MECH370: Modelling, Simulation and Analysis of Physical...

MECH370: Modelling, Simulation and Analysis of

Physical SystemsChapter 6

Electrical SystemsVariables and relationship with translational mechanical elementsElement lawsInterconnection lawsObtaining the system model

Lecture Notes on Lecture Notes on MECH 370MECH 370 –– ModellingModelling, Simulation and Analysis of Physical Systems, Simulation and Analysis of Physical SystemsChapter 6Chapter 6 2

Course OutlineCourse OutlineModellingModelling (Ch. 2,3,4,5,6,9,10,11,12) Simulation (4) Analysis (7(Ch. 2,3,4,5,6,9,10,11,12) Simulation (4) Analysis (7,8),8)

1. Definition and classification of dynamic systems (chapter 1)

2. Translational mechanical systems (chapter 2) 3. Standard forms for system models (chapter 3)4. Block diagrams and computer simulation with

Matlab/Simulink (chapter 4)5. Rotational mechanical systems (chapter 5)6. Electrical systems (chapter 6)7. Analysis and solution techniques for linear systems

(chapters 7 and 8)8. Developing a linear model (chapter 9)9. Electromechanical systems (chapter 10)10.Thermal and fluid systems (chapters 11, 12)


Electrical SystemsExample:

e - voltage (electric potential difference) in volts (V)

i - current in Amperes (A)

q - charge in Coulombs (C)

Variable:

Current is defined as the flow rate of charge

i


Electrical Systems (cont’d)

dtdqi =

Current has a direction and magnitude-i i

=

iii == 21

Charge cannot accumulate in a component

A voltage at a point a is the electric potential difference from an arbitrary reference called a ground.

Grounde0=0

i1 Circuitelement

i2

Circuitelement

ea -+e1 e2

iCircuitelement


Electrical Systems (cont’d):is element an across voltageThe ae

21 eeea −=

iep ⋅=

∫=1

0

t

tpdtE

The power supplied to an element is given by:

The energy supplied to the element is given by:

It has both a sign and magnitude-ea -+

=ea +-


Relationship Between Electrical and Translational Elements

Notation Variable Notation Variable

Position Charge

Velocity Current

Acceleration Change in current

Force Voltage

Newton’s 2nd law Inductor law

Spring lawCapacitor law

Damper law Resistor law

)( xv &=

x

)( xva &&& ==

F

MaF =

KxF =

BvF =

q

qi &=

qdtdi

&&=

)(or ve

dtdiLe =

duuiC

ete t )(1)0()( 0∫=−

qC

e 1=

Rie =


Relationship Between Electrical and Translational Elements(cont’d)

Notation Variable Notation Variable

Mass Inductance

Spring constant 1/Capacitance

Damping constant R Resistance

K

M

B

L

C1


Element Laws

iRe ⋅=

QReRieip &====

22

R is the resistance in Ohms (Ω)

The energy is converted into heat.

Ohm’s Law:

i

ReResistor:

A resistor dissipates energy:

i

+ -e

R

http://en.wikipedia.org/wiki/


Element Laws (cont’d)

Ceq =

∫+==t

tidτ

ctete

dtdeCi

0

1)()( , 0

Capacitor:

C is the capacitance in Farads (F)2

21 CeEp =

e

Cq

A capacitor stores energy in an electric field:

material dielectric theofty permittivi =⋅

= εε ;d

AC

i

+d

+

-A

-ee



Element Laws (cont’d)

∫ τ+==t

ted

Ltiti

dtdiLe

0

1)()( , 0

2

21 LiEp =

Inductor:

where L is the inductance in Henries (H).

e

i

+ -e

An inductor stores energy in a magnetic field:

dtdiLLi

dtddtde

==

=

)(

linkage)(flux i

+ -

L



Element Laws (cont’d)Notations:

Short circuit

Current source

Voltage source

Open circuit

circuit

e(t) circuit

i(t)

i(t)

i i=0

+ e=0e -


Interconnection Laws

Kirchhoff’s Voltage Law (KVL):

e3

+

-

e4 +-

e1

+

-

e2 +-

loopany around ,0=∑ je

e.g.

00

2143

4321

=−−+=−−+

eeeeeeee


Interconnection Laws (cont’d)

Kirchhoff’s Current Law (KCL):

Note: away from the node is +; into the node is –

nodeany around ,0=∑ jie.g.

0321 =+− iii

-

C

i1

A R

i2i3L


Electrical Systems

RiedtdiLe

dtdeCi

RL ==

=

and

Example 6.1

From KVL:

0)(1)0(

0)(

0=−τ+++∴

=−++

∫ teidC

eRidtdiL

teeee

i

e

t

ce

e

iCRL

C

R

L

44 344 21

Since:


Electrical Systems (cont’d)The above equation is not in state-variable or input-output form

)(1)0(0

teidC

eRidtdiL i

t

c =τ+++ ∫

: terms of ridget toateDifferenti ∫

dttdei

CdtdiR

dtidL i )(12

2

=++

Compare this equation with the one in Ex. 5.1 (textbook, pp.105) and in M-S-D system (next slide, or pp. 3-7)!

I/O form

It is interesting that the flow of current in this circuit is mathematically equivalent to motion of spring-mass system.

Lecture Notes on Lecture Notes on MECH 370MECH 370 –– ModellingModelling, Simulation and Analysis of Physical Systems, Simulation and Analysis of Physical SystemsChapter 6Chapter 6 16)(

)(

),(

)(

)(

2

2

tfKxxBxM

tfKxdtdxB

dtxdM

vdtdxtfKxBv

dtdvM

KxBvtf

fftfdtdvM

a

a

a

a

KBa

=++

=++

==++

−−=

−−=

&&&

Example:

Free body diagram (FBD):

Obtaining the System Model(Review)

Newton’s 2nd Law:

KM fa(t)

B x, v

M

fB

fK

fa(t)

fI

Force exerted by the dashpot

the applied force

Force exerted by the dashpot

the inertial force

x

y

D’Alembert’s Law(use the idea of an inertial force, fI)

)(0)(

)(0

tfaKxxBxMKxxBxMtfaffftfaf KBIi

=++⇒=−−−=

−−−==∑

&&&

&&&

dttdei

CdtdiR

dtidL i )(12

2

=++


Obtaining the System Model (cont’d) (Review)

Example 5.1: Find the input-output and state-variable equations for the Example given in pp. 5.

( ):0 Law sAlembert'D'by or Law 2nd sNewton' =iiτΣ

(t)τKθBωωJBωKθ(t)τωJ

a

a

=++−−=

&

&

or

Very important step towards to obtaining correct system model!

Are the above equations input-output equation format?

(t)τKθBωωJ(t)τKθBωωJ

a

a

=++=+−−−

&

&

or 0

(t)τKθBωωJ(t)τKθBωωJ

a

a

=++=−++

&

&

or 0

+ +


Obtaining the System Model (cont’d) (Review)

: iableoutput varfor equation output -Input θ

⎥⎦

⎤⎢⎣

⎡θω

=

τ⎥⎥

⎦

⎤

⎢⎢

⎣

⎡+⎥

⎦

⎤⎢⎣

⎡⎥⎥

⎦

⎤

⎢⎢

⎣

⎡ −−=⎥⎦

⎤⎢⎣

⎡

=

τ+−−=

1] [0

0

1

01

1

y

)t(Jθω

JK

JB

θω

ωθ

))t(KθBω(J

ω

a

a

&&

&

&

⎟⎠⎞

⎜⎝⎛ ====++ θω,θ

dtdθω(t)τKθθBθJ a

&&&&&&&

[ ] ⎥⎦

⎤⎢⎣

⎡=

τ⎥⎥⎦

⎤

⎢⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡

⎥⎥⎦

⎤

⎢⎢⎣

⎡−−=⎥

⎦

⎤⎢⎣

⎡

τ+−−=

=

ωθ

y

)t(Jω

θ

JB

JK

ωθ

)t(J

ωJBθ

JKω

ωθ

a

a

01

1010

1

&

&

&

&

State-variable equations:

or

or

(t)τKθBωωJBωKθ(t)τωJ

a

a

=++−−=

&

&

or


Electrical Systems (cont’d)(1))( or LLteeRi

dtdiL iC +−−=

From the element law for the capacitor

(2)LLidt

deC C =

(1) and (2) are in the state-variable form, or

1

)(11

iCdt

de

teL

eL

iLR

dtdi

C

iC

=

+−−=)(

0

1

01

1

teLei

C

LLR

dtdedtdi

iCC ⎥

⎥

⎦

⎤

⎢⎢

⎣

⎡+⎥

⎦

⎤⎢⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡ −−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

If we are interested in ec as output variable, then

[ ] ⎥⎦

⎤⎢⎣

⎡==

CC e

iey 10

[ ] ⎥⎦

⎤⎢⎣

⎡===

CRR e

iRRieye 0 then iable,output var as or

or


Obtain the Input-Output Model

e+

-i1 i2

There are two main approaches:1) Loop equation method (mesh analysis)

We assume a number of fictitious loop currents (i1, i2, …)

Then apply Kirchhoff’s voltage law and write the voltage in terms of the currents using element laws.2) Node equation method (node analysis)

i3

i2i1

e1 e2


Obtain the Input-Output Model (cont’d)• Select an arbitrary ground node

• Label the remaining node voltages (e1,e2, …)

• Label unknown currents (i1, i2, …)

• Apply Kirchhoff’s current law for nodes

• Express currents in terms of node voltages using element laws.

*We will focus on this method.


Obtain the Input-Output Model (cont’d)Example: Find I/O equation for input ei(t) and output eo.

KCL:

( )

( ) O) (Node 01)0(

A) (Node 0)(0 :O Node0 :A Node

0 0022

0

12

0

1

22

121

=τ+++−

−

=+−

+−

−∴

=++−=++−

∫t

LA

AAAi

LCR

CRR

deL

ieCR

ee

eCR

eeR

eteiiiiii

&

&

ei(t)+

-

R1 R2C1 L C2

ei(t) eA eO

A O


Obtain the Input-Output Model (cont’d)

O) (Node 0)0( (2)

A) (Node 0)(2 (1)get to1H ,F1 ,Ω1Let

0

2121

=−τ+++

=−−+=====

∫ A

t

OLOO

iOAA

edeiee

teeeeLCCRR

&

&

To get the input-output equation for ei, eO differentiate (2) to get

0 (3) =−++ AOOO eeee &&&&

2)(

(3)'0)1()3(

0)()2()1(2

+−

=∴

=−++⇒

=−−+⇒

petee

peepp

teeep

oiA

AO

iOA

LL

Use p operator to get



To get state-variable equations leave the inductor current

LOAO

iOAA

ieeeteeee

−−=⇒++−=⇒

&

&

(2))(2(1)

(2)0)0(

(1)0)(2

0LL

4434421&

LL&

=−τ+++

=−−+

∫ A

i

t

OLOO

iOAA

edeiee

teeee

Lfrom inductor element law

Substitute into (3)’ to get

)(223 )()223(

))(()1)(2(

))((2

)1

23

2

2

teeeeetpeeppp

etepeppp

etep

pep(p

iOOOO

iO

OiO

OiO

&&&&&&& =+++∴=+++

+=+++

++

=++

I/O form

OL e

dtdiL =


Obtain the Input-Output Model (cont’d)Example 6.4: Find the input-output equation for eO and ei(t).

KCL:

(2)0(1)0

22

231

LL

LL

=++=−++

CR

RRL

iiiiiii

(1) + (2): 0321 =++++ CRRRL iiiii



Differentiate (3) to get: )()( 12 teteee OL −+=∴

[ ] [ ] [ ] 01)(1)()(1)()(1

22

312\

112 =++++−++−+ OOOOO eCe

Rtee

Rtetee

Rtetee

L&&&&&&&&

Collecting terms gives:

[ ])()(1)(11)(11111212

311

1321

teteL

teRR

teR

eL

eRRR

eC OOO −+⎟⎟⎠

⎞⎜⎜⎝

⎛+−=+⎟⎟

⎠

⎞⎜⎜⎝

⎛+++ &&&&&

[ ] [ ]

[ ]

(3)01)(1

)()(1)()(1)0(

23

1

22

3

121

0 12

LL&

3214434421

444 3444 2144444 344444 21

=++++

+−++τ−++ ∫

C

RR

RL

iO

i

O

i

O

i

O

i

t

OL

eCeR

teeR

teteeR

dteteeL

i

0)()( 112 =−−+ teetee RO0)()( 12 =−−+ teetee LO


Obtain the State-Variable Model

:form followinghave willequations variable-state The . e.g. voltagescapacitor

of in terms voltagesnode expressingby avoided becan assuch termsCapacitor terms.integral avoid to variablestate as choosecan

one example, previous theoftion representa variable-state obtain the To

C

C

L

eCeC

i

&

&

LL

cc

eLdt

di

iC

e

1

1

=

=&

. )( outputsfor equations algebraic Write(5)

.1 :currentsinductor for equations state Write(4)

.1 :voltagescapacitor for equations state Write(3)

).,( inputs and , variables-state of in terms currentscapacitor for Solve (2)

. voltagescapacitor of in terms voltagesnode Write(1):Procedures

K

&

K

,,ie

eLdt

di

iC

e

(t) (t), ieiei

e

OO

LL

CC

ii

LCc

C

=

=


Obtain the State-Variable Model (cont’d)Example 6.7: Find the state-variable and output (eO) equations.

CLL

iCLC

iCLC

LCCi

eL

eLdt

di

tieR

iC

e

tieR

ii

ieR

iti

11

))(1(1

)(1

01)(

==

+−−=

+−−=

=+++−

&

or

[ ] ⎥⎦

⎤⎢⎣

⎡=

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡+⎥

⎦

⎤⎢⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡ −−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

L

C

iL

C

L

C

ie

e

tiCie

L

CRC

dtdidt

de

01

)(0

1

01

11

0

KCL:


Obtain the State-Variable Model (cont’d)Example 6.8: Find the state-variable equations

KCL:

B) (Node 01)(1

A) (Node 0)(1))((1

2221

2111

32

21

=+++−−

=−++−−

CCLCC

CCCCi

eR

iieeR

eeR

ieteR

2

2122

2111

11

11111

)(111111

32211

122111

CLL

CCLCC

iCCCC

eL

eLdt

di

eRR

eR

iC

iC

e

teR

eR

eRRC

iC

e

==

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+−+−==

⎥⎦

⎤⎢⎣

⎡++⎟⎟

⎠

⎞⎜⎜⎝

⎛+−==

&

&


Obtain the State-Variable Model (cont’d)or

)(00

1

010

11111

01111

11

2

1

232222

12211

2

1

teCR

iee

CRRCCR

CRRRC

dtdidt

dedt

de

i

B

L

C

C

A

L

C

C

32144444444 344444444 21

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

+⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−⎟⎟⎠

⎞⎜⎜⎝

⎛+−

⎟⎟⎠

⎞⎜⎜⎝

⎛+−

=

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

( )

( )

)(100

001

1111010

)(1

11y

1

2

1

1

212

1

23

212

2

1

2 te

Ri

ee

R

RRR

eteR

eR

ieeR

e

iie

i

L

C

C

Ai

CLCC

C

R

C

B

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

+⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−−−=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=


Obtain the State-Variable Model (cont’d)Example 6.9: Find the state-variable equations with output eO.

))(263(51

2 (2)

))(3(51

0222)((2)(1)(2)B node 02 (1)A node 0)(2))(( :KCL

0

teeii

eii

teeie

eieeteeeeii

ieeteee

iCLC

OLC

iCLO

OCCOiCO

LC

CCOiCO

+−=

+=⇒

+−−=

=++++−+⇒+∴=++−

=+++−+LL

LL


Obtain the State-Variable Model (cont’d)

)(11

))(263(101 C

DOLL

iCL

eeL

eLdt

di

teeie

−==

+−=∴ &

LDDL ieei5105:D nodefor KCL =⇒=+−

))(3(51

))(32(52

))(32(51

teei e

teeidtdi

teeie

iCLO

iCLL

iCLL

+−−=

+−−=

+−−=∴


Obtain the State-Variable Model (cont’d)Dependent state variables

0)(:KVL =−+ teee iBA 0)(:KCL =++− BAi iiti

State variable are not independent for these problems:In this case, we will get one state equation for each independent state variable.

Algebraic relationship

B



.1 ,

variable,-state theas choose figure hand-left above for theequation variable-state theFind

11

A CAiB

A

iC

eeee

e

=−= &

KCL: (Node A)

Example 6.10:

0))((11))((21

2 1=−+++−−

444 3444 21

&&

ABi

iAACAi teeR

eR

ieteC

⎥⎦

⎤⎢⎣

⎡++⎟⎟

⎠

⎞⎜⎜⎝

⎛+−−=

++⎟⎟⎠

⎞⎜⎜⎝

⎛+−−=∴

)(1)(111or

)(1)(11

22

212

1

22

2121

teR

teCeRR

eCC

e

teR

teCeRR

eCi

iiAAA

iiAAC

&&&

&&

Ae•

B



⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛+−+⎟⎟

⎠

⎞⎜⎜⎝

⎛+−

+=∴

+−=

⎥⎦

⎤⎢⎣

⎡+⎟⎟

⎠

⎞⎜⎜⎝

⎛+−

+=

+−

⎥⎦

⎤⎢⎣

⎡++⎟⎟

⎠

⎞⎜⎜⎝

⎛+−

+=

)(111111

)(let and

)(1111)(

)(1)(111

21

2

2122121

21

2

2212121

2

22

2121

teCC

CRRR

xRRCC

x

teCC

Cex

teR

eRRCC

teCC

Ce

teR

teCeRRCC

e

i

iA

iAiA

iiAA

&

&&

&&

Solve for

Note that one state variable is sufficient to represent the system dynamics.


Reading and ExerciseReading and Exercise• Reading

Chapter 6: 6.1-6.6

• Assignment #3Ex. 6.2, 6.19, 6.28, 7.6, 7.11, 7.15 (to be

handed in)

Due: Friday, 27/7/07 at lecture.

Ex. 6.3, 6.6, 6.29, 7.3, 7.20 (for your practice, no need to hand in)

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