Mechatronics Engineering - Biomechanics and Soft …softrobotics.buaa.edu.cn/PPT/Chapter 5 Matlab...

Li Wen

Mechatronics Engineering

Bio-inspired robot-DC motor drive

Mirko Kovac，EPFLUnstable system

Modeling and simulation

of the control system

Problems

1. Why we establish mathematical model of

the control system?

2. Modeling methods and procedures?

3. How to create a mathematical model of the

DC servo motor?

4. The simulation tool of the control system?

1Significance of control system simulation

analysis

2

3

4

Modeling methods and procedures

5

DC motor modeling examples

Analysis and correction for linear

motion unit closed-loop

simulation

Introduction to MATLAB /SIMULINK

Contents


analysis

2

3

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simulation


Contents

• By the specific physical problems, from a qualitative

understanding of engineering problems to rise to the precise

quantitative understanding of the key.

• Research and to analyze a mechanical control system, not

only to qualitatively understand the working principle and

characteristics of the system, but also quantitatively

describe the system dynamic performance.

Establish the significance of the mathematical model

• Mathematical description of the dynamic characteristics of

the system:

Because during the transition process, the system variable you want to

change over time, thus describing the system appears not only in

mathematical model of dynamic characteristics of the variable itself, but also

contain all order derivatives of these variables, so the system of dynamic

equations are differential equations, it is the most basic form representing

mathematical model of the system.

Basic concepts of mathematical models

Bio-inspired soft robot _ slow response

• First, analysis, starting from the physical or chemical

laws, establishing mathematical model and

experimental verification

• Two is an experimental method, by adding a certain

forms of input signals to the system or component,

evaluating output response for system or component,

building mathematical models.

• This lesson uses analysis

Control system modeling approach

• Inexact: Theoretically none can be absolutely accurate

mathematical expressions to describe a system because,

in theory, any system is nonlinear, time and distribution

parameters change, the random factors are present, the

more complex the system, the situation is more

complicated.

• Simplification: Ignore secondary factors, seize the

main problem for modeling, quantitative analysis.

Principle of establishing mathematical models

• Mathematical model is divided into:

– Time Domain Model

– Complex domain model

– Frequency domain model

• Time domain model:

Advantages: it is describes the control system in the time domain, and has the

advantage of an intuitive, accurate, all of the responses and can provide

the system time information.

Disadvantages: complex; hard to find system of structure parameters on

performance of control system of general rules, cannot find improvement

program is not easy on the system analysis and design.

Mathematical Model Types

Complex domain models:

It includes the transfer function and structure of the system. It demonstrates

its characteristic of the system and of the input signal;It not only characterize the

dynamic performance of the system, but can also affect the structure or the study of

changes in system parameters on system performance

Frequency domain model:

Describes the frequency characteristics of the system, with a clear

physical meaning, experimental methods are available to determine.


Relationship between the three commonly used mathematical

models

Linear Systems

Transfer

function

Differential

Equations

Frequency

Characteristics

Rumsfeld

TransformFourier

Transform


1. A linear system of equations:

① determine the input and output of the system

② The system is divided into several areas, from the input start signal is transmitted in the order, according to the laws of physics followed each variable (Newton's law, Kirchhoff's current and voltage law), etc., lists various aspects of linearization original equation;

2. For the establishment of differential equations, Laplace transform one by one, eliminating the intermediate variables, get the system transfer function model

Modeling steps

Biomimetic robot-DC drive

Mirko Kovac，EPFLSelf-stable system

Biomimetic robot-DC drive

Mirko Kovac，EPFLSelf-stable system

with steering


analysis

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simulation


Contents

19

Solution: armature controlled DC motor is essentially the work of the

input electrical energy into mechanical energy, which is the Input of

the armature voltage Ua (t) generated armature current Ia (t) in the

armature circuit, and then by the current Ia (t) and the excitation flux

generated by the interaction of electromagnetic torque Mm (t), to drag

the load movement. Therefore, the equation of motion of the DC motor

by the following three components.

– Armature circuit voltage balance equation

– Electromagnetic torque equation

– Turn the motor shaft from the balance equation

DC modeling analysis

(1) According to Kirchhoff's voltage law, the armature

winding voltage balance equation

（1）

Where, La and Ra were inductance (Henry) and the resistance of the

armature windings (Ohm)

aa a a a a

diu i R L E

dt


(2) When the rotation of the DC motor armature, the armature

windings produce anti potential, it is generally proportional to

the motor speed, i.e.,

（2）

Where, Ea is the back EMF (V), Ke is a scaling factor (V.rad / s)

ma e

dE K

dt


(3) the interaction between the armature current and the

magnetic field to produce an electromagnetic torque.

General electromagnetic torque is proportional to the armature current,

namely:

Where Mm is the electromagnetic torque (Nm), Ia is the armature current (A),

Km for the moment coefficient (Nm / A)


(3)m m aM K i

(4) for driving the electromagnetic torque to overcome the friction and load

torque, assuming only consider the viscous friction is proportional to the

speed, the DC motor torque balance equation

The formula: J m The total moment of inertia of the motor shaft

(Including the moment of inertia of the rotor and the load)

The angular displacement of the motor shaft (rad);

As a viscous friction coefficient of the motor shaft

The role of the applied load on the motor shaft torque

2

2( )m m

m m m c

d dM J B M t

dt dt

2.. 秒米牛

m

mB

秒/弧度/米.牛


( )cM t

To find the angular velocity and load control model of

the motor armature voltage U, namely the transfer function.

We assume zero initial conditions in these kinds of Laplace

Transform, respectively

m

( )

a a a a a a

a e m

m m a

m m m m c

U s L I s s R I s E s

E s K s

M s K I s

M s J S B s M s


Erasing the armature current ia, and then take the armature

voltage Ua is input, the angular velocity of the motor output shaft,

i.e.,

Whereby the DC motor can be controlled in the model, i.e., the

transfer function is:

2(s)

( )

m m

a a m a m m a a m e m

s K

U s L J s L B J R s R B K K

.m as s U s

m


DC servo motor Structure


Created in MATLAB using the Simulink simulation model

Li Wen et al, Beihang University


Open Simulink

Modeling Overview of Linear Motion Units

•Linear motion units has flexibility and mechanical friction,

etc., so it is virtually impossible to establish a precise

mathematical model.

•We usually use approximate model, assuming driver and

transmission is ideal rigid, and there is no elastic deformation

New Simulink model

The system block diagram of the various

modules and drag it to the model file

Pull in

(2) Linear motion unit components

Ball screw

Coupling

DC motorReducer

Modeling of linear motion unit control system

The system block diagram of the various

modules and drag it to the model file

The system block diagram of all the modules file and drag it to

the model and adjust the layout and orientation

Connection

Variable and label

Substituting parameters La，Ra，Jm，Bm，Km，Ke

• La = 0.001Hery；Ra = 1.2oum；Jm = 1e-5 kg.m2；Bm = 5e-4；Km

= 0.08N.m/A；Ke = 0.08V.S/rad；

Input signal:

• Amplitude of Input voltage Ua is 1V, the frequency of square wave is 1Hz

• Amplitude of the interference torque Mf 0.01Nm, frequency of sinusoidal signal 1Hz

Simulink block diagram to obtain arguments:


matlab m files for variable assignment

Generate input and load (or disturbance torque) signal

generator

Amplitude of Input voltage is Ua 1V, the frequency of square wave is 1Hz

Amplitude of the interference torque Mf 0.01Nm, frequency sinusoidal signal of 1Hz For example: motorSlip ring frictionProduce

Click on Run

Judging from the simulation curve, the response curve is a cycle curve, is in

response to a step input and Input of the linear superposition of the load cycle,

from the curve to see the system is still stable, which can be from the poles

and zeros of the transfer function are located to the left half plane verify get.


Necessary and sufficient conditions for stability of the

system is necessaryAll the roots of the characteristic equation must be negative real part, that is

all the roots in the complex plane of the left half-plane


0.08

G(s) = -------------------------------------------

1e-08 s^2 + 1.25e-05 s + 0.007

Continuous-time transfer function

1

2

625 556i

625 5.56i

Root system characteristics

2 ( )

m m

a a m a m m a a m e m

s C

U s L J s L B J R s R B C C

Jm=10-5

Jm=10-4


Jm=10-3

In three Jm, the system is stable, but smaller overshoot Jm more powerful;

Jm greater the longer the rise time of the system.

Overshoot

Jm Impact on system performance

Bm

=5X10-4


Bm

=2X10-3

Bm

=1X10-4

Bm impact on performance(Bm bigger (output / input) becomes smaller,

shorter adjustment time)

The greater the damping coefficient, the smaller the value of the unit step response

(speed / voltage value becomes smaller), the rise time becomes longer, but the time

to reach steady state becomes shorter.


Mc = 1Nm

Load impact on performance

Load increase reduce the system response (moving speed), larger changes in the

steady-state error of the system (open-loop steady-state error is large), the

adjustment time becomes longer, the rise time becomes long.

Mc = 2Nm Mc = 3Nm


analysis

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simulation


Contents

Model building

Specify the slider velocity (unit: mm/s) as the input, and the slider

actual speed (mm/s) as the output, establish a mathematic model for

the linear motion unit speed control system.

Modeling Overview of linear motion units

Rated voltage 24VBack-EMF constant

(Ke)0.0307v*s/rad

Reduction ratio (i) 29：1 Amplifier (Ka) 2.4

Motor resistor (Ra) 21.6欧 Torque constant (Km) 0.0307N*m/A

Motor inductance (La) 1.97mHRotor moment of

inertia (Jx)

Screw lead (p) 2mmEquivalent damping

(Bm)0.0005

Screw diameter (d) 11.5mm Speed gain (Ka) 0.0212v*s/rad

Screw Length (L) 540mm

Workbench mass (m) 0.315kg

(1) Linear motion unit system components and parameters


7 24.2 10 .kg m

1 1 2 2 2( ) ( )22 4 2 25.362 -7

2e kg m

m xi

d pd l m

J J

Equivalent moment of inertia of the motor shaft

1. The relationship between the motor and the screw speed

The actual relationship between the rotational speed of the motor shaft and the

screw speed: ( i is reduction ratio and the value is 29)

m ot i t

tm

to

Motor shaft speed

Actural speed of screw


2. Modeling of DC servo motor


2. Modeling of DC servo motor

Potential balance equation of armature windings:

Relations between the counter-electromotive force and speed

The relationship between the armature current and the armature

torque is:

Torque balance equation is:


3. Angular velocity feedback

To constitute the load shaft speed control system, there must be speed feedback

of load shaft, the error voltage can be obtained by velocity error:

a a n mu t k t t

n t is the input shaft speed of motor;ak is the speed feedback gain

n t

m t

au tak


• Laplace transform of the above formula :

This equation describes the relationship between the input

control voltage U and the rotational angular velocity of the

drive shaft.

m ot i t m os i s

( )m m aM K i N m

( ) aa e m a a a

diu k t L R i

dt

a a n mu t k t t

m m aM s K I s

a a a a a eU s L I s s R I s k s

(s)m m m m m cM s J s s B s M

a a n mu s k s s

(t)(t) J (t) (t)m

m m m m c

dM B M

dt


Specify the slider velocity (unit: mm/s) as the input, and the slider

actual speed (mm/s) as the output, establish a mathematic model for

the linear motion unit speed control system. The above notation is

expressed as the angular velocity, as the screw lead P is 2mm, we

can Build relationships between line speed and angular velocity.

P

2

o

o

sV

2

P

oo

Vs

o t is input speed

oV is slider speed


Build the system model in Simulink:


After determining the mathematical model of the

system, you can use several different methods to

analyze the dynamic performance and steady-state

performance of the control system.

Method:

• Time domain analysis

• Frequency domain analysis.

Simulation analysis of control systems

Dynamic performance and steady-state performance


Input step signal (amplitude is 1) to analyze the time-domain

response

From the results, we can get that rise time, peak time and settling time are

relatively small, although there is a certain system overshoot, but eventually

stabilized, but there is an error between the input and output. Therefore,

system stability and accuracy are not very well.



Analyze the system frequency response, draw bode plot

Conclusion: The system gain margin is infinite, phase margin

is 71.8 °, system stability is very good.


Analyze the system frequency response, draw Nyquist plot

6mm

Impacts on system when change parameters

8mm2mm

Impact on the performance of the screw lead

In the open-loop control, when lead increases, changes in steady-state error

is large and the system response increases, rise time get lower.

6mm

8mm

2mm


Impact on the performance of the screw lead

In the open-loop control, changes in lead will not change the dynamic

characteristic.

25 3515

Impact on the performance of the reduction ratio

In the open-loop control, when the reduction ratio increases, system

response time decreases


25

35

15

Impact on the performance of the reduction ratio


In the open-loop control, changes in reduction ratio will not change the

dynamic characteristic.


Changing the system structure parameters, and analyze its impact

on system performance

Bode plot


Open Simulink


analysis

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simulation


Contents

biorobotics_fast response

Rob Wood lab, Harvard University

X0.1

Active control

New Simulink model

According to the system block diagram, and drag various modules

to the model file

Drag

According to the system block diagram, and drag various modules

to the model file

Connection

Brought variables and mark

Brought parameters: La，Ra，Jm，Bm，Km，Ke

• La = 0.001H; Ra = 1.2Ω; Jm = 1e-5 kg.m2；Bm = 5e-4；Km = 0.08N.m/A; Ke = 0.08V.S/rad；

Input signal is：

• Square wave: Input voltage Ua is 1V，frequency is 1Hz;

• Sin signal: Disturbance torque Mf is 0.01N.m, frequency is 1Hz;

Simulink block diagram:


Variable assignment in matlab m file

Generate input and load (or disturbance torque) signal

using generator

Input voltage Ua: 1V, frequency 1Hz, of square wave.

Disturbance torque Mf is0.01N.m, frequency 1Hz, of sin signal.

Run

Judging from the simulation curve, the response curve is a cycle curve, which

is a linear superposition response to a step input and periodic load input.

From the curve to see the system is still stable, which can be verified from the

transfer function poles and zeros are in the left half plane.


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Mechatronics Engineering - Biomechanics and Soft …softrobotics.buaa.edu.cn/PPT/Chapter 5 Matlab...

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