ece 330 power circuits and electromechanics lecture …

Copyright © 2017 Hassan Sowidan

ECE 330

POWER CIRCUITS AND ELECTROMECHANICS

LECTURE 20

STABILITY OF ELECTROMECHANICAL

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.

11/6/2017


LINEARIZATION

• When we write the equations on the electrical and

mechanical side, we have complete dynamic

description of the system.

• Then we put them in the state space form as a set of

first-order equations for analysis

• Setting , we get the algebraic equations

, which may have several solutions

called static equilibrium solutions.

( , )x f x u

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0x

ˆ0 ( , )f x u

1 2, ,....e ex x


LINEARIZATION

• By integrating , we can obtain

the time domain response.

• a linearized analysis is helpful to determine if the

equilibrium point is stable or not by merely

computing the eigenvalues of a matrix.

• For large disturbances, a direct method using energy

functions can in some cases be used to assess

stability without time-domain simulations

( , )x f x u

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0(0)x x


LINEAR CIRCUIT MODELING

FIRST-ORDER EXAMPLE

• The circuit differential equation is:

• State space form

Solution:

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R

EL

t=0 +

V

-

i

1di Ri v

dt L L

( ) t

ssi t ae i

div i R L

dt

(0) 0, (0) 0 0, 0i v for t v E for t


FIRST-ORDER EXAMPLE

• DC steady state (equilibrium):

,

• How to solve for λ (eigenvalue)?

• How do we solve for a ? By using initial condition.

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10

Ri E

L L

| | | | | |R R R

A I I AL L L

ss

Ei

R

(0) 0

( )R

tL

E Ei a a

R R

E Ei t e

R R


SECOND-ORDER EXAMPLE

• The circuit differential equations are:

,

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R

EL

t=0 + v -i

C

1 1

1

di Ri v E

dt L L L

dvi

dt C

0

(0) 0

(0)

i

v V

diE i R L v

dt

dvi

dt



Then,

The solutions become:

The system is stable if Re{λ1, λ2}<0

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11

100

di R

idt L LEL

dv v

Cdt

1 2

1 2

( )

( )

t t

ss

t t

ss

i t ae be i

v t ce be v



At equilibrium:

Finding the eigenvalues:

The eigenvalues are stable since the real part is <0.

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1 1 10 , 0

Ri v E i

L L L C

0,ss ssi v E

2

1

1| | 0

1

R

RL LI A

L LC

C

2

1,2

1 4

2 2

R R

L L LC



Solve the following equations to find a, b, c, and d

,

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1 2

0

1 2

0

1 1(0) (0)

1(0)

di Ri v E a b

dt L L L

dvi c d

dt C

01 2

01 1 a

VEb

L L

0

1 2

1 1

0

c V E

d

00 : (0),At t a b i c d E


LINEARIZATION

A nonlinear system can be expressed as:

The power system may be subjected to small

disturbances which may lead to oscillations of power

in transmission lines. If not properly damped it may

lead to cascading outages and major disturbances

(faults).

( , )x f x u

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LINEARIZATION

• Linearization is applied around an equilibrium point

x e and an input .

• For a scalar x and u , the Taylor series expansion is:

ˆ

ex x x

u u u

u

ˆ ˆ( , ) ( , ) ( ) ( ) . . .e e

e e

x x

f ff x u f x u x x u u h o t

x u

ˆ. ., ,ei e x x x and u u u

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LINEARIZATION

• Then,

• For a second-order system:

e ex x

f fx x u

x u

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

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

1 1 1 2 2 2ˆ, , .e eLet x x x x x x u u u

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LINEARIZATION

• Linearizing around the equilibrium point, we get:

• For an n th- order system, a similar approach is

followed.

1 1 1

1 21 1

2 2 22 2

1 2

e e e

ee e

x x x

xx x

f f f

x x ux xu

x x ff f

ux x

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LINEARIZATION

• The eigenvalues of A determine the stability for

small disturbances around the equilibrium point, and

obtained by solving for .

• If all the eigenvalues lie in the left-half plane (i.e.,

real parts < 0 ), we say the system is stable. If an

eigenvalue lies in right-half plane, then the system is

unstable. If a pair of complex eigenvalues lies on the

imaginary axis, the system is marginally stable.

[ ] =| |= 0det A I A I

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EXAMPLE 1

• Linearize the following nonlinear system:

• X2e = 0, x1

e = sin-1(0.866) = 60o, 120o, etc. ; and:

• Eq. pt. 1: Unstable Eq. pt. 2: Stable

1 2

2 1 2sin( ) 0.866

x x

x x x

1 1

2 1 2

0 1

cos( ) 1

x x

x x x

1

1 2

10 1

, 1cos( ) 1 1

2

0.366 0, 1.366 0

A A Ix

1 2

1

11

2

0.5 0.5 , 0.5 0.5

A I

j j

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EXAMPLE 2

• Linearize the following nonlinear system:

• Equilibrium points

Linearization

1 2

2 1 26sin 4 3

x x

x x x

2

1

0

0 6sin 3

e

e

x

x

2

1

1

0

sin 0.5

5,

6 6

e

e

x

x

1 2

2 1 2( 6sin ) 4e

x x

x x x x

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EXAMPLE 2

Compute eigenvalues

Characteristic equation

0 1

6cos( ) 4eA

x

21

4 6cos6cos 4

e

eA I x

x

2 4 6cos 0ex

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EXAMPLE 2

Check stability of two equilibrium points:

1- 2-

24 14 24cos

2 2

ex

1 2, 06

e ex x

12 16 24cos

2 6

52

2j stable

1 2

5, 0

6

e ex x

1 52 16 24cos

2 6

382

2j unstable

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ece 330 power circuits and electromechanics lecture …

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