02_StateEstimation_R3.pdf

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1

State Estimation

A. J. Conejo

Univ. Castilla – La Mancha

2011



2

0. References

1. Introduction

2. Example

3. Formulation and solution

3.1 Linear case

3.2 Nonlinear case

4. Observability

5. Detection and identification of bad measurements

State Estimation



3

State Estimation0. Referencias

A. Gómez Expósito, A. J. Conejo, C. Cañizares.

“Electric Energy Systems. Analysis and Operation ”.

CRC Press, Boca Raton, Florida, 2008.

A. Abur, A. Gómez Expósito.

“Power System State Estimation. Theory and Implementation ”.Dekker, New York, 2004.



4

• SCADA (Supervisory Contron Adn Data Adquisition):

Data Adquisition

Data base maintenance

Issue warning

• EMS (Energy Management System)

State Estimation1. Introduction



5

Estimate the state variables using a redundant set of

measurements that include measurement errors.

Statica problem: all measurements are symultaneous.




6

• Reduncancy in measurements to improve theestimation

• Measurements:

1. P and Q flows.

2. P and Q injections.

3. Voltage magnitudes.

4. Current magnitues (uncommon).

5. Voltage and current angles (if PMUs)

• Other measurements: virstual (no error) and pseudo-

measurements (inexacts).




7

1. Measurement filtering

2. Topology checking

3. Observability analysis

4. Estimation

5. Detection and identification of bad measurements

Functions:




8

Online data

MeasurementFiltering

Topological Analysis

StateEstimation

Bad MeasurementDetection

Bad MeasurementSuppression

Man-machine interface

Fixed data

Observability Analysis

System state

Warning:Measurement errors

Unobserv

able

areas

Breaker status

urements




9

25.0X

4.0X

0.2X

MVA)100(Base

23

13

12

reference

Estimate voltage angles

Reactances in pu

State Estimation2. Example



10

Power flow meters: 321312 M,M ,M

One meter isredundant




11

Only two measurements (exact):

2 state variables:

2 meters:

)reference,0δ(,δ,δ 321

3213 M,M

3213 M,M

pu40.0Mδδx

1P

pu05.0Mδδx1P

pu4.0MW40M,pu05.0MW5M

3223

23

32

1331

13

13

3213




12

Exact solution:

0.02δ1 0.1δ2

0δ3




13

Solution using measurements:

0.024δ1 0.0925δ2

pu.370M,pu0.06M 3213

MW60MW25.58=M12

0δ3

M13

13 M

32M




14

0.0315δ1 0.0925δ2

pu.370M,pu0.62M 3212

MW5MW875.7M13

0δ3

M12 32M


Solution using measurements:



15

ηzz r m

Measurement

True value

Measurement error

Erroneous measurements:

ηh(x)zm

State vector is x, thus

State Estimation3. Formulation



16

• True values are unknown: they need to be estimated.

• Weighted least squares is a common approach for

estimation.




17

Quadratic errors

Weights

Weight for measurement i

Measurement number

Objective function

iiw

m

J(x)

N

1i

2

i

m

iii

x

(x)hzwJ(x)

J(x)Minimizar

Weighted least squares (WLS):




18

mm

22

11

mm

w

w

w

W

xhzWxhzJ(x)T

Measurement verctor

Functions to compute measurement

Weighting matrix

State variable (n)

mz

h(x)

W

x

1m

mm

1m




19

• is a measure of the measurement involved

• higher involves higher precision

21 21

2

m

2

2

2

1

1-

σ1

σ1

σ1

RW




20

First order optimality conditions:

x

xhxH :where

0xhzWx2H 0x

xJ mT




21

)E1( )x(h-zWxH ΔxxG kkTkk

WH(x)xHxG T

where the gain matrix is


Newton solution at iteration k



22

If H has full rank then G is definite positive an (E1) has a

single solution

kk1k Δxxx

1. Initialize x0 (flat start) (k=0).

2. Compute H(xk) and calculate G(xk).

3. Solve .

4. Update

5. Is Dx is small enough stot, otherwise go to 2

)h(x-zWxH ΔxxG kkTkk




23

• If h(x) is linear, everything is easier.

• Just one iteration:

observablenot :mnIf -

tsmeasuremenneeded jus t,zHx̂ :mnIf -

estimationOK,WzHGx̂ :mnIf -m1

mT1

mT1mT zWHGx0HxzW2H 0x

xJ




24

444

m

321312

101010w

0.370.060.62z

MMM

2 estate variables:

3 measurements:

21 δ,δ

321312 M,M,Mn < m

State EstimationExample



25

0,1

0,02δ

0,0943

0,0286δˆ

WzHGδˆ

0,37

0,060,62

z,

1000

01000010

W

WHH)x(G

40

02,555

H,

δ4

δ2,5δ5δ5

δh

REAL

mT1

m

4

4

4

T

2

1

21


S



26

464

m

321312

101010w

0.370.060.62z

MMM

Better meter

mT1

REAL

4

6

4

WzHGδ̂

0,1

0,02

δ0,0970

0,0241

δˆ

10000100

0010

W


S E i i



27

Optimization problem with no contraints

N

1i

2i

miii

x(x)hzwMinimize

1. P measurement

2. Q measurement

3. Voltage measurement

ijmij pp

ijmij qq

imi vv

-

(·)q-q(·)pp

i

m

i

i

m

i4. P, Q injections


St t E ti ti



28

Active Power flow:

jiijij

ji

ijij

2i

ij

δδθcosz

vvcosθ

z

vp

Reactive Power flow:

jiijij

ji

ijij

2i

ij δδθsenz

vvsenθ

z

vq


St t E ti ti



29

Estimation problem

2m

ijijiΩ jΩi

q

ij

2mijijiΩ jΩi

pij

2

Ωi

mii

vi

iδ,

iv

qqw

ppw

vvwMinimizar


St t E ti ti



30

2 buses, 1line with reactance x = 0.15 pu

Voltage measurements at buses 1 y 2 are,

respectively, 1.07 y 1.01 puV Active power flow measurements (1-2) are 0.83 y 0.81

puW

Reactive power flow measurements (1-2) are 0.73 y

0.58 puVar

Reference bus is 2


St t E ti ti



31

Optimization problem:

2

m

21

2

2121

2

m

21121

2

m

12121

2

1

2

m

12121

2m

22

2m

11δ,v,v

qvx

1cosδvvx

1psenδvvx

1

qcosδvvx

1v

x

1psenδvv

x

1vvvveMinimize121


St t E ti ti



32

20,582

2v

0.15

1

1cosδ

2v

1v

0.15

1-20.81

1senδ

2v

1v

0.15

1-

20,73

1cosδ

2v

1v

0.15

121

v0.15

120.83

1senδ

2v

1v

0.15

1

21.01

2v

21.07

1ve

Minimize1δ,

2v,

1v

---

--

Also


St t E ti ti



33

scalars

x Reactancia de la linea /0.15/*mediciones

v1m Tension medida en el nudo 1 /1.07/


p12m Potencia activa medida en el nudo 1 /0.83/

q12m Potencia reactiva medida en el nudo 1 /0.73/

p21m Potencia activa medida en el nudo 2 /0.81/

q21m Potencia reactiva medida en el nudo 2 /0.58/;

variablesv1 Modulo de la tension en el nudo 1. Valor exacto: 1.0966

v2 Modulo de la tension en el nudo 2. Valor exacto: 1.0000

d1 Argumento de la tension en el nudo 2. Valor exacto: 0.1097

e Error cuadratico medio;

equations

error Ecuacion del error cuadratico medio;

*Definicion de la ecuacion del error cuadratico medio

error .. e =e= sqr(v1-v1m) + sqr(v2-v2m) +

sqr((1/x)*v1*v2*sin(d1)-p12m) +

sqr((1/x)*(sqr(v1)-v1*v2*cos(d1))-q12m) +

sqr((1/x)*v1*v2*sin(d1)-p21m) +

sqr((1/x)*(-sqr(v2)+v1*v2*cos(d1))-q21m);

model estimo /all/;

solve estimo using nlp minimizing e;


St t E ti ti



34

LOWER LEVEL UPPER MARGINAL

---- VAR v1 -INF 1.088 +INF -1.185E-8

---- VAR v2 -INF 0.994 +INF 7.6054E-9

---- VAR d1 -INF 0.114 +INF -1.043E-8

---- VAR e -INF 7.9359E-4 +INF .



d1 Argumento de la tension en el nudo 1. Valor exacto: 0.1097

e Error cuadratico medio


St t E ti ti



35

Wood & Wollenberg, pg. 4776-bus network:


State Estimation



36

Wood & Wollenberg, pg. 478

Measurements


State Estimation



37

Wood & Wollenberg, pg. 480

Estimation:


State Estimation



38

Network

1

2

3


State Estimation



39

Data:

Measurement Value p.u. Variance

V1 0.95 10-3

V2 1.02 10-3

V3 1.03 10-3

P12 1 10-2

Q21 -0.4 10-2

P13 0.8 10-2

Q31 -0.2 10-2

X12 = X13 = X23 = 0.1j


State Estimation



40

Solution

• State vertor:

3

2

3

2

1

5

4

3

2

1

δ

δ

V

V

V

x

x

x

x

x

x

ii δV0δslack


State Estimation



41

• Measurements:

2.0

80.040.0

00.1

03.1

02.1

95.0

Q

PQ

P

V

V

V

z

31

13

21

12

3

2

1


State Estimation



42

• Weighting matrix

100000000

010000000001000000

000100000

0000100000

0000010000

000001000

W


State Estimation



43

• Functional measurement vector

1

2

3

1 2

4

2

1 4 2

1 3

5

3

3 1 5

sin( )

( )

cos( )

sin( )

cos( )

x

x

x

x x x

X

h x x

x x x X

x x x

X

x x x x

X


State Estimation



44

• Jacobian matrix, : j

iij

x

hH

)x(senX

xx0

X

x2xcosx0)xcos(

X

x

)xcos(X

xx0)x(sen

X

x0)x(sen

X

x0)x(senX

xx

0X

x2xcosx

)xcos(X

x

0)xcos(X

xx0)x(sen

X

x)x(sen

X

x00100

0001000001

)x(H

531351

53

531

51

53

4

21241

4

2

421

41

42


State Estimation



45

1. Initialization x0 :

2. Compute H(xk) and calculate G(xk).

-Funcitonal vector:

T0 00111x

T0 0000111)x(h


State Estimation



46

- Jacobian matrix:

0010010

100000

0001010

010000

00100

00010

00001

)x(H 0


State Estimation



47

- Gain matrix:

100000

010000

0011010

0001110

00101021

10WHH)x(G 3T0


State Estimation



48

- Residual vector :

2000.0

8000.0

4000.00000.1

0300.0

0200.0

0500.0

)x(hz)x(r

h(x)-zr(x)

00


State Estimation



49

3. Solution of the system of equations

Thus

D

0800.0

1000.0

0007.00184.0

0177.0

Wr HGx T10

)x(hzW)x(Hx)x(G kKTkk D


State Estimation



50

4. Update the state vector:

kk1k xxx D

D

0800.0

1000.0

0007.19816.0

0177.1

xxx 001


State Estimation



51

5. Repeat until residuals below 0.001


State Estimation



52

Solution:

J(x) V1 V2 V3

Matlab 7.8062 1.0216 0.9800 1.0013 -0.0996 -0.0781


State Estimation



53

Matlab solution:

1.0215 ; 0.9800 ; 1.0014 ; -0.0996 ; -0.0781x=

Residuals:

r 1 2 3 4

V1 -0.0500 -0.0677 -0.0714 -0.0715

V2 0.0200 0.0384 0.0401 0.0400

V3 0.0300 0.0293 0.0287 0.0286

P12 1.0000 0.0027 0.0043 0.0043

Q21 -0.4000 -0.0950 -0.0427 -0.0428

P13 0.8000 -0.0139 0.0024 0.0023

Q31 -0.2000 -0.0615 -0.0293 -0.0293


State Estimation



54

GAMS solution:scalars

x reactancias de todas las lineas /0.1/

*mediciones

v1m tension medida del nudo 1 /0.95/



p12m potencia activa medida en el nudo 1 /1.00/

q21m potencia reactiva medida en el nudo 2 /0.4/

p13m potencia activa medida en el nudo 1 /0.8/

q31m potencia reactiva medida en el nudo 3 /0.2/

sg1 varianza 1 /0.001/

sg2 varianza 2 /0.01/;

variables

v1 modulo de la tension en el nudo 1. Valor exacto:1.0216



d2 argumento de la tension en el nudo 2. Valor exacto:-0.0996

d3 argumneto de la tension en el nudo 3. Valor exacto:-0.0781

e error cuadratico medio;


State Estimation



55

equations

error ecuacion del error cuadratico medio;

error.. e =e= 1/sg1*(sqr(v1-v1m))+1/sg1*(sqr(v2-v2m))+

1/sg1*(sqr(v3-v3m))+

1/sg2*(sqr((1/x)*v1*v2*sin(d2)-p12m))+

1/sg2*(sqr((1/x)*v1*v3*sin(d3)-p13m))+

1/sg2*(sqr((1/x)*(-sqr(v2)+v1*v2*cos(d2))-q21m))+

1/sg2*(sqr((1/x)*(-sqr(v3)+v1*v3*cos(d3))-q31m));

model estimo /all/;

solve estimo using nlp minimizing e;


State Estimation



56


---- VAR v1 -INF 1.022 +INF 8.2392E-9---- VAR v2 -INF 0.980 +INF -3.839E-9

---- VAR v3 -INF 1.001 +INF -4.782E-9

---- VAR d2 -INF -0.100 +INF -2.068E-9

---- VAR d3 -INF -0.078 +INF -2.375E-9

---- VAR e -INF 7.806 +INF .

GAMS actual solution:


State Estimation



57

Comparison:

Variable Matlab GAMS Error

V1 1.0215p.u. 1.022p.u. 0.04%

V2 0.9800p.u. 0.980p.u. 0.00%

V3 1.0014p.u. 1.001p.u. 0.04%

δ2 -0.0996 -0.100 0.40%

δ3 -0.0781 -0.078 0.13%


State Estimation



58

Two alternatives:

1. Zero injections

modeled using largeweights.

2. Zero injection

equalities explicitly

modeled

Variable Value p.u Variance

V1 1.05 10-3

V2 1 10-3

V3 0.95 10-3

P3 1 10-2

Q3 0.2 10-2

One Two Three


State Estimation



59

• Matlab solution (case 1):

Matrix W:

2

2

6

6

3

3

3

10000000

01000000

00100000

00010000

00001000

0000010000000010

W


State Estimation



60

Z is:

P2 y Q2 are zero (exact!)

0.2

1

0

0

0.95

1

1.05

Q

P

Q

P

V

V

V

Z

3

3

2

2

3

2

1


State Estimation



61

Vector h(x):

x1

x2

x3

( )h x

-10((x1 x2 sin(x4))+(x2 x3 sin(x4-x5)))

10 x2((x1 cos(x4)-x2)-(x2-x3 cos(x4-x5)))

10x2 x3 sin(x4-x5)10x3 (x2 cos(x4-x5)-x3)


State Estimation



62

• Iteration:

• Tolerance for Δx below |0.00001|

kk1kT1k

k

T

xxxWr HGx)x(hz)x(r

WHHG

DD


State Estimation



63

Values per iteration:

Iteration 1 2 3 4

V1 1.025000 1.035781 1.035794 1.035793

V2 0.999999 0.996133 0.996001 0.995999

V3 0.974999 0.966660 0.966443 0.966440

2 -0.099999 -0.097648 -0.0977013 -0.097702

3 -0.199999 -0.202269 -0.202438 -0.202440


State Estimation



64

• Solution in 4 iterations

• Results (Caso 1):

Estimate Residual

V1 1.035794 0.014205

V2 0.996001 0.003998

V3 0.966443 -0.016443

2 -0.0977013 -0.006335

3-0.202438 -0.032906

State st at oExample

State Estimation



65

• GAMS (case 1) (NO weights):

scalars

x Reactancia de la linea /0.1/

*mediciones


v2m Tension medida en el nudo 2 /1/


p2m Potencia activa medida en el nudo 2 /0/

q2m Potencia reactiva medida en el nudo 2 /0/



variables


v2 Modulo de la tension en el nudo 2. Valor exacto: 1.0000v3 Modulo de la tension en el nudo 3. Valor exacto: 1.0000

d2 Argumento de la tension en el nudo 2. Valor exacto: -0.0100


Example

State Estimation



66


equations

error Ecuacion del error cuadratico medio;


error .. e =e= sqr(v1-v1m) + sqr(v2-v2m) + sqr(v3-v3m) +

sqr(((-1/x)*v1*v2*sin(d2)+(-1/x)*v2*v3*sin(d2-d3))-p2m) +sqr(((1/x)*v2*(v1*cos(d2)-v2)-((1/x)*v2*(v2-v3*cos(d2-d3))))-q2m) +

sqr((1/x)*v2*v3*sin(d2-d3)-p3m) +

sqr((1/x)*v3*(v2*cos(d2-d3)-v3)-q3m);

model estimo /all/;

SOLVE estimo USING nlp MINIMIZING e;

display v1.L,v2.L,v3.L,d2.L,d3.L;

Example

State Estimation



67

• GAMS (caso 1):

Solution is:

error Ecuación del error cuadrático medio


---- VAR v1 -INF 1.033 +INF EPS

---- VAR v2 -INF 0.996 +INF 1.8628E-8

---- VAR v3 -INF 0.970 +INF -1.837E-8

---- VAR d2 -INF -0.097 +INF -1.350E-8

---- VAR d3 -INF -0.201 +INF 9.2908E-9

Example

State Estimation



68

• Comparison case 1:

Matlab GAMSV1 1.035794 1.033

V2 0.996001 0.996

V3 0.966443 0.970

2-0.0977013 -0.097

3-0.202438 -0.201

Example

State Estimation



69

• Case 2

• Zero injections explicitly modeled as equalities.

• This generally avoid numerical issues.

Example

State Estimation



70

Vector h(x) is:

2.0

1

95.0

1

05.1

Q

P

V

V

V

Z

3

3

3

2

1

x3)-x5)-cosx4(x210x3

x5)-sin(x4x310x2x3

x2

x1

)x(h

Example

State Estimation



71

Constraint vector:

x5)))-cos(x4x3-(x2-x2)-cos(x4)((x1x210

x5)))-sin(x4x3(x2+sin(x4))x2((x110-c

Example

State Estimation



72

• System to be solved:

• C is the Jacobian matrix of the constraints and λ the

multipied vector of the constraints

D

)x(c

Wr Hx

0C

CWHHk

kTkTT

Example

State Estimation



73

• Values per iteration:

Iteration 1 2 3 4

V1 1.025000 1.035781 1.035794 1.035793

V2 1 0.996133 0.996001 0.995999

V3 0.975 0.966660 0.966443 0.966440

2 -0.1 -0.097648 -0.0977013 -0.097702

3 -0.2 -0.202269 -0.202438 -0.202440

Example

State Estimation



74

• Solution in 4 iterations

• Final results (Case 2):

Variable Estimate Residual

V1 1.035794 0.014205

V2 0.996001 0.003998

V3 0.966443 -0.016443

2 -0.0977013 -0.006335

3 -0.202438 -0.032906

Example

State Estimation



75

• GAMS (Cas2 2) (NO weights):scalars

x Reactancia de la linea /0.1/

*mediciones


v2m Tension medida en el nudo 2 /1/


p2m Potencia activa medida en el nudo 2 /0/q2m Potencia reactiva medida en el nudo 2 /0/



Variables







Example

State Estimation



76

equations

error Ecuacion del error cuadratico medio

R1 Restriccion 1

R2 Restriccion 2;


error .. e =e= sqr(v1-v1m) + sqr(v2-v2m) + sqr(v3-v3m) +

sqr((1/x)*v2*v3*sin(d2-d3)-p3m) +sqr((1/x)*v3*(v2*cos(d2-d3)-v3)-q3m);

R1 .. ((-1/x)*v1*v2*sin(d2)+(-1/x)*v2*v3*sin(d2-d3)-p2m)=E=0;

R2 .. ((1/x)*v2*(v1*cos(d2)-v2)-((1/x)*v2*(v2-v3*cos(d2-d3)))-q2m)=E=0;

model estimo /all/;

SOLVE estimo USING nlp MINIMIZING e;

display v1.L,v2.L,v3.L,d2.L,d3.L;

Example

State Estimation



77

• GAMS solution (caso 2):

error objective function

R1 Constraint 1

R2 Constraint 2


---- VAR v1 -INF 1.032 +INF -1.290E-8

---- VAR v2 -INF 0.996 +INF .---- VAR v3 -INF 0.970 +INF 1.3559E-8

---- VAR d2 -INF -0.097 +INF .

---- VAR d3 -INF -0.201 +INF -4.461E-9

Example

State Estimation



78

• Comparison:

CASE 1 CASE 2

Matlab GAMS Matlab GAMS

V1 1.035794 1.033 1.035794 1.032

V2 0.996001 0.996 0.996001 0.996

V3 0.966443 0.970 0.966443 0.970

2 -0.097701 -0.097 -0.097701 -0.097

3 -0.202438 -0.201 -0.202438 -0.201

Example

State Estimation



79

Do we have enough measurement to estimate

the state?

4. Observability

State Estimation



80

If the system as al whole is not observable, some

“islands” would be observable.

4. Observability

State Estimation



81

State variables: n

Measurements: M

1) m<n unboservable;

2) m=n no redundancy;

3) m>n generally observable.

Measurements have to be properly distributed.

4. Observability

State Estimation



82

It the system is unobservable, we use pseudo-

measurements

1) Reasonable values.

2) Previous estimates.

4. Observability

State Estimation



83

N bus network: n=2N-1 state variables, and m

measurements:

)x(h)x(H;

)x...,x,x(h

)x...,x,x(h

)x...,x,x(h

)x(h;

z

z

z

z x

n21m

n212

n211

m

2

1

4. Observability

State Estimation



84

For obervability purposes:

We also linearize. If the linear system is observable,the nonlinear system is observable too:

zH·x

z)x(h

4. Observability

State Estimation



85

H

zx·H

m

measurements

N

State variables

H

Linear system of

equations

4. Observability

State EstimationO



86

General solution:

·Nxxzx·H Par

where,

- xPar is a particular solution

- N is the null space of the system

- ρ is a coefficient vector

4. Observability

State Estimation4 Ob bili



87

For a single solution, the null space needs to be null:

j,i;0Nx ijSINGLEi

If so, the system is observable

4. Observability

State Estimation4 Ob bilit



88

Observability algorithm:

1) Compute H.

)x...,x,x(h

)x...,x,x(h

)x...,x,x(h

)x(h

n21m

n212

n211

4. Observability




89

To compute the Jacobian

a) Flat start:

b) Typical values:

i0,1V ii d

i0],,[]05.1,95.0[V iii dd

4. Observability




90

2) Compute the null space of H,

using MatLab N=null(H);

3) Chech NT columnwise A) Null column variable is observable.

B) No null column variable not observable.

C) Null matrix system observable.

4. Observability

State Estimation4 Ob bilit E l



91

System

V1

P13

2

V21

Q31

Q21

V33

Xij=0.1j j,i

4. Observability – Example

State Estimation4 Obser abilit E ample



92

Measurements:

Variable Value p.u. Variance

V1 0.95 10-3

V2 1.02 10-3

V3 1.03 10-3

P12 1 10-2

Q21 -0.4 10-2

P13 0.8 10-2

Q31 -0.2 10-2


State Estimation4 Observability Example



93

State variables 33322211 δVU,δVU,0VU

3

2

3

2

1

5

4

3

2

1

δ

δ

V

V

V

x

x

x

x

x

x

)xx(cos[xX

x

)xsen(X

xx

)xx(cos[xX

x

)xsen(X

xx

x

x

x

h(x)

35113

3

513

31

241

12

2

412

213

2

1





94

Jacobian matrix h(x): Hij=

j

i

xh

)x(senX

xx0

X

]x2)xcos(x[0)xcos(

X

x

)xcos(X

xx

0)x(senX

x

0)x(senX

x

0)x(senX

xx0

X

]x2)xcos(x[)xcos(

X

x

0)xcos(X

xx0)x(sen

X

x)x(sen

X

x00100

00010

00001

)x(H

5

13

31

13

3515

13

3

513

21

513

1

513

3

412

21

12

2414

12

2

4

12

214

12

14

12

2





95

1) H evaluated as a typical woring point and at the flat

start

Tfun ]3.02.098.098.001.1[x

T0 ]00111[x





96

Results:

Significant differences

0010010

100000

0001010

01000000100

00010

00001

)x(H 0

95.2086.9036.9

55.9001.3089.2

099.1060.961.9

080.9003.295.100100

00010

00001

)x(H fun





97

2) Compte the null space H(xfun) N

3) Check columnwise NT

MatLab

» N_t=null(H)'

N_t =

Empty matrix: 0-by-5

»

Null space is null.





98

Two redundant measurements

- We eliminate Q31

- State still observableV1

1

V3

V22

P13

P12

Q21

Q31TN Empty matrix: 0-by-5





99

- We eliminate P13

θ3 is no longer observable

This is consistent because the only

Information availbe at bus 3 is the voltage

magnitude.

V1

1

V3

2

P13

P12

Q21

Q31TN [ 0 0 0 0 1 ]

V2





100

Once observability is guarenteed we estimate the

state





101

)]x(hz[W)x(Hx)x(G kkTkk D

)x(WH)x(H)x(G kkTk

xxx k1k D





102

Tolerance is 10-3xD

Iteration 1 2 3 4

V1 1 1.0177 1.0214 1.0214 0.1005

V2 1 0.9816 0.9799 0.9799 0.1173

V3 1 1.0007 1.0013 1.0013 0.0995

Θ2 0 -0.1000 -0.0966 -0.0996 0.0209

Θ3 0 -0.0800 -0.0781 -0.0781 0.0014

34 10·x D





103

GAMS





104

Solution:





105

Comparison

Solution V1 V2 V3 Delta2(rad) Delta3(rad)

MatLab 1.0214 0.9799 1.0013 -0.0966 -0.0781

GAMS 1.022 0.980 1.001 -0.103 -0.078





106

Flat start

Flat start is good to copute the Jacobian because such Jacobian

involves teh minumum amount of “non-natural” couplings.

)x(H]00111[x 0T

0





107

V1

P13

2V2

1

Q31

Q21

V3 3

Xij=0.1j j,i


State Estimation4 Observability – Example



108

Natural coupling:

If P13 is eliminated, info pertaining to bus 3 is:

VQδP

313 QV





109

Jacobians

0010010

100000

0001010

010000

00100

00010

00001

)x(H0

7976.0084.9098.9

19.10079.0078.0

099.0043.975.9

095.9002.198.0

00100

00010

00001

)x(H fun

Row 5 not used (P13)





110

Null space is obtaine for both matrices

Flat start: [ 0 0 0 0 1 ]

variable θ3 NOT observable

Estimate: Empty matrix: 0-by-5

variable θ3 is observable

What do we do?





111

Using GAMS:

Phase of bus 3 is null no flow through line 1-3

Inconsistent result.





112

- System analytically observable since δ3 is computedusing Q13 y V3, which are measurements looselyrelated to δ3. The outcome is thus unreliable.

- Flat start result is more appropriate from a practicalviewpoint.

4. Observability Example

State Estimation5 Bad measurements



113

• Some bad measurements may scape the initial

screening.

• Bad measurement detection: Test x 2 .

• Bad measurement identification: Largest normalized

residual test.

5. Bad measurements




114

Chi-square with K degrees of freedom. :xJ

:nNK Number of measurement minus number of

state variables.

:KxJ

:2Kσ xJ

If J(x) large enough, bad measurement likely.

Average value.

Standard deviation.

5. Bad measurements




115

Specifically:

• If we fix α (e.g. 0.01), since J(x) is x 2 with K

degrees of freedom, we can compute t j

• If the value for J(x) from the optimization problme

is smaller than t j no bad measurements within a

confidence level of 1α

αtxJP j

5. Bad measurements




116

5. Bad measurements




117

1. Compute J(x’), where x’ is the state estimate

2. Considering a x 2 with K degrees of freedom and a

confidence level, compute t j

3. If J(x’) t j , bad measurments are present with a degree

of certitude of 1-α.

Algorithm:

αtx'JP j

5. Bad measurements




118

51K11n

62N

76.6t0.01α j

If no bad measurement with

donfidence 99 % 40.33x̂J

If bad measurements are detected, they need to

be identified.

If bad measurement present with

confidence 99 %

207.94x̂J

Six-bus case:

5. Bad measurements

State Estimation5. Bad measurements



119

5. Bad measurements




120

Identification:

1. Residuals, r = z-h(x ), are distributes following a

N(0,Ω), where Ω is the covariance matrix:

2. Normalized residual follow a N(0,1).

T11T HHRHHRΩ

ii

iNi

Ω

r r

5. Bad measurements




121

nir

nir PDF

ri

esti

min

iσ

hzr

Valor real

5. Bad measurements

Largest normalized residual test:




122

If r iN is N(0,1), for a condidence level of 95%:

• If |r iN| >1.6449 zi

m erroneous

• If |r iN| <1.6449 zi

m OK

5. Bad measurements

State Estimation5. Bad measurements - example



123

Convariance matrix:

6400.00501.0-3260.0-0326.06260.0-3043.03218.0

0501.0-0039.00255.00025.0-0490.00238.0-0252.0-3260.0-0255.06808.00680.0-3188.06355.0-3431.0

0326.00025.0-0680.0-0068.00319.0-0635.00343.0-

6260.0-0490.03188.00319.0-6124.02976.0-3148.0-

3043.00238.0-6355.0-0635.02976.0-5933.03203.0-3218.00252.0-3431.00343.0-3148.0-3203.0-6612.0

10Ω3

p




124

Residual and normalized residual verctor:

Since |r 1N| >1.6449, voltage measumrent at bus 1 is

erroneous.

0.0299-

0.0023

0.0433-

0.00430.0286

0.0400

0.0715-

r

1806.1-

1733.1

6599.1-

6611.11557.1

6406.1

7806.2-

Ω

r r

ii

iN

p




125

• Measuremnt V1 eliminated:

X12,X13,X23 = 0.1

V1

V3

V2

M13

M12

tres

dos

Magnitud Valor p.u. Varianza

V2 1.02 10-3

V3 1.03 10-3

P12 1 10-2

Q21 -0.4 10-2

P13 0.8 10-2

Q31 -0.2 10-2

p




126

3

2

3

2

1

5

4

3

2

1

δ

δ

V

V

V

x

x

x

x

x

x

p




127

• Measurement vector:

• Measurement V1 not used.

2.0

80.0

40.0

00.1

03.1

02.1

Q

P

Q

P

V

V

z

31

13

21

12

3

2

p




128

• Weighting matrix:

• Measurements as a function of state variables:

10000000

01000000

00100000

00010000

000010000

000001000

W

2

3

1 2

4

2

1 4 2

1 3

5

3

3 1 5

sin( )

( ) cos( )

sin( )

cos( )

x

x

x x x

X

xh x x x x

X

x x x

X

x x x x

X




129

• Jacobian: j

iij

x

hH

)x(senX

xx0

X

x2xcosx0)xcos(

X

x

)xcos(X

xx0)x(senXx0)x(sen

Xx

0)x(senX

xx0

X

x2xcosx)xcos(

X

x

0)xcos(X

xx0)x(sen

X

x)x(sen

X

x00100

00010

)x(H

531351

53

531

51

53

421241

42

421

41

42




130

1. Initialization:

2. H(xk) and G(xk):

T0 00111x

T0 000011)x(h




131

- Jacobian:

- Gain matriz:

0010010100000

0001010

010000

00100

00010

)x(H 0

100000

010000

00110100001110

00101020

10WHH)x(G 3T0




132

- Residual vector:

2000.0

8000.0

4000.0

0000.1

0300.0

0200.0

)x(hz)x(r

h(x)-zr(x)

00




133

3. Solving

4. New state:

D

0800.0

1000.0

0345.0

0155.0

0550.0

Wr HGx T10

)x(hzW)x(Hx)x(GkKTkk

D

kk1k xxx D

D

0800.0

1000.0

0345.1

0155.1

0550.1

xxx 001




134

5. We iterate until Dx is smaller than 0.001.




135

Solution:

J(x) V1

V2

V3

Matlab 0.0644 1.0581 1.0147 1.0355 -0.0932 -0.0731

Gams 0.0645 1.0581 1.0147 1.0355 -0.0932 -0.0731

2δ

3δ




136

Checking again:

J(x) = 0.0645 < 3.8415, no bad measurements within

a confidence level of 95%.

8415.3t95.0α105.0α




• Residual and normalized residual vectors:

Since |r iN| <1.6449, no bad measurements.

0.0052

0.0001-

0.0058-

0.0007

0.0055-

0.0053

r

2432.0

0814.0-

2650.0-

3616.0

2536.0-

2540.0

Ω

r r

ii

iN

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