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

Q1: The transfer function of a phase lead compensator is given by
The frequency (in rad/sec), at which ∠D(jω) is maximum, is   (2019)
(a)
(b)
(c)
(d)
Ans: (b)
Sol: Frequency at which ∠T(jω) is maximum,


Q2: The transfer function C(s) of a compensator is given below.
The frequency range in which the phase (lead) introduced by the compensator reaches the maximum is   (SET-2(2017))
(a) $0.1<\omega <1$0.1 < ω < 1  
(b) $1<\omega <10$1 < ω < 10
(c) 10 < ω < 100
(d) ω > 100
Ans: (a)
Sol: Pole zero diagram of compensator transfer function is shown below.
Maximum phase lead is between 0.1 and 1.
0.1 < ω < 1

Q3: For the network shown in the figure below, the frequency (in rad/s) at which the maximum phase lag occurs is, ___________.   (SET-2(2016))
(a) 0.15
(b) 0.32
(c) 0.66
(d) 0.92
Ans: (b)
Sol: Assuming
We can write,

Q4: The transfer function of a compensator is given asThe phase of the above lead compensator is maximum at   (2012)
(a) √2 rad/s
(b) √3 rad/s
(c) √6 rad/s
(d) 1/√3 rad/s
Ans: (a)
Sol: 
Q5: The transfer function of a compensator is given as
G_c(s) is a lead compensator if   (2012)
(a) a = 1, b = 2
(b) a = 3, b = 2
(c) a =- 3, b =- 1
(d) a = 3, b = 1
Ans: (a)
Sol: For G_c(s) to be a lead compensator
Option (A) Satisfies the above equation.

Q6: The transfer functions of two compensators are given below :
Which one of the following statements is correct ?   (2008)
(a) $C1$C₁ is a lead compensator and C₂ is a lag compensator
(b) C₁ is a lag compensator and C₂ is a lead compensator
(c) Both C₁ and C₂ are lead compensator
(d) Both C₁ and C₂ are lag compensator
Ans: (a)
Sol: Zero at s = -1
Pole at s = -10
As zero is closer origin, zero dominates pole. Hence  C₁ is lead compensator.
Zero at s = -10
Pole at s = -1
As pole is closer to origin, pole dominates zero. Hence C₂ is lag compensator.

Q7: The system  is to be compensated such that its gain-crossover frequency becomes same as its uncompensated phase crossover frequency and provides a 45° phase margin. To achieve this, one may use   (2007)
(a) a lag compensator that provides an attenuation of 20 dB and a phase lag of 45° at the frequency of 3 √3 rad/s
(b) a lead compensator that provides an amplification of 20 dB and a phase lead of 45° at the frequency of 3 rad/s
(c) a lag-lead compensator that provides an attenuation of 20 dB and phase lag of 45° at the frequency of √3 rad/s
(d) a lag-lead compensator that provides an attenuation of 20 dB and phase lead of 45° at the frequency of 3 rad/s
Ans: (d)
Sol: Let uncompensated syatem,
Phase crossover frequency of uncompensated system = (ω_pc), at this frequency phase of T(jω) is −180°
Put s = jω in T(s)
Gain cross frequency of compensated system, (ω_gc)₂ = phase cross frequency of uncompensated system, (ω_pc)₁
phase angle of ∠T(jω) is −135° and phase of uncompensated systen is −180° at 3 rad/sec.  
Therefore, the compensator provides phase lead of 45° at the frequency of 3 rad/sec.
Let X dB is the gain provided by the compensator, so at gain cross frequency, ∣T(jω)∣_com=1 or 0 dB. Gain of uncompensated system  Gain of compensated system,
So, the compensator provides an attenuation od 20dB. Hence option (D) is correct.

Q8: A lead compensator used for a closed loop controller has the following transfer function  For such a lead compensator   (2003)
(a) $a<b$a<b
(b) $a>b$a>b
(c) a>Kb
(d) $a<Kb$a<Kb
Ans: (a)
Sol: Transfer function Zero of TF = -a
Pole of TF = -b
For a lead-compensator, the zero is nearer to origin as compared to pole, hence the effect of zero is dominant, therefore, the lead-compensator when introduced in series with forward path of the trnasfer function the phase shift is increased.
So, from pole-zero configuration of the compensator a < b.