Previous Year Questions- Feedback Characteristics of Control Systems | Control Systems - Electrical Engineering (EE) PDF Download

Q1: Consider the cascaded system as shown in the figure. Neglecting the faster component of the trasient response, which one of the following options is a first-order pole-only approximation such that the steady-state values of the unit step responses of the original and the approximated system are same?   (2024)
(a)
(b)
(c)
(d)
Ans: (b)
Sol: Transfer function
Answer = 2/(s+1)

$Q2:\; =\frac{2}{s}$Q2: Consider a unity-gain negative feedback system consisting of the plant G(s) (given below) and a proportional-integral controller. Let the proportional gain and integral gain be 3 and 1, respectively. For a unit step reference input, the final values of the controller output and the plant output, respectively, are   (2023)
G(s) = 1/(s-1)
(a) ∞, ∞
(b) 1, 0
(c) 1, -1
(d) -1, 1
Ans: (d)
Sol: Given plant:
So,
Closed loop transfer function,
Final value of plant,
From plant,
∴ Final value of controller,
x(∞) = LimsX_s→0 s(s) = −1

Q3: The block diagram of a system is shown in the figure
If the desired transfer function of the system is  then G(s) is   (SET-3 (2014))
(a) 1
(b) s
(c) 1/s
(d)
Ans: (b)
Sol: For all given values of G(s) in options, G(s) does not satisfy the desired condition. But option (B) is official answer given by IIT.

Q4: The closed-loop transfer function of a system is  The steady state error due to unit step input is _____.   (SET-2  (2014))
(a) 0
(b) 0.15
(c) 0.75
(d) 1
Ans: (a)
Sol: Closed loop transfer function,
= Open llop transfer function ( for unity feedback system)
Given, input = u(t) = r(t)
∴ R(s) = 1/s
Steady state error
∴ Steady state error for step input,  e_ss = 0.

Q5: The open-loop transfer function of a dc motor is given as  When connected in feedback as shown below, the approximate value of K_a that will reduce the time constant of the closed loop system by one hundred times as compared to that of the open-loop system is   (2013)
(a) 1
(b) 5
(c) 10
(d) 100
Ans: (c)
Sol: Closed loop systems,
According to quetion,

Q6: As shown in the figure, a negative feedback system has an amplifier of gain 100 with ±10% tolerance in the forward path, and an attenuator of value 9/100 in the feedback path. The overall system gain is approximately :   (2010)
(a) 10 ± 1%
(b) $10\pm 2$10 ± 2%
(c) $10\pm 5$10 ± 5%
(d) 10 ± 10%
Ans: (a)
Sol: