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Unit-3

Time Response Analysis

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Lecture 2

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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 𝐢(𝑠)

𝑅(𝑠) = πœ”π‘›2

𝑠2 + 2πœ‰πœ”π‘›π‘  + πœ”π‘›2

Where, πœ‰ = damping ratio, πœ”π‘› = natural frequency

Fig. 1 Block Diagram of a second-order system

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β€’ If πœ‰=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œ‰ < 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œ‰ > 1, the system is called overdamped. The poles are real, negative and unequal and the output rises towards final value in a sluggish manner.

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β€’ If πœ‰=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œ‰ < 1 (except robotic applications) to have high response speed.

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Characteristic equation:

𝑠2 + 2πœ‰πœ”π‘›π‘  + πœ”π‘›2 = 0

Eq. (7) is called the characteristic equation of the system.

The roots of this equation are given by

𝑠1, 𝑠2 = βˆ’πœ‰πœ”π‘› Β± π‘—πœ”π‘› 1 βˆ’ πœ‰2 = βˆ’πœ‰πœ”π‘› Β± π‘—πœ”π‘‘

πœ”π‘‘ = πœ”π‘› 1 βˆ’ πœ‰2 , πœ”π‘‘ is called the damped natural frequency.

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Step response of Second-Order Systems

𝐢(𝑠)

𝑅(𝑠) = πœ”π‘›2

𝑠2 + 2πœ‰πœ”π‘›π‘  + πœ”π‘›2

β€’ For a unit-step input, R s = 1s 𝐢(𝑠)

𝑅(𝑠) = πœ”π‘›2

𝑠(𝑠2 + 2πœ‰πœ”π‘›π‘  + πœ”π‘›2) = 1

𝑠 βˆ’ 𝑠 + 2πœ‰πœ”π‘›

𝑠2 + 2πœ‰πœ”π‘›π‘  + πœ”π‘›2

= 1

𝑠 βˆ’ 𝑠 + 2πœ‰πœ”π‘›

[(𝑠 + πœ‰πœ”π‘›)2+πœ”π‘›2 βˆ’ πœ”π‘›2πœ‰2]

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= 1

𝑠 βˆ’ 𝑠 + 2πœ‰πœ”π‘›

[(𝑠 + πœ‰πœ”π‘›)2+πœ”π‘‘2] βˆ’ πœ‰πœ”π‘›

πœ”π‘‘ . πœ”π‘‘

𝑠 + πœ‰πœ”π‘› 2 + πœ”π‘‘2

= 1

𝑠 βˆ’ 𝑠 + 2πœ‰πœ”π‘›

[(𝑠 + πœ‰πœ”π‘›)2+πœ”π‘‘2] βˆ’ πœ‰

1 βˆ’ πœ‰2 . πœ”π‘‘

𝑠 + πœ‰πœ”π‘› 2 + πœ”π‘‘2 Taking the inverse Laplace transform of eq. (8),

𝑐 𝑑 = 1 βˆ’ π‘’βˆ’πœ‰πœ”π‘›π‘‘π‘π‘œπ‘ πœ”π‘‘π‘‘ βˆ’ πœ‰π‘’βˆ’πœ‰πœ”π‘›π‘‘

1 βˆ’ πœ‰2 π‘ π‘–π‘›πœ”π‘‘π‘‘

= 1 βˆ’ π‘’βˆ’πœ‰πœ”π‘›π‘‘

1βˆ’πœ‰2 ( 1 βˆ’ πœ‰2 π‘π‘œπ‘ πœ”π‘‘π‘‘ + πœ‰π‘ π‘–π‘›πœ”π‘‘π‘‘)

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= 1 βˆ’ π‘’βˆ’πœ‰πœ”π‘›π‘‘

1 βˆ’ πœ‰2 π‘ π‘–π‘›πœƒπ‘π‘œπ‘ πœ”π‘‘π‘‘ + π‘π‘œπ‘ πœƒπ‘ π‘–π‘›πœ”π‘‘π‘‘

= 1 βˆ’ π‘’βˆ’πœ‰πœ”π‘›π‘‘

1βˆ’πœ‰2 𝑠𝑖𝑛(πœ”π‘‘π‘‘ + πœƒ)

= 1 βˆ’ π‘’βˆ’πœ‰πœ”π‘›π‘‘

1βˆ’πœ‰2 𝑠𝑖𝑛 πœ”π‘› 1 βˆ’ πœ‰2𝑑 + π‘‘π‘Žπ‘›βˆ’1 1βˆ’πœ‰2

πœ‰ π‘“π‘œπ‘Ÿ 𝑑 β‰₯ 0.

The error signal,

𝑒 𝑑 = π‘Ÿ 𝑑 βˆ’ 𝑐 𝑑 = π‘’βˆ’πœ‰πœ”π‘›π‘‘(π‘π‘œπ‘ πœ”π‘‘π‘‘ + πœ‰

1 βˆ’ πœ‰2 π‘ π‘–π‘›πœ”π‘‘π‘‘)

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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

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β€’ Fig.3 shows the time response for various values of πœ‰.

β€’ As πœ‰ is increased, the response becomes progressively less oscillatory till it becomes critically damped for πœ‰=1 and becomes overdamped for πœ‰ > 1.

Fig. 3 Step response curves of a second-order system

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Time response Specifications

Fig. 4 Time Response Specifications

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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

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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.

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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 = 𝑐 𝑑𝑝 βˆ’π‘(∞)

𝑐(∞) x 100%

5. Settling Time(ts): The time required for transient’s 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.

𝑒𝑠𝑠 = lim

π‘‘β†’βˆž[π‘Ÿ 𝑑 βˆ’ 𝑐 𝑑 ]

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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,

𝑐 𝑑 = 1 βˆ’ π‘’βˆ’πœ‰πœ”π‘›π‘‘

1 βˆ’ πœ‰2 𝑠𝑖𝑛(πœ”π‘‘π‘‘ + πœƒ)

Since rise time is the time taken by the output to reach from 0 to 100% of the final value, therefore, at t = tr, 𝑐 𝑑 = 1

Therefore, from eq.(9), 1 = 1 βˆ’ π‘’βˆ’πœ‰πœ”π‘›π‘‘π‘Ÿ

1βˆ’πœ‰2 𝑠𝑖𝑛(πœ”π‘‘π‘‘π‘Ÿ + πœƒ)

(9)

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or, π‘’βˆ’πœ‰πœ”π‘›π‘‘π‘Ÿ

1βˆ’πœ‰2 𝑠𝑖 𝑛 πœ”π‘‘π‘‘π‘Ÿ + πœƒ = 0 Since, π‘’βˆ’πœ‰πœ”π‘›π‘‘π‘Ÿ

1βˆ’πœ‰2 β‰  0,𝑠𝑖 𝑛 πœ”π‘‘π‘‘π‘Ÿ + πœƒ must be equal to zero.

Therefore,𝑠𝑖 𝑛 πœ”π‘‘π‘‘π‘Ÿ + πœƒ = 0 = π‘ π‘–π‘›πœ‹

πœ”π‘‘π‘‘π‘Ÿ + πœƒ = πœ‹ or πœ”π‘‘π‘‘π‘Ÿ = πœ‹ βˆ’ πœƒ

Therefore, rise time

π‘‘π‘Ÿ = πœ‹ βˆ’ πœƒ

πœ”π‘‘ = πœ‹ βˆ’

π‘‘π‘Žπ‘›βˆ’1 1 βˆ’ πœ‰2 πœ‰

πœ”π‘› 1 βˆ’ πœ‰2

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2. Peak Time: As obtained earlier, the output of a second order underdamped system excited by unit-step input is,

𝑐 𝑑 = 1 βˆ’ π‘’βˆ’πœ‰πœ”π‘›π‘‘

1 βˆ’ πœ‰2 𝑠𝑖𝑛(πœ”π‘‘π‘‘ + πœƒ)

Since peak time is the time taken by the output to reach the maximum value of magnitude, therefore, at t = tp, the slope of 𝑐 𝑑 should be zero.

Therefore, from eq.(10), 𝑑𝑐(𝑑)

𝑑𝑑 𝑑=𝑑𝑝 = βˆ’ π‘’βˆ’πœ‰πœ”π‘›π‘‘

1βˆ’πœ‰2 π‘π‘œ 𝑠 πœ”π‘‘π‘‘ + πœƒ . πœ”π‘‘ βˆ’ 𝑠𝑖 𝑛 πœ”π‘‘π‘‘ + πœƒ π‘’βˆ’πœ‰πœ”π‘›π‘‘

1βˆ’πœ‰2 βˆ’πœ‰πœ”π‘› 𝑑=𝑑𝑝 = 0

(10)

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∴ πœ‰πœ”π‘›π‘ π‘– 𝑛 πœ”π‘‘π‘‘π‘ + πœƒ βˆ’ πœ”π‘› 1 βˆ’ πœ‰2π‘π‘œπ‘  πœ”π‘‘π‘‘π‘ + πœƒ = 0

= πœ‰π‘ π‘– 𝑛 πœ”π‘‘π‘‘π‘ + πœƒ βˆ’ 1 βˆ’ πœ‰2π‘π‘œπ‘  πœ”π‘‘π‘‘π‘ + πœƒ = 0

= π‘π‘œπ‘ πœƒπ‘ π‘– 𝑛 πœ”π‘‘π‘‘π‘ + πœƒ βˆ’ π‘ π‘–π‘›πœƒπ‘π‘œπ‘  πœ”π‘‘π‘‘π‘ + πœƒ = 0

= 𝑠𝑖 𝑛 πœ”π‘‘π‘‘π‘ + πœƒ βˆ’ πœƒ = 0 = π‘ π‘–π‘›πœ‹

∴ πœ”π‘‘π‘‘π‘ = πœ‹ Therefore, peak time,

𝑑𝑝 = πœ‹

πœ”π‘‘ = πœ‹

πœ”π‘› 1 βˆ’ πœ‰2

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3. Peak Overshoot: As obtained earlier, the output of a second order underdamped system excited by unit-step input is,

𝑐 𝑑 = 1 βˆ’ π‘’βˆ’πœ‰πœ”π‘›π‘‘

1 βˆ’ πœ‰2 𝑠𝑖𝑛(πœ”π‘‘π‘‘ + πœƒ)

The peak overshoot is the difference between the maximum value and the reference input.

Therefore, from eq.(11),

𝑀𝑝 = 𝑐 𝑑𝑝 βˆ’ 1 = 1 βˆ’ π‘’βˆ’πœ‰πœ”π‘›π‘‘π‘

1 βˆ’ πœ‰2 𝑠𝑖 𝑛 πœ”π‘‘π‘‘π‘ + πœƒ βˆ’ 1

= βˆ’ π‘’βˆ’πœ‰πœ”π‘›π‘‘

1 βˆ’ πœ‰2 𝑠𝑖 𝑛 πœ”π‘‘π‘‘π‘ + πœƒ

(11)

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Substituting the expression of peak time,

𝑀𝑝 = βˆ’ π‘’βˆ’πœ‰πœ”π‘›

πœ‹

πœ”π‘› 1βˆ’πœ‰2

1 βˆ’ πœ‰2 𝑠𝑖 𝑛 πœ”π‘‘ πœ‹

πœ”π‘‘ + πœƒ

= βˆ’ 𝑒

βˆ’πœ‰πœ‹ 1βˆ’πœ‰2

1 βˆ’ πœ‰2 (βˆ’π‘ π‘– 𝑛 πœƒ) = 𝑒

βˆ’πœ‰πœ‹ 1βˆ’πœ‰2

1 βˆ’ πœ‰2 1 βˆ’ πœ‰2

= 𝑒

βˆ’πœ‰πœ‹ 1βˆ’πœ‰2

Percentage Overshoot, = 100 x 𝑒

βˆ’πœ‰πœ‹

1βˆ’πœ‰2%

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4. Settling Time: It is given by,

𝑑𝑠 = 4𝑇 = 4 πœ‰πœ”π‘›

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.

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Example 1 – The closed loop transfer functions of second order unity feedback systems are given below. Determine the type of damping in the systems.

a 𝐢(𝑠)

𝑅(𝑠) = 8

𝑠2 + 3𝑠 + 8 𝑏 𝐢(𝑠)

𝑅(𝑠) = 2

𝑠2 + 4𝑠 + 2 𝑐 𝐢(𝑠)

𝑅(𝑠) = 2

𝑠2 + 2𝑠 + 1

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Solution –

(a) Comparing the given transfer function with the standard form of a second order system transfer function

𝐢(𝑠)

𝑅(𝑠) = 8

𝑠2 + 3𝑠 + 8 = πœ”π‘›2

𝑠2 + 2πœ‰πœ”π‘›π‘  + πœ”π‘›2

∴ πœ”π‘›2 = 8 𝑖. 𝑒. πœ”π‘› = 2.82

2πœ‰πœ”π‘› = 3 𝑖. 𝑒. πœ‰ = 3

2πœ”π‘› = 3

2 x 2.82 = 0.53 Since πœ‰ < 1, it is an underdamped system.

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(b) Comparing the given transfer function with the standard form of a second order system transfer function

𝐢(𝑠)

𝑅(𝑠) = 2

𝑠2 + 4𝑠 + 2 = πœ”π‘›2

𝑠2 + 2πœ‰πœ”π‘›π‘  + πœ”π‘›2

∴ πœ”π‘›2 = 2 𝑖. 𝑒. πœ”π‘› = 1.414

2πœ‰πœ”π‘› = 4 𝑖. 𝑒. πœ‰ = 4

2πœ”π‘› = 4

2 x 1.414 = 1.41 Since πœ‰ > 1, it is an overdamped system.

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(c) Comparing the given transfer function with the standard form of a second order system transfer function

𝐢(𝑠)

𝑅(𝑠) = 2

𝑠2 + 2𝑠 + 1 = πœ”π‘›2

𝑠2 + 2πœ‰πœ”π‘›π‘  + πœ”π‘›2

∴ πœ”π‘›2 = 1 𝑖. 𝑒. πœ”π‘› = 1

2πœ‰πœ”π‘› = 2 𝑖. 𝑒. πœ‰ = 2

2πœ”π‘› = 2

2 x 1 = 1 Since πœ‰ = 1, it is a crtically-damped system.

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Example-2 The open loop transfer function of a unity feedback system is

𝐺(𝑠) = 4

𝑠(𝑠 + 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 – Given, 𝐺(𝑠) = 𝑠(𝑠+1)4

The closed loop transfer function is given by, 𝐢(𝑠)

𝑅(𝑠) =

𝑠(𝑠 + 1)4 1 + 4

𝑠(𝑠 + 1)

= 4

𝑠2 + 𝑠 + 4

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Comparing it with the standard form of closed loop transfer function of a second order system,

𝐢(𝑠)

𝑅(𝑠) = 4

𝑠2 + 𝑠 + 4 = πœ”π‘›2

𝑠2 + 2πœ‰πœ”π‘›π‘  + πœ”π‘›2

∴ πœ”π‘›2 = 4 𝑖. 𝑒. πœ”π‘› = 2, πœ‰ = 2πœ”1

𝑛 = 2 x 21 = 0.25 Since πœ‰ < 1, it is an underdamped system.

Therefore, πœ”π‘‘ = πœ”π‘› 1 βˆ’ πœ‰2 = 2 x 1 βˆ’ 0.252 = 1.936 π‘Ÿπ‘Žπ‘‘

𝑠

πœƒ = π‘‘π‘Žπ‘›βˆ’1 1 βˆ’ πœ‰2

πœ‰ = π‘‘π‘Žπ‘›βˆ’1 1 βˆ’ 0.252

0.25 = 1.310 π‘Ÿπ‘Žπ‘‘

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πœƒ = π‘‘π‘Žπ‘›βˆ’1 1 βˆ’ πœ‰2

πœ‰ = π‘‘π‘Žπ‘›βˆ’1 1 βˆ’ 0.252

0.25 = 1.310 π‘Ÿπ‘Žπ‘‘ The rise time, π‘‘π‘Ÿ = πœ‹βˆ’πœƒ

πœ”π‘‘ = 3.141βˆ’1.310

1.936 = 0.945 𝑠 The peak time, 𝑑𝑝 = πœ‹

πœ”π‘‘ = 3.141

1.936 = 1.622

The peak overshoot, 𝑀𝑝 = 𝑒

βˆ’πœ‰πœ‹

1βˆ’πœ‰2 = 0.4326

Percentage overshoot = 𝑀𝑝 x 100 % = 43.26 %

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Solve the following problems:

1. Obtain the response of a unity feedback system whose open loop transfer function is

𝐺(𝑠) = 3

𝑠(𝑠 + 4) for a unit step input.

2. The open loop transfer function of a unity feedback system is 𝐺(𝑠) = 10

𝑠(𝑠 + 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.

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3. The open loop transfer function of a unity feedback system is given by

𝐺(𝑠) = 𝐾

𝑠(𝑇𝑠 + 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%.

(32)

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.

References

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