IC8451 - Control Systems - Unit-2 Two marks

Subject Name: Control Systems

Subject Code: IC8451

B.E. EEE/EIE/ICE Dept - semester 04

Anna University Regulation 2017 B.E. EEE/EIE/ICE Dept 4th semester IC8451(Control Systems) 2 marks for 2nd unit is provided below. IC8451 all units 2 marks, Control Systems all units notes & syllabus are listed down for students to utilize it for scoring maximum marks with our study materials.

1.How to determine breakaway point of a root locus ?

Answer

Steps to determine the co-ordinates of breakaway points are, Step 1 : Construct the characteristic equation 1 + G(s)H(s) = 0 of the system. Step 2 : From this equation, separate the terms involving 'K' and terms involving 's'. Write the value of K in terms of s i.e. K in terms of F(s). Step 3 : Differentiate above equation w.r.t. 's', equate it to zero i.e. dK/ds = 0. Step 4 : Roots of the equation dK/ds = 0 gives us the breakaway points.

2.State the rules for obtaining the breakaway point in root locus.

Answer

1) If there are adjacently placed poles on the real axis and the real axis between them is a part of the root locus then there exists minimum one breakaway point in between adjacently placed poles. 2) If there are two adjacently placed zeros on real axis and section of real axis in between them is a part of the root locus then there exists minimum one breakaway point in between adjacently placed zeros. 3) If there is a zero on the real axis and to the left of that zero there is no pole or zero existing on the real axis and complete real axis to the left of this zero is a part of the root locus then there exists minimum one breakaway point to the left of that zero.

3.State the various standard test inputs used in control systems.

Answer

The various standard test inputs used in control systems are step, ramp, parabolic, impulse and sinusoidal.

4.What is steady state error ?

Answer

The difference between the desired output i.e. reference input and the actual output of the system is called steady state error which is denoted as eₛₛ.

5.What are the properties of a good control system ?

Answer

The properties of a good control system are less peak overshoot, less rise time indicating fast system, less settling time, less steady state error and absolute stability.

6.List out any four time-domain specifications.

Answer

The various time domain specifications are delay time, rise time, peak time, peak overshoot, settling time and steady state error.

7.How the transient response of a system with feedback differ from that without feedback ?

Answer

The feedback reduces the time constant of the system. Lesser the time constant, faster is the response. Thus the feedback makes the system fast. The speed of the transient response increases due to the feedback.

8.Mention the characteristics of PI controller.

Answer

The PI controller has following characteristics : i) It increases order of the system. ii) It increases TYPE of the system. iii) Design of Kᵢ must be proper to maintain stability of system. So it makes system relatively less stable. iv) Steady state error reduces tremendously for same type of inputs.

9.Define time response of a system.

Answer

The response given by the system which is function of the time, to the applied excitation is called time response of a control system.

10.Define transient response and steady state response of a system.

Answer

The output variation during the time it takes to achieve its final steady value is called transient response. Steady state response is that part of the time response which remains after complete transient response vanishes from the system output.

11.State the factors on which the steady state error of a system depends.

Answer

The steady state error of a system depends on, i) R(s) i.e. reference input, its type and magnitude. ii) G(s)H(s) i.e. open loop transfer function. iii) Dominant nonlinearities present if any.

12.State the factors on which transient response of the system depends.

Answer

The transient response of a system depends on, i) Poles of the closed loop T.F. and ii) Location of the poles of the closed loop T.F.

13.How a control system is classified depending on the value of damping ?

Answer

A control system is classified depending on the value of damping depending on the value of the damping ratio ξ as, i) 0 < ξ < 1 : Underdamped system ii) ξ = 1 : Critically damped system iii) ξ > 1 : Overdamped system iv) ξ = 0 : Undamped system

14.What is the effect of PD controller on the performance of system ?

Answer

The PD controller has following characteristics, i) It increases damping ratio. ii) 'ωₙ' for system remains unchanged. iii) 'TYPE' of the system remains unchanged. iv) It reduces peak overshoot. v) It reduces settling time. vi) Steady state error remains unchanged.

15.Why derivative controller is not used in the control system ?

Answer

The derivative controller acts only during transient period when the error varies with time and does not produce any corrective action for a constant error as derivative of a constant error is zero. Hence the derivative controller is never used alone but always used along with some other type of controller.

16.What is the effect of adding a pole to a second order system ?

Answer

The second order system is generally stable. If a pole is added to it, it becomes third order due to which it becomes less stable in nature. It increases peak overshoot and settling time.

17.Give two disadvantages of feedback in control.

Answer

i) The system stability must be observed strictly if feedback is used. ii) The system becomes complicated. iii) The system becomes oscillatory. iv) The system becomes sensitive to parameter variations in the feedback.

18.Why might a designer wish to put the derivative term in the feedback rather than in the error path ?

Answer

The derivative term if put in the error path produces zero output for zero error which is not desirable. It adds with proportional signal to increase noise. If it is in the feedack path, it gets compared with proportional signal and reduces the noise.

19.State the advantages of generalised error coefficients.

Answer

i) It gives variation of error as a function of time. ii) For any input other than the standard inputs, error can be determined. iii) As variation of error as a function of time is available, the design of system becomes easy and optimum.

20.Define adaptive control system.

Answer

Adaptive control system has a property or an automatic control system having a property of automatically changing the characteristics, system parameters and/or structure of the controller during the normal operation with the aim of maintaining a given standard of performance under varying inputs and varying arbitrary, conditions of operation is called as adaptive or self adjusting system.

21.What is the function of controller ?

Answer

A controller accepts error as its input and manipulates the error according to the requirement of the system and gives output to the plant or the process to be controlled.

22.What will happen to damping factor ξ and natural frequency of oscillations ωₙ if gain K of a second order system is increased.

Answer

As the value of gain K increases the natural frequency of oscillations also increases and the value of damping ratio ξ decreases. Due to this system starts becoming more oscillatory in nature.

23.How can the maximum overshoot of a system be decreased without affecting the steady state error ?

Answer

With the use of PD i.e. proportional plus derivative controller, it can be observed that transient response and value of damping ratio ξ increases without affecting steady state error. As ξ increases, the maximum overshoot decreases. So using PD controller it is possible to decrease maximum overshoot without affecting the steady state error.

24.Explain the concept of damping.

Answer

Every system has a tendency to pass through the transients before achieving steady state. Such transients may be osicllatory. The system oppose to cause oscillations naturally on its own. This opposition by the system to cause oscillations is called damping.

25.What is meant by time constant of the system ?

Answer

The time required by the system output to reach to 63.2 % of its final value during the first attempt is called one time constant of the system.

26.What is root locus ?

Answer

The locus of the closed loop poles obtained when system gain 'K' is varied from —∞ to +∞ is called Root Locus. When 'K' is varied from 0 to +∞ , the plot is called Direct root locus while when 'K' is varied from —∞ to 0, the plot obtained is called Inverse root locus.

27.What is angle criterion for root locus ?

Answer

Angle condition can be stated as, ∠G(s)H(s) | for any value of 's' which is the root of equation [1 + G(s)H(s) = 0] is = ±(2q + 1) 180° where q = 0, 1, 2 ... = Odd multiple of 180°

28.State the magnitude condition for the root locus.

Answer

The magniude condition is, | G(s)H(s) | at a point in s-plane which is on root locus = 1.

29.State the rule for finding out the root loci on the real axis.

Answer

A point on the real axis lies on the root locus if the sum of the number of open loop poles and the open loop zeros, on the real axis, to the right hand side of this point is odd.

30.Define breakaway point of a root locus.

Answer

Breakaway point is a point on the root locus where multiple roots of the characteristic equation occurs, for a particular value of K.

31.How to find the intersection of root locus with imaginary axis ?

Answer

The intersection of root locus with imaginary axis can be obtained as, Step 1 : Consider characteristic equation 1 + G(s)H(s) = 0 as obtained in Rule 6. Step 2 : Construct Routh’s array in terms of "K". Step 3 : Determine Kmarginal i.e. value of K which creates one of the rows of Routh’s array as row of zeros, except the row of s⁰. Step 4 : Construct auxiliary equation A(s) = 0 by using coefficients of a row which is just above the row of zeros. Step 5 : Roots of auxiliary equation A(s) = 0 for K = Kₘₐᵣ are nothing but the intersection points of the root locus with imaginary axis.

32.State the effect of addition of poles on a root locus and the system.

Answer

Effects of addition of open loop poles in a root locus can be stated as i) Root locus shifts towards imaginary axis. ii) System stability relatively decreases. iii) System becomes more oscillatory in nature. iv) Range of operating values of 'K' for stability of the system decreases.

33.What are the applications of root locus method ?

Answer

The various applications of foot locus method are, i) The absolute stability of the system can be predicted from the locations of the roots in the s-plane. ii) Limiting range of the values of the system gain 'K' can be decided for absolute stability of the system. iii) Marginal value of the system gain 'K' which makes the system marginally stable can be determined and the corresponding value of the frequency of oscillations can be determined from intersection of root locus with imaginary axis. iv) Using root locus, value of system gain 'K' for any point on the root locus can be determined, by using magnitude condition. v) For particular damping ratio of the system, gain 'K' can be determined which helps to design system more correctly.

34.What is root contour ?

Answer

The locus of roots of characteristic equation obtained by varying parameter of the system other than K from 0 to ∞ is called root contour of the system.

35.At breakaway point in root locus several branches meet. Why ?

Answer

A breakaway point is a point where multiple roots of characteristic equation exists for a certain value of K. Thus multiple roots have same value at breakaway point hence several branches meet at breakaway point indicating occurrence of multiple roots at one point.

36.State the basics properties of root locus.

Answer

1. Root locus is symmetrical about real axis. 2. The number of branches of the root locus equals the number of closed loop poles. 3. The branch of a root locus starts from open loop pole and terminates at open look zero. If open loop zero does not exist, branch terminates at ∞. 4. If a point on the s-plane, satisfies the angle condition ∠G(s)H(s) | At point (2q+1)180° then that point lies on the root locus.

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