Unit-3
Time Response Analysis
Lecture 2
Time response of Second-Order Systems
β’ A second-order system is shown in Fig. 1 below.
β’ The transfer function of the system can be written as πΆ(π )
π (π ) = ππ2
π 2 + 2ππππ + ππ2
Where, π = damping ratio, ππ = natural frequency
Fig. 1 Block Diagram of a second-order system
β’ If π=0, the system is called undamped. In this case, the poles are purely imaginary and the transient response is purely oscillatory i.e. it does not die out.
β’ If 0 < π < 1, the closed loop poles lie in the left half plane of the s-plane and they are complex conjugates. In this case, the system is called underdamped. The transient response shows oscillations about the final value (also called overshoot) before finally settling down.
β’ If π > 1, the system is called overdamped. The poles are real, negative and unequal and the output rises towards final value in a sluggish manner.
β’ If π=1, the system is called critically damped. In this case, the poles are real, negative and equal. The response rises slowly and reaches the final value.
β’ Critically damped responses are the fastest possible without any overshoot (shown by underdamped systems).
β’ Most control systems are designed with damping ratio π < 1 (except robotic applications) to have high response speed.
Characteristic equation:
π 2 + 2ππππ + ππ2 = 0
Eq. (7) is called the characteristic equation of the system.
The roots of this equation are given by
π 1, π 2 = βπππ Β± πππ 1 β π2 = βπππ Β± πππ
ππ = ππ 1 β π2 , ππ is called the damped natural frequency.
(7)
Step response of Second-Order Systems
πΆ(π )
π (π ) = ππ2
π 2 + 2ππππ + ππ2
β’ For a unit-step input, R s = 1s πΆ(π )
π (π ) = ππ2
π (π 2 + 2ππππ + ππ2) = 1
π β π + 2πππ
π 2 + 2ππππ + ππ2
= 1
π β π + 2πππ
[(π + πππ)2+ππ2 β ππ2π2]
= 1
π β π + 2πππ
[(π + πππ)2+ππ2] β πππ
ππ . ππ
π + πππ 2 + ππ2
= 1
π β π + 2πππ
[(π + πππ)2+ππ2] β π
1 β π2 . ππ
π + πππ 2 + ππ2 Taking the inverse Laplace transform of eq. (8),
π π‘ = 1 β πβππππ‘πππ πππ‘ β ππβππππ‘
1 β π2 π πππππ‘
= 1 β πβππππ‘
1βπ2 ( 1 β π2 πππ πππ‘ + ππ πππππ‘)
(8)
= 1 β πβππππ‘
1 β π2 π ππππππ πππ‘ + πππ ππ πππππ‘
= 1 β πβππππ‘
1βπ2 π ππ(πππ‘ + π)
= 1 β πβππππ‘
1βπ2 π ππ ππ 1 β π2π‘ + π‘ππβ1 1βπ2
π πππ π‘ β₯ 0.
The error signal,
π π‘ = π π‘ β π π‘ = πβππππ‘(πππ πππ‘ + π
1 β π2 π πππππ‘)
β’ The error signal shows a damped sinusoidal oscillation. At steady state, there is no error exists between the output and the input.
β’ The time response of an underdamped second order system is shown in Fig.2
Fig. 2 Step response of a second-order underdamped system
β’ Fig.3 shows the time response for various values of π.
β’ As π is increased, the response becomes progressively less oscillatory till it becomes critically damped for π=1 and becomes overdamped for π > 1.
Fig. 3 Step response curves of a second-order system
Time response Specifications
Fig. 4 Time Response Specifications
Time response Specifications
The time response specifications are as follows:
1. Rise time 2. Peak Time 3. Delay time
4. Maximum overshoot 5. Settling time
6. Steady state error
1. Rise Time(tr): It is the time required for the response to rise from zero to final value (i.e 0% to 100%) for an under damped system. If the signal is over damped, then rise time is defined as the time required by the response to rise from 10% to 90%
of its final value.
2. Peak Time(tp): It is defined as the time required by response to reach its first peak of the overshoot.
3. Delay Time(tp): The time required to reach 50% of its final value by the time response for the first time is termed as delay time.
4. Maximum (or peak) Overshoot (Mp): The amount that the waveform overshoots the steady state or final value at the peak time is known as maximum or peak overshoot.
Maximum percent overshoot = π π‘π βπ(β)
π(β) x 100%
5. Settling Time(ts): The time required for transientβs damped oscillations to reach and stay within Β±2% of the steady state value.
6. Steady State Error (ess): It indicates the error between actual output and desired output as time (t) tends to infinity.
ππ π = lim
π‘ββ[π π‘ β π π‘ ]
Expressions for time response parameters
1. Rise Time: As obtained earlier, the output of a second order underdamped system excited by unit-step input is,
π π‘ = 1 β πβππππ‘
1 β π2 π ππ(πππ‘ + π)
Since rise time is the time taken by the output to reach from 0 to 100% of the final value, therefore, at t = tr, π π‘ = 1
Therefore, from eq.(9), 1 = 1 β πβππππ‘π
1βπ2 π ππ(πππ‘π + π)
(9)
or, πβππππ‘π
1βπ2 π π π πππ‘π + π = 0 Since, πβππππ‘π
1βπ2 β 0,π π π πππ‘π + π must be equal to zero.
Therefore,π π π πππ‘π + π = 0 = π πππ
πππ‘π + π = π or πππ‘π = π β π
Therefore, rise time
π‘π = π β π
ππ = π β
π‘ππβ1 1 β π2 π
ππ 1 β π2
2. Peak Time: As obtained earlier, the output of a second order underdamped system excited by unit-step input is,
π π‘ = 1 β πβππππ‘
1 β π2 π ππ(πππ‘ + π)
Since peak time is the time taken by the output to reach the maximum value of magnitude, therefore, at t = tp, the slope of π π‘ should be zero.
Therefore, from eq.(10), ππ(π‘)
ππ‘ π‘=π‘π = β πβππππ‘
1βπ2 ππ π πππ‘ + π . ππ β π π π πππ‘ + π πβππππ‘
1βπ2 βπππ π‘=π‘π = 0
(10)
β΄ ππππ π π πππ‘π + π β ππ 1 β π2πππ πππ‘π + π = 0
= ππ π π πππ‘π + π β 1 β π2πππ πππ‘π + π = 0
= πππ ππ π π πππ‘π + π β π ππππππ πππ‘π + π = 0
= π π π πππ‘π + π β π = 0 = π πππ
β΄ πππ‘π = π Therefore, peak time,
π‘π = π
ππ = π
ππ 1 β π2
3. Peak Overshoot: As obtained earlier, the output of a second order underdamped system excited by unit-step input is,
π π‘ = 1 β πβππππ‘
1 β π2 π ππ(πππ‘ + π)
The peak overshoot is the difference between the maximum value and the reference input.
Therefore, from eq.(11),
ππ = π π‘π β 1 = 1 β πβππππ‘π
1 β π2 π π π πππ‘π + π β 1
= β πβππππ‘
1 β π2 π π π πππ‘π + π
(11)
Substituting the expression of peak time,
ππ = β πβπππ
π
ππ 1βπ2
1 β π2 π π π ππ π
ππ + π
= β π
βππ 1βπ2
1 β π2 (βπ π π π) = π
βππ 1βπ2
1 β π2 1 β π2
= π
βππ 1βπ2
Percentage Overshoot, = 100 x π
βππ
1βπ2%
4. Settling Time: It is given by,
π‘π = 4π = 4 πππ
Selection of damping ratio for industrial control applications requires a trade off between relative stability.
In most cases, it is desirable that the transient response be fast with sufficient damping. Thus, for a desirable transient response, the damping ratio must be between 0.4 and 0.8. for large values of damping ratio, the response becomes sluggish.
Example 1 β The closed loop transfer functions of second order unity feedback systems are given below. Determine the type of damping in the systems.
a πΆ(π )
π (π ) = 8
π 2 + 3π + 8 π πΆ(π )
π (π ) = 2
π 2 + 4π + 2 π πΆ(π )
π (π ) = 2
π 2 + 2π + 1
Solution β
(a) Comparing the given transfer function with the standard form of a second order system transfer function
πΆ(π )
π (π ) = 8
π 2 + 3π + 8 = ππ2
π 2 + 2ππππ + ππ2
β΄ ππ2 = 8 π. π. ππ = 2.82
2πππ = 3 π. π. π = 3
2ππ = 3
2 x 2.82 = 0.53 Since π < 1, it is an underdamped system.
(b) Comparing the given transfer function with the standard form of a second order system transfer function
πΆ(π )
π (π ) = 2
π 2 + 4π + 2 = ππ2
π 2 + 2ππππ + ππ2
β΄ ππ2 = 2 π. π. ππ = 1.414
2πππ = 4 π. π. π = 4
2ππ = 4
2 x 1.414 = 1.41 Since π > 1, it is an overdamped system.
(c) Comparing the given transfer function with the standard form of a second order system transfer function
πΆ(π )
π (π ) = 2
π 2 + 2π + 1 = ππ2
π 2 + 2ππππ + ππ2
β΄ ππ2 = 1 π. π. ππ = 1
2πππ = 2 π. π. π = 2
2ππ = 2
2 x 1 = 1 Since π = 1, it is a crtically-damped system.
Example-2 The open loop transfer function of a unity feedback system is
πΊ(π ) = 4
π (π + 1)
Determine the nature of response of the closed-loop system for a unit step input. also, determine the rise time, peak time, peak overshoot and settling time.
Solution β Given, πΊ(π ) = π (π +1)4
The closed loop transfer function is given by, πΆ(π )
π (π ) =
π (π + 1)4 1 + 4
π (π + 1)
= 4
π 2 + π + 4
Comparing it with the standard form of closed loop transfer function of a second order system,
πΆ(π )
π (π ) = 4
π 2 + π + 4 = ππ2
π 2 + 2ππππ + ππ2
β΄ ππ2 = 4 π. π. ππ = 2, π = 2π1
π = 2 x 21 = 0.25 Since π < 1, it is an underdamped system.
Therefore, ππ = ππ 1 β π2 = 2 x 1 β 0.252 = 1.936 πππ
π
π = π‘ππβ1 1 β π2
π = π‘ππβ1 1 β 0.252
0.25 = 1.310 πππ
π = π‘ππβ1 1 β π2
π = π‘ππβ1 1 β 0.252
0.25 = 1.310 πππ The rise time, π‘π = πβπ
ππ = 3.141β1.310
1.936 = 0.945 π The peak time, π‘π = π
ππ = 3.141
1.936 = 1.622
The peak overshoot, ππ = π
βππ
1βπ2 = 0.4326
Percentage overshoot = ππ x 100 % = 43.26 %
Solve the following problems:
1. Obtain the response of a unity feedback system whose open loop transfer function is
πΊ(π ) = 3
π (π + 4) for a unit step input.
2. The open loop transfer function of a unity feedback system is πΊ(π ) = 10
π (π + 4)
Determine the nature of response of the closed loop system for a unit-step input. Also, determine the rise time, peak time, peak overshoot and settling time.
3. The open loop transfer function of a unity feedback system is given by
πΊ(π ) = πΎ
π (ππ + 1)
Where K and T are positive constants. By what factor, the amplifier gain K be reduced so that peak overshoot of unit step response of the system is reduced from 60% to 30%.
References
1. Control Systems by A. Anand Kumar.
2. Modern Control Engineering by Katsuhiko Ogata.
3. Control Systems by M. Gopal
4. NPTEL Video Lectures on Control Engineering (https://nptel.ac.in/courses/108/106/108106098/)
5. Control Systems Engineering by Norman S. Nise.