1

Electromechanical System Dynamics, energy Conversion, and Electromechanical Analogies

Modeling of Dynamic SystemsModeling of dynamic systems may be done in several ways:

Use the standard equation of motion (Newton’s Law) for mechanical systems

Use circuits theorems (Ohm’s law and Kirchhoff’s laws: KCL and KVL)

Another approach utilizes the notation of energy to model the dynamic system (Lagrange model).

2

Mathematical Modeling and System DynamicsNewtonian Mechanics: Translational Motion

• The equations of motion of mechanical systems can be found using Newton’s second law of motion. F is the vector sum of all forces applied to the body; a is the vector of acceleration of the body with respect to an inertial reference frame; and m is the mass of the body.

• To apply Newton’s law, the free-body diagram in the coordinate system used should be studied.

∑ = aF m

3

Translational Motion in Electromechanical Systems

• Consideration of friction is essential for understanding the operation of electromechanical systems.

• Friction is a very complex nonlinear phenomenon and is very difficult to model friction.

• The classical Coulomb friction is a retarding frictional force (for translational motion) or torque (for rotational motion) that changes its sign with the reversal of the direction of motion, and the amplitude of the frictional force or torque are constant.

• Viscous friction is a retarding force or torque that is a linear function of linear or angular velocity.

Fcoulomb :Force

4

Newtonian Mechanics: Translational Motion

• For one-dimensional rotational systems, Newton’s second law of motion is expressed as the following equation. M is the sum of all moments about the center of mass of a body (N-m); J is the moment of inertial about its center of mass (kg/m2); and α is the angular acceleration of the body (rad/s2).

αjM =

5

The Lagrange Equations of Motion• Although Newton’s laws of motion form the fundamental foundation for the study

of mechanical systems, they can be straightforwardly used to derive the dynamics of electromechanical motion devices because electromagnetic and circuitry transients behavior must be considered. This means, the circuit dynamics must be incorporated to find augmented models.

• This can be performed by integrating torsional-mechanical dynamics and sensor/actuator circuitry equations, which can be derived using Kirchhoff’s laws.

• Lagrange concept allows one to integrate the dynamics of mechanical and electrical components. It employs the scalar concept rather the vector concept used in Newton’s law of motion to analyze much wider range of systems than F=ma.

• With Lagrange dynamics, focus is on the entire system rather than individual components.

• Γ, D, Π are the total kinetic, dissipation, and potential energies of the system. qiand Qi are the generalized coordinates and the generalized applied forces (input).

ii

ii

i

Qdqd

qd

dDdqd

qd

ddtd

=Π

++Γ

−

Γ

..

6

Electrical and Mechanical Counterparts

ResistorRi2

Damper / Friction0.5 Bv2

Dissipative

Capacitor0.5 Cv2

Gravity: mghSpring: 0.5 kx2

Potential

Inductor0.5 Li2

Mass / Inertia0.5 mv2 / 0.5 jω2

Kinetic

ElectricalMechanicalEnergy

7

Mathematical Model for a Simple Pendulum

( )θ

θ

cos1 :isenergy potential The21

21 :is bob pendulum theofenergy kinetic The

2.2

−==Π

==Γ

mglmgh

lmmv

y0

x

y

x

mgmg cosθ

l

θ

Ta, θ

8

Electrical Conversion

InputElectrical Energy

OutputMechanical

Energy

CouplingElectromagnetic

Field

Irreversible Energy Conversion

Energy Losses

Energy Transfer in Electromechanical Systems

flux. theis where;di WWhere

),(),( :is nt,displacemeangular andcurrent of

function a as torque,neticelectromag themotion, rotationalFor

ic ψψ

θθ

θ

∫=

=didW

iT ce

9

Electromechanical Analogies

• From Newton’s law or using Lagrange equations of motions, the second-order differential equations of translational-dynamics and torsional-dynamics are found as

dynamics) (Torsional )(

dynamics) onal(Translati )(

2

2

2

2

tTkdtdB

dtdj

tFxkdtdxB

dtxdm

asm

asv

=++

=++

θθθ

10

For a series RLC circuit, find the characteristic equation and define the analytical relationships between the characteristic roots and circuitry parameters.

LCLR

LRs

LCLR

LRs

LCs

LRs

dtdv

Li

LCdtdi

LR

dtid a

122

122

are roots sticcharacteri The

01

11

2

2

2

1

2

2

2

−

+−=

−

−−=

=++

=++

11

Resistance, R (ohm)

v(t) R

i(t)

)(1)(

)()()(Current

)( voltageAppied

tvR

ti

tRitvti

tv

=

=

12

Inductance, L (H)

v(t) L

i(t)

∫=

=

t

tdttv

Lti

dttdiLtv

titv

0

)(1)(

)()(

)(Current )( voltageAppied

13

Capacitance, C (F)

v(t) C

i(t)

dttdvCti

dttiC

tv

titv

t

t

)()(

)(1)(

)(Current )( voltageAppied

0

=

= ∫

14

Translational Damper, Bv (N-sec)

Fa(t)

x(t)

∫=

==

=

=

t

ta

v

mma

am

ma

dttFB

tx

dttdxBtvBtF

tFB

tv

tvBtFtxtvtF

0

)(1)(

)()()(

)(1)(

)()((m) )(position Linear (m/sec) )(ocity Linear vel

Newtonin )( force Appied a

Bm

15

Translational Spring, k (N)

Fa(t)

x(t)

∫=

==

=

=

t

tsa

a

s

as

sa

dttvktF

dttdF

kdttdxtv

tFk

tx

txktFtxtvtF

0

)()(

)(1)()(

)(1)(

)()((m) )(position Linear (m/sec) )(ocity Linear vel

Newtonin )( force Appied a

16

Rotational Damper, Bm (N-m-sec/rad)

Fa(t)

θ (t)

∫=

==

=

=

t

ta

m

mma

am

ma

dttTB

t

dttdBtBtT

tTB

t

tBtTt

ttT

0

a

)(1)(

)()()(

)(1)(

)()((rad) )(nt displacemeAngular

(rad/sec) )(locity Angular vem)-(N )( torqueAppied

θ

θω

ω

ωθ

ω

ω (t)

Bm

17

Rotational Spring, ks (N-m-sec/rad)

Fa(t)

θ (t)

∫=

==

=

=

t

tsa

a

s

as

ma

dttktT

dttdT

kdttdt

tTk

t

tBtTt

ttT

0

)()(

)(1)()(

)(1)(

)()((rad) )(nt displacemeAngular

(rad/sec) )(locity Angular vem)-(N )( torqueAppied a

ω

θω

θ

θθ

ω

ω (t)

ks

18

Mass Grounded, m (kg)

Fa(t)

x (t)

∫=

==

t

ta

a

dttFm

tv

dttxdm

dtdvmtF

txtvtT

0

)(1)(

)()(

(m) )(position Linear (m/sec) )(ocity Linear vel

m)-(N )( torqueAppied

2

2

a

v (t)

m

19

Mass Grounded, m (kg)

Fa(t)

θ (t)

∫=

==

t

ta

a

dttTJ

t

dttdJ

dtdJtT

tt

tT

0

2

2

a

)(1)(

)()(

(rad) )( nt displacemeAngular (rad/sec) )( locity Angular ve

m)-(N )( torqueAppied

ω

θω

θω

ω (t)

m

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Steady-State Analysis• State: The state of a dynamic system is the smallest set of variables

(called state variables) so that the knowledge of these variables at t= t0, together with the knowledge of the input for t ≥ t0, determines the behavior of the system for any time t ≥ t0.

• State Variables: The state variables of a dynamic system are the variables making up the smallest set of variables that determine the state of the dynamic system.

• State Vector: If n state variables are needed to describe the behavior of a given system, then the n state variables can be considered the n components of a vector x. Such vector is called a state vector.

• State Space: The n-dimensional space whose coordinates axes consist of the x1 axis, x2 axis, .., xn axis, where x1, x2, .., xn are state variables, is called a state space.

• State-Space Equations: In state-space analysis we are concerned with three types of variables that are involved in the modeling of dynamic system: input variables, output variables, and state variables.

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State Variables of a Dynamic System

Dynamic SystemState x(t)

u(t) Input y(t) Output

x(0) initial condition

dynamics thedescribing equations theand inputs, excitation thestate,present given the

system, a of response future thedescribe variablesstate The

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Electrical Example: An RLC Circuit

C

L

Ru(t)

iL

iC

vC

( ) ( )

Lc

c

cL

LC

itudtdv

Ci

txtxCvLi

tixtvx

−+==

+=

==

)(

junction at the KCL USEnetwork theofenergy

initial total theis )( and )(2/12/1

)( );(

0201

2221

ξ

23

The State Differential Equation

Equation) alDifferenti (StateBu Axx.

+=

Equation)(Output Du Cxy +=

+

=

++++++=

++++++=

++++++=

mnmn

m

nnnnn

n

n

n

mnmnnnnnnn

mmnn

mmnn

u

u

bb

bb

x

xx

aaa

aaaaaa

x

xx

dtd

ububxaxaxax

ububxaxaxax

ububxaxaxax

.....

................

.

. . .

.

......

......

......

1

1

1112

1

21

22221

11211

2

1

112211

.212122221212

.111112121111

.

State Vector

matrix nsmission direct tra :D matrix;Output :Cmatrixinput :B matrix; State :A

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The Output Equation

Equation) alDifferenti (StateBu Axx.

+=

Equation)(Output Du Hxy +=

x

x

xx

hhh

hhhhhh

y

yy

y

xaxaxayxhxhxhyxhxhxhy

nbnbb

n

n

b

nbnbbb

nn

nn

H.

. . .

.

...

......

2

1

21

22221

11211

2

1

2211

22221212

12121111

=

=

=

+++=+++=+++=

matrix nsmission direct tra :D matrix;Output :Hmatrixinput :B matrix; State :A

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Example 1: Consider the given series RLC circuit. Derive the differential equations that map the circuitry dynamics.

V(t)

R

L

C

i(t)

( ))(1

1

)(

tvRivLdt

di

iCdt

dv

tvRivdtdiL

idtdv

C

c

c

c

c

+−−=

=

+−−=

=

26

Example 2: Using the state-space concept, find the state-space model and analyze the transient dynamics of the series RLC circuit.

BuAxvL

iv

LR

L

C

dtdidtdv

dtdxdtdx

dtdx

tvtitxtvtx

tvRivLdt

di

iCdt

tdv

ac

c

c

c

+=

+

−=

=

=

==

+−−=

=

10

- 1

1 0

control theis )(states) theare (These )()( );()(

))((1

1)(

2

1

21

27

Continue with Values..• Assume R = 2 ohm, L = 0.1 H,

and C = 0.5 F, find the following coefficients.

• The initial conditions are assumed to be vc(t0)=vc0=15 V; and I (t0) = i0 = 5 A.

• Let the voltage across the capacitor be the output; y(t)= vc(t). The output equation will be

• The expanded output equation in y

• The circuit response depends on the value of v (t)

=

=

100

B and 20- 10-2 0

A

=

=

515

20

100 x

xx

[ ] [ ]0 1H ;H 0 1 ==

= x

iv

y c

[ ] [ ] DuHxviv

y ac +=+

= 0 0 1