ECEN 667

Power System Stability

1

Lecture 22: Power System Stabilizers

Prof. Tom Overbye

Dept. of Electrical and Computer Engineering

Texas A&M University, overbye@tamu.edu

Special Guest Lecturer: TA Iyke Idehen

Announcements

• Read Chapter 8

• Homework 7 is posted; due on Tuesday Nov 28

• Final is as per TAMU schedule. That is, Friday Dec 8

from 3 to 5pm

• We’ll be doing power system stabilizers today, and will

pickup with modal analysis next time

2

Overview of a Power System

Stabilizer (PSS)

• A PSS adds a signal to the excitation system to improve

the rotor damping

– A common signal is proportional to speed deviation; other

inputs, such as like power, voltage or acceleration, can be used

– Signal is usually generated locally (e.g. from the shaft)

• Both local mode and inter-area mode can be damped.

When oscillation is observed on a system or a planning

study reveals poorly damped oscillations, use of

participation factors helps in identifying the machine(s)

where PSS has to be located

• Tuning of PSS regularly is important

3

Block Diagram of System with a PSS

4Image Source: Kundur, Power System Stability and Control

Power System Stabilizer Basics

• Stabilizers can be motivated by considering a classical

model supplying an infinite bus

• Assume internal voltage has an additional component

• This will add additional damping if sin(d) is positive

• In a real system there is delay, which requires

compensation 5

dts

dd

0

0

2sins

Md ep

E VH dT D

dt X X

d

orgE E K

Example PSS

• An example single input stabilizer is shown below

(IEEEST)

– The input is usually the generator shaft speed deviation, but it

could also be the bus frequency deviation, generator electric

power or voltage magnitude

6

VST is an

input into

the exciter

The model can be

simplified by setting

parameters to zero

Example PSS

• Below is an example of a dual input PSS (PSS2A)

– Combining shaft speed deviation with generator electric

power is common

– Both inputs have washout filters to remove low frequency

components of the input signals

7

IEEE Std 421.5 describes the common stabilizers

Washout Parameter Variation

• The PSS2A is the most common stabilizer in both the

2015 EI and WECC cases. Plots show the variation in

TW1 for EI (left) and WECC cases (right); for both the

x-axis is the number of PSS2A stabilizers sorted by

TW1, and the y-axis is TW1 in seconds

8

PSS Tuning: Basic Approach

• The PSS parameters need to be selected to achieve the

desired damping through a process known as tuning

• The next several slides present a basic method using a

single machine, infinite bus (SMIB) representation

• Start with the linearized differential, algebraic model

with controls u added to the states

• If D is invertible then

9

Δx = AΔx BΔy EΔu

0 = CΔx+DΔy

-1

sysΔx = A BD C Δx EΔu A Δx EΔu

PSS Tuning: Basic Approach

• Low frequency oscillations are considered the

following approach

• A SMIB system is setup to analyze the local mode of

oscillation (in the 1 to 3 Hz range)

– A flux decay model is used with Efd as the input

• Then, a fast-acting exciter is added between the input

voltage and Efd

• Certain constants, K1 to K6, are identified and used to

tune a power system stabilizer

10

SMIB System (Flux Decay Model)

• SMIB with a flux decay machine model and a fast

exciter

11

~

)2/()( d j

qd ejII 0V

ejXeR

jteV

' '

' ' '

'

[ ( ( ) )]

( ( ) )

pu s

pu M q q d d d q pu

q q d d d fd

do

1T E I X X I I D

2H

1E E X X I E

T

d

Stator Equations

• Assume Rs=0, then the stator algebraic equations are:

' '

( /2)

( /2)

sin( ) 0 (1)

cos( ) 0 (2)

( ) (3)

(4)

Expand the right hand side of (4)

sin( ) cos( ) (5)

sin( ) cos( )

(1) (2)

q q t

q t d d

j j

d q t

j j

d q t

d q t t

d t q t

X I V

E V X I

V jV e V e

V jV V e e

V jV V jV

V V and V V

and beco

d

d

d

d

d d

d d

' '

0 (6)

0 (7)

q q d

q q d d

me X I V

E V X I

12

Network Equations

• The network equation is (assuming zero angle at the infinite bus and no local load)

( / )

( / )

( / )

( )( )

( )( )

j 2

d qj 2

d q

e e

j 2

d q

d q

e e

V jV e V 0I jI e

R jX

V jV V eI jI

R jX

d

d

d

Simplifying,

~

)2/()( d j

qd ejII 0V

eXeR

jteV

sin

cos

e d e q d

e d e q q

R I X I V V

X I R I V V

d

d

13

Complete SMIB Model

' ' '

'

' '

' '

( ( ) )

note in the book

[ ( ( ) )]

sin

cos

q q d d d fd

do

pu s pu

s

pu M q q q d d q pu

q q d

q q d d

e d e q d

e d e q q

1E E X X I E

T

1T E I X X I I D

2H

X I V 0

E V X I 0

R I X I V V

X I R I V V

d

d

d

Stator equations

Network equations

~

)2/()( d j

qd ejII

0V

ejXeR

jteV

Machine equations

14

Linearization of SMIB Model

''

Step 1: Linearize the stator equations

Step 2: Linearize the network equations

cos

sin

Step 3: Equate the

d dq

q q qd

d de e

q qe e

V I 00 X

V I EX 0

V IR X V

V IX R V

dd

d

''

righthand sides of the above equations

( ) cos

( ) sin

de e q

q qe d e

I 0R X X V

I EX X R V

dd

d

Notice this is equivalent to a generator at the infinite

bus with modified resistance and reactance values15

Linearization (contd)

• Final Steps involve

1. Linearizing Machine Equations

2. Substitute (1) in the linearized equations of (2).

'

'

'

Solve for ,

cos sin ( )( )

sin cos ( )

The determinant is ( )( )

d q

d e q e q e q

q e e d e

2

e e q e d

I I

I X X R V V X X E11

I R R V V X X

R X X X X

d d

d d d

( , , , , ) (2)d q fd MI I E Tx f x

16

Linearization of Machine

Equations

• Substitute (3) in (2) to get the linearized model.

Symbolically we have

'

' '

' '

'

' ' '

( )

( )

( ) ( )

do

oq s

do do

1

Tq q

s

I Dpu pu

2H 2H

1 1d dT T

d fd

q Mo o o1 1 1 1q d q d q q q2H 2H 2H 2H

0 0E E

0 0 1

0

X X 0 0I E

0 0 0 0I T

I X X X X I E 0

d d

( )

( )

d q

d q

I 2

3

x A x B u C

I D x

17

Linearized SMIB Model

• Excitation system is not yet included.

• K1 – K4 constants are defined on next slide

' '

' ' '

'

4q q fd

3 do do do

s pu

s2 1pu q pu M

K1 1E E E

K T T T

DK K 1E T

2H 2H 2H 2H

d

d

d

18

K1 – K4 Constants

• K1 – K4 only involve machine and not the exciter.

}]sincos)}{(){(

}cossin)){(([1

)())((1

cossin)()(

))((1

1

'''

'

1

'''

2

'

4

'

3

dd

dd

dd

eed

o

q

o

dqd

eeqqd

o

q

o

qe

o

dqdeeqqd

o

q

o

q

eeqdd

eqdd

RXXEIXXV

RXXXXVIK

ERIXXRXXXXIIK

RXXXXV

K

XXXX

K

19

Manual Control

• Without an exciter, the machine is on manual control.

• The previous matrix will tend to have two complex

eigenvalues, corresponding to the electromechanical

mode, and one real eigenvalue corresponding to the flux

decay

• Changes in the operating point can push the real

eigenvalue into the right-hand plane

– This is demonstrated in the following example

20

Numerical Example

• Consider an SMIB system with Zeq = j0.5, Vinf = 1.05,

in which in the power flow the generator has

S = 54.34 – j2.85 MVA with a Vt of 115

– Machine is modeled with a flux decay model with (pu, 100

MVA) H=3.2, T'do=9.6, Xd=2.5, Xq=2.1, X'd=0.39, Rs=0, D=0

21

Saved as case B2_PSS_Flux

Initial Conditions

• The initial conditions are

22

( / ) .( ) .

.

( ) ( ) .

( . )( . )

. .

j j 2

G d q

j j

t s q G

1 15 1 05 0I e I jI e 0 5443 18

j0 5

0 angle of E where E V e R jX I e

E 1 15 j2 1 0 5443 18

1 4788 65 5

d

d

( / )

( / )

. . ,

. , . .

.

Hence, . , .

j j 2

d q G

d q

j j 2

d q

d q

I jI I e e 0 5443 42 48

I 0 4014 and I 0 3676

V jV Ve e 1 39 48

V 0 7718 V 0 6358

d

d

SMIB Eigenvalues

• With the initial condition of Q = -2.85 Mvar the SMIB

eigenvalues are -0.223 j7.374, -0.0764

• Changing the Q to 20 Mvar gives eigenvalues of

-0.181 j7.432, -0.171

• Change the Q to –20 Mvar gives eigenvalues of

-0.256 j6.981, +0.018

23

This condition

is unstable

in Eqp

Including Terminal Voltage

• The change in the terminal voltage magnitude also needs

to be include since it is an input into the exciter

q

t

o

q

d

t

o

dt

q

o

qd

o

dtt

qdt

qdt

VV

VV

V

VV

VVVVVV

VVV

VVV

222

222

22

refVrefV

tV

tV

Differentiating

24

Computing

• While linearizing the stator algebraic equations, we had

• Substitute this in expression for Vt to get

'

''

d q q

q qd

V 00 X E

V EX 0 d

'

'

'

'

[ sin cos ( )]

[ ( cos ( )sin )]

( )

t 5 6 q

o

d5 q e d e

t

o

q

d e q e

t

o ooq qd

6 q e d q e

t t t

V K K E

V1K X R V V X X

V

VX R V V X X

V

V VV1K X R X X X

V V V

d

d d

d d

25

Heffron–Phillips Model

(from 1952 and 1969 )

• Add a fast exciter with a single differential equation

• Linearize

• This is then combined with the previous three differential equations to give

( )A fd fd A ref tT E E K V V

'

( )

( )

A fd fd A ref t

A fd fd A ref A 5 6 q

T E E K V V

T E E K V K K K Ed

[ ]

sys

T

M refT V

x A x BΔu

ΔuSMIB Model

26

Block Diagram

• K1 to K6 are affected by system loading

27

On diagram

v is used to

indicate what

we've been

calling pu

Add an EXST1 Exciter Model

• Set the parameters to KA = 400, TA=0.2, all others zero

with no limits and no compensation

• Hence this simplified exciter is represented by a single

differential equation

28

( )A fd fd A REF t sT E E K V V V

Vs is the input

from the stabilizer,

with an initial

value of zero

Initial Conditions (contd)

• From the stator algebraic equation,' '

. ( . )( . ) .

q q d dE V X I

0 63581 0 39 0 4014 0 7924

' '

' '

( )

0.7924 (2.5 0.39)0.4014 1.6394

1.63941 1.0041

400

377, ( )

(0.7924)(0.3676) (2.1 0.39)(0.4014)(0.3676)

0.5436 ( sin ) /

fd q d d d

fd

REF t

A

s M q q q d d q

t e

E E X X I

EV V

K

T E I X X I I

checks with VV X

29

SMIB Results

• Doing the SMIB gives a matrix that closely matches the

book’s matrix from Example 8.7

– The variable order is different; the entries in the w column are

different because of the speed dependence in the swing

equation and the Norton equivalent current injection

30

The values are

different from the

book because of

speed dependence

included in the

stator voltage

and swing

equations

Computation of K1 – K6 Constants

• Calculating the values with the formulas gives

'

'

'

' ' '

1 5 6

( )( ) .

( )( ).

.

( )[( )sin cos ] .

[ ( )( ) ( ) ] .

Similarly K ,K ,and K are calculat

2

e e q e d

d d q e

3

3

d d4 q e e

o o o o

2 q q d q q e e d q d e q

R X X X X 2 314

X X X X11 3 3707

K

K 0 296667

V X XK X X R 2 26555

1K I I X X X X R X X I R E 1 0739

d d

ed as

. , . , .1 5 6K 0 9224 K 0 005 K 0 3572

31

Damping of Electromechanical Modes

• Damping can be considered using either state-space

analysis or frequency analysis

• With state-space analysis the equations can be written

as

The change in Efd comes from the exciter

32

' '

4

3 do do doq q

s fd

pu s pu2 1

K1 10

K T T TE E

0 0 0 E

D 0K K

2H 2H 2H

d d

State-Space Analysis

• With no exciter there

are three loops

– Top complex pair of

eigenvalues loop

– Bottom loop through

Eq’ contributes positive

damping

• Adding the exciter gives

33

5 6

1fd fd A A q A ref

A

E E K K K K E K VT

d

B2_PSS_Flux Example

• The SMIB can be used to plot how the eigenvalues

change as the parameters (like KA) are varied

34

''

4

3 do do doqq

s

refs2 1pupu

A

fdfdA 6 A 5 A

A A A

K1 10

0K T T TEE

00 0 0

V0DK K0

K2H 2H 2HEE

K K K K 1 T0

T T T

dd

Values for K1 to K6

can be

verified for

previous case

using the

PowerWorld

SMIB matrix

Verified K Values

• The below equations verify the results provided on the

previous slide SMIB matrix with the earlier values

35

11

22

3

3

44

55

65

0.1441 0.2882 3.2 0.9222

0.1677 0.3354 3.2 1.0732

1 10.3511 0.297

0.3511 9.6

0.236 0.236 9.6 2.266

10.069 0.210.069 0.005

400

714.5 0.714.5

do

do

A

A

A

A

KK

H

KK

H

KK T

KK

T

K KK

T

K KK

T

20.357

400

Root Locus for Varying KA

• This can be considered a feedback system, with the

general root locus as shown

36

Frequency Domain Analysis

• Alternatively we could use frequency domain analysis

• Ignoring TA gives

• If K5 is positive, then the

response is similar to the

case without an exciter

• If K5 is negative, then with

a sufficiently large KA the

electromechanical modes

can become unstable

37

'

( )

( )

q 3 4 A 5

A 3 6 3 do

E s K K K K

s 1 K K K sK Td

Frequency Domain Analysis

• Thus a fast acting exciter can be bad for damping, but

has over benefits, such as minimizing voltage

deviations

• Including TA

• Including the torque angle gives

38

'

( )

( )

3 4 A A 5q

A 3 6 3 do A

K K 1 sT K KE s

s K K K 1 sK T 1 sTd

( )( )( )

( ) ( )

qe2

E sT sK H s

s sd d

Torque-Angle Loop

• The torque-angle loop is as given in Figure 8.11;

with damping neglected it has two imaginary

eigenvalues

39

11,2

2

sKj

H

Torque-Angle Loop with

Other Dynamics

• The other dynamics can be included as in Figure 8.12

40

2

1

2( ) 0

s

Hs K H sd d d

H(s) contributes

both synchronizing

torque and

damping torque

Torque-Angle Loop with

Other Dynamics

• Previously we had

• If we neglect K4 (which has little effect at 1 to 3 Hz)

and divide by K3 we get

41

'

( )( )

( )

2 3 4 A A 5e

A 3 6 3 do A

K K K 1 sT K KT sH s

s K K K 1 sK T 1 sTd

( )( )

( )

e 2 A 5

2AA 6 do do A

3 3

T s K K KH s

s T1K K s T s T T

K K

d

Synchronizing Torque

• The synchronizing torque is determined by looking at

the response at low frequencies (0)

• With high KA, this is approximately

42

(j ) 2 A 5

A 6

3

K K KH 0

1K K

K

(j ) 2 5

6

K KH 0

K

2

1

2( ) 0

s

Hs K H sd d d

The synchronizing

torque is dominated

by K1; it is enhanced

if K5 is negative