ECE 330 POWER CIRCUITS AND …...POWER CIRCUITS AND ELECTROMECHANICS LECTURE 19 DYNAMICS OF LUMPED MECHANICAL SYSTEMS Acknowledgment-These handouts and lecture notes given in class

Copyright © 2017 Hassan Sowidan

ECE 330

POWER CIRCUITS AND ELECTROMECHANICS

LECTURE 19

DYNAMICS OF LUMPED MECHANICAL SYSTEMS

Acknowledgment-These handouts and lecture notes given in class are based on material from Prof. Peter

Sauer’s ECE 330 lecture notes. Some slides are taken from Ali Bazi’s presentations

Disclaimer- These handouts only provide highlights and should not be used to replace the course textbook.

10/24/2017



So far we have studied the electrical subsystem and its

interaction with the lossless magnetic field system, which

resulted in the force of electric origin acting on the

mechanical system. We next study the dynamics of the

mechanical system and finally get the overall dynamic

model.

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Lumped elements of mechanical systems:

• Masses: store kinetic energy(KE)

• Springs: store potential energy(PE)

• Dashpots: the dissipative elements

Elements of electrical systems:

- Capacitance C, inductance L: store energies

- Resistance R: dissipate energy

Instead of KCL and KVL, we have Newton’s law to write

the equations of motion in a mechanical system.

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MASS-SPRING SYSTEM

For a mass suspended from a spring of stiffness K.

The gravitational force in equilibrium position is

is the stretching of the spring due to the weight W

W Mg is balanced by spring force K

=W

Mg

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MASS-SPRING SYSTEM

Displacement x downward causes a force in the spring to

pull the mass upward.

Newton’s law: acceleration force in the positive x direction

is equal to the algebraic sum of all the forces acting on the

mass in the positive x direction.

= = 0Mx Kx or Mx Kx

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MASS-SPRING SYSTEM WITH DISSIPATIVE ELEMENT

The same example with dashpot arrangement.

f(t) is the applied force.

Application of Newton’s law

1 2= ( ) K K BMx f t f f f

= BB

dxf B

dt

1 2= ( )dx

Mx f t K x K x Bdt

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EXAMPLE

The mechanical equations for the two degrees of freedom

system.x1 and x2 are measured from the equilibrium position

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EXAMPLE

The free body diagram

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EXAMPLE

Define

Applying Newton’s law to each of the two masses, we get

2 1 =x x x

1 1 1 2 2 1

2 2 1 1 1 1 1

= ( ) ( )

( )

M x f t K x x

B x x B x K x

2 2 2 2 2 1

2 2 1 3 2 3 2

= ( ) ( )

( )

M x f t B x x

K x x B x K x

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EXAMPLE

Suppose f1(t) and f2(t) are unit step functions of magnitudes

and respectively.

In steady state the derivatives of and are zero.

1F̂2F̂

1 1 21 2 10 = ( )ss ss ssK x K x x F

2 1 23 2 20 = ( )ss ss ssK x K x x F

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1x 2x


EXAMPLE

We solve for and which will be the new

equilibrium positions

1 2 2 11

2 2 3 2 2

ˆ=

ˆ

ss

ss

K K K Fx

K K K x F

1

ssx2

ssx

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STATE SPACE MODELS

The electrical and mechanical equations are coupled.

State space models are writing the coupled electrical and

mechanical equations in the form of a set of first order

deferential equations.

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EXAMPLE

Write the electrical and mechanical equations of the motion

for the system and put them in the state space form.

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1 2= ( )dx

Mx f t K x K x Bdt


EXAMPLE

We have derived the flux linkages in example 4.7:

Note and

cc

A

0

2( )g

xx

A

2

=( )c g

N i

x

2

0

=2c

N i

x

A A

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EXAMPLE

The co-energy is:

Where,

The force of electrical origin:

=e mWf

x

2 2 2 2

0

=2( ( )) 2 ( )

i

m

c g

N i N iW di

R R x R x

( ) = ( )c gR x R R x

2 2

2

0

0

=2

e

c

N if

xA R

A

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EXAMPLE

Equations for the electrical side:

=s

dv i R

dt

2 2

2

0

0 0

2=

2 2s

c c

N di N i dxv iR

dt A dtx xR RA A

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EXAMPLE

Mechanical equation:

Let be the static equilibrium position of the moving

member. This is the same as the upstretched length of the

spring with . We assume , i.e., the two pieces

do not touch each other

= 0i > 0

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EXAMPLE

If we measure the position of the moving member from the

equilibrium position, then the mechanical equations have

the variable

( )x 2 2

2 2

( )=

d x d x

dt dt

( )=

d x dx

dt dt

2 2 2

22

0

0

( ) = =2

e

c

d x dx N iM K x B f

dt dt xA R

A

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EXAMPLE

State space model:

A set of three first-order differential equations. Define the

state variables as

,

, ( ( ))dx

x denoted by velocity and idt

= ...................(1)dx

dt

2

2=

d d x

dt dt

2 2

2

0

0

1= ( ) ..........(2)

2c

d N iK x B

dt M xA R

A

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2

0

( ) =2

c

NL x

xR

A

EXAMPLE

2

2

0

0

1 2= .........(3)

( ) 2s

c

di N iiR

dt L x AxR

A

Where

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EXAMPLE

Equations 1,2,3 are first-order differential equations and are

called the state space equations of electromechanical system.

The initial conditions are , , and .

is a forcing function. Generally, we can also redefine

the state variables mathematically as , ,

, and

(0) =x (0) = 0 (0) = 0i

( )sv t

1=x x2= x

3=i x =sv u

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EXAMPLE

The state space equations are then of the form

1 1 1 2 3= ( , , )x f x x x

2 2 1 2 3= ( , , )x f x x x

3 3 1 2 3= ( , , , )x f x x x u

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EXAMPLE

In the matrix form

Where

are the initial conditions. In this case ,

. Given we can

integrate to obtain the time domain response.

1

2

3

(0)

= ( , ), (0) = (0) , = ,

(0)

x

x f x u x x u u a scalar

x

1 1

2 2

3 3

= , =

x f

x x f f

x f

(0)x 1(0) =x

2 3(0) = (0) = 0x x (t) and knowing x(0),u

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EQUILIBRIUM POINTS

Consider the equation , if ,

by setting we get algebraic equations

This may have several solutions for x denoted , ,….

are called static equilibrium solutions.

( , )x f x u

ˆ0 ( , )f x u

ˆ.u is const u

= 0x

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1

ex 2

ex


EQUILIBRIUM POINTS

Graphically, setting derivatives equal to zero, ,

Since , x is the only unknown

Plot and

= 0e

2 2

2

0

0

( )( ) = = ( , )

2

ee e

c

N iK x f i x

xA R

A

e svi

R

( )K x ( , )e ef i x

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NUMERICAL INTEGRATION

There are two kinds of methods: a) explicit and b) implicit.

Euler’s method is an explicit one. Euler’s method is easier

to implement for small systems.

For large systems, the implicit method is better because it is

numerically stable.

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EXPLICIT METHOD(EULER’S METHOD)

The equation is

are vectors. We split the time interval into

equally spaced time instants

From

0= ( , ), (0) =x f x u x x

, and ux f

0 1 2 1, , , , , ,n nt t t t t

1n nt to t

1 1( ) = ( , )

t tn n

t tn n

x t dt f x u dt

1 1( ) ( ) = ( ) ( ( ), ( ))n n n n n nx t x t t t f x t u t

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EXPLICIT METHOD (EULER’S METHOD)

Where

is the integration step size

are known from the previous time instant

1( ) = ( ) [ ( ( ), ( ))]n n n nx t x t t f x t u t

1= = , = 0,1,2n nt t t constant n

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

n nx x x x

t

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APPLICATION FOR HIGHER-ORDER SYSTEMS

The same procedure is applicable.

Let

Application of Euler’s method gives

Superscript notation is used to denote the time instant

,

1 1 1 2 3= ( , , , )x f x x x t

2 2 1 2 3= ( , , , )x f x x x t

3 3 1 2 3= ( , , , )x f x x x t

( 1) ( ) ( ) ( ) ( )

1 1 1 1 2 3= ( ( , , , ))n n n n n

nx x t f x x x t

( 1) ( ) ( ) ( ) ( )

2 2 2 1 2 3= ( ( , , , ))n n n n n

nx x t f x x x t

( 1) ( ) ( ) ( ) ( )

3 3 3 1 2 3= ( ( , , , ))n n n n n

nx x t f x x x t

( ) = nn n t t 1( 1) ( 1) = nn n t t

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ECE 330 POWER CIRCUITS AND …...POWER CIRCUITS AND ELECTROMECHANICS LECTURE 19 DYNAMICS OF LUMPED MECHANICAL SYSTEMS Acknowledgment-These handouts and lecture notes given in class