Ch.04 Time Response

of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Learning Outcome After completing this chapter, the student will be able to • Use poles and zeros of transfer functions t

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04 Time Response

System Dynamics and Control 4.01 Time Response

HCM City Univ of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien

Learning Outcome After completing this chapter, the student will be able to

• Use poles and zeros of transfer functions to determine the time response of a control system

• Describe quantitatively the transient response of 1st-order systems

• Write the general response of second-order systems given the pole location

• Find the damping ratio and natural frequency of a second-order system

• Find the settling time, peak time, percent overshoot, and rise time for an underdamped second-order system

• Approximate higher-order systems and systems with zeros as first- or second-order systems

• Describe the effects of nonlinearities on the system time response

• Find the time response from the state-space representation

§1.Introduction

- This chapter

• Analyze of system transient response

• Demonstrates applications of the system representation by

evaluating the transient response from the system model

• Evaluate the response of a subsystem prior to inserting it into

the closed-loop system

• Describe a valuable analysis and design tool, poles and zeros

• Analyze the models to find the step response of first- and

second-order systems

§2.Poles, Zeros, and System Response

- The output response of a system is the sum of two responses

• The forced response

• The natural response

- Output response of a system depends on the positions of the poles and zeros and their relationships

⟹ study

• the concept of poles and zeros

• fundamental to the analysis and design of control systems

• simplifies the evaluation of a system’s response

- Transfer function

𝐺 𝑠 =𝑌 𝑠

𝑈 𝑠 =

𝑏0𝑠𝑚+ 𝑏1𝑠𝑚−1+ ⋯ + 𝑏𝑚−1𝑠 + 𝑏𝑚

𝑎0𝑠𝑛+ 𝑎1𝑠𝑛−1+ ⋯ + 𝑎𝑛−1𝑠 + 𝑎𝑛

• The poles of a transfer function are roots of

𝑎0𝑠𝑛+ 𝑎1𝑠𝑛−1+ ⋯ + 𝑎𝑛−1𝑠 + 𝑎𝑛= 0

• The zeros of a transfer function are roots of

𝑏0𝑠𝑚+ 𝑏1𝑠𝑚−1+ ⋯ + 𝑏𝑚−1𝑠 + 𝑏𝑚= 0

- Example of poles and zeros of a first order system

System

𝐶 𝑠 =1

𝑠×

𝑠 + 2

𝑠 + 5=

𝑠 + 2 𝑠(𝑠 + 5)

𝐴 = 𝑠 + 2 (𝑠 + 5)𝑠→0=

2

5, 𝐵 =

𝑠 + 2

𝑠 𝑠→−5=

2 5

𝑐 𝑡 =2

5+

3

5𝑒

−5𝑡

=𝐴

𝑠+

𝐵

𝑠 + 5=

2/5

𝑠 +

3/5

𝑠 + 5

Response

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Evolution of a system response

§2.Poles, Zeros, and System Response Conclusions

• A pole of the input function generates the form of the forced response (that is, the pole at the origin generated a step function at the output)

• A pole of the transfer function generates the form of the natural response (that is, the pole at5 generated 𝑒5𝑡)

• A pole on the real axis generates an exponential response of the form𝑒𝑎𝑡, where𝑎 is the pole location on the real axis

Thus, the farther to the left a pole is on the negative real axis, the faster the exponential transient response will decay to zero

• The zeros and poles generate the amplitudes for both the forced and natural responses (this can be seen from the calculation of𝐴 and 𝐵)

Write the output,𝑐(𝑡), in general terms, for the given system

Specify the forced and natural parts of the solution

Solution

• Each system pole generates an exponential as part of the

natural response

• The input’s pole generates the forced response

𝐶 𝑠 = 𝐾1

𝐾2

𝑠 + 2+

𝐾3

𝑠 + 4+

𝐾4

𝑠 + 5

⟹ 𝑐 𝑡 = 𝐾1 + 𝐾2𝑒−2𝑡+ 𝐾3𝑒−4𝑡+ 𝐾4𝑒−5𝑡

Thus

§3.First Order Systems

- A first-order system without zeros can be described by the transfer function

- If the input is step

𝐶 𝑠 = 𝑅 𝑠 𝐺 𝑠 =1

𝑠×

𝑎

𝑠 + 𝑎=

𝑎 𝑠(𝑠 + 𝑎)=

1

𝑠−

1

𝑠 + 𝑎

- Taking the inverse transform, the step response is given by

𝑐 𝑡 = 𝑐𝑓𝑡 + 𝑐𝑛𝑡 = 1 − 𝑒−𝑎𝑡

• input pole, 0, generates the forced response 𝑐𝑓𝑡 = 1

• system pole, −𝑎, generates the natural response 𝑐𝑛𝑡 = −𝑒−𝑎𝑡

𝐶 𝑠 = 𝑅 𝑠 𝐺 𝑠 =1

𝑠×

𝑎

𝑠 + 𝑎=

𝑎 𝑠(𝑠 + 𝑎)= 1

𝑠− 1

𝑠 + 𝑎⟹ 𝑐 𝑡 = 𝑐𝑓 𝑡 + 𝑐 𝑛 𝑡 = 1 − 𝑒 −𝑎𝑡

1.Time Constant𝑇𝑐

- Time constant𝑇𝑐= 1/𝑎

𝑒−𝑎𝑡

𝑡=1𝑎= 𝑒−1= 0.37

𝑐(𝑡)

𝑡=1𝑎= 1 −𝑒−𝑎𝑡

𝑡=1𝑎= 0.63

- The time constant is the time it takes for the step response to

rise to63% of its final value

First-order system response to a unit step

§3.First Order Systems 𝑑

𝑑𝑡𝑒−𝑎𝑡

𝑡=0

=−𝑎𝑒−𝑎𝑡 𝑡=0= −𝑎 𝑎: the initial rate of change

of the exponential at𝑡 = 0

- Transient response specifications

𝑎 : time constant

𝑇𝑟: rise time

𝑇𝑠: settling time

-𝑎 relates to the speed at which the system responds to a step input: the farther the pole from the imaginary axis, the faster the transient response

-𝑎 has the units 1/𝑠, or frequency ⟹ theexponential frequency

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2.Rise Time𝑇𝑟

- The time for the waveform to go

from0.1 to 0.9 of its final value

- Rise time is found by solving

𝑐 𝑡 = 1 − 𝑒−𝑎𝑡

𝑐 𝑡1 = 0.1

1 − 𝑒−𝑎𝑡 1= 0.1 ⟹ 𝑡1= −𝑙𝑛0.9

𝑎

𝑐 𝑡2 = 0.9

1 − 𝑒−𝑎𝑡 2= 0.9 ⟹ 𝑡2= −𝑙𝑛0.1

𝑎

⟹ 𝑡2− 𝑡1= −𝑙𝑛0.1

𝑙𝑛0.9

0.11

2.31

𝑎 ⟹𝑇𝑟=

2.2 𝑎

§3.First Order Systems 3.Settling Time𝑇𝑠

- The time for the response to reach, and stay within,2% of its final value

- Settling time is found by solving

𝑐 𝑡 = 1 − 𝑒−𝑎𝑡

𝑐 𝑇𝑠 = 0.98

1 − 𝑒−𝑎𝑇 𝑠= 0.98 ⟹ 𝑇𝑠= −𝑙𝑛0.02

𝑎

⟹𝑇𝑠=4 𝑎

4.First-Order Transfer Functions via Testing

- Consider a simple first-order system with the step input

𝐶 𝑠 =1

𝑠×

𝐾

𝑠 + 𝑎=

𝐾/𝑎

𝑠 + 𝑎

If we can identify𝐾 and 𝑎 from laboratory testing, we can

obtain the transfer function of the system

§3.First Order Systems Example using the plot of

From the plot

• Steady state value 𝑐𝑠𝑠= 0.72

• Time constant 𝑇𝑐= 0.13𝑠 Therefore

𝑇𝑐=1

𝑎⟹ 𝑎 = 1/0.13 = 7.7 𝐾

𝑎= 𝑐𝑠𝑠⟹ 𝐾 = 𝑎𝑐𝑠𝑠= 7.7 × 0.72 = 5.54

The transfer function of the plot Note: 1storder system has no overshoot and nonzero initial slope

𝑠 + 7

𝐺 𝑠 = 5.54

𝑠 + 7.7

Skill-Assessment Ex.4.2

Problem Find the time constant,𝑇𝑐, rising time,𝑇𝑟, and settling

time,𝑇𝑠, of the system with the following TF

𝐺 𝑠 = 50

𝑠 + 50 Solution From the TF

𝑎 = 50

𝑇𝑐=1

𝑎=

1

50= 0.02𝑠

𝑇𝑟=2.2

𝑎 =

2.2

50= 0.044𝑠

𝑇𝑠=4

𝑎=

4

50= 0.08𝑠

Time constant

Rising time

Settling time

§4.Second Order Systems: Introduction

- The general second order system

- Depending on component values, the second order system exhibits a wide range of responses such as

• first-order system

• damped oscillation

• pure oscillation

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[r,p,k]=residue([9],[1,9,9,0]), 𝐶 𝑠 = 1/𝑠 → 𝑐(𝑡) = 1, 𝐶 𝑠 = 1/(𝑠 + 𝑎) → 𝑐(𝑡) = 𝑒 −𝑎𝑡

- Second-order systems, pole plots, and step responses

• Overdamped

𝑠(𝑠 + 1.146)(𝑠 + 7.854)

⟹𝑐 𝑡 = 1 + 0.171𝑒−7.854𝑡− 0.171𝑒−1.146𝑡

Poles: two real at−𝜎1and−𝜎2

Natural response: 𝑐𝑛𝑡 = 𝐾1𝑒−𝜎 1 𝑡+ 𝐾2𝑒−𝜎 2 𝑡

• Underdamped

𝑠(𝑠2+ 2𝑠 + 9)

𝑠 𝑠 − −1 − 𝑗 8 𝑠 − −1 + 𝑗 8

⟹𝑐 𝑡 = 1 − 1.06𝑒−𝑡cos( 8𝑡 − 19.470) Poles: two complex at−𝜎𝑑± 𝑗𝜔𝑑 Natural response: 𝑐𝑛𝑡 = 𝐴𝑒−𝜎 𝑑 𝑡𝑐𝑜𝑠 𝜔𝑑𝑡 − 𝜙

• Undamped

𝑠(𝑠2+ 9)=

1

𝑠−

𝑠

𝑠2+ 32

⟹𝑐 𝑡 = 1 − 𝑐𝑜𝑠3𝑡 Poles: two imaginary at±𝑗𝜔1 Natural response:𝑐𝑛𝑡 = 𝐴𝑐𝑜𝑠 𝜔1𝑡 − 𝜙

𝐶 𝑠 = 𝑠/(𝑠 2 + 𝑎 2 ) → 𝑐(𝑡) = 𝑐𝑜𝑠𝑎𝑡

𝐶 𝑠 = 𝑎/(𝑠 + 𝑎) 2 → 𝑐(𝑡) = 𝑎𝑡𝑒 −𝑎𝑡

• Critically damped

𝑠(𝑠2+ 6𝑠 + 9)

=1

𝑠−

1

𝑠 + 3−

3

𝑠 + 32

⟹𝑐 𝑡 = 1 − 3𝑡𝑒−3𝑡− 𝑒−3𝑡

Natural response:𝑐𝑛𝑡 = 𝐾1𝑒−𝜎 1 𝑡+ 𝐾2𝑡𝑒−𝜎 1 𝑡

• Step responses for second-order system damping cases

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Problem For each of the following transfer functions, write, by

inspection, the general form of the step response

𝑠2+ 12𝑠 + 400

𝑠2+ 90𝑠 + 900

𝑠2+ 30𝑠 + 225

𝑠2+ 625

a

b

c

d

𝐺(𝑠) has poles at −𝜎 𝑑 ± 𝑗𝜔 𝑑 ⟹ natural response 𝑐 𝑛 𝑡 = 𝐴𝑒 −𝜎 𝑑 𝑡 𝑐𝑜𝑠 𝜔 𝑑 𝑡 − 𝜙

𝐺(𝑠) has poles at −𝜎 1 , −𝜎 2 ⟹ natural response 𝑐 𝑛 𝑡 = 𝐾 1 𝑒 −𝜎 1 𝑡 + 𝐾 2 𝑒 −𝜎 2 𝑡

𝑠2+ 12𝑠 + 400

𝑠 − −6 − 𝑗19.08 𝑠 − −6 + 𝑗19.08 The system has two complex poles at−6 ± 𝑗19.08 The general form of the step response

𝑐 𝑡 = 𝐴 + 𝐵𝑒−6𝑡𝑐𝑜𝑠 19.08𝑡 + 𝜙

𝑠2+ 90𝑠 + 900=

900

𝑠 + 78.54 𝑠 + 11.46 The system has two real poles at−78.54 and −11.46 The general form of the step response

𝑐 𝑡 = 𝐴 + 𝐵𝑒−78.54𝑡+ 𝐶𝑒−11.46𝑡

Solution a

b

𝐺(𝑠) has poles at −𝜎 1 , −𝜎 1 ⟹ natural response 𝑐 𝑛 𝑡 = 𝐾 1 𝑒 −𝜎 1 𝑡 + 𝐾 2 𝑡𝑒 −𝜎 1 𝑡

𝐺(𝑠) has poles at ±𝑗𝜔 1 ⟹ natural response 𝑐 𝑛 𝑡 = 𝐴𝑐𝑜𝑠 𝜔 1 𝑡 − 𝜙

𝑠2+ 30𝑠 + 225=

400

𝑠 + 152

The system has two repeated real poles at−15

The general form of the step response

𝑐 𝑡 = 𝐴 + 𝐵𝑒−15𝑡+ 𝐶𝑡𝑒−15𝑡

𝑠2+ 625=

625

𝑠2+ 252

The system has two imaginary poles at±𝑗25

The general form of the step response

𝑐 𝑡 = 𝐴 + 𝐵𝑐𝑜𝑠(25𝑡 + 𝜙)

c

d

§5.The General Second Order Systems 1.Natural frequency𝜔𝑛

- The natural frequency of a second-order system is the frequency of oscillation of the system without damping 2.Damping ratio𝜁

- The natural frequency of a second-order system is the frequency of oscillation of the system without damping

𝜁 =exponentialdeca𝑦frequency natural frequency =

1 2𝜋

natural period exponentialtimeconstant

- Consider the general second order system

𝑠2+ 𝑎𝑠 + 𝑏≡

𝜔𝑛2

𝑠2+ 2𝜁𝜔𝑛𝑠 + 𝜔𝑛2

𝜁 = 𝑎

§5.The General Second Order Systems

- Ex.4.5 Finding𝜁 and 𝜔𝑛For a Second-Order System

Given the transfer function of Eq (4.23), find𝜁and 𝜔𝑛

𝑠2+ 4.2𝑠 + 36

Solution

From observation

• The natural frequency of the given system

𝜔𝑛= 𝑏 = 36 = 6

• The damping ratio of the given system

𝜁 = 𝑎

2 𝑏=

4.2

2 × 6= 0.35

(4.23)

- The poles of the transfer function

2

𝑠2+ 2𝜁𝜔𝑛𝑠 + 𝜔𝑛2⟹ 𝑝1,2= −𝜁𝜔𝑛± 𝜔𝑛 𝜁2− 1

⟹ Second-order response as a function of damping ratio

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- Ex.4.4 Characterizing Response from the Value of𝜁

Find the value of 𝜁 and report the kind of response expected

Solution

𝜁 = 𝑎

2 𝑏=

8

2 12= 1.16

⟹ overdamped

𝜁 = 𝑎

2 𝑏=

8

2 16= 1

⟹ critically damped

𝜁 = 𝑎

2 𝑏=

8

2 20= 0.89

⟹ underdamped

ℒ 𝑒 −𝑎𝑡 𝑐𝑜𝑠𝜔𝑡 =(𝑠+𝑎)𝑠+𝑎2 +𝜔 2 , ℒ 𝑒 −𝑎𝑡 𝑠𝑖𝑛𝜔𝑡 =(𝑠+𝑎)𝜔2 +𝜔 2 (2.34-35)

§6.Underdamped Second Order Systems

- The step response for the general second-order system

2

𝑠 𝑠2+ 2𝜁𝜔𝑛𝑠 + 𝜔𝑛2

=1

𝑠−

𝑠 + 𝜁𝜔𝑛

𝑠+𝜁𝜔𝑛2+ 𝜔𝑛 1−𝜁2

1−𝜁2

𝜔𝑛 1 − 𝜁2

𝑠+𝜁𝜔𝑛2+ 𝜔𝑛 1−𝜁2

⟹ 𝑐 𝑡 = 1 − 𝑒−𝜁𝜔 𝑛 𝑡 𝑐𝑜𝑠𝜔𝑛 1 − 𝜁2𝑡 + 𝜁

1 − 𝜁2𝑠𝑖𝑛𝜔𝑛 1 − 𝜁2𝑡 𝑐(𝑡) = 1 − 1

1 − 𝜁2𝑒−𝜁𝜔 𝑛 𝑡𝑐𝑜𝑠 𝜔𝑛 1 − 𝜁2𝑡 − 𝜙

𝜙 = 𝑡𝑎𝑛−1 𝜁/ 1 − 𝜁2

(4.28)

- Second-order underdamped responses for damping ratio

values

- Second-order underdamped response specifications

• Rise time,𝑇𝑟

𝑇𝑟= 𝑡

0.9𝑐 𝑓𝑖𝑛𝑎𝑙

− 𝑡

0.1𝑐 𝑓𝑖𝑛𝑎𝑙

• Peak time,𝑇𝑝

𝑇𝑝= 𝑡

𝑐 𝑚𝑎𝑥

• Percent overshoot,%𝑂𝑆

%𝑂𝑆 =𝑐𝑚𝑎𝑥− 𝑐𝑓𝑖𝑛𝑎𝑙

𝑐𝑓𝑖𝑛𝑎𝑙 × 100

• Settling time,𝑇𝑠 The time required for thetransient’s damped oscillations to reach and stay within2% of the steady-state value

𝑐(𝑡) = 1 − 1

1−𝜁 2 𝑒 −𝜁𝜔 𝑛 𝑡 𝑐𝑜𝑠 𝜔 𝑛 1 − 𝜁 2 𝑡 − 𝜙 (4.28)

1.Evaluation of𝑇𝑝

𝑇𝑝= 𝑡

𝑐 𝑚𝑎𝑥

= 𝑡

𝑐=0

𝑐 𝑡 = 𝜔𝑛

1 − 𝜁2𝑒−𝜁𝜔 𝑛 𝑡𝑠𝑖𝑛𝜔𝑛 1 − 𝜁2𝑡

𝑐 𝑡 = 0 ⟹ 𝜔𝑛 1 − 𝜁2𝑡 = 𝑛𝜋

𝜔𝑛 1 − 𝜁2

(4.28)⟹

(4.34)

𝑐(𝑡) = 1 − 1

1−𝜁 2 𝑒 −𝜁𝜔 𝑛 𝑡 𝑐𝑜𝑠 𝜔 𝑛 1 − 𝜁 2 𝑡 − 𝜙 (4.28)

§6.Underdamped Second Order Systems 2.Evaluation of%𝑂𝑆

%𝑂𝑆 =𝑐𝑚𝑎𝑥− 𝑐𝑓𝑖𝑛𝑎𝑙

𝑐𝑓𝑖𝑛𝑎𝑙 × 100 (4.28) ⟹ 𝑐𝑚𝑎𝑥= 𝑐(𝑇𝑝)

= 1 − 𝑒−𝜁𝜋/ 1−𝜁 2

𝑐𝑜𝑠𝜋 + 𝜁

1 − 𝜁2𝑠𝑖𝑛𝜋

= 1 + 𝑒−𝜁𝜋/ 1−𝜁 2

For the unit step𝑐𝑓𝑖𝑛𝑎𝑙= 1

⟹%𝑂𝑆 = 𝑒−𝜁𝜋/ 1−𝜁 2

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%𝑂𝑆 = 𝑒 −𝜁𝜋/ 1−𝜁 2

𝜁 can be found from given %𝑂𝑆

(4.38) ⟹𝜁 = − 𝑙𝑛(%𝑂𝑆/100)

𝜋2+ 𝑙𝑛2(%𝑂𝑆/100) Percent overshoot versus damping ratio

(4.39)

𝑐(𝑡) = 1 − 1

1−𝜁 2 𝑒 −𝜁𝜔 𝑛 𝑡 𝑐𝑜𝑠 𝜔 𝑛 1 − 𝜁 2 𝑡 − 𝜙 (4.28)

§6.Underdamped Second Order Systems 3.Evaluation of𝑇𝑠

- To find the settling time⟹ to find the time it takes for the amplitude of the decaying sinusoid in Eq (4.28) to reach0.02 1

1 − 𝜁2𝑒−𝜁𝜔 𝑛 𝑡= 0.02

⟹ 𝑇𝑠=−𝑙𝑛(0.02 1 − 𝜁

2)

𝜁𝜔𝑛

- An approximation for the settling time that will be used for all values of𝜉

𝑇𝑠= 4

4.Evaluation of𝑇𝑟

- A precise analytical relationship between rise time and

damping ratio,𝜁, cannot be found ⟹ using numerical method

- Ex Normalized rise time versus damping ratio for a

second-order underdamped response

𝜔𝑛𝑇𝑟≈ 1.76𝜁3− 0.417𝜁2+ 1.039𝜁 + 1

- Ex.4.5 Finding𝑇𝑝,%𝑂𝑆, 𝑇𝑠, and𝑇𝑟from a Transfer Function Finding𝑇𝑝,%𝑂𝑆, 𝑇𝑠, and𝑇𝑟from given transfer function

𝑠2+ 15𝑠 + 100 Solution

From observation

𝜔𝑛= 𝑏 = 100 = 10

𝜁 = 𝑎

2 𝑏=

15

2 100= 0.75

𝜔𝑛 1 − 𝜁2= 𝜋

10 1 − 0.752= 0.475

𝑇𝑠= 4

0.75 × 10= 0.533

%𝑂𝑆 = 𝑒−𝜁𝜋/ 1−𝜁 2

× 100 = 𝑒−0.75𝜋/ 1−0.75 2

× 100 = 2.84

- The relation between peak time, percent overshoot, and settling

time to the location of the poles

• Pole plot for an underdamped second-order system

𝜔𝑑≡ 𝑗𝜔𝑛 1 − 𝜁2

𝜎𝑑≡ 𝜁𝜔𝑛

𝑐𝑜𝑠𝜃 = 𝜁

𝜔𝑑 : damped frequency of oscillation

𝜎𝑑 : exponential damping frequency

- The peak time and settling time in terms of the pole location

𝜔𝑛 1 − 𝜁2= 𝜋

𝜔𝑑

𝑇𝑠= 4

𝜁𝜔𝑛=

4

(4.44)

- Lines of constant peak time,𝑇𝑝, settling time,𝑇𝑠, and percent overshoot,%𝑂𝑆 (𝑇𝑠1< 𝑇𝑠2,𝑇𝑝1< 𝑇𝑝2,%𝑂𝑆1< %𝑂𝑆2)

Trang 8

- Step responses of the

2nd-order underdamped

systems as poles move with

• constant imaginary part

• constant real part

• constant damping ratio

- Ex.4.6 Finding𝑇𝑝,%𝑂𝑆, and 𝑇𝑠from Pole Location Given the pole plot, find𝜁; 𝜔𝑛;𝑇𝑝;%𝑂𝑆, and 𝑇𝑠

Solution The damping ratio

𝜁 = 𝑐𝑜𝑠𝜃 = 𝑐𝑜𝑠 𝑡𝑎𝑛−17/3 = 0.394 The natural frequency

𝜔𝑛= 72+ 32= 7.616 The peak time

𝑇𝑝= 𝜋/𝜔𝑑= 𝜋/7 = 0.449𝑠 The percent overshot

%𝑂𝑆 = 𝑒−𝜁𝜋 1−𝜁 2

× 100 = 26%

The approximate settling time

𝑇𝑠= 4/𝜎𝑑= 4/3 = 1.333𝑠

- Ex.4.7 Transient Response Through Component Design

Find𝐽 and 𝐷 to yield 20% overshoot and a settling time of2𝑠 for a step input of torque𝑇(𝑡)

Solution

The rotational mechanical system

Requirement

𝑇𝑠= 4/𝜁𝜔𝑛= 2 ⟹ 𝜔𝑛= 2/0.456 = 4.386

Observation𝜔𝑛= 𝐾/𝐽 = 4.386

2𝜁𝜔𝑛= 𝐷/𝐽 = 2 × 0.456 × 4.386 = 38.474

Therefore 𝐷 = 1.04𝑁𝑚𝑠/𝑟𝑎𝑑, 𝐽 = 0.26𝑘𝑔𝑚2

𝐺 𝑠 =

1 𝐽

𝑠2+𝐷𝐽𝑠 +𝐾𝐽

%𝑂𝑆 = 20 ⟹ 𝜁 = −𝑙𝑛(%𝑂𝑆/100)

𝜋2+ 𝑙𝑛2(%𝑂𝑆/100)= 0.456

§6.Underdamped Second Order Systems 5.Second-Order Transfer Functions via Testing

- Consider a simple first-order system with the step input

From testing data in laboratory

• The response curve for percent overshoot, %𝑂𝑆 , and settling time,𝑇𝑠

%𝑂𝑆, 𝑇𝑠⟹ 𝜁, 𝜔𝑛

• The expected steady-state values, 𝑐𝑓𝑖𝑛𝑎𝑙

𝑐𝑓𝑖𝑛𝑎𝑙 ⟹ 𝐾

- A problem at the end of the chapter illustrates the estimation

of a second-order transfer function from the step response

𝜔 𝑛 𝑇 𝑟 ≈ 1.76𝜁 3 − 0.417𝜁 2 + 1.039𝜁 + 1

Problem Find𝜁, 𝜔𝑛,𝑇𝑠,𝑇𝑝,𝑇𝑟,%𝑂𝑆 for

Solution𝜔𝑛= 361 = 19

2𝜁𝜔𝑛= 16 ⟹ 𝜁 = 16

2𝜔𝑛=

16

2 × 19= 0.421

𝑇𝑠= 4

𝜁𝜔𝑛=

4

8= 0.5𝑠

𝜔𝑛 1 − 𝜁2= 𝜋

19 1 − 0.4212= 0.182𝑠

𝜔𝑛𝑇𝑟= 1.4998 ⟹ 𝑇𝑟=1.4998

𝜔𝑛 =1.4998

19 = 0.079𝑠

%𝑂𝑆 = 𝑒

−𝜁𝜋

1−𝜁 2

× 100% = 𝑒

−0.421𝜋 1−0.421 2× 100% = 23.3%

𝑠2+ 16𝑠 + 361

Matlab numg=361;

deng=[1 16 361];

omegan=sqrt(deng(3)/deng(1)) zeta=(deng(2)/deng(1))/(2*omegan) Ts=4/(zeta*omegan)

Tp=pi/(omegan*sqrt(1-zeta^2)) pos=100*exp(-zeta*pi/sqrt(1-zeta^2)) Tr=(1.768*zeta^3

-0.417*zeta^2+1.039*zeta+1)/omegan Result

𝜁 = 0.421, 𝜔𝑛= 19, 𝑇𝑠= 0.5𝑠, 𝑇𝑝= 0.182𝑠,

𝑇𝑟= 0.079𝑠, %𝑂𝑆 = 23.3%

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§7.Systems Response with Additional Poles

- The formulas describing percent overshoot, settling time, and

peak time were derivedonly forasystem with two complex

poles and no zeros

⟹ for a system with more than two poles or with zeros ?

- Before apply the derived formulas, a system with more than two

poles or with zerosmust be approximatedas a second-order

system that has just two complex dominant poles

- In this section, we investigate the effect of an additional pole on

the second-order response In the next section, we analyze the

effect of adding a zero to a two-pole system

- Let us now look at the conditions that would have to exist in order to approximate the behavior of a three-pole system as that of a two-pole system

Consider a three-pole system with complex poles, −𝜁𝜔𝑛±

𝑗𝜔𝑛 1 − 𝜁2, and a third pole on the real axis,−𝛼𝑟

The output transform

𝐶 𝑠 =𝐴

𝑠+

𝐵 𝑠 + 𝜁𝜔𝑛 + 𝐶𝜔𝑑

𝑠 + 𝜁𝜔𝑛2+ 𝜔𝑑2 + 𝐷

𝑠 + 𝛼𝑟

In time domain

𝑐 𝑡 = 𝐴𝑢 𝑡 + 𝑒−𝜁𝜔 𝑛 𝑡𝐵𝑐𝑜𝑠𝜔𝑑𝑡 + 𝐶𝑠𝑖𝑛𝜔𝑑𝑡 + 𝐷𝑒−𝛼 𝑟 𝑡

- Component responses of a three-pole system

Case I: Nondominant pole is

near dominant 2 nd -order pair

Case II: Nondominant Case III: Nondominant

pole is at infinity

- Ex.4.8 Comparing Responses of Three-Pole Systems Find the step response of each of the TFs and compare them

𝑇1𝑠 = 24.542

𝑠2+ 4𝑠 + 24.542, 𝑇2𝑠 =

1

𝑠 + 10𝑇1𝑠 , 𝑇3𝑠 =

1

𝑠 + 3𝑇1𝑠 Solution

𝑇1𝑠 = 24.542

𝑠2+ 4𝑠 + 24.542

⟹ 𝑐1𝑡 = 1 − 1.09𝑒−2𝑡cos(4.532𝑡 − 23.80)

(𝑠 + 10)(𝑠2+ 4𝑠 + 24.542)

⟹ 𝑐2 𝑡 = 1 − 0.29𝑒−10𝑡− 1.189𝑒−2𝑡cos(4.532𝑡 − 53.340)

(𝑠 + 3)(𝑠2+ 4𝑠 + 24.542)

⟹ 𝑐3 𝑡 = 1 − 1.14𝑒−3𝑡+ 0.707𝑒−2𝑡cos(4.532𝑡 + 78.630)

𝑇1 𝑠 = 24.542

𝑠2+ 4𝑠 + 24.542

𝑠 − (−2 − 𝑗4.532) 𝑠 − (−2 + 𝑗4.532)

⟹ 𝑐1𝑡 = 1 − 1.09𝑒−2𝑡cos(4.532𝑡 − 23.80)

𝑐1 𝑡 : the pure second-order system response

(𝑠 + 10)(𝑠2+ 4𝑠 + 24.542)

(𝑠 + 10) 𝑠 − (−2 − 𝑗4.532) 𝑠 − (−2 + 𝑗4.532)

⟹ 𝑐2 𝑡 = 1 − 0.29𝑒−10𝑡− 1.189𝑒−2𝑡cos(4.532𝑡 − 53.340)

𝑐2𝑡 : with the third pole at −10 and farthest from the dominant poles, is the better approximation of𝑐1𝑡

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(𝑠 + 3)(𝑠2+ 4𝑠 + 24.542)

(𝑠 + 3) 𝑠 − (−2 − 𝑗4.532) 𝑠 − (−2 + 𝑗4.532)

⟹ 𝑐3𝑡 = 1 − 1.14𝑒−3𝑡+ 0.707𝑒−2𝑡cos(4.532𝑡 + 78.630)

𝑐3𝑡 : with the third pole at −3, close to the dominant poles,

yields the most error

§7.Systems Response with Additional Poles Run ch4p2 in Appendix B Learn how to use MATLAB to

• generate a step response for a transfer function

• plot the response directly or collect the points for future use

• solve Ex.4.8

System responses can alternately be obtained using

Simulink Simulink is a software package that is

integrated with MATLAB to provide a graphical user

interface (GUI) for defining systems and generating

responses

The reader is encouraged to study Appendix C, which

contains a tutorial on Simulink as well as some

examples One of the illustrative examples, Ex.C.1,

solves Ex.4.8 using Simulink

§7.Systems Response with Additional Poles Another method to obtain systems responses is through the use ofMATLAB’s LTI Viewer An advantage

of the LTI Viewer is that it displays the values of settling time, peak time, rise time, maximum response, and the final value on the step response plot

The reader is encouraged to study Appendix E, which contains a tutorial on the LTI Viewer as well as some examples Ex.E.1 solves Ex.4.8 using the LTI Viewer

Problem Determine the validity of a second-order approximation

for each of these two transfer functions

(𝑠 + 15)(𝑠2+ 4𝑠 + 100)

(𝑠 + 4)(𝑠2+ 2𝑠 + 90)

a

b

§7.Systems Response with Additional Poles Solution

[𝑠 − (−15)] 𝑠 − (−2 − 𝑗9.798) 𝑠 − (−2 + 𝑗9.798) The dominant poles have a real part of−2 and the higher-order pole is at−15, i.e more than five-times further

⟹ The second-order approximation is valid

[𝑠 − (−4)] 𝑠 − (−1 − 𝑗9.434) 𝑠 − (−1 + 𝑗9.434) The dominant poles have a real part of−1 and the higher-order pole is at−4, i.e not more than five-times further

⟹ The second-order approximation is not valid

b

a

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