principle of automatic control chapter 2 writing...

1

Principle of Automatic Control

浙江大学控制科学与工程学系

By Chunyue Song

Qinmin Yang

CHAPTER 2Writing System Equations ----Modeling

1

2

Outline of this chapter

Introduction Electric Circuits and Components Basic Linear Matrix Algebra & State Concepts Transfer Function and Block Diagram Mechanical Translation Systems Analogous Circuits Other Mathematic Modeling Examples

Mechanical Rotational SystemsThermal SystemsHydraulic Linear ActuatorLiquid-Level System……

Determination of the overall transfer function Linearization

2

Mechanical Translation Systems

（机械传递系统）

4

Mechanical systems obey Newton’s law (inertial frame 惯性坐标系)

that the sum of the forces equals to the momentum; that is, the

sum of the applied forces minus the reactive force.


xMDMaMDvMvdt

dp

dt

dF

ii

2)( where is the sum of the external forces, M is the mass, v is the inertial velocity and p is the momentum, a is the acceleration, and x is the displacement.

i

iF


The three elements considered in a mechanical translation system are mass (质量块), elastance (spring 弹簧), and damping (阻尼). Analysis is carried out based on linear systems, and other nonlinear terms are not considered.

3

5

Mechanical Translation Systems: basic elements

a

M

b

c

K

d

e

B

f

Fig. 2.9 Network elements of mechanical translation system

(a) (b) (c)

The network representation of mass is shown in Fig.2.9a.

xMDMDvMafM2


The elastance, or stiffness, K provides a restoring force as represented by a spring, which is shown in Fig.2.9b. (胡克定律)

)( dcK xxKf cK Kxf 0 dxif

The damping, or viscous, friction B characterizes the element that absorbs energy. The damping force is proportional to the difference in velocity of two bodies, which is shown in Fig.2.9c.

( ) ( )B e f e f ef B v v B Dx Dx BDx

6

Mechanical Translation Systems: dashpot construction

A dashpot (阻尼器) construction is shown in Fig.2.10. (See P36)

)()( fefeB DxDxBvvBf

f(t)

Fig. 2.10 Dashpot (阻尼器、缓冲器) construction

Filled with incompressible fluid

ba

Housing

Piston

If a force f is applied to the shaft, what happens to the dashpot?

What does damping B depend on?

The force equation is written for each node or position by equating the sum of the forces at each position to zero.

The equation is similar to the node equations in an electric circuit, force analogous to current, velocity analogous to voltage (See Table 2.3, P37).


4

7

Mechanical Translation Systems: dashpot construction


8

Simple Mechanical Translation System

The system shown in Fig.2.11 is initially at rest.

A force f applied at the end of the spring must be balanced by a compression of the spring. The same force is also transmitted through the spring and acts on point xb.

Figure. 2.11

reference

M

xb

K B

(a) Simple mass-spring-damping mechanical system

xa

f(t)


5

9

Simple Mechanical Translation SystemSet the displacements xa and xb as nodes. According to

Newton’s second law, at each node the sum of the forces must equal to zero.

)( baK xxKff node a: (2.58)

According the concept of transfer function, it is possible to obtain equations relating xa to f , xb to xa , or xb to f by combining Eqs. (2.58) and (2.59).

node b: bbBMK BDxxMDfff 2 (2.59)

output input


Figure. 2.11

reference

M

xb

K B


xa

f(t)

10

Then the transfer functions can be obtained by combining Eqs.

)( baK xxKff (2.58)

bbBMK BDxxMDfff 2(2.59)

fKBDMDxBDMDK a )()( 22 xa—f :

)( 2

2

1 BDMDK

KBDMD

f

xG a

xb—xa : ab KxxKBDMD )( 2

KBDMD

K

x

xG

a

b

22

xb—f : fxBDMD b )( 2

BDMDf

xG b

2

1

K

fxx ab



Figure. 2.11

reference

M

xb

K B


xa

f(t)

6

11

)()()( 2 sBsXsXMssF bb Take 2.59 L’ transformation

the transfer functions of Laplace form can be obtainedeasily.

Thus:

Take 2.58 L’ transformation ))()(()( sXsXKsF ba

)()(

)()(

2

2

1BsMsK

KBsMs

sF

sXsG a

KBsMs

K

sX

sXsG

a

b

22 )(

)()(

BsMssF

sXsG b

2

1

)(

)()(

Comparing these Eqs. with…….

fKBDMDxBDMDK a )()( 22 xa—f :

)( 2

2

1 BDMDK

KBDMD

f

xG a

xb—xa : ab KxxKBDMD )( 2

KBDMD

K

x

xG

a

b

22

xb—f : fxBDMD b )( 2

BDMDf

xG b

2

1


Simple Mechanical Translation System)( baK xxKff (2.58)

bbBMK BDxxMDfff 2(2.59)

12

Note that the last equation is equal to the product of the first two

)()(

)()(

2

2

1BsMsK

KBsMs

sF

sXsG a

KBsMs

K

sX

sXsG

a

b

22 )(

)()(

BsMssX

sX

sF

sXsGsGsG

a

ba

221

1

)(

)(

)(

)()()()(

G1(s) G2(s)f xbxa

in cascade

G (s)f xb

input output

overall



Figure. 2.11

reference

M

xb

K B


xa

f(t)

7

13

Let x1= xb=y and x2= vb = Dx1, then the state and output equation is

1 2

1 1

2 2

0 1 0

x x u

x xu Ax buK B K

x xM M M

cxxy 01 (2.69)

ab KxxKBDMD )( 2 (2.61)

Eq. (2.61) involves two energy-storage elements, K and M, whose energy variables are xb and vb respectively. Note that xa is input.

KuKxBxMDx 122



Determine the state space equations for Eq.(2.61)

Figure. 2.11

reference

M

xb

K B


xa

f(t)

14

State Concepts (P28-34)State Concepts

14

8

15

Multiple-Element Mechanical Translation System

reference

M2

xb

K2

B3

Fig. 2.13(a) Multiple-element mechanical system

xa

f(t)

M1K1

B1 B2

A force f is applied to a mass M1 with friction B1.

The system equations can be written in terms of the two displacements xaand xb.


The sum of the forces at each node must equal to zero. The equations are written according to the rules for node equation.

fxDBxKDBDBDM ba )()( 31312

1node a: (2.70)

node b: 0)()( 2322

23 ba xKDBDBDMxDB (2.71)

Each term in the Eq. must be a force.

16

There are four energy-storage elements, thus the four assigned state variables are xa, xb, Dxa, Dxb. f is input.

Determine the state space equations for Fig.2.13, where xb is output.

21 Kfor ,bxx 212 Mfor ,bvxx

13 Kfor ,axx 134 Mfor ,avxx

1 , xxyfu b

cxy

buAxx

(2.72)



reference

M2

xb

K2

B3


xa

f(t)

M1K1

B1 B2

9

17


1node a: (2.70)

node b: 0)()( 2322




,1 bxx bvxx 12

,3 axx avxx 34

1 , xxyfu b

2331434141 xBxKxBxBfDxM

1223224322 xKxBxBxBDxM

1xy reference

M2

xb

K2

B3


xa

f(t)

M1K1

B1 B2

18


Multiple-Element Mechanical Translation System2331434141 xBxKxBxBfDxM

1223224322 xKxBxBxBDxM

reference

M2

xb

K2

B3


xa

f(t)

M1K1

B1 B2

10

19

Ex. When a external force f(t) is applied to the mass, a displacement y(t) is present. Write the system equation.

2

1 2 2

d( ) ( ) ( )

d

yf t f t f t M

t

where f1(t) — damper resistance forcef2(t) — stiffness spring force.

1

d ( )( )

d

y tf t B

t where B — damping coefficient

First step: Write the overall system dynamics.

Second step: Write corresponding subsystemequations.

f1

f2input

output

Example: Spring-Mass-Damper System


20

Assume the spring characteristics is linear, i.e. K is a constant value.

f2 (t) = Ky(t)

Third step: Replace f1(t) and f2(t) in the original equation

2

2

d ( ) d ( )( ) ( )

d d

y t y tM B Ky t f t

t t

2

2

d ( ) d ( ) 1( ) ( )

d d

M y t B y ty t f t

K t K t K

or

Let

B

BT

K 2

M

MT

Kand



11

21

Then we get a differential model, which describes the system dynamics.

22

2

d ( ) d ( ) 1( ) ( )

d dM B

y t y tT T y t f t

t t K

where TM and TB are called the time constants of the system. 1/K

is a ratio of the system’s output and input at steady condition.

列写系统的微分方程式（建模）时，输出量及其各阶导数项列写在方

程式左端，输入项列写在右端。由于一般物理系统均有质量、惯性或储

能元件，左端的导数阶次一般来说要比右端的高。要注意的是，二阶系

统中的时间常数已经不具备像一阶系统中时间常数的物理意义。二阶系

统有它自己特殊重要的参数表征其系统特性。







22

12

Analogous Circuits

（相似电路）

Example: the system shown in Fig.2.11.

M

xb

KB

(b) Corresponding mechanical network

xa

f(t)

referenceFigure. 2.11

reference

M

xb

K B


xa

f(t)

Analogous Circuits

Analogous Circuits

)( baK xxKff node a: (2.58)


24

13

Look at the 2 figures below, and compare them graphically.

Node a (2.58-1))(1

ba xxL

i

Node b (2.59-1)R

DxxCDi b

b 2

Compare them again, this time mathematically.

M

xb

KB


xa

f(t)

reference

va vb

Reference(b’)

L

Ri(t)Ci i2

i1

Analogous Circuits

Analogous Circuits)( baK xxKff node a: (2.58)


25

M

xb

KB


xa

f(t)

reference

ua ub

Reference(b’)

L

Ri(t)Ci i2

i1

Circuit Fig.(b’) resembles mechanical translation system in Fig.(b) byletting force f=i (current), velocity v=Dx (displacement)=Voltage,Mass=C, K=1/L, and B=1/R (See P41 table 2.4). These systems are calledanalogous systems, which means they have the same form.

(2.58-2)1

( )a bi u uL

Node a )( ba xxKf

(2.59-2)2 bb

Dui CD u

R Node b bb BDxxMDf 2

Analogous Circuits

Analogous Circuits

26

14

Analogous Circuits

Analogous Circuits

27

The force f and the current i are analogs and are classified as “through” variables.（测量：ammeter or force indicator placed in serie, point of reference not needed）

The velocity v and the voltage v are analogs and are classified as “across” variables.（测量：across the system, point of reference needed）

Analogous Circuits: Example: S-M-D System

f1

f2input

output

(a)(a)

y

M

Reference

K Bf(t)

(b)

1 2 3( ) ( ) ( ) ( )f t f t f t f t

2

2

d ( ) d ( )( ) ( )

d d

y t y tM B Ky t f t

t t

Analogous Circuits

28

2

1 2 32

d d ( )( ) , ( ) ( ), ( )

d d

y y tf t M f t Ky t f t B

t t

15

y

M

Reference

K Bf(t)

(b)

Dy

Reference(b’)

L Ri(t)C

i0 i2i1

Write Eq. of circuit Fig.(b’)

DyR

i1

2

yL

DyLD

i11

1

yCDi 20 iiii 210

iyL

DyR

yCD 112

fKyBDyyMD 2

Note !!!

Analogous Circuits

Analogous Circuits: Example:S-M-D System

2929

30


reference

M2

xb

K2

B3


xa

f(t)

M1K1

B1 B2

The system equations can be written in terms of the two displacements xaand xb.

The mechanical network is drawn as the following figure.

xb

K1

Fig.2.13(b) Corresponding mechanical network

xa

f(t) B2M2

reference

M1 K2

B3

B1


16

31

By using Kirchhoff’s laws,


1node a: (2.70)

node b: 0)()( 2322


xb

K1

Fig.2.13(b) Corresponding mechanical network

xa

f(t) B2M2

reference

M1 K2

B3

B1




32




17

Other Mathematic Modeling Examples

-----Mechanical Rotational Systems

（机械旋转系统）

Mechanical Rotational Systems (P41)


a

J

b

c

K

d

e

B

f

Fig. 2.14 Network elements of mechanical rotational system

(a) (b) (c)

The applied torque (转矩) is equal to the sum of the reaction torques. (Newton’s Second Law)

The three elements in a rotational system are inertia (惯量), the spring, and the dashpot. The mechanical-network representation of these elements is shown in Fig.2.14.

Mechanical rotational systems are comprised of a group of mechanical elements that only rotate. The equation characterizing rotational systems are similar to those for translation systems, where the displacement, velocity, and acceleration terms are now angular quantities.

Comparing Fig.2.14 with Fig.2.9.! -------Almost same?

34

18

a

J

b

c

K

d

e

B

f


(a) (b) (c)

The torque applied to a body having a moment of inertia J produces an angular acceleration.

When a torque is applied to a spring, the spring is twisted by an angle .

2JDJDJaTJ (2.73)

)( dcK KT (2.74)

To produce motion of the body, a torque must be applied to overcome the reaction damping torque. The damping torque

The torque equation is written for each node by equating the sum of the torque at each node to zero.

)()( fefeB DDBBT (2.75)



35

a

J

b

c

K

d

e

B

f


(a) (b) (c)

Compare: Mechanical rotational and mechanical translation system

2JDJDJaTJ

)( dcK KT

)()( fefeB DDBBT

a

M

b

c

K

d

e

B

f

Fig. 2.9 Network elements of mechanical translation system

(a) (b) (c)

xMDMDvMafM2

)( dcK xxKf cK Kxf 0 dxif

)()( fefeB DxDxBvvBf



36

19

Mechanical Rotational Systems

There is one node having a displacement , therefore only one equation is necessary. An electrical analog can be obtained, thus torque is analog of current, and the moment of inertia (转动惯量）is the analog of capacitance.

Fig.2.15 (a) Simple rotational system

torqueT(t)

Fluid K

B

Wire

K

2.15 (b) Corresponding mechanical network

T(t) BJ

)(2 tTKBDJD (2.76)

The system shown in Fig.2.15 has a mass, with a moment of inertia J, immersed in a fluid. A torque T is applied to the mass.


37


1.（10%）图1为转动物体，J表示转动惯量，B表示粘滞系数。若输入为转矩，输出为轴角位移，求传递函数。


图1 转动物体

解：由转矩方程得：

设初始条件为零，对上式取拉氏变换得：

mechanical network

T(t) BJ

B

T

)(2

2

tTdt

dB

dt

dJ

2 ( ) ( ) ( )

( ) 1

( ) ( )

Js s Bs s T s

s

T s s Js B

38

20

Multiple-element mechanical rotational systemThe system represented by Fig.2.16a has two disks that have damping between

them and also between each of them and the frame. The corresponding mechanical network is drawn in Fig.2.16b.

K1

Fig.2.16(b) Rotational system ‘s corresponding mechanical network

2

T(t) B2J1

3

J2 K2

B3

B1

1


39


Multiple-element mechanical rotational systemThe system represented by Fig.2.16a has two disks that have damping between

them and also between each of them and the frame. The corresponding mechanical network is drawn in Fig.2.16b.

)(:1 Node 2111 tTKK (2.77)

0])([:2 Node 3321312

111 DBKDBBDJK (2.78)

0])([:3 Node 32322

223 KDBBDJDB (2.79)


40

Mechanical Rotational SystemsK1

Fig.2.16(b) Rotational system’s corresponding mechanical network

2

T(t) B2J1

3

J2 K2

B3

B1

1

21

Multiple-element mechanical rotational system

These three equations can be solved simultaneously for 1, 2, and 3 as a function of the applied torque.

If 3 is the system output, the transfer function of each unit is respectively

TDG 1

1 )(

1

22 )(

DG2

33 )(

DG

(2.80)TT

GGGG 3

2

3

1

21321

Figure 2.17 Detailed and overall representations of Fig.2.16

G1(D)T(t) 3

G2(D) G3(D)21

G (D)T(t) 3

And the overall transfer function of the system is

Question 2.17: State equation for this system?


41


More mechanical rotational system examples are similar, for example

Effective moment of inertia and damping of a gear trainshown in Fig.2.18a in P.45


42


22

Effective moment of inertia and damping of a gear train。


43



44


23


----- Thermal Systems（热力系统）

(1) Simple mercury thermometer

(2) Direct steam heater

Thermal Systems(P46): Basic Elements

R

Fig. 2.19 Network elements of thermal systems

C

(a) (b)

The necessary condition of equilibrium requires that the heat added to the system equals to the heat stored plus the heat carried away, which results in the increase of temperature of the system. This requirement can also be expressed in terms of rate of heat flow within unit time.

A limited number of thermal systems can be represented by differential equations. The basic requirement is that the temperature of a body be considered uniform.


The symbols shown in table 2.6 are used for thermal systems.

A thermal system network is drawn by thermal capacitance and thermal resistance.

46

24

The additional heat stored in a body whose temperature is raised from 1 to 2 is given by

2 1heat energy: ( )h C (2.84)

In terms of rate of heat flow 2 1( )d h

q C Dd t

C: The thermal capacitance (热容) determines the amount of heat stored in a body,-----like a capacitor in an electric circuit.

Rate of heat flow through a body in terms of the two boundary temperatures 3 and 4 is

Rq 43 (2.86)

R: The thermal resistance (热电阻) determines the rate of heat flow through the body, -----like a resistor in an electric circuit.


Thermal Systems(P46): Basic Elements

47

Thermal Systems: Simple Mercury Thermometer

Consider a thin glass-walled thermometer (having a capacitance Cand a resistance R) filled with mercury that has stabilized at a temperature 1 (initial condition). It is plunged into a bath of temperatures 0 at t=0. The mercury temperature at time t is m(t).

The flow rate of heat is R

q m 0

The heat entering the thermometer stored in the C, is given by )( 1 mC

D

qh

(2.88)

These equations can be combined to form

)( 10

mm C

RDh

Differentiating Eq.(2.88) and rearranging terms gives

0 mmRCD (2.90)


48

25

0 mmRCD (2.90)

1

1)(

0

RCDDG m

or1

1

)(

)()(

0

RCss

ssG m

The thermal network is drawn in Fig. 2.20. The node equation for this circuit, with the temperature considered as a voltage, gives Eq.(2.89) directly. Then, the transfer function G= m/0 may be obtained.

+

mR

C

0

-0

Fig.2.20 Simple network representation of thermometer

Let x1=m, u=0, the state equation is uRC

xRC

x11

11 (2.91)


49

Thermal Systems: Simple Mercury Thermometer

Thermal Systems: Direct Steam Heater

cold-liquid

steam W

qc, c

qa, a

hot-liquid

environment temperature

Objective: Heat cold-liquid to the temperature a

Step 1: Input and output?

Output (controlled variable): a

Input (control variable) can be: W, qc , c and , etc.. The most suitable variable is W. Others are as disturb variables.

Step 2: Assumptions

(1) Lumped model

(2) Good heat preservation

(3) qc, c, are approximately constant


50

26

cold-liquid

steam W

qc, c

qa, a

hot-liquid


qs

Step 3: Develop mathematics model • Let Q denote quantity of heat

According to Conservation of Energy Law in unit time:

i n p u t o u tp u t

d QQ Q

d t

Firstly, consider steady state, i.e.

0dt

dQ

1QQQQ asc

asc QQQ

From assumption (2), Q1=0, then

Q1


Thermal Systems: Direct Steam Heater

51

cold-liquid

steam W

qc, c

qa, a

hot-liquid


qs

Q1

, , c c c c s a a a aQ q c Q W H Q q c

where H is enthalpy of steam.

Assuming uniform volume of heat:

000 Wcq

H

aca

ccc ac

cca qWqq

where W is very small.

This is the system’s steady statemodel – algebraic equation.


Thermal Systems: Direct steam heater

asc QQQ

y a b u Note:52

27

cold-liquid

steam W

qc, c

qa, a

hot-liquid


qs

Q1

Secondly, consider system dynamic model, which is more important.

WHcqcqdt

dC ccaa

a

asc QQQdt

dQ

aQ V c where

and V is available volume,

is fluid’s density, so

a ad dd QV c C

d t d t d t

where C is called capacity coefficient. It represents the capability of the tank to store heat energy.


Thermal Systems: Direct steam heaterasc QQQ

53

cold-liquid

steam W

qc, c

qa, a

hot-liquid


qs

Q1

Let cq

Ra

1 R represents the resistance to prevent heat energy

from departing the tank. It is called heat resistance.

Then

aa c

dR C K W

d t

cca qWqq

1 1aa c

dC W H

d t R R

where T=RC is the time constant, and K=HR is the amplificatory coefficient, also called steady state gain of a system. c is a disturbance variable, which is another uncontrolled input.

aa c

dT K W

d t

or **


Thermal Systems: Direct steam heatera

a a c c

dC q c q c WH

dt

54

28

cold-liquid

steam W

qc, c

qa, a

hot-liquid


qs

KWdt

dT ca

a The system’s differential equation:

The system’s transfer function

Regulating path

1)(

)()(

Ts

K

sW

ssG a

o

Disturbance path

1

1

)(

)()(

Tss

ssG

c

ad

If there is delay

1)(

)()(

Ts

Ke

sW

ssG

sa

o


Thermal Systems: Direct steam heater

55任意一个输入到任意一个输出都构成一个path，也都存在一个传递函数。

Increment form differential equation

In process control, we usually consider increment equation of variables,

e.g.

Subtract the steady equation from the dynamic one, where a= a0+ a

000 KWKWdt

dT ccaa

a

KWdt

dT ca

a **System dynamic equation

System steady equation000 W

cq

H

aca ---operation point.

We obtain increment equation WKdt

dT ca

a

Except for increment symbol , the equation is same as the dynamic equation. For LTIS, symbol is often omitted.


56

29

First-order systems

KWdt

dT ca

a **

Considering (**), which is a first-order differential equation, it can represent different systems, e.g. RC circuit and the heater.

ieedt

deT 0

0 ***

First-order systemControl path

Disturbed pathoutput

1)(

)(

Ts

K

sW

sa(W)

1

1

)(

)(

Tss

s

c

a

(c)

Laplace )()1)((0 sETssE i

1

1

)(

)(0

TssE

sE

i


57

The first-order system is represented by first-order differential equation.

1

1

)(

)()(

TssU

sYsG

First-order system

Control path

Disturbed pathoutputInput

A

t

u

t

y

A

T1

0.632A

T2

( ) (1 )tTy t A e

First-order system’s step response

)()()(

tutydt

tdyT


58

30


----- Liquid-Level System （液位系统）

(1) System-1(P53)

(2) System-2(supplementary)

Liquid level system

h1

A1

u

R1

qin

Tank 1

Disturb input

qout

Single tank with a valve

q = rate of flow of fluid,

h = height of fluid level

R = flow resistance

A = cross-sectional tank areas


60

1outq h K f

R

近似的流量方程：

液位差引起的流量

变化

阀门开度引起的流

量变化

31

Liquid level system-1 (P53)

h1

A1A2

R1

q1

qin

Tank 1

Disturb input

h2

R2qout

Tank 2 Control input

q2+qf

Two-tank liquid-level control system consists of two first-order dependent plants that are connected in series. Note that the heights h1 and h2 of the tanks are coupled（耦合）(see P53 Fig.2.23).



R = flow resistance


Keep R1 still, and adjust R2.

Objective: To keep h2 on a constant level, which is related to qout and qin.


61

h

1

A1A2

R1

q1

qin

Tank 1

Disturb input

h

2

R2qout


q2+qf

Tank 1 11

1 qqdt

dhA in where )(

121

11 hh

Rq

Tank 2outqq

dt

dhA 1

22 fout qh

Rq 2

2

1where



Objective: To hold h2 constant, which is related to qout and qin

The first item depends on h2, while the second one depends on R2.

62设R1开度不变；R2为调节阀，因开度变化引起流量变化为qf。

32

h1

A1A2

R1

q1

qin

Tank 1

Disturb input

h2

R2qout


q2+qf

Eliminate internal variables q1, h1, qout, etc. Then we get

dt

dqRTqRqRh

dt

dhRATT

dt

hdTT f

fin 212222

212122

2

21 )(



此为输入输出模型，可以反映输出h2与输入qin与qf之间的关系，但无法反映变

量h1与其他变量的关系，从而损失了系统内部变量h1的信息。解决办法？？？

63设R1开度不变；R2为调节阀，因开度变化引起流量变化为qf。

1 1 1

2 2 2

T A R

T A R

Liquid level system-1: transfer function

Disturbance input

h1

A1A2

R1

q1

qin

Tank 1

h2

R2

qout


q2+qf

The system’s transfer function is

dt

dqRTqRqRh

dt

dhRATT

dt

hdTT f

fin 212222

212122

2

21 )(

1)()(

)(

)(

)()(

21212

21

2122

11

DRATTDTT

DRTR

tq

th

tu

tyDG

f

control path

1)()(

)(

)(

)()(

21212

21

22

22

DRATTDTT

R

tq

th

tu

tyDG

in

disturbance path


64

33

Liquid level system-1: state equation

h

1

A1A2

R1

q1

qin

Tank 1

Disturb input

h

2

R2qout


q2+qf

Assigned state variables as 122

1

h

hx

and input variables as Tfin qqu

Let output variables are 122

1

h

hy

Then the state Eq. is 1 1 11 1

2 2

1 2 1 2 2 2

1 1 10

1 1 1 10

in

f

qT T Ax x

qx x

R A R A T A

and the output Eq. is

2

1

10

01

x

xy

Tank 1 11

1 qqdt

dhA in where )(

121

11 hh

Rq

Tank 2outqq

dt

dhA 1

22 fout qh

Rq 2

2

1where


65

Liquid level system-2

Two-tank liquid level control system as shown below with definitions



R = flow resistance


Objective: hold h2 constant, which is related to qout and qin

means: control qout(---- qf)

h1

A1

A2

h2R2

R1

q1

qout

qin

Tank 2

Tank 1

This system consists of two first-order independent plants that are cascaded.

Note that the heights h1

and h2 here are uncoupled.

q = rate of flow of fluid


R = flow resistance



66

34

Tank 1 11

1 qqdt

dhA in

11

1

1h

Rq where

After eliminating internalvariables q1, h1, etc., we get

h1

A1

A2

h2R2

R1

q1

qout

qin

Tank 2

Tank 1

output

Control input

Tank 2outqq

dt

dhA 1

22

wherefout qh

Rq 2

2

1

Disturbance input

different from Sys. 1


Liquid level system-2

)(1

211

1 hhR

q

67

h1

A1

A2

h2R2

R1

q1

qout

qin

Tank 2

Tank 1

Disturbance input

output

Control input

dt

dqRTqRqRh

dt

dhTT

dt

hdTT f

fin 212222

2122

2

21 )(

The transfer functions of the system is

1)(

)(

)(

)(

)()(

212

21

212

2

11

DTTDTT

DRTR

tq

th

tu

tyDG

f

1)(

)(

)(

)(

)()(

212

21

2

2

22

DTTDTT

R

tq

th

tu

tyDG

in

调节（控制）通道传递函数

干扰通道传递函数

Liquid level system-2: transfer function


68

35

h1

A1

A2

h2R2

R1

q1

qout

qin

Tank 2

Tank 1

Disturbance input

output

Control input

Since there are two energy-storage elements, select them as the state variables. Or you can judge from the order of the ODE.

Assign state variables as

122

1

h

hx

input variables12

f

in

q

qu

output variable 2hy

Tank 1

1

11

11 R

hqqq

dt

dhA inin

Tank 2

f

out

qR

h

R

h

qqdt

dhA

2

2

1

1

12

2



69

h1

A1

A2

h2R2

R1

q1

qout

qin

Tank 2

Tank 1

Disturb input

output

Control input

Then the state equation is 1 11 1

2 2

1 2 2 2

1 10 0

1 1 10

in

f

qT Ax x

qx x

R A T A

and the output equation

22

110 hh

hy

If the output variables are

11 hy 22 hy

the output equation

2

1

10

01

x

xy



70

36

The state equation 1 11 1

2 2

1 2 2 2

1 10 0

1 1 10

in

f

qT Ax x

qx x

R A T A

The state equation1 1 11 1

2 2

1 2 1 2 2 2

1 1 10

1 1 1 10

in

f

qT T Ax x

qx x

R A R A T A

Compare it with

Difference?!

Liquid level System-2:

Liquid level System-1:



71思考题：对于两个液位系统，qin为干扰，分别分析qf对于h1/h2的“可控”能力(提示：从状态方程模型出发)

System Modeling

Differential equation

physical laws: electrical, mechanical, thermal, liquid-level

State space model

basic matrix A, B, C; state variable

state equation and output equation

The third form: transfer function and block diagram

72


37

Ex: Modeling by block diagram--direct steam heater

steam

W,P

qa, T

cold-liquidqc, Tc

TTTC

Ref. Ts

Objective: T, by control W

hot-liquid

What is the controlled process (plant)? Controlled variable? Output/Input variable? Control (manipulation) variable?

G0 (s)W T

Tc

73



steam

W,P

qa, T

cold-liquidqc, Tc

TTTCRef. Ts


hot-liquid

Consider each component of the system as follows

Control path W T’

G01(s)W T’

Disturbance path Td

G02(s)Tc Td

Define T’+Td=T

74


38

steam

W,P

qa, T

cold-liquidqc, Tc

TTTCRef. Ts


hot-liquid

G01(s)T

G02(s)Tc

W


+ +

Controller_1-TC (main loop)

GTC(s)u1

GTT(s) T

Ts +

75


steam

W,P

qa, T

cold-liquidqc, Tc

TTTCRef. Ts


hot-liquid

Since the steam pressure is usually changing, an additional disturbance like path is added. Thus make it to be a T plus G cascaded controller. The main disturbance is Tc.

GC GT2 paths that causing W


G03(s)W

G04(s)Steam P

Gv(s)control U

u valve W

++

76


39

Controller_2-GC (slave loop)

GGC(s)u

GGT(s) W

u1

environment temperature Te

steam

W,P

qa, T

hot-liquid

cold-liquidqc, Tc

TTTC

GC GT

Ref. Ts


+

77


Overall system block diagram model is as follows

GTT(s)

T

T’GTC(s)Ts

GGC(s)u

GGT(s)

u1G03(s)

W

G04(s)Steam P

Gv(s) G01(s)

G02(s)Tc

Td


How to get the overall transfer function of the multi-loop block diagram?

78


40

79

一．1．求理想运算放大器的传递函数，结构图如下：(2003年)

2003年第1题示意图

其它对象的建模也是一样的方法，如无源放大器。

80


)()(

)()()()()()(

)(

)(:得

)(消去

0

)(

)(

)(

)()(

)(

)()(

)(

)()(

)(

)()(

即：

41

424332

423

21

423

21

sZsZ

sZsZsZsZsZsZ

sU

sU

sU

U

sZ

sU

sZ

sUsU

sZ

sUsU

sZ

sUsU

sZ

sUsU

iii

ii

i

O

F

B

FFBOF

FBBi

考虑：如

果这里用

方块图建

模方法，

你行否？

41


UF(s)

81

1

1

Z oUiU

0BU 33 ZIUU Fo

42

42

4242423

)(

ZZ

UZZ

Z

U

Z

U

Z

U

Z

UUIII FFFFBF

22 ZIU F

2Z

2I

112 Z

UII i

42

42

ZZ

ZZ FU

3I3Z

)()(

)()()()()()(

)(

)(

41

424332

sZsZ

sZsZsZsZsZsZ

sU

sU

i

O

82

一．R-L-C网络如图所示，信号源内阻为零，Ur(t)为输入变量，U0(t)为

输出变量，试求电网络的状态变量表达式。 (2002年)

－

ur(t)u0(t)

R2


L

R1

＋

e(t)

i1 i2

i3

C

42

83

－

ur(t) u0(t)R2


L

R1

＋

e(t)

i1 i2

i3

C

rcc u

RRL

RCRR

i

u

RRL

RR

RRL

RCRR

R

CRRi

ux

)(

)(

1

)()(

)()(

1

21

2

21

3

21

21

21

1

21

1

21

3

rc u

RR

R

i

u

RR

RR

RR

Ry

21

2

321

21

21

2

84

一．运算放大器的电路如附图2-1所示，写出该电路的传递函数。

(2000年)

UB

-KU1

Uo

R2

C

2000年第1题

R1i1

i2

i3

43

85

UB

-KU1

Uo

R2

C

2000年第1题

R1i1

i2

i3

1 2 3

0 01

1 2

2

1 1 2

( ) ( ) ( ) ( )( ) ( )1( )

0

( ):

( ) ( 1)

B BB

B

O

i i i

U s U s U s U sU s U s

R s RC s

U

U s R

U s R R C s

即：

得

principle of automatic control chapter 2 writing...

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