Previous Year Questions- Time Response Analysis - 2 - Control Systems - Electrical

Q16: A two loop position control system is shown below
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: G(s) = 1/s(s+1+k) and H(s) = 1
Characteristic equation 1 + G(s)H(s) = 0
⇒ s(s + 1 + k) + 1 = 0
⇒ s² + (R + 1)s + 1 = 0
Comapring with,
Natural frequency ω_n = 1 remains constant and does not depend on k.
2ξω_n = k + 1
Damping ratio depends on k
Peak overshoot
Since, m_p depends 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)
(a) 0
(b) 0.1
(c) 1
(d) 10
Ans: (a)
Sol: Let the system is represnted as
H(s) = 1(unity feedback)
Error = E(s) =  X(s)/1+G(s)
Steady state error for X(s) = 1/s, e_ss = 0.1
When input,
X(t) = 10[u(t) − u(t − 1)]

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: C(t) = L⁻¹[C(s)]=2[−1 − e^−t]
Final value of C(t) = C_ss = 2
98% of C_ss = 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)
(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
e_ss = 1 − 0.75 = 0.25
Steady state erro using find value theorem,
⇒ k = 6

Q20: The transfer function of a system is given as 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:  Comparing with standard form,
∴ 2ξω_n = 20  
ω²_n = 100
ξ = 1
ω_n = 10
∴ The system is critically damped.

Q21: The transfer function of a linear time invariant system is given as 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
Steady state value of output, using final value theoram,
C_ss = lim_s→0 sC(s)  

Q22: Consider the R-L-C circuit shown in figure
 If the above step response is to be observed on a non-storage CRO, then it would be best have the e_i 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: Settling time For a square wave T/2 should be grater than t_s
For f₁ = 50Hz
For f₂ = 300Hz
For f₃ = 2kHz
Therefore, it would be best to have e_i as a square wave of 300Hz.  

Q23: Consider the R-L-C circuit shown in figure
For a step-input e_i, the overshoot in the output e_o will be  (2007)
(a) 0, since the system is not under damped
(b) 5%
(c) 16%
(d) 48%
Ans: (c)
Sol: Taking Laplace transform,
Characteristic equation,
Comparing with,
Overshot 
Q24: Consider the feedback system shown below which is subjected to a unit step input. The system is stable and has following parameters k_p = 4, k_i = 10, ω = 500 and ξ = 0.7. The steady state value of z is  (2007)
(a) 1
(b) 0.25
(c) 0.1
(d) 0
Ans: (a)
Sol: Step−input ⇒ R(s) = 1/s
and H(s) = 1
Steady state value of Z,

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 ω_n of 5 rad/sec, are respectively equal to   (2004)
 (a) 20 and 0.3
(b) 20 and 0.2
(c) 25 and 0.3
(d) 25 and 0.2
Ans: (d)
Sol: Closed-loop transfer function,
So, characteristic equation = s² + (2 + kP)s + k
Comparing with standard equation  (where ω_n = undamped natueal frequency)  
2ξω_n = 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)
(a) 25%
(b) 0.75%
(c) 6%
(d) 33%
Ans: (a)
Sol: Open loop transfer function

The system is type-0.
Steady state error to unit-step input
where,
 k_p = Position error constant = lim_⁡s→0 G(s)H(s)