Models for Dynamic Systems and Systems Similaritynhuttho/me584/Chapter 2 Models for Dynamic... · Models for Dynamic Systems and Systems Similarity Dr. Nhut Ho ME584 Chap2 1. ...

Models for Dynamic Systems

and Systems Similarity

Dr. Nhut Ho

ME584

Chap2 1

Agenda

• Formulation of Models for Engineering

Systems

• Review of Solution of Differential

Equations

• Engineering Systems Similarity

• Simulation with Matlab

• Active learning: Pair-share questions,

Concept questions

Chap2 2

Formulation of Models for

Engineering Systems

Chap2 3

Conservation Laws of

Engineering Systems

• Linear momentum: translational

momentum is equal to applied impulse

(rigid body an or fluid)

• Angular momentum: angular impulse is

equal to time rate of change of angular

momentum

Chap2 4

mvdt

dFnet

Jdt

dTnet


Engineering Systems

• Conservation of Charge/Kirchhoff’s law:

sum of all currents at a node is equal to

rate of storing charge

• Conservation of Mass: net mass flowrate

at a location equal to time rate of change

of mass that that location

Chap2 5

dt

deC

dt

dQinode

VVVdt

dmnet


Engineering Systems

• Conservation of Energy

– Energy in a system is constant

– Reduces to Bernoulli equation: energy is not

exchanged or stored

Chap2 6

Review of Solution of

Differential Equations

Chap2 7

Input Signals

Chap2 8

Step input Ramp input

Input Signals

Chap2 9

Pulse input Sinusoidal input

Classical Differential EquationsChap2 10

)(]1[,,

.00

:1

1

01

001

tGuxDa

a

dt

dD

a

bGLet

aandaifstableisSystem

ubxadt

dxaOrder

oo

o

st

Example Classical Differential Equations

• Use Table E.3 to find the solution, and find

steady state value. Is this system stable?

Chap2 11

1,0;226 oxuuxx

0)(

)(lim

13

1

26

2

)(

)()(2)(2)(6)2

0)()1

2:

3..00

32

6,1

2

2,2,2,6

:

0

3/

3//

1

11

sU

sXsx

sssU

sXsUsXssX

etxx

methodsvaluestateSteady

eexxETablestableisSystemaanda

a

a

a

bGbaa

Solution

sss

ss

tt

oo

oo

ooo

Classical Differential Equations:

Pair-Share Exercise

• Use Table E.3 to find solution, and find steady state value for

this system. Is this system stable?

Chap2 12

.

.)(lim)(

)(lim

;1

)(;210

10

)(

)(

)]1(5[5)]1([

52

10,5

2

10,2,10,10

:

00

5///

11

unstableisSystem

sXsU

sXsx

ssU

ssU

sX

etetGuexx

a

a

a

bGaab

Solution

ssss

tt

o

t

o

oo

ooo

0),(1;10210 oo xinputstepuuuxx

Classical Differential EquationsChap2 13

)(:1

)(:1

)(:1

,,2

,1

,

.0,0,0

:2

12

2

012

0012

2

2

responseinonoscilliatiNocasedampedCritically

responseinonoscilliatiNocaseOverdamped

responseinonOscilliaticasedUnderdampe

dt

dDand

a

a

a

a

a

bGLet

aandaaifstableisSystem

ubxadt

dxa

dt

xdaOrder

ooo

o

nd

nn

Chap2 14

Classical Differential Equations

Classical Differential Equations

• Example: Using Table E.4 to find solution for this

system. Is this system stable?

Chap2 15

)866.0sin(866.0

5.)866.0cos(

)sin()sin(1

)cos()(

866.01

:1

,5.12

,111

,1

,1.,1,1,1

:

5.

2

2

12

2

012

tte

tex

ttextx

casedUnderdampe

a

a

a

a

a

bG

bstableisSystemaaa

Solution

t

tot

o

ooo

o

o

d

n

d

dd

n

nd

n

n

n

0,1,0; oo xxuuxxx

State Space Description

• State variables describe present configuration of

a system and can be used to determine future

response, given inputs and dynamic equations

• Number of state variables

– Should be as small as possible, but a minimum

number of state variables must be selected as

components of the state vector (sufficient to describe

completely the state of the system)

– Usually equal to order of system’s differential

equation or number of energy-storage elements

• Choice of state variables is not unique

Chap2 16

State Space Representation

Chap2 17

d matrixfeedforwarD

rixoutput matC

ixinput matrB

rixsystem matA

orinput vectu

ectorhe state vative of ttime derivx

toroutput vecor ystate vectx

DuCxy

BuAxx


• Example: Express in State

Space format

• Is this system stable?

Chap2 18

fz

x

z

x

fxzfxxzx

xz

xz

fxxx

1

0

23

10

3232

32

fxxx 32

Yes

State Space Representation:

Pair-Share Exercise

• Example: Express in state space format

• Is this system stable?

Chap2 19

f

y

z

x

y

z

x

fxzy

fxxxxzy

xzy

xz

1

0

0

234

100

010

432

432

fxxxx 432

Yes


• Example: develop a state space model with output vL

• Ldi/dt+Ri+(1/c)∫idt=v(t)

• i(t)=dq/dt, i’=q”

• Lq”+Rq’+(1/c)q=v(t)

• To make two 1st-order equations– dq/dt=i

– di/dt=-(1/LC)q-R/Li+v(t)/L

• Output: vL=Ldi/dt

Chap2 20


Concept Question• Which variables can be chosen to describe the state of the system?

A. Only i

B. Only q

C. i and q

D. Only VR

E. Only VC

F. L, R, or C

G. Other possibilities

• How many state variables are sufficient to describe the state of the

system?

• Some possible pairs

– i and q

– VR and VC

– ……

Chap2 21


Chap2 22

1D ,R-c

1-C ),(

v(t)u , 10

B ,

110

A , /

/

)(1

Lc

1

:

tvy

DuCxy

equationsoutput

Li

qx

L

R

Lcdtdi

dtdqx

tvL

iL

Rq

dt

dii

dt

dq

BuAxx

iandqChoose

L


Pair-Share Exercise

Chap2 23

Form state space model with vR and vC

• vL=Ldi/dt=(L/R) dvR/dt

• As vL+vR+vC=v(t)

• (L/R) dvR/dt+vR+vC= v(t)

• dvR/dt=-(R/L) vR-(R/L) vC+(R/L)v(t)

• vC=(1/c) ∫(vR/R)dt

• Differentiate

• dvC/dt=(1/RC)vR

• New state equations

State Space

Graphical Representation

• q,i were selected

as state variables

• Graphical

representation

of state space

and a state vector

Chap2 24

Function of time t

q

i


• Example: RLC Electrical circuit

Chap2 25

L

C

L

C

L

C

C

CL

LCC

i

v

R

tv

Li

v

L

CRCi

v

vR

output

tvL

vLdt

di

iC

vRCdt

dv

01

i

)(10

01

11

1i

)(11

11

R

R

Output iR(t)

State Space Solution

• Solution for state space equation

Chap2 26

:)(

.)(,)(),0(),(

,)()(

)()]()0()()(

)()()0()()(

)()0()()(

)()()0()(

1

0

11

matrixtransitionstatet

ynumericallevaluatedorsolvedbecantxknownuandxtWith

AsIoftransformLaplaceistwhere

dButxttx

sBUAsIXAsIsX

sBUXsXAsI

sBUsAXXssX

BuAxx

t

State Space Solution

• Example

Chap2 27

tt

tt

tttt

tttt

ee

eexttx

eeee

eeeet

transformLaplaceinverseTake

ss

s

ss

ssss

s

s

s

sssAsI

uandxxx

2

2

22

22

1

1

22

2)0()()(

)2()22(

)()2()(

,

)2)(1()2)(1(

2

)2)(1(

1

)2)(1(

3

2

13

)2)(1(

1

32

10

10

01)(

.00

1)0(,

32

10

State Space Solution:

Pair-Share exercise

• Find solution for this system

Chap2 28

.01

1)0(,

31

20

uandxxx

t

t

tttt

tttt

e

exttx

eeee

eeeet

transformLaplaceinverseTake

s

s

sssAsI

2

2

22

22

1

1

)0()()(

)2()(

)22()2()(

,

1

23

)2)(1(

1

31

20

10

01)(

Engineering Systems

Similarity

Chap2 29

System Similarity

• Models in different engineering systems

have same or similar differential equations

• Similar basis elements

– Resistance

– Capacitance

– Inductance

• Dynamic behaviors have different time

constants, influencing modeling choices

and assumptions

Chap2 30

System Similarity

Chap2 31

Effort and Flow Variables

• Variables used for differential equations are

either effort or flow variables

– Effort variables: express effort placed on a

component, e.g., force, voltage

– Flow variables: express flow or time rate of change of

a system variable, e.g., velocity, current

• Impedance: is any transfer function between a

flow variable as input and effort variable as

output

– Effort = impedance x flow

– Impedance can be static (resistance) or dynamic

(capacitance/inductance)

Chap2 32

Chap2 33

General Model Structure for

Mechanical System

Chap2 34

General Model Structure for

Electrical System

Chap2 35

Simulation with Matlab

Chap2 36

1st Order System: Differential

Equation and Transfer Function

Free body diagram: Sliding block with friction

• f(t) - m v' - b v = 0

• L{ v(t) } = V(s), L{ v’(t) } = sV(s) – v(0), L{ f(t) } = F(s)

• Assume v(0) = 0,

• m[sV(s)] + b[V(s)] – F(s) = 0

• Rearranging we get: [ms + b]V(s) = F(s)

• The transfer function is: V(s)/F(s) = 1/[ms+b]

Chap2 37

1st Order System: TF in Matlab

• Let m = 20,000kg, b = 500kg/s

• Matlab commands:

>> m = 20000;

>> b = 500;

>> num = [ 1 ];

>> den = [ m b ];

>> first_tf = tf(num, den)

Transfer function:

1

-------------

20000 s + 500

Chap2 38

1st Order System: Step Response

>> u = 1;

>> step(u * first_tf)

Chap2 39

>> u = 20000;

>> step(u * first_tf)

1st Order System: Impulse

Response

>> u = 1;

>> impulse(u * first_tf)

Chap2 40

>> u = 20000;

>> impulse(u * first_tf)

1st Order System: State Space

x' = Ax + Bu

y = Cx + Du

>> [A, B, C, D] = tf2ss(num,den)

A =

-0.0250

B =

1

C =

5.0000e-005

D =

0

Chap2 41

>> first_ss = ss(first_tf)

a = x1

x1 -0.025

b = u1

x1 0.007813

c = x1

y1 0.0064

d = u1

y1 0

2nd Order System: Differential

Equation and Transfer Function• Sum of all torques

• kq + bq' + Iq" = 0

• L{ q(t) } = Q(s)

• L{ q’(t) } = sQ(s) – q(0)

• L{ q"(t) } = s2Q(s) – sq(0) – q'(0)

• k[Q(s)] + b[sQ(s)] + I[s2Q(s) - w(0)] = 0

• Or [Is2 + bs + k]Q(s) = I*w(0)

• Q(s) = [ 1/[Is2 + bs + k] ] * I*w(0)

• The transfer function is:

• G(s) = 1/[Is2 + bs + k]

Chap2 42

bq’

kqIq”

a rotational mass inertia I,

fiction oefficient b, spring

coefficient k

2nd Order System:

I = 1,500,000kgm2

b = 200,000Nm/s

k = 4,000,000Nm

>> I = 1500000;

>> b = 200000;

>> k = 4000000;

>> num = [ 1 ];

>> den = [ I b k ];

>> second_tf = tf(num, den)

1

------------------------------

1.5e006 s^2 + 200000 s + 4e006

Chap2 43

• >> u = 0.002 * 1.5e6;

>> impulse(u * second_tf)

Matlab Simulation Exercise

• Find the transfer function, state space

model, and simulate the impulse and step

response for this system

Chap2 44

0)0(

0)0(

32

x

x

uxxx

Homework 2: chapter 2

• 2.14

• 2.15

• 2.24

• 2.25

Chap2 45

References

• Woods, R. L., and Lawrence, K., Modeling

and Simulation of Dynamic Systems,

Prentice Hall, 1997.

• Palm, W. J., Modeling, Analysis, and

Control of Dynamic Systems

• http://www.me.cmu.edu/ctms/index.html

Chap2 46

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