1
GATE ECE 2024
MCQ (Single Correct Answer)
+1
-0.33

In the feedback control system shown in the figure below $G(s) = \dfrac{6}{s(s+1)(s+2)}$.

$R(s), Y(s),$ and $E(s)$ are the Laplace transforms of $r(t), y(t),$ and $e(t)$, respectively. If the input $r(t)$ is a unit step function, then __________

A

$\lim\limits_{t \to \infty} e(t) = 0$

B

$\lim\limits_{t \to \infty} e(t) = \dfrac{1}{3}$

C

$\lim\limits_{t \to \infty} e(t) = \dfrac{1}{4}$

D

$\lim\limits_{t \to \infty} e(t)$ does not exist, $e(t)$ is oscillatory

2
GATE ECE 2024
MCQ (Single Correct Answer)
+2
-1.33

Consider a unity negative feedback control system with forward path gain $G(s) = \frac{K}{(s + 1)(s + 2)(s + 3)}$ as shown.

The impulse response of the closed-loop system decays faster than $e^{-t}$ if ________.

A

$1 \leq K \leq 5$

B

$7 \leq K \leq 21$

C

$-4 \leq K \leq -1$

D

$-24 \leq K \leq -6$

3
GATE ECE 2024
MCQ (Single Correct Answer)
+2
-1.33

A satellite attitude control system, as shown below, has a plant with transfer function $G(s) = \frac{1}{s^2}$ cascaded with a compensator $C(s) = \frac{K(s +\alpha)}{s + 4}$, where $K$ and $\alpha$ are positive real constants.

In order for the closed-loop system to have poles at $-1 \pm j \sqrt{3}$, the value of $\alpha$ must be ______.

A

0

B

1

C

2

D

3

4
GATE ECE 2024
MCQ (More than One Correct Answer)
+2
-1.33

Consider a system $S$ represented in state space as

$$\frac{dx}{dt} = \begin{bmatrix} 0 & -2 \\ 1 & -3 \end{bmatrix}x + \begin{bmatrix} 1 \\ 0 \end{bmatrix}r , \quad y = \begin{bmatrix} 2 & -5 \end{bmatrix}x.$$

Which of the state space representations given below has/have the same transfer function as that of $S$?

A

$$\frac{dx}{dt} = \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix}x + \begin{bmatrix} 0 \\ 1 \end{bmatrix}r , \quad y = \begin{bmatrix} 1 & 2 \end{bmatrix}x.$$

B

$$\frac{dx}{dt} = \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix}x + \begin{bmatrix} 1 \\ 0 \end{bmatrix}r , \quad y = \begin{bmatrix} 0 & 2 \end{bmatrix}x.$$

C

$$\frac{dx}{dt} = \begin{bmatrix} -1 & 0 \\ 0 & -2 \end{bmatrix}x + \begin{bmatrix} -1 \\ 3 \end{bmatrix}r , \quad y = \begin{bmatrix} 1 & 1 \end{bmatrix}x.$$

D

$$\frac{dx}{dt} = \begin{bmatrix} -1 & 0 \\ 0 & -2 \end{bmatrix}x + \begin{bmatrix} 1 \\ 1 \end{bmatrix}r , \quad y = \begin{bmatrix} 1 & 2 \end{bmatrix}x.$$

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