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1
GATE ECE 2025
MCQ (Single Correct Answer)
+1
-0.33

Consider the unity-negative-feedback system shown in Figure (i) below, where gain $K \geq 0$. The root locus of this system is shown in Figure (ii) below. For what value(s) of $K$ will the system in Figure (i) have a pole at $-1+j 1$ ?

GATE ECE 2025 Control Systems - Root Locus Diagram Question 3 English 1 GATE ECE 2025 Control Systems - Root Locus Diagram Question 3 English 2
A
$K=5$
B
$K=\frac{1}{5}$
C
For no positive value of $K$
D
For all positive values of $K$
2
GATE ECE 2022
MCQ (Single Correct Answer)
+1
-0.33

The root-locus plot of a closed-loop system with unity negative feedback and transfer function KG(s) in the forward path is shown in the figure. Note that K is varied from 0 to $$\infty$$.

Select the transfer function G(s) that results in the root-locus plot of the closed-loop system as shown in the figure.

GATE ECE 2022 Control Systems - Root Locus Diagram Question 4 English

A
$$G(s) = {1 \over {{{(s + 1)}^5}}}$$
B
$$G(s) = {1 \over {{s^5} + 1}}$$
C
$$G(s) = {{s - 1} \over {{{(s + 1)}^6}}}$$
D
$$G(s) = {{s + 1} \over {{s^6} + 1}}$$
3
GATE ECE 2020
MCQ (Single Correct Answer)
+1
-0.33

The pole-zero map of a rational function $G(s)$ is shown below. When the closed contour $\Gamma$ is mapped into $G(s)$-plane, then the mapping encircles

GATE ECE 2020 Control Systems - Root Locus Diagram Question 2 English
A

the point $-1+j 0$ of the $G(s)$-plane once in clockwise direction.

B

the origin of $G(s)$-plane once in counter - clockwise direction.

C

the point $-1+j 0$ of the $G(s)$-plane once in counter - clockwise direction.

D

the origin of $G(s)$-plane once in clockwise direction.

4
GATE ECE 2011
MCQ (Single Correct Answer)
+1
-0.3
The root locus plot for a system is given below. The open loop transfer function corresponding to this plot is given by GATE ECE 2011 Control Systems - Root Locus Diagram Question 21 English
A
$$G\left(s\right)H\left(s\right)=k\frac{s\left(s+1\right)}{\left(s+2\right)\left(s+3\right)}$$
B
$$G\left(s\right)H\left(s\right)=k\frac{\left(s+1\right)}{s\left(s+2\right)\left(s+3\right)^2}$$
C
$$G\left(s\right)H\left(s\right)=k\frac1{s\left(s-1\right)\left(s+2\right)\left(s+3\right)}$$
D
$$G\left(s\right)H\left(s\right)=k\frac{\left(s+1\right)}{s\left(s+2\right)\left(s+3\right)}$$
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