ECE 576 – Power System
Dynamics and Stability
Lecture 28: Power System Stabilizer
Prof. Tom Overbye
University of Illinois at Urbana-Champaign
overbye@illinois.edu
1
Announcements
•
•
•
•
Read Chapters 8 and 9
Homework 8 should be completed before final but need
not be turned in
Final is Wednesday May 14 at 7 to 10pm
Key papers for book's approach on stabilizers are
– F.P. DeMello and C. Concordia, "Concepts of Synchronous
Machine Stability as Affected by Excitation Control, IEEE
Trans. Power Apparatus and Systems, vol. PAS-88, April
1969, pp. 316-329
– W.G. Heffron and R.A. Philips, "Effects of Modern Amplidyne
Voltage Regulator in Underexcited Operation of Large Turbine
Generators," AIEE, PAS-71, August 1952, pp. 692-697
2
Functions of a PSS
•
•
•
•
PSS adds a signal to the excitation system proportional
to speed deviation. This adds positive damping
– Other inputs, like power, voltage or acceleration, can be used
– Signal is generated locally 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.
Modal Analysis technique forms the basis for multimachine systems.
3
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
VST is an
input into
the exciter
4
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
IEEE Std 421.5 describes the common stabilizers
5
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
Δx = AΔx  BΔy  EΔu
0 = CΔx + DΔy
•
If D is invertible then
Δx =  A  BD-1C  Δx  EΔu  A sys Δx  EΔu
6
SMIB System (Flux Decay Model)
•
Process is introduced using an SMIB with a flux decay
machine model and a fast exciter
Vt e j
Re
~
( I d  jIq )e j (  / 2)
jX e
V 0
   pus
1
 pu 
[TM  ( Eq' I q  ( X d  X d' ) I d I q  D pu )]
2H
1
'
Eq  ' ( Eq'  ( X d  X d' ) I d  E fd )
Tdo
7
Stator Equations
•
Assume Rs=0, then the stator algebraic equations are:
X q I q  Vt sin(    )  0
(1)
Eq'  Vt cos(   )  X d' I d  0
( 2)
(Vd  jVq )e j (  / 2 )  Vt e j
(3)
 Vd  jVq  Vt e j e  j (  / 2 )
(4)
Expand right hand side of (4)
Vd  jVq  Vt sin(    )  jVt cos(   )
(5)
 Vd  Vt sin(    ) and Vq  Vt cos(   )
X q I q  Vd  0
( 6)
Eq'  Vq  X d' I d  0
(7 )
(1) and (2) become
8
Network Equations
•
The network equation is (assuming zero angle at the
infinite bus and no local load)
( I d  jI q )e j (  / 2 ) 
( I d  jI q ) 
(Vd  jVq )e j (  / 2 )  V 0
Re  jX e
(Vd  jVq )  V e  j (  / 2 )
Re  jX e
Re I d  X e I q  Vd  V sin 
Simplifying,
X e I d  Re I q  Vq  V cos 
Vt e j
~
Re
( I d  jIq )e j (  / 2)
Xe
V 0
9
Complete SMIB Model
1
E  ' ( Eq'  ( X d  X d' ) I d  E fd )
Tdo
'
q
   pus
Machine equations
1
 pu 
[TM  ( Eq' I q  ( X q  X d' ) I d I q  D pu )]
2H
X q I q  Vd  0
Stator equations
Eq'  Vq  X d' I d  0
Re I d  X e I q  Vd  V sin 
Network equations
X e I d  Re I q  Vq  V cos 
Vt e
~
j
Re
( I d  jIq )e j (  / 2)
V 0
jX e
10
Linearization of SMIB Model
Step 1: Linearize the stator equations
X q   I d   0 
 Vd   0
 V    '
 I    E ' 

 q   X d 0   q   q 
Step 2: Linearize the network equations
 Vd   Re  X e   I d   V cos   

 V   
 I   


 q   X e Re   q   V sin  
Step 3: Equate the righthand sides of the above equations
Re
( X e  X q )   I d   0   V cos  


 I    E '   
( X  X ' )


Re
d
 e
  q   q   V sin   
Notice this is equivalent to a generator at the infinite
bus with modified resistance and reactance values
11
Linearization (contd)
Solve for I d , I q
 ReV cos    V sin  ( X q  X e )   E ' 
 I d  1 ( X e  X q )
q



 I 

 (1)
'
Re
ReV sin    V cos  ( X d  X e )    
 q   
The determinant is   Re2  ( X e  X q )( X e  X d' )
•
1.
2.
Final Steps involve
Linearizing Machine Equations
.
x  f ( x, I d , I q , E fd , TM )
(2)
Substitute (1) in the linearized equations of (2).
12
Linearization of Machine
Equations
1
 Eq'   Tdo'

 
     0
  pu    I qo

  2 H
0   E ' 
 q 
0 s     (1)

Ds   
0  2 H   pu 
0
 T'1 ( X d  X d' )
 do

0
 1 o '
 2 H I q ( X d  X q )
0
0
1
2H
( X d'  X q ) I qo  21H

 T'1
  I d   do
0


I q  
'o  
Eq 
0



0
  E fd 
0 


T
M 
1 
2H 

Symbolically we have
x  Ax  Bu  C I d  q
(2 )
I d  q  Dx
( 3)
•
Substitute (3) in (2) to get linearized model.
13
Linearized SMIB Model
K4
1
1
'
E  
Eq  '   ' E fd
'
K 3Tdo
Tdo
Tdo
'
q
  s  pu
 pu
•
•
Ds
K2
K1
1
'

Eq 
 
 pu 
TM
2H
2H
2H
2H
Excitation system is yet to be included.
K1 – K4 constants are defined on next slide
14
K1 – K4 Constants
1
 1
K3
( X d  X d' )( X q  X e )



V ( X d  X d' )
K4 
( X q  X e ) sin    Re cos  

1 o
K 2  I q   I qo ( X d'  X q )( X q  X e )  Re ( X d'  X q ) I do  Re Eq'o

1 o
K1   [ I q V ( X d'  X q ){( X q  X e ) sin    Re cos  }

 V {( X d'  X q ) I do  Eq'o }{( X d'  X e ) cos    Re sin  }]

•

K1 – K4 only involve machine and not the exciter.
15
Including Terminal Voltage
•
The change in the terminal voltage magnitude also needs
to be include since it is an input into the exciter
Vref
Vref




Vt
Vt
Vt  Vd2  Vq2
Differentiating
Vt 2  Vd2  Vq2
2Vt Vt  2Vdo Vd  2Vqo Vq
Vt 
o
d
Vqo
V
Vd 
Vq
Vt
Vt
16
Computing
•
•
While linearizing the stator algebraic equations, we had
 Vd   0
 V   
'

X
 q  d
X q   Eq'   0 

   E ' 

0      q 
Substitute this in expression for Vt to get
Vt  K 5   K6 Eq'
1 Vdo
K 5   X q [ ReV sin    V cos  ( X d'  X e )]
  Vt

 [ X ( ReV cos    V ( X q  X e ) sin  )]
Vt

Vqo
'
d
o
o
o


V
V
1 V
 q
q
K6   d X q Re 
X d' ( X q  X e )  
  Vt
Vt
 Vt
17
Heffron–Phillips Model
•
Add a fast exciter with a single differential equation
TA E fd   E fd  K A (Vref  Vt )
•
Linearize
TA E fd  E fd  K A (Vref  Vt )
TA E fd  E fd  K A Vref  K A ( K 5   K 6 Eq' )
•
This is then combined with the previous three
differential equations to give
x  A sys x  BΔu
Δu  [TM
Vref ]
T
SMIB Model
18
Block Diagram
•
K1 – K6 are affected by system loading and Xe
19
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 115
– 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
Saved as case B2_PSS_Flux (but only available in v18)
20
Initial Conditions
•
The initial conditions are
j
I G e  ( I d  jI q )e
j (  / 2 )
115  1.050

 0.544318
j0.5
 (0)  angle of E where E  Vt e j  ( Rs  jX q ) I G e j .
E  115  ( j2.1)(0.544318)
 1.4788 65.5
I d  jI q  I G e j e  j (  / 2 )  0.544342.48,
I d  0.4014, and I q  0.3676.
Vd  jVq  Ve j e  j (  / 2 )  139.48
Hence, Vd  0.7718, Vq  0.6358
21
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
V is the input
TA E fd   E fd  K A (VREF  Vt  Vs )
s
from the stabilizer,
with an initial
value of zero
22
Initial Conditions (contd)
•
From the stator algebraic equation,
Eq'  Vq  X d' I d
 0.63581  (0.39 )(0.4014 )  0.7924
E fd  Eq'  ( X d  X d' ) I d
 0.7924  (2.5  0.39)0.4014  1.6394
VREF
E fd
1.6394
 Vt 
 1
 1.0041
KA
400
s  377, TM  Eq' I q  ( X q  X d' ) I d I q
 (0.7924)(0.3676)  (2.1  0.39)(0.4014)(0.3676)
 0.5436
 checks with (VtV sin  ) / X e 
23
Computation of K1 – K6 Constants
•
The formulas are used.
  Re2  ( X e  X q )( X e  X d' )  2.314
1
 1
K3
( X d  X d' )( X q  X e )

 3.3707
K 3  0.296667
V ( X d  X d' )
K4 
[( X q  X e ) sin    Re cos  ]  2.26555

1
K 2  [ I qo   I qo ( X d'  X q )( X q  X e )  Re ( X d'  X q ) I do  Re Eq' o ]  1.0739

Similarly K1 , K 5 , and K 6 are calculated as
K1  0.9224 , K 5  0.005 , K 6  0.3572
24
Effect of Field Flux on System Stability
•
With constant field voltage, i.e. Efd=0, from the block
diagram
Te  K 2 K3 K 4


1 K3Tdo' s
– K2, K3, and K4 are usually positive.
On diagram
v is used to
indicate what
we've been
calling pu
25
Effect of Field Flux on System Stability
•
For low frequencies and in steady-state, as s=j goes to
zero, then
Te  K 2 K 3 K 4

 Te   K 2 K 3 K 4 
'

1  K 3Tdo s
•
Hence field flux variation due to  feedback (armature
reaction) introduces negative synchronizing torque
26
Effect of Field Flux on System Stability
•
At higher frequencies the denominator is dominated
by the K3T'dos term so
 K2 K3 K4
K2 K4
Te 
 
j 
'
'
jTdo K 3
Tdo
•
•
This is a phase with (j)
Hence Efd results in a positive damping component.
27
Effect of Field Flux on System Stability
•
In general, positive damping torque and negative
synchronizing torque due to Efd at typical oscillation
frequencies (1-3 Hz).

Te
Ts
TD
 K3 (E fd  K 4  ) 
K 2 E  K 2 

'
(
1

K
T
s
)
3 do


'
q

28
Effect of Field Flux
•
•
•
•
There are situations when K4 can become negative.
V ( X d  X d' )
K4 
( X q  X e ) sin    Re cos  






may become negative
Case1. A hydraulic generator without a damper
winding, light load ( is small) and connected to a line
with high Re/Xe ratio and a large system.
Case 2. Machine is connected to a large local load,
supplied partly by generator and partly by
remote large system.
If K4 < 0 then the damping due to Efd is negative
29
Effect of Excitation System
•
•
•
•
•
Ignore dynamics of
AVR & KA >> 0
Gain through T'do
is -(K4+KAK5)K3
If K5 < 0 then gain
becomes positive.
VREF = 0
Large enough K5 may
make system unstable
Te
 ( K 4  K A K5 ) K3 K 2

 (1  sK 3'Tdo'  K A K3 K 6 )
30
Numerical Example (Effect of K5)
•
Test system 1
Test System 2
K A  50, K 5  0
K A  50, K 5  0
K1  3.7585 K 2  3.6816
K1  0.9813
K 3  0.2162 K 4  2.6582
K 3  0.3864 K 4  1.4746
K 5  0.0544 K 6  0.3616
K 5  0.1103 K 6  0.4477
Tdo'  5 sec
Tdo'  5 sec
H  6 sec
TA  0.2 sec
K 2  1.093
H  6 sec
TA  0.2 sec
Eigen Values
Test System 1
Test System 2
 0.353  j10.946
0.015  j 5.38
 2.61  j 3.22
 2.77  j 2.88
unstable
31
Addition of PSS Loop
•
•
The impact of a PSS can now be considered in which
the shaft speed signal is fed through the PSS transfer
function G(s) to the excitation system.
To analyze effect of PSS signal assume , VREF=0
32
PSS Contribution to Torque
 pu
TPSS
G (s ) PSS
K2
K3
1  sK 3Tdo'
KA
1  sTA
vs



Vref  0
K6
TPSS
G ( s) K 2 K A K 3

v
K A K 3 K 6  (1  sK 3Tdo' )(1  sTA )
G ( s) K 2 K A
 1
 G ( s )GEP( s )
TA
'
2 '
K 3  K A K 6  s K 3  Tdo  s TdoTA
(for usual range
G ( s) K 2 K A
of constants)
 1
'
33

K
K
1

s
T
A 6
do / K A K 6 1  sT A 
K3

 

 


PSS Contribution to Torque
•
For high gain KA
TPSS K 2
G ( s)

v
K 6 1  s Tdo' / K A K 6 1  sTA 
 
•

If PSS were to provide pure damping (in phase with
pu), then ideally,
 

G ( s)  K PSS 1  s Tdo' / K A K 6 1  sTA 
•
K PSS  Gain of PSS
i.e. pure lead function
Practically G(s) is a combination of lead and lag
blocks.
34
PSS Block Diagram
ymax
 pu
sTw
1  sTw
1  sT1
1  sT2
1  sT3
1  sT4
K PSS
v s
ymin
G ( s )  K PSS
•
•
•
•
(1  sT1 ) (1  sT3 ) sTw
 K PSS G1 ( s )
(1  sT2 ) (1  sT4 ) (1  sTw )
Provide damping over the range of frequencies (0.1-2 Hz)
“Signal wash out” (Tw) is a high pass filter
– TW is in the range of 5 to 20 seconds
Allows oscillation frequencies to pass unchanged.
Without it, steady state changes in speed would modify
terminal voltage.
35
Criteria for Setting PSS Parameters
•
•
•
•
KPSS determines damping introduced by PSS.
T1, T2, T3, T4 determine phase compensation for the
phase lag present with no PSS.
A typical technique is to compensate for the phase lag
in the absence of PSS such that the net phase lag is:
– Between 0 to 45 from 0.3 to 1 Hz.
– Less than 90 up to 3 Hz.
Typical values of the parameters are:
–
–
–
–
–
KPSS is in the range of 0.1 to 50
T1 is the lead time constant, 0.2 to 1.5 sec
T2 is the lag time constant, 0.02 to 0.15 sec
T3 is the lead time constant, 0.2 to 1.5 sec
T4 is the lag time constant, 0.02 to 0.15 sec
36
Design Procedure
•
•
•
•
•
The desired stabilizer gain is obtained by finding the
gain at which the system becomes unstable by root
locus study.
The washout time constant TW is set at 10 sec.
KPSS is set at (1/3)K*PSS, where (1/3)K*PSS is the gain
at which the system becomes unstable.
There are many ways of adjusting the values of T1,
T2, T3, T4
Frequency response methods are recommended.
37
Design Procedure Using the
Frequency-Domain Method
•
TPSS
K 2 K A K3
 GEP( s )G ( s ) , GEP( s ) 
v
K A K 3 K 6  (1  sTdo' K 3 )(1  sTA )
Step 1: Neglecting the damping due to all other sources, find the
undamped natural frequency  n in rad/sec of the
torque - angle loop from:
2H 2
s  K1  0, i.e. s1, 2   j n
K1
s
Te

s


2Hs

1
s
where  n 

D
TM  0
 Te
Step 2: Find the phase lag of

2H
s
1
s 2  Ds  K1
K1 s
2H

GEP(s) at s  j n
38
PSS Design (contd)
•
•
Step 3:
Adjust the phase lead of G(s) such that
G ( s ) s  j n GEP( s ) s  j n  0
•
•
 1  sT1 

Let G ( s )  K PSS 
 1  sT2 
k
Ignoring the washout filter whose net phase
contribution is approximately zero.
k  1 or 2 with T1  T2 . Thus, k  1 :
1  j nT1  1  j nT2  GEP( j n )
39
PSS Design (contd)
•
•
Knowing n and GEP(j n) we can select T1. T2
can be chosen as some value between 0.02 to 0.15 sec.
Step 4: Compute K*PSS, the gain at which the system
becomes unstable, using the root locus. Then have
*
K PSS  13 K PSS
.
Alternative Procedure
The char. equation is
X
  n
n 1  2
2H
cos 1 
s
S 2  Ds  K1  0 (1)
The damping ratio is
X
Damping ratio
  12 D / MK1
where M  2H /  s
40
PSS Design (contd)
•
The characteristic roots are:
s1,2 
 MD 
 
D
M
2
 4MK1
2
2
2
K1  D 
4K1
D
D

j

 if   
2M
M  2M 
M
M 
 n  jn 1   2
Note that n 
K1
M
. Therefore

Verify that
D
D

2M n 2M
M
D

K1 2 K1 M
K1 D 2
 D 
2 2
n   .


M 4K1 M  2M 
2
41
PSS Design (contd)
•
Since the phase lead of G(s) cancels phase lag due to GEP(s) at
the oscillatory frequency, the contribution of PSS through GEP(s)
is a pure damping torque with a damping coefficient DPSS

DPSS  K PSS GEP( s )
s  jn
G1 ( s )
s  jn
The characteristic equation is
DPSS
K1
s 
s
0
M
M
i.e., s 2  2n M  0
2
DPSS  2n M  K PSS GEP ( jn ) G1 ( jn )
We can thus find K PSS , knowing ωn and the desired ξ.
A reasonable choice for ξ is between 0.1 and 0.3.
42
PSS Design (contd)
•
•
•
•
The washout filter should not have any effect on
phase shift or gain at the oscillating frequency.
A large value of TW is chosen so that sTW is much
larger than unity.
Choose Gw ( j n )  1
Hence, its phase contribution is close to zero. The
PSS will not have any effect on the steady state
behavior of of the system since in steady state
pu=0
43
Example
•
•
•
•
Without the PSS, the A matrix is
0 0.104 
 0.3511 0.2236
 0

0
377
0


 0.1678 0.144
0
0 



714
.
4

10
0

5


The eigenvalues are l1,2  0.0875  j7.11, l3,4  2.588  j8.495.
The electromechanical mode l1,2 is poorly damped.
Instead of a two-stage lag lead compensator, assume a
single-stage lag-lead PSS. Assume that the damping D in
the torque-angle loop is zero.
The input to the stabilizer is pu. An extra state equation
will be added. The washout stage is omitted.
44
Example (contd)
Vref
 pu
K PSS
(1  sT1 )
(1  sT2 )
PSS
•

Vs


KA
1  sTA
E fd
Vt
The added differential equation is then
K PSS
T1
1
Vs   Vs 
 pu  K PSS  pu
T2
T2
T2
K PSS
T1   K 2
K1
1

'

Vs 
 pu  K PSS 
Eq 
 
T2
T2
T2  2H
2H

45
Example (contd)
With a choice of K PSS  0.5, T1  0.5, T2  0.1, the new matrix is
0 0.104
0 
 0.3511 0.236
 0

0
377
0
0


 0.1678 0.144 0
0
0 



714
.
4

10
0

5
2000


 0.42
0.36
5
0
10 
The eigenvalues are l1,2  0.8612  j7.7042 l3,4  1.6314  j8.5504,
l5  10.3661.
Note the improvement of the electromechanical mode l1,2
46
Example (industry)
•
CONVERT TO SMIB (RETAIN G16)
47
Example (contd)
Eigenvalue
-0.18929
-0.91111
-0.017097  j3.3105
-10.145
-8.0319  j16.785
-24.531
-29.031
Frequency Hz Damping Ratio
0
0
0.52688
0.0051643
0
2.6714
0.43165
0
0
E.M. Mode has f = 0.5688 Hz
The participator for this mode is
•
State Number
1

2

3
E'q

E fd
4
5
E'd
E'q
6
E fd
7
tg1
8
9
tg 2
10 tg 3
Participation Factor
0.45177  j0.027549
0.50613  j0.028591
-0.00011  j0.011599
0.00012  j0.00011
-0.004354  j0.01354
-0.00034  j7.033e-6
-0.00177  j0.00075
0.005614  j0.016166
0.040345  j0.023404
0.0025854  j0.0092191
*
*
Poorly damped
E.M . Mode   0.005
Exciter mode
well damped
Convert these into
magnitudes
} Turbine Governor States
Largest Participation by angle and speed states.
48
Root Locus
•
If G(s) (Transfer function of PSS) is a real gain, it
will add pure damping Root Locus Study (with KPSS
= 10).
49
Phase Compensation
GEP( s ) s  j
Ideal phase lead.
PSS phase lead.
G( s) 
10s (1  .03s)(1  .04s)
1  10s (1  .01s)(1  .01s)
Washout does not add
much phase lead.
50
Root Locus with PSS
51
Eigenvalues with PSS
Eigenvalue
-0.10078
-0.18895
-0.90803
-1.8476
-0.40649  j3.2883
-10.151
-7.668  j16.996
-24.531
-29.031
-95.914
-104.05
Frequency Hz
0
0
0
0
0.52334
0
2.705
0
0
0
0
Damping Ratio
1
1
1
1
0.12268
1
0.41125
1
1
1
1
Note Improvement in Damping Ratio of E.M Mode
from 0.05 to 0.123
52