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Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE) PDF Download

Q16: A two loop position control system is shown below
Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)The gain K of the Tacho-generator influences mainly the  (2011)
(a) Peak overshoot
(b) Natural frequency of oscillation
(c) Phase shift of the closed loop transfer function at very low frequencies (ω → 0)
(d) Phase shift of the closed loop transfer function at very high frequencies (ω → ∞)
Ans:
(a)
Sol: Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)G(s) = 1/s(s+1+k) and H(s) = 1
Characteristic equation 1 + G(s)H(s) = 0
Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)⇒ s(s + 1 + k) + 1 = 0
⇒ s+ (R + 1)s + 1 = 0
Comapring with,
Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Natural frequency ωn = 1 remains constant and does not depend on k.
2ξωn = k + 1
Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Damping ratio depends on k
Peak overshoot Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)
Since, mdepends on ξ which depends on k. Hence, peak overshoot is influenced by gain (k) of techo-generator.

Q17: The steady state error of a unity feedback linear system for a unit step input is 0.1. The steady state error of the same system, for a pulse input r(t) having a magnitude of 10 and a duration of one second, as shown in the figure is  (2011)
Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)(a) 0
(b) 0.1
(c) 1
(d) 10
Ans:
(a)
Sol: Let the system is represnted as
Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)H(s) = 1(unity feedback)
Error = E(s) =  X(s)/1+G(s)
Steady state error for X(s) = 1/s, ess = 0.1
Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)When input,
Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)X(t) = 10[u(t) − u(t − 1)]
Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)
Q18: For the system 2/(s + 1), the approximate time taken for a step response to reach 98% of the final value is  (2010)
(a) 1s
(b) 2s
(c) 4s
(d) 8s
Ans:
(c)
Sol: Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)C(t) = L−1[C(s)]=2[−1 − e−t]
Final value of C(t) = Css = 2
98% of Css = 0.98 × 2 = 1.96
Let, t = T, the response reaches 98% of its final values.  
1.96 = 2[1 − e−T]
T ≈ 4sec

Q19: The unit-step response of a unity feed back system with open loop transfer function G(s) = K/((s + 1) (s + 2)) is shown in the figure. The value of K is  (2009)
Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)(a) 0.5
(b) 2
(c) 4
(d) 6
Ans:
(d)
Sol: Steady state value of response = 0.75
Input is unit step
So steady state error
ess = 1 − 0.75 = 0.25
Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Steady state erro using find value theorem,
Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)⇒ k = 6

Q20: The transfer function of a system is given as Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)This system is  (2008)
(a) An over damped system
(b) An under damped system
(c) A critically damped system
(d) An unstable system
Ans:
(c)
Sol: Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE) Comparing with standard form,
Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)∴ 2ξωn = 20  
ω2n = 100
ξ = 1
ωn = 10
∴ The system is critically damped.

Q21: The transfer function of a linear time invariant system is given as Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)The steady state value of the output of the system for a unit impulse input applied at time instant t = 1 will be  (2008)
(a) 0
(b) 0.5
(c) 1
(d) 2
Ans:
(a)
Sol: r(t) = unit impulse applied at t = 1
= δ(t − 1)
R(s) = 1[r(t)] = e−s
Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Steady state value of output, using final value theoram,
Css = lims→0 sC(s)  
Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)
Q22: Consider the R-L-C circuit shown in figure
Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE) If the above step response is to be observed on a non-storage CRO, then it would be best have the ei as a  (2007)
(a) Step function
(b) Square wave of 50 Hz
(c) Square wave of 300 Hz
(d) Square wave of 2.0 KHz
Ans:
(c)
Sol: Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Settling time Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)For a square wave T/2 should be grater than ts
For f1 = 50Hz
Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)For f2 = 300Hz
Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)For f3 = 2kHz
Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Therefore, it would be best to have ei as a square wave of 300Hz.  

Q23: Consider the R-L-C circuit shown in figure
Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)For a step-input ei, the overshoot in the output ewill be  (2007)
(a) 0, since the system is not under damped
(b) 5%
(c) 16%
(d) 48%
Ans:
(c)
Sol: Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Taking Laplace transform,
Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Characteristic equation,
Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Comparing with,
Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Overshot Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)
Q24: Consider the feedback system shown below which is subjected to a unit step input. The system is stable and has following parameters k= 4, k= 10, ω = 500 and ξ = 0.7. The steady state value of z is  (2007)
Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)(a) 1
(b) 0.25
(c) 0.1
(d) 0
Ans:
(a)
Sol: Step−input ⇒ R(s) = 1/s
Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)and H(s) = 1
Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Steady state value of Z,
Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)
Q25: The block diagram of a closed loop control system is given by figure. The values of K and P such that the system has a damping ratio of 0.7 and an undamped natural frequency ωn of 5 rad/sec, are respectively equal to   (2004)
Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE) (a) 20 and 0.3
(b) 20 and 0.2
(c) 25 and 0.3
(d) 25 and 0.2
Ans: 
(d)
Sol: Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Closed-loop transfer function,
Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)So, characteristic equation = s+ (2 + kP)s + k
Comparing with standard equation  Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)(where ωn = undamped natueal frequency)  
2ξωn = 2 + kP
⇒ 2 × 0.7 × 5 = 2 + 25P
(where ξ = damping ratio)
P = 0.2

Q26: The block diagram shown in figure gives a unity feedback closed loop control system. The steady state error in the response of the above system to unit step input is  (2003)
Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)(a) 25%
(b) 0.75%
(c) 6%
(d) 33%
Ans: 
(a)
Sol: Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)Open loop transfer function
Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)

The system is type-0.
Steady state error to unit-step input
Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)where,
 kp = Position error constant = lim⁡s→0 G(s)H(s)
Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE)

The document Previous Year Questions- Time Response Analysis - 2 | Control Systems - Electrical Engineering (EE) is a part of the Electrical Engineering (EE) Course Control Systems.
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FAQs on Previous Year Questions- Time Response Analysis - 2 - Control Systems - Electrical Engineering (EE)

1. What is time response analysis?
Ans. Time response analysis is a method used in control systems engineering to study the behavior of a system in the time domain. It involves analyzing how a system responds to different inputs over time.
2. What are the key parameters to consider in time response analysis?
Ans. The key parameters to consider in time response analysis include rise time, settling time, peak time, peak overshoot, and steady-state error. These parameters help in evaluating the performance of a control system.
3. How is the settling time of a system determined in time response analysis?
Ans. The settling time of a system is the time it takes for the system's response to reach and stay within a specified range around its final value without oscillating. It is typically measured as the time taken for the response to settle within a certain percentage (e.g., 5%) of the final value.
4. What is the significance of peak time in time response analysis?
Ans. Peak time is the time taken for the response of a system to reach the peak value or overshoot before settling down. It is an important parameter as it indicates how quickly the system responds to a change in input.
5. How does time response analysis help in designing and analyzing control systems?
Ans. Time response analysis helps in designing and analyzing control systems by providing insights into the system's dynamic behavior, stability, and performance. It allows engineers to optimize the system's response to achieve desired specifications and improve overall system efficiency.
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