6} = \frac{3s + 5}{s^2 + 4s + 6}

\begin{aligned} \frac{Y(s)}{U(s)} &= \frac{1.5(2s) + 0.625(8)}{s^2 + 4s + 6} \\ &= \frac{3s + 5}{s^2 + 4s + 6} \end{aligned}

Option (C).

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Full question on page 56 of the official paper (PDF)

GATE 2026, Q.59

Control Systems2 marksNAT

Unity negative feedback is applied around $G(s) = \dfrac{K}{s(s+7)(s+11)}$ . Find the gain $K > 0$ at which the system is marginally stable.

Block diagram: R(s) to a summing junction, then K/(s(s + 7)(s + 11)) to C(s), with unity negative feedback. — Figure from the official GATE 2026 EC paper, IIT Guwahati

Answer format: Integer

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Key idea: For $s^3 + as^2 + bs + K$ , the Routh array hits the stability edge at $K = ab$ .

Official answer: 1386

The characteristic equation is

s(s+7)(s+11) + K = s^3 + 18s^2 + 77s + K = 0

\begin{aligned} &s(s+7)(s+11) + K \\ &\quad = s^3 + 18s^2 + 77s + K \\ &\quad = 0 \end{aligned}

For a cubic $s^3 + a s^2 + b s + c$ , the Routh array's $s^1$ row is $(ab - c)/a$ . It reaches zero, the edge of stability, when

K = 18 \times 77 = 1386

At that gain the auxiliary equation $18s^2 + 1386 = 0$ gives poles at $s = \pm j\sqrt{77} \approx \pm j8.77$ rad/s, on the imaginary axis. $K = 1386$ .

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Full question on page 15 of the official paper (PDF)

GATE 2025, Q.16

Control Systems1 markMCQ

The figure shows a Nyquist plot with the unit circle dashed. $\omega_P$ , $\omega_Q$ , $\omega_R$ and $\omega_S$ are the positive frequencies at the points P, Q, R and S. Which statement is TRUE?

A Nyquist plot. The positive-frequency branch starts at omega = 0+ far below the real axis, rises through point S on the dashed unit circle, then through point R, crosses the negative real axis at point Q inside the unit circle, and loops through point P to the origin at omega = infinity. Dashed lines from the origin at 45 degrees above and below the negative real axis pass through P and R. The mirror-image branch for negative frequencies starts at omega = 0- above the real axis. — Figure from the official GATE 2025 EC paper, IIT Roorkee

Option A: $\omega_S$ is the gain crossover frequency and $\omega_P$ is the phase crossover frequency
Option B: $\omega_Q$ is the gain crossover frequency and $\omega_R$ is the phase crossover frequency
Option C: $\omega_Q$ is the gain crossover frequency and $\omega_S$ is the phase crossover frequency
Option D: $\omega_S$ is the gain crossover frequency and $\omega_Q$ is the phase crossover frequency

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Key idea: Gain crossover is where the plot meets the unit circle; phase crossover is where it meets the negative real axis.

Official answer: D

At the gain crossover frequency $|G(j\omega)H(j\omega)| = 1$ , so the plot is on the unit circle. Of the four points, only S lies on the dashed unit circle.

At the phase crossover frequency the phase is $-180°$ , so the plot is on the negative real axis. That is Q, where the branch crosses the axis inside the circle.

P and R are marked where the plot meets the $45°$ lines through the origin; they are neither crossover. So $\omega_S$ is the gain crossover and $\omega_Q$ the phase crossover. Option (D).

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Full question on page 25 of the official paper (PDF)

GATE 2025, Q.26

Control Systems1 markMSQ

The unity negative-feedback system in the figure has gain $K \ge 0$ in front of $G(s)$ . Its root locus, also shown, has open-loop poles at $-1$ and $-2$ and zeros at $-3$ and $-4$ . For what value or values of $K$ does the closed loop have a pole at $-1 + j1$ ?

Figure (i): a unity negative-feedback loop with R(s) into a summing junction, then a gain K, then G(s), giving Y(s). Figure (ii): the root locus, with open-loop poles at -1 and -2 and open-loop zeros at -3 and -4 on the real axis. The branches leave the poles, meet between -2 and -1, travel round a circle centred near -2.5 that reaches just under j1 and -j1, and return to the real axis between -4 and -3 to end at the zeros. — Figure from the official GATE 2025 EC paper, IIT Roorkee

One or more options are correct.

Option A: $K = 5$
Option B: $K = 1/5$
Option C: For no positive value of $K$
Option D: For all positive values of $K$

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Key idea: A point is a closed-loop pole for some $K > 0$ only if it lies on the root locus, which the angle condition decides.

Official answer: C

From the figure, $G(s) = \dfrac{(s+3)(s+4)}{(s+1)(s+2)}$ up to a positive constant. At $s = -1 + j$ , the angle of $G$ is

\angle(2 + j) + \angle(3 + j) - \angle(j) - \angle(1 + j) = 26.6° + 18.4° - 90° - 45° = -90°

\begin{aligned} &\angle(2 + j) \\ &\qquad + \angle(3 + j) \\ &\qquad - \angle(j) \\ &\qquad - \angle(1 + j) \\ &\quad = 26.6° \\ &\qquad + 18.4° \\ &\qquad - 90° \\ &\qquad - 45° \\ &\quad = -90° \end{aligned}

A point on the root locus needs $\pm 180°$ , so $-1 + j$ is not on it. Solving for $K$ directly confirms it:

K = -\frac{(s+1)(s+2)}{(s+3)(s+4)}\bigg|_{s=-1+j} = -\frac{-1 + j}{5 + 5j} = -0.2j

\begin{aligned} K &= -\frac{(s+1)(s+2)}{(s+3)(s+4)}\bigg|_{s=-1+j} \\ &= -\frac{-1 + j}{5 + 5j} \\ &= -0.2j \end{aligned}

That isn't a real number, so no $K$ works. The figure agrees: the circle only reaches about $\pm j0.87$ . Option (C). Option (B) is the size of that complex value, $0.2$ , taken as if it were a gain.

Full question on page 31 of the official paper (PDF)

GATE 2025, Q.36

Control Systems2 marksMCQ

$G(s) = \dfrac{1}{10s^2}$ is controlled by $M(s)$ in a unity negative-feedback loop, as in the figure. (i) No controller $M(s) = K_I/s$ with $K_I > 0$ makes the closed loop stable. (ii) At least one controller $M(s) = K_P + sK_D$ with $K_P, K_D > 0$ makes it stable. Which is correct?

A unity negative-feedback loop: R(s) into a summing junction, then M(s), then G(s), giving Y(s), with Y(s) fed back to the junction's minus input. — Figure from the official GATE 2025 EC paper, IIT Roorkee

Option A: (i) is TRUE and (ii) is FALSE
Option B: (i) is FALSE and (ii) is TRUE
Option C: Both (i) and (ii) are FALSE
Option D: Both (i) and (ii) are TRUE

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Key idea: Write the closed-loop characteristic polynomial: a missing power of $s$ rules out stability.

Official answer: D

The closed-loop poles are the roots of $1 + M(s)G(s) = 0$ .

(i) With $M = K_I/s$ : $10s^3 + K_I = 0$ . The $s^2$ and $s$ terms are missing, so some root is in the right half-plane (or on the axis) for every $K_I > 0$ . No such controller works: (i) is TRUE.
(ii) With $M = K_P + sK_D$ : $10s^2 + K_D s + K_P = 0$ . A second-order polynomial with all coefficients positive has both roots in the left half-plane, so any positive $K_P$ and $K_D$ work: (ii) is TRUE.

Option (D).

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Full question on page 32 of the official paper (PDF)

GATE 2025, Q.37

Control Systems2 marksMCQ

For $p(s) = s^5 + 7s^4 + 3s^3 - 33s^2 + 2s - 40$ , let $L$ , $I$ and $R$ be the numbers of roots with negative real parts, purely imaginary roots, and roots with positive real parts. Which is correct?

Option A: $L = 2$ , $I = 2$ , $R = 1$
Option B: $L = 3$ , $I = 2$ , $R = 0$
Option C: $L = 1$ , $I = 2$ , $R = 2$
Option D: $L = 0$ , $I = 4$ , $R = 1$

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Key idea: Every option has imaginary roots, so look for a factor $s^2 + a$ and divide it out.

Official answer: A

Try $s = j$ , so $s^2 = -1$ , $s^3 = -j$ , $s^4 = 1$ , $s^5 = j$ :

j + 7 - 3j + 33 + 2j - 40 = 0

So $s^2 + 1$ is a factor, and dividing it out leaves a cubic that factors further:

p(s) = (s^2 + 1)(s^3 + 7s^2 + 2s - 40) = (s^2 + 1)(s - 2)(s + 4)(s + 5)

\begin{aligned} p(s) &= (s^2 + 1)(s^3 + 7s^2 + 2s - 40) \\ &= (s^2 + 1)(s - 2)(s + 4)(s + 5) \end{aligned}

The roots are $\pm j$ , $2$ , $-4$ and $-5$ : $L = 2$ , $I = 2$ , $R = 1$ . Option (A). A Routh array gets the same counts, with a row of zeros at the $s^2 + 1$ factor.

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Full question on page 46 of the official paper (PDF)

GATE 2025, Q.53

Control Systems2 marksMSQ

A system has states $x_1(t)$ , $x_2(t)$ and $x_3(t)$ and input $u(t)$ , with $\dfrac{d}{dt}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 2 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} + \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} u(t)$ . Which statements are TRUE?

One or more options are correct.

Option A: The signals $x_1(t)$ , $x_2(t)$ and $x_3(t)$ are bounded for all bounded inputs
Option B: There exists a bounded input such that at least one of $x_1(t)$ , $x_2(t)$ and $x_3(t)$ is unbounded
Option C: There exists a bounded input such that $x_1(t)$ , $x_2(t)$ and $x_3(t)$ are all unbounded
Option D: The signals $x_1(t)$ , $x_2(t)$ and $x_3(t)$ are unbounded for all bounded inputs

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Key idea: The matrix is diagonal, so each state is a separate first-order system, and each one's pole decides what a bounded input can do to it.

Official answer: B

Each state obeys its own equation (from rest):

$\dot{x}_1 = 2x_1 + u$ : pole at $+2$ . A constant $u = 1$ gives $x_1 = (e^{2t} - 1)/2$ , unbounded.
$\dot{x}_2 = -2x_2 + u$ : pole at $-2$ . For any input with $|u| \le U$ , $|x_2| \le U/2$ , so $x_2$ is always bounded.
$\dot{x}_3 = u$ : an integrator. $u = 1$ gives $x_3 = t$ , unbounded.

So $u = 1$ makes $x_1$ unbounded: (B) is TRUE and (A) is FALSE. $x_2$ can never be unbounded, so (C) is FALSE, and $u = 0$ keeps everything at zero, so (D) is FALSE. Option (B).

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Full question on page 48 of the official paper (PDF)

GATE 2025, Q.56

Control Systems2 marksMSQ

Which of the signal flow graphs (A) to (D) in the figure represent the block diagram shown above them? In the block diagram, the error $E = R - G_1 E$ is formed by feeding back only $G_1$ 's output, and $Y = G_1 E + G_2 E$ .

A block diagram and four signal flow graphs. Block diagram: R(s) enters a summing junction (plus); its output E goes through G1 and also, above, through G2; G1's output is fed back to the junction's minus input and also goes to a second summing junction, which adds G2's output to give Y(s). Graph A: G1 and G2 both run between the same two nodes, with a -1 branch back from the second node to the first. Graph B: G1 runs from the first node to a middle node, which has a -1 branch back to the first node and a branch of 1 to a third node; G2 runs from the first node straight to the third node, which feeds Y(s). Graph C: like A but the return branch is +1. Graph D: G1 and G2 run between the same two nodes, and a -1 branch returns from a later node, after both paths have joined, to the first node. — Figure from the official GATE 2025 EC paper, IIT Roorkee

One or more options are correct.

Option A: Graph (A)
Option B: Graph (B)
Option C: Graph (C)
Option D: Graph (D)

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Key idea: Find where the feedback is taken: here it is after $G_1$ but before $G_2$ 's path joins, so the loop gain is $-G_1$ alone.

Official answer: B

From the block diagram, $E = R - G_1 E$ , so $E = R/(1 + G_1)$ , and

\frac{Y}{R} = \frac{G_1 + G_2}{1 + G_1}

Graph (B) has a single loop, $G_1$ and the $-1$ branch, with forward paths $G_1$ and $G_2$ that both touch it, so Mason's rule gives $(G_1 + G_2)/(1 + G_1)$ . It matches.

In (A) and (D) the $-1$ branch returns from a node where $G_2$ has already joined, so the loop gain is $-(G_1 + G_2)$ and the result is $(G_1 + G_2)/(1 + G_1 + G_2)$ . Graph (C) has positive feedback. Option (B) only.

Full question on page 10 of the official paper (PDF)

GATE 2024, Q.12

Control Systems1 markMCQ

In a Bode magnitude plot, a slope of 40 dB/decade is the same as

Option A: 12 dB/octave
Option B: 6 dB/octave
Option C: 20 dB/octave
Option D: 10 dB/octave

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Key idea: A decade multiplies frequency by 10 and an octave by 2, so scale the slope by $\log_{10} 2$ .

Official answer: A

40\ \text{dB/decade} \times \log_{10} 2 = 40 \times 0.301 \approx 12\ \text{dB/octave}

\begin{aligned} &40\ \text{dB/decade} \\ &\qquad \times \log_{10} 2 \\ &\quad = 40 \times 0.301 \\ &\quad \approx 12\ \text{dB/octave} \end{aligned}

Option (A). It is the familiar 20 dB/decade, or 6 dB/octave, of one pole, doubled.

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Full question on page 11 of the official paper (PDF)

GATE 2024, Q.13

Control Systems1 markMCQ

In the unity negative-feedback loop in the figure, $G(s) = \dfrac{6}{s(s+1)(s+2)}$ . If $r(t)$ is a unit step, what happens to the error $e(t)$ ?

A unity negative-feedback loop: R(s) enters a summing junction (plus), whose output E(s) feeds G(s), giving Y(s), which is fed back to the junction's minus input. — Figure from the official GATE 2024 EC paper, IISc Bengaluru

Option A: $\lim_{t\to\infty} e(t) = 0$
Option B: $\lim_{t\to\infty} e(t) = \tfrac{1}{3}$
Option C: $\lim_{t\to\infty} e(t) = \tfrac{1}{4}$
Option D: $\lim_{t\to\infty} e(t)$ does not exist; $e(t)$ is oscillatory

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Key idea: The final value theorem only applies to a stable loop, so check the closed-loop poles before using it.

Official answer: D

The closed-loop characteristic equation is $1 + G(s) = 0$ :

s^3 + 3s^2 + 2s + 6 = (s + 3)(s^2 + 2) = 0

\begin{aligned} &s^3 + 3s^2 + 2s + 6 \\ &\quad = (s + 3)(s^2 + 2) \\ &\quad = 0 \end{aligned}

Two poles sit on the imaginary axis at $\pm j\sqrt{2}$ , so the loop is marginally stable and $e(t)$ keeps oscillating. The limit doesn't exist. Option (D).

The final value theorem would give 0 for this type 1 system, option (A), but it doesn't apply when there are poles on the imaginary axis.

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Full question on page 28 of the official paper (PDF)

GATE 2024, Q.37

Control Systems2 marksMCQ

A unity negative-feedback system has forward gain $G(s) = \dfrac{K}{(s+1)(s+2)(s+3)}$ . The closed-loop impulse response decays faster than $e^{-t}$ if

Option A: $1 < K < 5$
Option B: $7 < K < 21$
Option C: $-4 < K < -1$
Option D: $-24 < K < -6$

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Key idea: Decaying faster than $e^{-t}$ means every closed-loop pole is left of $s = -1$ ; shift the axis there and apply Routh.

Official answer: A

The closed-loop poles solve $(s+1)(s+2)(s+3) + K = 0$ . Put $s = p - 1$ , so that $p$ in the left half-plane means $s$ left of $-1$ :

p(p + 1)(p + 2) + K = p^3 + 3p^2 + 2p + K = 0

\begin{aligned} &p(p + 1)(p + 2) + K \\ &\quad = p^3 + 3p^2 + 2p + K \\ &\quad = 0 \end{aligned}

For a cubic $p^3 + a p^2 + b p + c$ , Routh needs every coefficient positive and $ab > c$ : here $K > 0$ and $3 \times 2 > K$ . So $0 < K < 6$ , and only the range $1 < K < 5$ lies inside it. Option (A).

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Full question on page 29 of the official paper (PDF)

GATE 2024, Q.38

Control Systems2 marksMCQ

A satellite attitude loop has plant $G(s) = \dfrac{1}{s^2}$ in cascade with a compensator $C(s) = \dfrac{K(s + \alpha)}{s + 4}$ , $K, \alpha > 0$ , under unity negative feedback. For closed-loop poles at $-1 \pm j\sqrt{3}$ , what must $\alpha$ be?

Option A: 0
Option B: 1
Option C: 2
Option D: 3

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Key idea: Match the closed-loop cubic to the required quadratic factor times an unknown real pole.

Official answer: B

1 + C(s)G(s) = 0 \;\Rightarrow\; s^3 + 4s^2 + Ks + K\alpha = 0

\begin{aligned} 1 + C(s)G(s) &= 0 \\ &\Rightarrow\; s^3 + 4s^2 + Ks + K\alpha \\ &\quad = 0 \end{aligned}

Poles at $-1 \pm j\sqrt{3}$ mean a factor $s^2 + 2s + 4$ . With a third pole at $-c$ :

(s^2 + 2s + 4)(s + c) = s^3 + (2 + c)s^2 + (4 + 2c)s + 4c

\begin{aligned} &(s^2 + 2s + 4)(s + c) \\ &\quad = s^3 \\ &\qquad + (2 + c)s^2 \\ &\qquad + (4 + 2c)s \\ &\qquad + 4c \end{aligned}

Matching gives $c = 2$ , then $K = 8$ and $K\alpha = 8$ , so $\alpha = 1$ . Option (B).

Full question on page 41 of the official paper (PDF)

GATE 2024, Q.53

Control Systems2 marksMSQ

System S is $\dot{x} = \begin{bmatrix} 0 & -2 \\ 1 & -3 \end{bmatrix}x + \begin{bmatrix} 1 \\ 0 \end{bmatrix}r$ , $y = \begin{bmatrix} 2 & -5 \end{bmatrix}x$ . Which of these have the same transfer function?

One or more options are correct.

Option A: $\dot{x} = \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix}x + \begin{bmatrix} 0 \\ 1 \end{bmatrix}r$ , $y = \begin{bmatrix} 1 & 2 \end{bmatrix}x$
Option B: $\dot{x} = \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix}x + \begin{bmatrix} 1 \\ 0 \end{bmatrix}r$ , $y = \begin{bmatrix} 0 & 2 \end{bmatrix}x$
Option C: $\dot{x} = \begin{bmatrix} -1 & 0 \\ 0 & -2 \end{bmatrix}x + \begin{bmatrix} -1 \\ 3 \end{bmatrix}r$ , $y = \begin{bmatrix} 1 & 1 \end{bmatrix}x$
Option D: $\dot{x} = \begin{bmatrix} -1 & 0 \\ 0 & -2 \end{bmatrix}x + \begin{bmatrix} 1 \\ 1 \end{bmatrix}r$ , $y = \begin{bmatrix} 1 & 2 \end{bmatrix}x$

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Key idea: Compute $C(sI - A)^{-1}B$ for each; different realisations of one system share it.

Official answer: A and C

G_S(s) = \begin{bmatrix} 2 & -5 \end{bmatrix}\frac{1}{s^2 + 3s + 2}\begin{bmatrix} s + 3 \\ 1 \end{bmatrix} = \frac{2s + 1}{s^2 + 3s + 2}

\begin{aligned} G_S(s) &= \begin{bmatrix} 2 & -5 \end{bmatrix}\frac{1}{s^2 + 3s + 2}\begin{bmatrix} s + 3 \\ 1 \end{bmatrix} \\ &= \frac{2s + 1}{s^2 + 3s + 2} \end{aligned}

(A) is the controllable canonical form of $\frac{2s + 1}{s^2 + 3s + 2}$ : the same.
(B) gives $\frac{-4}{s^2 + 3s + 2}$ : different.
(C) gives $\frac{-1}{s + 1} + \frac{3}{s + 2} = \frac{2s + 1}{(s + 1)(s + 2)}$ : the same.
(D) gives $\frac{1}{s + 1} + \frac{2}{s + 2} = \frac{3s + 4}{(s + 1)(s + 2)}$ : different.

Options (A) and (C).

Full question on page 18 of the official paper (PDF)

GATE 2023, Q.25

Control Systems1 markMCQ

The open-loop transfer function of a unity negative feedback system is $G(s) = \dfrac{k}{s(1 + sT_1)(1 + sT_2)}$ , where $k$ , $T_1$ and $T_2$ are positive constants. What is the phase crossover frequency, in rad/s?

Option A: $\dfrac{1}{\sqrt{T_1 T_2}}$
Option B: $\dfrac{1}{T_1 T_2}$
Option C: $\dfrac{1}{T_1\sqrt{T_2}}$
Option D: $\dfrac{1}{T_2\sqrt{T_1}}$

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Key idea: At the phase crossover frequency the phase of $G(j\omega)$ is $-180°$ .

Official answer: A

-90^\circ - \tan^{-1}(\omega T_1) - \tan^{-1}(\omega T_2) = -180^\circ

\begin{aligned} &-90^\circ \\ &\qquad - \tan^{-1}(\omega T_1) \\ &\qquad - \tan^{-1}(\omega T_2) \\ &\quad = -180^\circ \end{aligned}

So $\tan^{-1}(\omega T_1) + \tan^{-1}(\omega T_2) = 90°$ , which needs $\omega T_1 \cdot \omega T_2 = 1$ :

\omega_{pc} = \frac{1}{\sqrt{T_1 T_2}}

Option (A).

Full question on page 28 of the official paper (PDF)

GATE 2023, Q.41

Control Systems2 marksMCQ

The closed-loop system in the figure has unity negative feedback round $G(s) = \dfrac{k}{s(s + 2)}$ , where $k > 0$ and $\alpha > 0$ . What is the steady-state error due to a ramp input, $R(s) = \alpha/s^2$ ?

Block diagram: R(s) = alpha/s^2 into a summing junction (plus), then E(s) into a block k/(s(s+2)) giving Y(s), which is fed back to the junction with a minus sign. — Figure from the official GATE 2023 EC paper, IIT Kanpur

Option A: $\dfrac{2\alpha}{k}$
Option B: $\dfrac{\alpha}{k}$
Option C: $\dfrac{\alpha}{2k}$
Option D: $\dfrac{\alpha}{4k}$

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Key idea: A type-1 loop follows a ramp with a constant error, $\alpha / K_v$ , where $K_v = \lim_{s \to 0} sG(s)$ .

Official answer: A

K_v = \lim_{s \to 0} s \cdot \frac{k}{s(s + 2)} = \frac{k}{2}, \qquad e_{ss} = \frac{\alpha}{K_v} = \frac{2\alpha}{k}

\begin{aligned} K_v &= \lim_{s \to 0} s \cdot \frac{k}{s(s + 2)} \\ &= \frac{k}{2}, \\ e_{ss} &= \frac{\alpha}{K_v} \\ &= \frac{2\alpha}{k} \end{aligned}

This holds when the loop is stable, which it is for every $k > 0$ : the characteristic equation $s^2 + 2s + k = 0$ has both roots in the left half-plane. Option (A).

Full question on page 29 of the official paper (PDF)

GATE 2023, Q.42

Control Systems2 marksMCQ

In the block diagram in the figure, $R(s)$ and $D(s)$ are two inputs, and the output is $Y(s) = G_1(s)R(s) + G_2(s)D(s)$ . What are $G_1(s)$ and $G_2(s)$ ?

Block diagram: R(s) enters a first summing junction (plus), whose output enters a second junction (plus) together with D(s) (plus). The second junction feeds G(s), whose output is Y(s). Y(s) is fed back through H(s) to the second junction with a minus sign, and directly to the first junction with a minus sign. — Figure from the official GATE 2023 EC paper, IIT Kanpur

Option A: $G_1 = \dfrac{G}{1 + G + GH}$ and $G_2 = \dfrac{G}{1 + G + GH}$
Option B: $G_1 = \dfrac{G}{1 + G + H}$ and $G_2 = \dfrac{G}{1 + G + H}$
Option C: $G_1 = \dfrac{G}{1 + G + H}$ and $G_2 = \dfrac{G}{1 + G + GH}$
Option D: $G_1 = \dfrac{G}{1 + G + GH}$ and $G_2 = \dfrac{G}{1 + G + H}$

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Key idea: Write the signal into $G$ and solve once for $Y$ ; both inputs enter at the same point, so they share a denominator.

Official answer: A

The second junction adds $D$ to the first junction's output, $R - Y$ , and subtracts $HY$ :

Y = G\left[(R - Y) + D - HY\right]

Collecting the $Y$ terms:

Y(1 + G + GH) = G R + G D

So $G_1 = G_2 = G/(1 + G + GH)$ . Option (A).

Full question on page 30 of the official paper (PDF)

GATE 2023, Q.43

Control Systems2 marksMCQ

The state equation of a second-order system is $\dot{\mathbf{x}}(t) = A\mathbf{x}(t)$ , with initial condition $\mathbf{x}(0)$ . $\lambda_1$ and $\lambda_2$ are two distinct eigenvalues of $A$ , and $\mathbf{v}_1$ and $\mathbf{v}_2$ the corresponding eigenvectors. For constants $\alpha_1$ and $\alpha_2$ , what is the solution $\mathbf{x}(t)$ ?

Option A: $\sum_{i=1}^{2} \alpha_i e^{\lambda_i t} \mathbf{v}_i$
Option B: $\sum_{i=1}^{2} \alpha_i e^{2\lambda_i t} \mathbf{v}_i$
Option C: $\sum_{i=1}^{2} \alpha_i e^{3\lambda_i t} \mathbf{v}_i$
Option D: $\sum_{i=1}^{2} \alpha_i e^{4\lambda_i t} \mathbf{v}_i$

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Key idea: Try $\mathbf{x} = e^{\lambda t}\mathbf{v}$ : it solves the equation exactly when $A\mathbf{v} = \lambda\mathbf{v}$ .

Official answer: A

\frac{d}{dt}\left(e^{\lambda_i t}\mathbf{v}_i\right) = \lambda_i e^{\lambda_i t}\mathbf{v}_i = A\left(e^{\lambda_i t}\mathbf{v}_i\right)

\begin{aligned} &\frac{d}{dt}\left(e^{\lambda_i t}\mathbf{v}_i\right) \\ &\quad = \lambda_i e^{\lambda_i t}\mathbf{v}_i \\ &\quad = A\left(e^{\lambda_i t}\mathbf{v}_i\right) \end{aligned}

Distinct eigenvalues give independent eigenvectors, so every solution is $\alpha_1 e^{\lambda_1 t}\mathbf{v}_1 + \alpha_2 e^{\lambda_2 t}\mathbf{v}_2$ , with $\alpha_1, \alpha_2$ fixed by $\mathbf{x}(0)$ . In (B) to (D), $\dot{\mathbf{x}}$ would be $k\lambda_i$ times each term, not $A\mathbf{x}$ . Option (A).

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Full question on page 41 of the official paper (PDF)

GATE 2023, Q.57

Control Systems2 marksNAT

The figure shows the asymptotic magnitude Bode plot of a minimum-phase system: $-20$ dB/decade up to $\omega_1$ , flat from $\omega_1$ to $\omega_2$ , and $-40$ dB/decade beyond $\omega_2$ . Its transfer function is $G(s) = \dfrac{k(s + z)^a}{s^b(s + p)^c}$ , where $k$ , $z$ , $p$ , $a$ , $b$ and $c$ are positive constants. What is $a + b + c$ ?

Asymptotic Bode magnitude plot: a -20 dB/decade line falling to omega_1, a flat line from omega_1 to omega_2, and a -40 dB/decade line beyond omega_2. — Figure from the official GATE 2023 EC paper, IIT Kanpur

Answer format: rounded off to the nearest integer

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Key idea: Each pole at the origin sets $-20$ dB/decade from the start; a zero adds $+20$ and a pole $-20$ dB/decade at its corner.

Official answer: 4

$-20$ dB/decade from the start: one pole at the origin, $b = 1$ .
At $\omega_1$ the slope rises by $20$ to 0: a single zero, $a = 1$ , $z = \omega_1$ .
At $\omega_2$ the slope falls by $40$ : a double pole, $c = 2$ , $p = \omega_2$ .

So $a + b + c = 1 + 1 + 2 = 4$ .

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