Ch.03 Modeling in Time Domain

of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien Chapter Objectives After completing this chapter, the student will be able to representation, for a linear, time invarian

Trang 1

03 Modeling in Time Domain

System Dynamics and Control 3.01 Modeling in Time Domain

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

Chapter Objectives After completing this chapter, the student will be able to

representation, for a linear, time invariant system

• Model electrical and mechanical systems in state space

• Convert a transfer function to state space

• Convert a state-space representation to a transfer function

• Linearize a state-space representation 𝐶

§1.Introduction

Two approaches are available for the analysis and design of

feedback control systems

- The classical, or frequency-domain, technique

Major disadvantage: can be applied only to linear,

time-invariant systems or systems that can be approximated as such

Major advantage: rapidly provide stability and transient

response information

§1.Introduction

- The modern, or time domain, state-space technique

• A unified method for modeling, analyzing, and designing a wide range of systems

• Can be used to represent nonlinear systems that have backlash, saturation, and dead zone

• Can handle, conveniently, systems with nonzero initial conditions

• Can be used to represent time-varying systems, (for example, missiles with varying fuel levels or lift in an aircraft flying through a wide range of altitudes)

• Can be compactly represented in state space for multiple-input, multiple-output systems

• Can be used to represent systems with a digital computer in the loop or to model systems for digital simulation

§1.Introduction

• With a simulated system, system response can be obtained

for changes in system parameters - an important design tool

• The state space approach is also attractive because of the

availability of numerous state-space software packages for

the personal computer

§2.Some Observations

Select the current𝑖(𝑡) as a variable, write the loop equation

𝐿𝑑𝑖(𝑡)

⟹𝑑𝑖(𝑡)

𝑅

𝐿𝑖 𝑡 + 𝑣(𝑡)

Trang 2

Select the current𝑖(𝑡) as a variable, write the loop equation

𝑖(𝑡) =𝑑𝑞(𝑡)

𝑑𝑡

𝐿𝑑𝑖(𝑡)

1

𝐶 𝑖𝑑𝑡 = 𝑣(𝑡)

𝑑𝑡 = 𝑖(𝑡)

𝑑𝑖(𝑡)

1

𝐿𝐶𝑞 𝑡 −

𝑅

𝐿𝑖 𝑡 +

1

𝐿𝑣(𝑡)

𝑑𝑞 𝑡

𝑑𝑖(𝑡)

1

𝑅

𝐿𝑖 𝑡 +

1

𝐿𝑣(𝑡) The state equation can be written in vector-matrix form

𝒙 = 𝑨𝒙 + 𝑩𝑢

𝑥 = 𝑞(𝑡)𝑖(𝑡) , 𝑨 =

𝑅 𝐿 , 𝑩 =

0 1 𝐿 , 𝑢 = 𝑣(𝑡) The output equation can be written in vector-matrix form

𝑦 = 𝑪𝒙 + 𝑫𝑢

𝑦 = 𝑣𝐿𝑡 , 𝑪 = −1

𝑞 𝑡

𝑖 𝑡 , 𝐷 = 1, 𝑢 = 𝑣(𝑡)

where,

§3.The General State-Space Representation

Review

-Linear combination: A linear combination of𝑛 variables, 𝑥𝑖, for

𝑖 = 1 ÷ 𝑛, is given by the following sum, 𝑆

𝑆 = 𝐾𝑛𝑥𝑛+ 𝐾𝑛−1𝑥𝑛−1+ ⋯ + 𝐾1𝑥1, where 𝐾𝑖is constant

-Linear independence: A set of variables is said to be linearly

independent if none of the variables can be written as a linear

combination of the others

-System variable: Any variable that responds to an input or initial

conditions in a system

-State variables: The smallest set of linearly independent

system variables such that the values of the members of the

set at time𝑡0 along with known forcing functions completely

determine the value of all system variables for allt ≥ 𝑡0

-State vector: A vector whose elements are the state variables

-State space: The 𝑛 -dimensional space whose axes are the state variables

In the figure

• the state variables: 𝑣𝑅and𝑣𝐶

• the state trajectory can be thought

of as being mapped out by the state vector,𝑥(𝑡), for a range of 𝑡

-State equations: A set of𝑛 simultaneous, first-order differential equations with𝑛 variables, where the 𝑛 variables to be solved are the state variables

-Output equation: The algebraic equation that expresses the output variables of a system as linear combinations of the state variables and the inputs

- A system is represented in state space by the following

equations

𝒙 = 𝑨𝒙 + 𝑩𝒖

𝒚 = 𝑪𝒙 + 𝑫𝒖 for𝑡 ≥ 𝑡0and initial conditions,𝒙(𝑡0), where

𝒙 : state vector

𝒙 : derivative of the state vector with respect to time

𝒚 : output vector

𝒖 : input or control vector

𝑨 : system matrix

𝑩: input matrix

𝑪 : output matrix

§4.Applying the State-Space Representation

In this section, we apply the state-space formulation to the representation of more complicated physical systems The first step in representing a system is to select the state vector, which must be chosen according to the following considerations 1.A minimum number of state variables must be selected as components of the state vector This minimum number of state variables is sufficient to describe completely the state of the system

2.The components of the state vector (that is, this minimum number of state variables)must be linearly independent

Trang 3

Given the electrical network, find a state-space representation if the output

is the current through the resistor Solution

Step 1 Label all of the branch currents in the network These

include𝑖𝐿,𝑖𝑅, and𝑖𝐶

Step 2 Select the state variables by writing the derivative

equation for all energy storage elements,𝐿 and 𝐶

𝐶𝑑𝑣𝐶

𝑑𝑡 = 𝑖𝐶

𝐿𝑑𝑖𝐿

𝑑𝑡 = 𝑣𝐿

(3.22) (3.23)

§4.Applying the State-Space Representation Step 3Apply network theory, such asKirchhoff’s voltage and

current laws, to obtain 𝑖𝐶 and 𝑣𝐿 in terms of the state variables,𝑣𝐶and𝑖𝐿

At Node 1,

𝑖𝐶= −𝑖𝑅+ 𝑖𝐿= −1

𝑅𝑣𝐶+ 𝑖𝐿 which yields𝑖𝐶in terms of the state variables,𝑣𝐶and𝑖𝐿 Around the outer loop,

which yields𝑣𝐿in terms of the state variable,𝑣𝐶, and the source,𝑣(𝑡)

(3.24)

𝐶𝑑𝑣𝐶

𝑑𝑡 = 𝑖 𝐶 (3.22), 𝐿𝑑𝑖𝐿

𝑑𝑡 = 𝑣 𝐿 (3.23), 𝑖 𝐶 = −1

𝑅 𝑣 𝐶 + 𝑖 𝐿 (3.24), 𝑣 𝐿 = −𝑣 𝐶 + 𝑣(𝑡) (3.25)

Step 4 Substitute the results of Eqs (3.24) and (3.25) into Eqs

(3.22) and (3.23) to obtain the following state equations

𝑑𝑣𝐶

1

𝑅𝐶𝑣𝐶+

1

𝐶𝑖𝐿

𝑑𝑖𝐿

1

𝐿𝑣(𝑡)

(3.27.a) (3.27.b)

𝑑𝑣 𝐶

𝑑𝑡 = −1

𝑅𝐶 𝑣 𝐶 +1𝑖 𝐿 (3.27.a), 𝑑𝑖𝐿

𝑑𝑡 = −1𝑣 𝐶 +1𝑣(𝑡) (3.27.b)

§4.Applying the State-Space Representation Step 5 Find the output equation Since the output is 𝑖𝑅(𝑡)

𝑖𝑅=1

𝑅𝑣𝐶 The state-space representation is found by representing Eqs (3.27) and (3.28) in vector-matrix form

𝑣𝐶

𝐿 = −1/𝑅𝐶−1/𝐿 1/𝐶0 𝑣𝑖𝐶

𝐿 + 1/𝐿0 𝑣(𝑡)

𝐿

(3.28)

- Ex.3.2 Representing an Electrical Network with a Dependent Source

network if the output vector is

𝑦 = 𝑣𝑅 2 𝑖𝑅 2 𝑇

Solution

Step 1 Label all of the branch currents in the network

Step 2 Select the state variables by listing the voltage-current

relationships for all of the energy-storage elements

𝐿𝑑𝑖𝐿

𝑑𝑡 = 𝑣𝐿

𝐶𝑑𝑣𝐶

𝑑𝑡 = 𝑖𝐶 Select the state variables:𝑥1= 𝑖𝐿and𝑥2= 𝑣𝐶 (3.31)

(3.30.a) (3.30.b)

§4.Applying the State-Space Representation Step 3 UsingKirchhoff’s voltage and current laws to find 𝑖𝐿,𝑣𝐶

in terms of the state variables

𝑣𝐿= 𝑣𝐶+ 𝑣𝑅 2= 𝑣𝐶+ 𝑖𝑅 2𝑅2

At node 2,𝑖𝑅2= 𝑖𝐶+ 4𝑣𝐿

𝑣𝐿= 𝑣𝐶+ (𝑖𝐶+ 4𝑣𝐿)𝑅2= 1

1 − 4𝑅2(𝑣𝐶+ 𝑖𝐶𝑅2)

At node 1

𝑖𝐶= 𝑖 − 𝑖𝑅 1− 𝑖𝐿= 𝑖 −𝑣𝑅1

𝑅1− 𝑖𝐿= 𝑖 −𝑣𝐿

𝑅1− 𝑖𝐿

(3.35)

(3.36)

Trang 4

𝑣 𝐿 =1−4𝑅1

2 (𝑣 𝐶 + 𝑖 𝐶 𝑅 2 ) (3.35), 𝑖 𝐶 = 𝑖 −𝑣𝐿

𝑅1− 𝑖 𝐿 (3.36)

Rewriting Eqs (3.35) and (3.36)

1 − 4𝑅2𝑣𝐿− 𝑅2𝑖𝐶= 𝑣𝐶

−(1/𝑅1)𝑣𝐿 − 𝑖𝐶= 𝑖𝐿− 𝑖

Writing the result in vector-matrix form

𝑣𝐿=1

∆ 𝑅2𝑖𝐿− 𝑣𝐶− 𝑅2𝑖

𝑖𝐶=1

∆ 1 − 4𝑅2𝑖𝐿+

1

𝑅1𝑣𝐶− 1 − 4𝑅2𝑖

𝑅1

𝐿

𝑣

𝐶 = 1 − 4𝑅𝑅2/(𝐿∆) −1/(𝐿∆)

2/(𝐶∆) 1/(𝑅1𝐶∆)

𝑖𝐿

𝑣𝐶 +

−𝑅2/(𝐿∆)

1 − 4𝑅2/(𝐶∆)𝑖 where,

(3.38) (3.39)

𝑣𝐿=1𝑅2𝑖𝐿− 𝑣𝐶− 𝑅2𝑖 (3.38), 𝑖𝐶=1 1 − 4𝑅2𝑖𝐿+1

𝑅 1 𝑣𝐶− 1 − 4𝑅2𝑖 (3.39)

§4.Applying the State-Space Representation Step 4 Derive the output equation

Since the specified output variables are𝑣𝑅 2 and𝑖𝑅 2, note that around the mesh containing𝐶, 𝐿, and 𝑅2

Substituting Eqs (3.38) and (3.39) into Eq.(3.42)

𝑣𝑅2

𝑖𝑅2 =

𝑅2/∆ −(1 + 1/∆) 1/∆ 1 − 4𝑅1/(𝑅1∆)

𝑖𝐿

𝑣𝐶 +

−𝑅2/∆

−1/∆ 𝑖

Find the state equations for the translational mechanical system

Solution

Find the Laplace-transformed equations of motion

−𝐾𝑋1+ 𝑀2𝑠2+ 𝐾 𝑋2= 𝐹

§4.Applying the State-Space Representation Take the inverse Laplace transform assuming zero initial conditions

𝑀1𝑠2+ 𝐷𝑠 + 𝐾 𝑋1−𝐾𝑋2= 0 ⟹ 𝑀1

𝑑2𝑥1

𝑑𝑡2+ 𝐷𝑑𝑥1

𝑑𝑡+ 𝐾(𝑥1− 𝑥2) = 0

−𝐾𝑋1+ 𝑀2𝑠2+ 𝐾 𝑋2= 𝐹 ⟹ −𝐾𝑥1+ 𝑀2

𝑑2𝑥2

𝑑𝑡2 + 𝐾𝑥2= 𝑓(𝑡)

𝑑𝑥1

𝑑𝑡 ≡ 𝑣1,

𝑑𝑥2

𝑑2𝑥1

𝑑𝑡2 =𝑑𝑣1

𝑑𝑡 ,

𝑑2𝑥2

𝑑𝑡2 =𝑑𝑣2 𝑑𝑡

𝑑𝑥1

𝑑𝑣1

𝐾

𝑀1𝑥1−𝐷

𝑀1𝑣1+𝐾

𝑀1𝑥2

𝑑𝑥2

𝑑𝑣2

𝐾

𝑀2𝑥1−𝐾

𝑀2𝑥2+ 1

𝑀2𝑓(𝑡)

(3.44) (3.45) Let

State equations

𝑑𝑥1

𝑑𝑣1

𝐾

𝑀1𝑥1−𝐷

𝑀1𝑣1+𝐾

𝑀1𝑥2

𝑑𝑥2

𝑑𝑣2

𝐾

𝑀2𝑓(𝑡)

In vector matrix form

𝑥

1

𝑣

1

𝑥2

𝑣

2

=

𝑀1

𝐾

+

0 0 0 1

𝑀2 𝑓(𝑡)

§4.Applying the State-Space Representation Note The equations of motion (3.44) and (3.45) can be derived

motion

𝑀1

𝑑2𝑥1

𝑑𝑡2 + 𝐷𝑑𝑥1

𝑑𝑡 + 𝐾(𝑥1− 𝑥2) = 0

𝑀2𝑑

2𝑥2

𝑑𝑡2 + 𝐾(𝑥2− 𝑥1) = 𝑓(𝑡)

2𝑥1

𝑑𝑡2 + 𝐷𝑑𝑥1

𝑑𝑡 + 𝐾(𝑥1− 𝑥2) = 0

−𝐾𝑥1+ 𝑀2

𝑑2𝑥2

𝑑𝑡2 + 𝐾𝑥2= 𝑓(𝑡)

(3.44) (3.45)

Trang 5

Skill-Assessment Ex.3.1

Problem Find the state-space representation of the electrical

network with the output is𝑣𝑜(𝑡) Solution

Identifying appropriate variables on the circuit yields

Writing the derivative relations

𝐶1𝑑𝑣𝐶1

𝑑𝑡 = 𝑖𝐶 1

𝐿𝑑𝑖𝐿

𝑑𝑡 = 𝑣𝐿

𝐶2

𝑑𝑣𝐶2

𝑑𝑡 = 𝑖𝐶2 System Dynamics and Control 3.25 Modeling in Time Domain

§4.Applying the State-Space Representation UsingKirchhoff’s current and voltage laws

𝑖𝐶 1= 𝑖𝐿+ 𝑖𝑅= 𝑖𝐿+1

𝑅(𝑣𝐿− 𝑣𝐶2)

𝑣𝐿= −𝑣𝐶 1+ 𝑣𝑖

𝑖𝐶2= 𝑖𝑅=1

𝑅(𝑣𝐿− 𝑣𝐶2) Substituting and rearranging

𝑑𝑣𝐶 1

1

𝑅𝐶1𝑣𝐶 1+1

𝐶1𝑖𝐿− 1

𝑅𝐶1𝑣𝐶 2+ 1

𝑅𝐶1𝑣𝑖

𝑑𝑖𝐿

1

𝐿 𝐶1+1

𝑑𝑣𝐶 2

1

𝑅𝐶2

𝑣𝐶1− 1

𝑅𝐶2

𝑣𝐶2+ 1

𝑅𝐶2

𝑣𝑖 The output 𝑣𝑜= 𝑣𝐶2

𝑑𝑣𝐶1

1

𝑅𝐶1𝑣𝐶 1+1

𝐶1𝑖𝐿− 1

𝑅𝐶1𝑣𝑖

𝑑𝑖𝐿

1

𝐿 𝐶 1+1

𝑑𝑣𝐶2

1

𝑅𝐶2𝑣𝐶 1− 1

𝑅𝐶2𝑣𝑖

The output 𝑣𝑜= 𝑣𝐶 2, the equations in vector-matrix form

𝒙 =

𝑣

𝐶 1

𝐿

𝑣

𝐶 2

=

𝑅𝐶1

1

𝑅𝐶1

𝑅𝐶2

+

1

𝑅𝐶1

1 𝐿 1

𝑅𝐶2

𝑣𝑖

𝑦 = 0 0 1 𝒙

§4.Applying the State-Space Representation Skill-Assessment Ex.3.2

Problem Represent the translational mechanical system in

state-space, where𝑥3(𝑡) is the output

Solution Writing the equations of motion

−𝑋2+ 𝑠2+ 𝑠 + 1 𝑋3= 0 Taking the inverse Laplace transform and simplifying 𝑥

1= − 𝑥1− 𝑥1+ 𝑥2+ 𝑓 𝑥

2= + 𝑥1− 𝑥2− 𝑥2+ 𝑥3

𝑥

3= − 𝑥3− 𝑥3+ 𝑥2

𝑥 1 = − 𝑥 1 − 𝑥 1 + 𝑥 2 + 𝑓, 𝑥 2 = + 𝑥 1 − 𝑥 2 − 𝑥 2 + 𝑥 3 , 𝑥 3 = − 𝑥 3 − 𝑥 3 + 𝑥 2

Defining state variables

𝑧1= 𝑥1, 𝑧2= 𝑥1, 𝑧3= 𝑥2, 𝑧4= 𝑥2, 𝑧5= 𝑥3, 𝑧6= 𝑥3

Write the state equations

1= 𝑧2

2= 𝑥1= − 𝑥1− 𝑥1+ 𝑥2+ 𝑓

= −𝑧2− 𝑧1+ 𝑧4+ 𝑓

3= 𝑥2= 𝑧4

4= 𝑥2= 𝑥1− 𝑥2− 𝑥2+ 𝑥3

= 𝑧2− 𝑧4− 𝑧3+ 𝑧5

5= 𝑥3

= 𝑧6

6= 𝑥3= − 𝑥3− 𝑥3+ 𝑥2

= −𝑧6− 𝑧5+ 𝑧3

1= 𝑧2

2= −𝑧2− 𝑧1+ 𝑧4+ 𝑓

3= 𝑧4

4= 𝑧2− 𝑧4− 𝑧3+ 𝑧5

5= 𝑧6

6= −𝑧6− 𝑧5+ 𝑧3

The output 𝑦 = 𝑧5, the equations in vector-matrix form

𝒛 =

𝒛 +

0 1 0 0 0 0 𝑓(𝑡)

𝑦 = 0 0 0 0 1 0 𝒛

Trang 6

§5.Converting a Transfer Function to State-Space

Consider the differential equation

𝑑𝑛𝑦

𝑑𝑡𝑛+ 𝑎𝑛−1𝑑𝑛−1𝑦

𝑑𝑡𝑛−1+ ⋯ + 𝑎1

𝑑𝑦

𝑑𝑡+ 𝑎0𝑦 = 𝑏0𝑢

Choose the output,𝑦, and its derivatives as the state variables

𝑥1= 𝑦, 𝑥2=𝑑𝑦

𝑑𝑡, 𝑥3=𝑑

2𝑦

𝑑𝑡2, ⋯ , 𝑥𝑛=𝑑

𝑛−1𝑦

𝑑𝑡𝑛−1

Differentiating both sides yields

𝑥1=𝑑𝑦

𝑑𝑡, 𝑥2=

𝑑2𝑦

𝑑𝑡2, 𝑥3=𝑑

3𝑦

𝑑𝑡3, ⋯ , 𝑥𝑛=𝑑

𝑛𝑦

𝑑𝑡𝑛

Differentiating both sides yields

𝑥1= 𝑥2, 𝑥2= 𝑥3, 𝑥3= 𝑥4, ⋯ , 𝑥𝑛−1= 𝑥𝑛

𝑥𝑛= −𝑎0𝑥1− 𝑎1𝑥2⋯ − 𝑎𝑛−1𝑥𝑛+ 𝑏0𝑢

§5.Converting a Transfer Function to State-Space The phase-variable form of the state equations 𝑥

1

𝑥

2

𝑥

3

⋮ 𝑥

𝑛−1

𝑥

𝑛

=

𝑥1

𝑥2

𝑥3

⋮

𝑥𝑛−1

𝑥𝑛 +

0 0 0

⋮ 0

𝑏0

𝑢

𝑦 = 1 0 0 ⋯ 0 0

𝑥1

𝑥2

𝑥3

⋮

𝑥𝑛−1

𝑥𝑛

Find the state-space representation

in phase-variable form for the TF Solution

Step 1 Find the associated differential equation

𝐶(𝑠)

24

𝑠3+ 9𝑠2+ 26𝑠 + 24

Take the inverse Laplace transform, assuming zero

initial conditions

𝑐 + 9 𝑐 + 26 𝑐 + 24𝑐 = 24𝑟

§5.Converting a Transfer Function to State-Space Step 2 Select the state variables

Choosing the state variables𝑥1= 𝑐, 𝑥2= 𝑐, 𝑥3= 𝑐

𝑥3= −24𝑥1− 26𝑥2− 9𝑥3+ 24𝑟

𝑦 = 𝑐 = 𝑥1

In vector-matrix form 𝑥

1

𝑥

2

𝑥

3

=

−24 −26 −9

𝑥1

𝑥2

𝑥3

+

0 0 24 𝑟

𝑦 = 1 0 0

𝑥1

𝑥2

𝑥3

An equivalent block diagram of the system

𝑥

1

𝑥

2

𝑥

3

=

−24 −26 −9

𝑥1

𝑥2

𝑥3 +

0 0 24 𝑟

𝑦 = 1 0 0

𝑥1

𝑥2

𝑥3

§5.Converting a Transfer Function to State-Space Run ch3p1 through ch3p4 in Appendix B Learn how to use MATLAB to

• represent the system matrix A, the input matrix B, and the output matrix C

• convert a transfer function to the state-space representation in phase-variable form

• solve Ex.3.4

Trang 7

- If a TF has a polynomial in𝑠 in the numerator that is of order

less than the polynomial in the denominator

the numerator and denominator can be handled separately:

separate the transfer function into two cascaded TFs

• The first TF with just the denominator is converted to the phase-variable representation in state space, phase variable

𝑥1is the output, and the rest of the phase variables are the internal variables of the first block

• The second TF with just the numerator yields

𝑌 𝑠 = 𝐶 𝑠 = (𝑏2𝑠2+ 𝑏1𝑠 + 𝑏0)𝑋1(𝑠) Taking the inverse Laplace transform with zeros initial conditions

𝑦 𝑡 = 𝑏2

𝑑2𝑥1

𝑑𝑡2 + 𝑏1

𝑑𝑥1

𝑑𝑡 + 𝑏0𝑥1

= 𝑏0𝑥1+ 𝑏1𝑥2+ 𝑏2𝑥3

Hence, the second block simply forms a specified linear combination of the state variables developed in the first block

Find the state-space representation

of the TF Solution

Step 1 Separate the system into two cascaded blocks

Step 2 Find the state equations for the block containing the

denominator

𝑥

1

𝑥

2

𝑥

3

=

−24 −26 −9

𝑥1

𝑥2

𝑥3

+

0 0 1 𝑟

§5.Converting a Transfer Function to State-Space Step 3 Introduce the effect of the block with the numerator The second block states that

𝐶 𝑠 = 𝑏2𝑠2+ 𝑏1𝑠 + 𝑏0𝑋1 𝑠

= 𝑠2+ 7𝑠 + 2 𝑋1 𝑠 Taking the inverse Laplace transform with zero initial conditions

𝑐 = 𝑥1+ 7 𝑥1+ 2𝑥1

= 𝑥3+ 7𝑥2+ 2𝑥1 The output equation

𝑦 = 𝑏0 𝑏1 𝑏2

𝑥1

𝑥2

𝑥3

= 2 7 1

𝑥1

𝑥2

𝑥3

An equivalent block diagram of the system

𝑥

1

𝑥2

𝑥3

=

−24 −26 −9

𝑥1

𝑥2

𝑥3 +

0 0 1

𝑥1

𝑥2

𝑥3

§5.Converting a Transfer Function to State-Space Skill-Assessment Ex.3.3

Problem Find the state equations and output equation for the phase-variable representation of the TF

𝑠2+ 7𝑠 + 9 Solution

First TF 𝑋(𝑠)

1

Taking inverse Laplace transform with zero initial conditions

𝑥 + 7 𝑥 + 9𝑥 = 𝑟 Defining the state variables as𝑥1= 𝑥, 𝑥2= 𝑥 𝑥

1= 𝑥2 𝑥

2= 𝑥 = −7 𝑥 − 𝑥 + 𝑟 = −9𝑥1− 7𝑥2+ 𝑟

Trang 8

𝑥2= −9𝑥1− 7𝑥2+ 𝑟, 𝑐 = 𝑥1+ 2𝑥2

Second TF

The output equation

𝑐 = 2 𝑥 + 𝑥 = 𝑥1+ 2𝑥2

Putting all equation in vector-matrix form

0

𝑐 = 1 2 𝒙

§6.Converting from State Space to a Transfer Function

- Given the state and output equations

𝒙 = 𝑨𝒙 + 𝑩𝒖

𝒚 = 𝑪𝒙 + 𝑫𝒖

- Take the Laplace transform assuming zero initial conditions

𝑠𝑿 𝑠 = 𝑨𝑿 𝑠 + 𝑩𝑼(𝑠)

𝒀 𝑠 = 𝑪𝑿 𝑠 + 𝑫𝑼(𝑠)

- After some arrangement

𝑿 𝑠 = (𝑠𝑰 − 𝑨)−1𝑩𝑼(𝑠)

The matrix𝑪 𝑠𝑰 − 𝑨−1𝑩𝑼 𝑠 + 𝑫: the transfer function matrix

- If𝑼 𝑠 = 𝑈(𝑠) and 𝒀 𝑠 = 𝑌(𝑠) are scalars, the TF

𝑈(𝑠)= 𝑪 𝑠𝑰 − 𝑨

Find the TF,𝑇 𝑠 = 𝑌(𝑠)/𝑈(𝑠), for given the system

𝒙 = 00 10 01

−1 −2 −3

𝒙 +

10 0 0

Solution

Find(𝑠𝑰 − 𝑨)−1

𝑠𝑰 − 𝑨 =

−

−1 −2 −3

=

(𝑠𝑰 − 𝑨)−1=adj(𝑠𝑰 − 𝑨)

det(𝑠𝑰 − 𝑨)=

𝑠3+ 3𝑠2+ 2𝑠 + 1

𝒙 = 00 10 01

−1 −2 −3

𝒙 + 10

0 𝑢, 𝑦 = 1 0 0 𝒙

𝑠3+ 3𝑠2+ 2𝑠 + 1 Then

= 1 0 0

𝑠3+ 3𝑠2+ 2𝑠 + 1

10 0 0 + 0

2+ 3𝑠 + 2)

𝑠3+ 3𝑠2+ 2𝑠 + 1

Run ch3p5 in Appendix B

Learn how to use MATLAB to

• convert a state-space representation to a transfer

function

• solve Ex.3.6

§6.Converting from State Space to a Transfer Function Run ch3sp1 in Appendix F

Learn how to use the Symbolic Math Toolbox to

• write matrices and vectors

• solve Ex.3.6

Trang 9

Problem Convert the state and output equations to a TF

2

Solution

−4 −1.5

𝑠 + 4 1.5

4 𝑠 + 4

𝑠2+ 4𝑠 + 6

4 𝑠 + 4

𝑠2+ 4𝑠 + 6

2 0

𝑠2+ 4𝑠 + 6

(3.78)

2

𝑦 = 1.5 0.625 𝒙 Matlab A=[-4 -1.5; 4 0]; B=[2 0]';

C=[1.5 0.625]; D=0;

T=ss(A,B,C,D); T=tf(T)

3 s + 5 -s^2 + 4 s + 6 Continuous-time transfer function

(3.78)

§7.Linearization

Walking robots, such as Hannibal shown here, can be used

to explore hostile environments and rough terrain, such as

that found on other planets or inside volcanoes

§7.Linearization

First represent the simple pendulum in state space:𝑀𝑔 is the weight, 𝑇 is an applied torque

in the 𝜃 direction, and 𝐿 is the length of the

distributed, with the center of mass at𝐿/2 Then

position with zero angular velocity Solution

Drawing the free body diagram Summing the torques

𝐽𝑑

2𝜃

𝑑𝑡2+𝑀𝑔𝐿

§7.Linearization

𝐽𝑑

2𝜃

𝑑𝑡2+𝑀𝑔𝐿

Letting𝑥1= 𝜃, 𝑥2= 𝑑𝜃/𝑑𝑡, the state equation

𝑥2= −𝑀𝑔𝐿

2𝐽 𝑠𝑖𝑛𝑥1+𝑇

𝐽 The nonlinear Eq (3.80) represent a valid and complete model

of the pendulum in state space even under nonzero initial

conditions and even if parameters are time varying

To apply classical techniques and convert these state equations

to a transfer function⟹ The nonlinear must be linearized

(3.80.b)

𝑓 𝑥 − 𝑓(𝑥 0 ) ≈𝑑𝑓𝑑𝑥 𝑥=𝑥

0

(𝑥 − 𝑥 0 ) (2.182), 𝑥 1 = 𝑥 2 (3.80.a), 𝑥 2 = −𝑀𝑔𝐿2𝐽𝑠𝑖𝑛𝑥 1 +𝑇𝐽(3.80.b)

§7.Linearization Linearize the equation about the equilibrium point,𝑥1= 𝜃 = 0,

equilibrium point, or

𝑥1= 0 + 𝛿𝑥1

𝑥2= 0 + 𝛿𝑥2 Using Eqs (2.182) 𝑠𝑖𝑛𝑥1− 𝑠𝑖𝑛0 =𝑑(𝑠𝑖𝑛𝑥1)

𝑑𝑥1

𝑥 1 =0

𝛿𝑥1= 𝛿𝑥1⟹ 𝑠𝑖𝑛𝑥1= 𝛿𝑥1

The state equations now become

𝛿𝑥1= 𝛿𝑥2

2𝐽 𝛿𝑥1+

𝑇 𝐽

Trang 10

Problem Represent the translational mechanical system in state

space about the equilibrium displacement The spring is

nonlinear,𝑓𝑠𝑡 = 2𝑥𝑠(𝑡) The applied force is 𝑓 𝑡 =

10 + 𝛿𝑓(𝑡), where 𝛿𝑓(𝑡) is a small force about the 10𝑁

displacement of the mass,𝑥(𝑡)

Solution

The equation of motion

𝑑2𝑥

𝑑𝑡2+ 2𝑥2= 10 + 𝛿𝑓(𝑡)

Letting𝑥 = 𝑥0+ 𝛿𝑥

𝑑2(𝑥0+ 𝛿𝑥)

𝑑𝑡2 + 2(𝑥0+ 𝛿𝑥)2= 10 + 𝛿𝑓(𝑡)

(1)

(2)

𝑓 𝑥 − 𝑓(𝑥 0 ) ≈𝑑𝑓𝑑𝑥 𝑥=𝑥

0

(𝑥 − 𝑥 0 ) (2.182), 𝑑𝑑𝑡2𝑥2 + 2𝑥 2 = 10 + 𝛿𝑓(𝑡) (1)

§7.Linearization Linearize𝑥2at𝑥0

(𝑥0+ 𝛿𝑥)2−𝑥0=𝑑 𝑥

2

𝑑𝑥

𝑥 0

𝛿𝑥 = 2𝑥0𝛿𝑥

Substituting Eq.(3) into Eq.(1)

𝑑2𝛿𝑥

𝑑𝑡2 + 4𝑥0𝛿𝑥 = −2𝑥0+ 10 + 𝛿𝑓(𝑡) The force of the spring at equilibrium𝐹 = 10 = 2𝑥0⟹ 𝑥0= 5 Substituting this value of𝑥0into Eq.(4)

𝑑2𝛿𝑥

(4)

§7.Linearization

𝑑2𝛿𝑥

Selecting the state variables𝑥1= 𝛿𝑥, 𝑥2= 𝛿𝑥

The state and output equations

𝑥1= 𝑥2

𝑥2= 𝛿𝑥 = −4 5𝑥1+ 𝛿𝑓 𝑡

𝑦 = 𝑥1

Converting to vector-matrix form

0

𝑦 = 1 0 𝒙

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