Dr R Jegatheesan

Professor, EEE Dept.SRM University

PS2106

POWER SYSTEM STATE ESTIMATION

UNIT I - INTRODUCTION TO STATE ESTIMATION (9 hours) Need for state estimation - Comparison of power flow and state estimation

problems - Measurements - Redundancy - Noise - Measurement functions -

Weighted least square errors - Measurement Jacobian matrix - Weights - Gain

matrix - State estimation applied to DC networks - Bad data deduction and

removal.

UNIT II - POWER SYSTEM STATE ESTIMATION (9 hours) Building model for power system components - Mean - Standard deviation -

Maximum likelihood estimation - Measurement equations - WLS State Estimation

- Measurement Jacobian matrix - Gain matrix - Cholesky decomposition -

Forward and backward substitutions - State estimation algorithm - Decoupled

formulation of WLS state estimation - DC State estimation.

UNIT III - NETWORK OBSERVABILITY ANALYSIS (9 hours) Networks and graphs - Methods of observability analysis - Numerical method

based on nodal variables - Determining the unobservable branches -

Identification of observable islands - Measurement placement to restore

observability.

UNIT IV - BAD DATA DETECTION AND IDENTIFICATION (9 hours) Properties of measurement residuals - Classification of measurements - Bad data

deduction using Chi-square distribution - use of normalized residuals for bad

data deduction - Bad data identification by largest normalized residual test.

Hypothesis testing identification.

UNIT V - PHASOR MEASUREMENT UNITS (9 hours) Basics of Phasor Measurement Unit (PMU) - Optimal placement of PMUs -

Methods to reduce the number of PMUs. State estimation including PMU -

Optimal placement of PMUs in large scale systems - State estimation including

FACTS devices.

REFERENCES 1. Ali Abur and Antonio Gomez Exposito, “Power System State Estimation

Theory and Implementation”, Marcel Dekker. Inc., New York, Basel 2004. 2. J J Grainger and W D Stevension, “Power System Analysis”, McGraw-Hill,

Inc., 1994. 3. A Monticelli, “State Estimation in Electric Power Systems”, Kluwer

Academic Publishers, 1999. 4. Mukhtar Ahmad, “Power System State Estimation”, LAP Lambert Acad

Publishers, 2013. 5. Naim Logic. “Power System State Estimation”, LAP Lambert Acad

Publishers, 2010.

Chapter 1

INTRODUCTION TO STATE ESTIMATION

Power systems are composed of generation, transmission, sub-transmission and

distribution systems. Power is injected into the system by the generators and

consumed by the loads. In between the generation system and loads there exists

large complex transmission and distribution system.

The output voltages of the generators are in the range of 11 to 25 kV.

Transformers are used to increase the voltage levels to levels ranging from 220

kV all the way up to 765 kV at the generating stations for efficient power

transmission. High voltage is preferred at the transmission system for different

reasons one of which is to minimize the copper losses that are proportional to the

current flows in the lines.

At the receiving end, the transmission systems are connected to the sub-

transmission and distribution systems which are operated at lower voltage levels

ranging from 132 kV to 0.44 kV. Distribution systems are typically configured to

operate in a radial configuration, where the feeders stretch from distribution

substations and form a tree structure with their roots at the substation and

branches spreading over the distribution area.

1.1 OPERATING STATES OF A POWER SYSTEM

Knowing the network model and the complex phasor voltages at every

system bus the operating conditions of a power system at a given

point in time can be determined. Since the set of complex phasor

voltages fully specifies the system, it is referred as the static state of

the system.

As operating conditions change, the system may move into one of the

five possible states namely, normal, alert, emergency, in extremis and

restorative. Figure 1.1 shows the different operating states of a power

system and the transitions between them.

Restorative

Alert

In extremis Emergency

Normal

Fig. 1.1 Power system operating states

Fault clearing Excitation control Fast valving Load shedding

Load shedding System separation

Generation shifting Increase reserve

Emergency

NORMAL STATE:

In the normal state, all system variables are within the normal range and no

component is being overloaded. The system operates in a secure manner and is

able to withstand a contingency without violating any of the constraints.

ALERT STATE:

The system enters the alert state if the security level falls below a certain limit of

adequacy, or if the possibility of disturbance increases because of adverse

weather conditions such as the approach of severe storms. In this state, all the

system variables are still within the acceptable range and all the constraints are

satisfied. However, the system has been weakened to a level where a contingency

may cause an overloading of a component that places the system in an

emergency state. If the disturbance is very severe, the in extremis state may result

directly from the alert state.

RESTORATIVE STATE:

Preventive action, such as generation shifting or increased reserve, can be taken

to restore the system to the normal state. If the restorative steps do not succeed,

the system will remain in the alert state.

EMERGENCY STATE:

The system enters the emergency state, if a sufficiently severe disturbance

occurs when the system is in the alert state. In this state, voltages at many buses

are low and / or component loadings exceed the short-term emergency ratings.

The system is still intact and may be restored to the alert state by initiating

emergency control actions such as fault clearing, excitation control, fast-valving

and load curtailment. If the above measures are not applied or are ineffective, the

system will move to in extremis state.

IN EXTREMIS STATE:

If the control action taken during the emergency state is insufficient, then the

system enters into in extremis state. The result is cascading outages and

possibly a shut-down of major portion of the system. Control actions, such as

load shedding and controlled system separation, are aimed at saving as much

system as possible from a widespread blackout.

In case the control action taken are effective, the system moves to restorative

state in which further action is being taken to reconnect all the facilities and to

restore the system load. The system transits from the restorative state to either

alert state or normal state, depending on the system conditions.

1.2 POWER SYSTEM SECURITY ANALYSIS

Power systems are operated by system operators from the area control centers.

The main goal of the system operator is to maintain the system in a normal

secure state as the operating conditions vary during the daily operation.

Following are required to achieve this goal requires

i) Continuous monitoring of the system conditions

ii) Identification of the operating state and

iii) Determination of the necessary preventive action in the case of state found

to be insecure.

This sequence of operation is referred as the security analysis of the system.

The first step of security analysis is to determine the current state of the system.

This involves acquisition of measurements from all parts of the system.

The measurement may be both of analog and digital type.

Substations are equipped with devices called Remote Terminal Unit (RTU) which

collect various types of measurements from the field and transmit them to the

control center.

More recently, the so-called Intelligent Electronic Device (IED) are replacing or

complementing the existing RTUs.

Once the data are collected at the they are processed at the control center in

order to determine the system state.

It is possible to have a mixture of these devices (RTUs and IEDs)

connected to a Local Area Network (LAN) along with Supervisory

Control And Data Acquisition (SCADA) front end computer, which

supports the communication of the collected measurements to the

host computer at the control center.

The SCADA host computer at the control center receives

measurements from all SCADA systems installed at the monitored

substations’ via one of many possible types of communication links

such as fiber optic, satellite, microwave, etc.

Fig.1.2 shows the configuration of EMS / SCADA system for a typical

power system.

CONTROL CENTER Local Area Network

Monitored Devices Substation

Fig. 1.2 EMS / SCADA system configuration

Communications Network

CORPORATE OFFICES

Fig.1.2 shows the configuration of EMS / SCADA system for a typical power

system.

Planning and

analysis functions

SCADA host computer

Energy Management System (EMS) Application Functions

State Estimator

SCADA Front end computer

RTU RTU IED RTU IED

Measurements received at the control center will include line power flows, bus

powers, bus voltage magnitudes, line current magnitudes, generator output

powers, loads powers, circuit breaker and switch status information, transformer

tap positions and switchable capacitor bank values.

These raw data and measurements are processed by State Estimator (SE) in

order to filter the measurement noise and also detect gross errors if any. Thus,

State estimator solution will provide reliable estimate of the system state based

on the available measurements and on the assumed system model.

The information pertaining to the system state will then be passed on to all the

Energy Management System (EMS) application functions such as the

contingency analysis, automatic generation control, automatic load frequency

control, economic load dispatching, load forecasting and optimal power flow, etc.

The same information will also be available via a LAN connection to the corporate

offices where other planning and analysis functions can be executed off-line.

Development of EMS that encompass State Estimation

Initially, power systems were monitored only by supervisory control systems

which essentially monitor and control the status of circuit breakers at the

substations. Generator outputs and the system frequency were also monitored

for purpose of Automatic Generation Control (AGC) and Economic Dispatch (ED).

These supervisory control systems were later augmented by real time data

acquisition capabilities. This allows the control centers to gather all sorts of

analog measurements and circuit breaker status data from the power system.

This led to the establishment of Supervisory Control and Data Acquisition

(SCADA) systems.

The main motivation behind the development of SCADA system was the

facilitation of security analysis. Various operating functions can be executed only

by knowing the real-time operating conditions of the system.

However, the information provided by the SCADA system may not always be

reliable due to the errors in the measurements, telemetry failures, communication

noise, etc.

Furthermore, the collected set of measurements may not allow direct extraction

of the corresponding A.C. operating state of the system. For instance, bus voltage

phase angles are not typically measured and not all the transmission line flows

are available.

Besides, it may not be economically feasible to telemeter all possible

measurements even if they are available.

1.3 STATE ESTIMATION

The idea of state estimation in power systems was first recognized in the year

1970 and subsequently addressed by Fred Schweppe.

Introduction of state estimation function broadened the capabilities of the SCADA

system computers, leading to the establishment of Energy Management System

(EMS).

State estimators provide a reliable real-time data base of the system, including the

existing state, based on which necessary security assessment functions can be

deployed in order to analyze contingencies, and to determine any required

corrective actions.

The state estimators typically include the following functions

Topology processor: Gathers status data about the circuit breakers and switches

and configures the one-line diagram of the system.

Observability analysis: Determines if a state estimation solution for the entire

system can be obtained using the available set of measurements. Identifies the

unobservable branches and the observable islands in the system if any exist.

State estimation solution: Determines the optimal estimate for the system state,

which is composed of complex bus voltages in the entire power system, based

on the network model and the gathered measurements from the system.

Bad data processing: Detect the existence of gross errors in the measurement

set. Identifies and eliminates bad measurements provided that there is enough

redundancy in the measurement configuration.

Parameter and structural error processing: Estimates various network

parameters such as transmission line model parameters, tap changing

transformer parameters, shunt capacitor or reactor parameters. Detects

structural errors in the network configuration and identifies the erroneous

breaker status.

APPLICATION FUNCTIONS

RAW MEASUREMENTS

RELIABLE STATE

Thus Power System State Estimator constitutes the core of

the on-line security analysis function.

It acts as a filter between the raw measurements received

from the system and all the application functions that require

the most reliable data base pertaining to the current state of

the system.

STATE ESTIMATOR APPLICATION FUCTIONS

DIFFERENCE BETWEEN POWER FLOW AND STATE ESTIMATION

• Power system networks are large and complex.

• For proper operation of power system, several control actions are necessary.

• To decide about the control actions, system status need to be determined first.

• System state can be decided by conducting Power flow analysis.

POWER FLOW ANALYSIS

If even one of the inputs is unavailable, power flow solution can not be obtained.

Gross errors in one or more input quantities can cause power flow results to become useless.

POWER SYSTEM NETWORK

P and Q injections

Bus voltages

V δand

STATE ESTIMATION

POWER SYSTEM NETWORK

Bus voltages

V δ

P and Q injectionsBus voltages

P and Q line flows

V

Conveniently

measurable

and

Number of inputs > Number of state variables.State variables are estimated.

Bad data if any can be detected and removed.

Let us say that we like to determine the area of a rectangular field.

Area A = L x B

We can measure L and B and compute the area.

But both the measurements may have certain errors and hence area calculated is

not accurate.

To overcome this we can measure

1. Length L

2. Breadth B

3. Diagonal D

4. Perimeter P

From these four measurements area A can be estimated filtering out the errors

and the area obtained in this way will be more accurate and reliable.

The following are the features of State Estimation

1. The techniques developed provide an estimate of the

system state.

2. Further they provide a quantitative measure of how good

the estimate is before it is used for real-time power

flow calculations.

3. The inputs are easily measurable P and Q line flows, P

and Q bus power injections and voltage magnitudes ІVІ.

4. The unavoidable errors of the measurements are

assigned statistical properties and the estimates of the

states are subjected to statistical testing before being

accepted as satisfactory.

THE METHOD OF WEIGHTED LEAST SQUARES

In the power system, meters are used to measure real power,

reactive power, voltages and currents.

The analog quantities measured at substations and other

strategical points, pass through transducers and analog-to-

digital converters and the digital outputs are then

telemetered to energy control center.

The data received at energy center are processed by the

computer.

The acquired data always contain inaccuracies.

The errors can be quantified in a statistical sense and the

estimated values are then either accepted as reasonable or

rejected based on the accuracy.

Because of noise, the true values of physical quantities are

never known.

The method of least squares is often used to “best fit”

measured data relating two or more quantities.

First, we apply the method to a simple dc circuit which

contains measurement errors.

Later we extend the estimation procedures to the ac power

system.

The best estimates are chosen as those which minimize the

weighted sum of squares of the measurement errors.

This method is known as Weighted Least Square ( WLS )

method.

Consider the simple dc circuit of Fig. 1 below.1z

Here and are the unknown values to be estimated.

They are represented by the state variables

, , and are the four measurements made.

In terms of network parameters and the state variables, true values of measurements can be written as (using Super position theorem)

1V 2V

1z

.2xand1x

V

A

V

A1 Ω

1 Ω 1 Ω

1 Ω 1 Ω

-

+ +

+ +

++

-

- -

--1V 2V4z

3z

2z

3z2z 4z

Each measurement will contain some error . Thus measurements are given by

(1)

(2)

21true4

21true3

21true2

21true1

x8

3x

8

1z

x8

1x

8

3z

x8

5x

8

1z

x8

1x

8

5z

4214

3213

2212

1211

ex8

3x

8

1z

ex8

1x

8

3z

ex8

5x

8

1z

ex8

1x

8

5z

iz ie

The above equation can be written as

(3)i.e.

(4)

In a compact form z = H x + e

Error e = z -

= z - H x

(5)

(6)

True values of can not be determined.

However, we can calculate the estimates of .

truez

21 x andx

21 xandx

4true442421414

3true332321313

2true222221212

1true112121111

ezexhxhz

ezexhxhz

ezexhxhz

ezexhxhz

4

3

2

1

2

1

4241

3231

2221

1211

4

3

2

1

e

e

e

e

x

x

hh

hh

hh

hh

z

z

z

z

e = z - H x (6) Just as truez = H x we can write,

z = H

x

where

x is the estimated state values and

z is the estimated measurements.

Estimated errors are obtained as

z-ze

xH-z (7) Substituting z = H x + e in eq. (7)

e H x + e - H

x = e - H (

x - x ) (8) PROCEDURE

First find the estimated state

x

From this compute estimated measurement

z using

z = H

x

Finally find estimated error

e using

z-ze

xH-z It is to be noted that

x ,

z and

e are related.

We need to find the state variables for which weighted sum of squares of errors is minimum. i.e. it is required to minimize

244

233

222

211

2j

4

1jj ewewewewewf

(9)

Noting that error is related to and necessary conditions for f to be minimum are

x = 0

]x

eew

x

eew

x

eew

x

eew[2

x

f

21

444

2

333

2

222

2

111

2

]x

eew

x

eew

x

eew

x

eew[2

x

f

1

444

1

333

1

222

1

111

1

x

x

x = 0

(10)

(11)

0

0

e

e

e

e

w

w

w

w

x

e

x

e

x

e

x

ex

e

x

e

x

e

x

e

4

3

2

1

4

3

2

1

2

4

2

3

2

2

2

1

1

4

1

3

1

2

1

1

(12)

x

ie 1x 2x

2

4

1

4

2

3

1

3

2

2

1

2

2

1

1

1

x

e

x

ex

e

x

ex

e

x

ex

e

x

e

= -

4241

3231

2221

1211

hh

hh

hh

hh

= - H where matrix H =

4241

3231

2221

1211

hh

hh

hh

hh

Note that the elements of H matrix are independent of state variables x1 and x2. This is true only

when the measurements are LINEARLY related to state variables as shown in eq. (1)

24214144

23213133

22212122

21211111

xhxhze

xhxhze

xhxhze

xhxhze

(3) eq. From

T

Using the above result in eq.(12) we have

4241

3231

2221

1211

hh

hh

hh

hh

0

0

e

e

e

e

w

w

w

w

4

3

2

1

4

3

2

1

(13)

We are more interested on

x rather than

e . We can replace

e by

x .

Using compact notation

MATRIXGAINtheascalledisG

HWHGwhere

(14) zWHGxTherefore

zWHxHWHThus

0)xHz(WHi.e.0)zz(WHi.e.0eWH

T

T1

TT

TTT

We expect

X to be close to the true values of X. Let us now develop an expression for

X - X.

Substituting z = H X + e in eq.(14) namely zWHGx T1

(15) eWHGXXThus

eWHGXeWHGXHWHG)eXH(WHGX

T1

T1T1T1T1

For the example circuit

4

3

2

1

22

11

e

e

e

e

****

****

xx

xx (16)

This means that any one or more of the four errors 4321 eande,e,e can influence the

difference between each of the state estimate and its true value. i.e. WLS calculation spreads the effect of the error in any one measurement, to some or all the estimates.

Similarly, we can compute the estimated error as (using Eq. 15)

(17) e)WHGHI(

e)WHG(He)XX(HeXHeXHzze

T1

T1

Use of eqs. (15) and (17) will be given after introducing the statistical properties of errors. Summary of formulas: z = H X + e e = z - H X

Using WLS method, zWHGX T1

where G = HWHT

z = H

x

)XX(HeXHeXHzze

Procedure for DC State Estimation:

1. Generally a set of measurements and weighting factors of errors will be the

input data.

2. Relate each measurement to the state variables and thereby find the H

matrix.

3. Compute the gain matrix G from G = HT W H

4. Determine estimated state variables from zWHGx T1

5. Compute

x Hz

6. Calculate the estimated errors from

z - ze

Example 1 In the dc circuit of Fig.1, the meter readings are 1z = 9.01 A, 2z = 3.02 A,

3z = 6.98 V and 4z = 5.01 V. Assuming that the ammeters are more

accurate than the voltmeters, let us assign the measurement weight 100w 1 , 50wand50w100,w 432 respectively. Determine WLS

estimates of the voltage sources 21 VandV . Also calculate the estimated measurements and the estimated errors in the measurements. Solution

Given ;

50

50

100

100

W;

5.01

6.98

3.02

9.01

z

it is required to find

eandz,x .

For the DC circuit shown in Fig. 1, measurements are related to the state

variables as

Thus

0.3750.125

0.1250.375

0.6250.125

0.1250.625

H

4214

3213

2212

1211

ex8

3x

8

1z

ex8

1x

8

3z

ex8

5x

8

1z

ex8

1x

8

5z

Gain matrix G = HWHT

WHT =

0.3750.1250.6250.125

0.1250.3750.1250.625

50

50

100

100

=

18.756.2562.5012.50

6.2518.7512.5062.50

Gain matrix G =

18.756.2562.5012.50

6.2518.7512.5062.50

0.3750.125

0.1250.375

0.6250.125

0.1250.625

=

48.437510.9375

10.937548.4375

Its inverse 1G =

0.02180.0049

0.00490.0218

Estimated state vector

zWHGX T1

=

0.02180.0049

0.00490.0218

18.756.2562.5012.50

6.2518.7512.5062.50

5.01

6.98

3.02

9.01

=

0.43860.22811.29820.0351

0.22810.43860.03511.2982

5.01

6.98

3.02

9.01

=

V8.0261

V16.0072

G-1 HT W

Estimated measurements

XHZ =

0.3750.125

0.1250.375

0.6250.125

0.1250.625

8.0261

16.0072 =

5.01070

7.00596

3.01544

9.00123

Estimated errors

zze =

5.01

6.98

3.02

9.01

-

5.01070

7.00596

3.01544

9.00123

=

V0.00070-

V0.02596-

A0.00456

A0.0877

In the proceeding example the state of the system is estimated using WLS method. The results are of little values if we cannot measure, by some means, how good the estimate is. Consider for instance, that the voltmeter reading 4z of Fig.1 is 4.40 V rather than 5.01 V used in Example 1. If other three meter readings are

unchanged, we can determine

X ,

z and

e as follows.

zWHGX T1

=

0.43860.22811.29820.0351

0.22810.43860.03511.2982

4.40

6.98

3.02

9.01

=

V7.75860

V15.86807

XHz =

0.3750.125

0.1250.375

0.6250.125

0.1250.625

7.75860

15.86807 =

4.89298

6.92035

2.86562

8.94772

zze =

4.40

6.98

3.02

9.01

-

4.89298

6.92035

2.86562

8.94772

=

V0.49298-

V0.05965

A0.15438

A0.06228

It is meaningless to compare the results obtained in the above two cases, because at any one time, only one set of measurement will be available. Given a set of measurements, how we can decide if the corresponding results should be accepted as good estimates of the true values? What criterion for acceptance is reasonable? If a grossly erroneous meter reading is present, can we detect that fact and identify the BAD DATA ?

STATISTICS, ERRORS AND ESTIMATES

Generally unavoidable random noise enters into measurement process and distorts the results from the true values.

Repeated measurements of the same quantity reveals certain statistical properties from which the true value can be estimated.

If the measured values of a random variable are plotted, taking the measured values along the horizontal axis and their relative frequency of occurrence along the vertical axis, a bell shaped function is obtained. This function is called the Gaussian or normal probability density function and has the formula

(18)2)

σ

μx(

2

1

ε2πσ

1f(x)

where x: randam variable

μ: mean or expected value of x = E(x)

σ: standard deviation of x

This function will have the maximum value when x = μ

The probability of x to lie between μ to μ + σ is 0.342

Area under the curve gives the probability with the corresponding intervals of the horizontal axis.

f(x) has its maximum value when x equals μ, which is the expected value of x, denoted by E(x) and defined by

dx)(xfx)(xEμ

The expected value μ is often called the MEAN.

(19)

The degree to which the curve f(x) spreads out about μ depends on the variance of x defined by2σ

dx)x(f)μx(])μx([Eσ 222

(20)

The positive square root of the variance is the standard deviation σ of x.

The total area of the curve is one.

We assume that the noise terms 4321 eande,e,e are independent Gaussian

random variables with zero means and the respective variances are

.σandσ,σ,σ 24

23

22

21

Two random variables ji eande are independent when E ( ji ee ) = 0 for ji .

Later we encounter the product vector, e and its transpose Te . This is given by

e Te =

4

3

2

1

e

e

e

e

24342414

43232313

42322212

41312121

4321

eeeeeee

eeeeeee

eeeeeee

eeeeeee

eeee (21)

The expected values of diagonal elements are nonzero and correspond to

the variance. Then the expected value of Eq.(21) becomes

E ( e Te ) = R =

24

23

22

21

σ000

0σ00

00σ0

000σ

(22)

We know 1true11 ezz

The shift of e1 by z1 true in no way alter the shape and spread of the

probability density function of e1. Hence, z1 also has a Gaussian

probability density function with a mean value μ1 equal to the true value

z1 true and variance σ12 equal to that of e1. Similar remarks are applicable to

other three measurements also.

Meters with smaller error variances have narrower curves and provide

more accurate measurements. In formulating the objective function f of

Eq. (9), preferential weighting is given to more accurate measurements by

choosing the weight wi as the reciprocal of the corresponding variance σi2.

Henceforth, we specify the weighting matrix W of Eqs. (15) and (17) as

24

23

22

21

1

σ

1σ

1σ

1σ

1

RW (23)

and the gain matrix G then becomes

HRHG 1T (24) In Example 1, the weight for the first ammeter is taken as 100. This means

that its variance is 1/100 = 0.01. Thus its standard deviation is 0.1. This

means that there is 99% probability for this ammeter, when functioning

properly, to give readings within 0.3 A (3σ = 0.3 A) deviation from the true

values of its measured current. Similar interpretation can be given for the

other meters.

Let us see how close the estimated value is to the true value.

From Eq. (19),

which defines the expected value the random variable x, we can show that

E[ax + b] = a E[x] + b if a and b are constants. Therefore, taking the

expected value of both side of Eq.(15) namely

(15)eWHGXX T1

gives

]xx[E

]xx[E

2

^

2

1

^

1 =

2

^

2

1

^

1

xx[E

xx[E

]

] = G-1 HT R-1

]e[E

]e[E

]e[E

][eE

4

3

2

1

=

0

0 (25)

From the above equation it follows that

E [ i

^

i x]x (26)

which implies that the WLS estimate of each state variable has an expected

value equal to the true value.

dx)(xfx)(xEμ

Similarly we can find how close estimated measurement is to the true

measurement.

Consider (17) e)WHGHI(zze T1

Then E

^

4

^

3

^

2

^

1

e

e

e

e

= E

^

44

^

33

^

22

^

11

zz

zz

zz

zz

= [ I - H G-1 HT W ] E

4

3

2

1

e

e

e

e

=

0

0

0

0

(27)

From the above

E [^

jz ] = E [ zj ] = E [ zj true + ej ] = zj true (28)

Equations (26) and (28) state that the WLS estimates of the state variables

and the measured quantities, on the average, are equal to their true values -

an obviously desirable property, and the estimates are said to be unbiased.

The estimated values can be accepted as unbiased estimates of the true

values provided no bad measurement data are present.

TEST FOR BAD DATA

When the system model is correct and the measurements are accurate,

there is good reason to accept the state estimates calculated by the WLS

estimator. But if a measurement is grossly erroneous or bad, it should be

detected and then identified so that it can be removed from estimator

calculations. The statistical properties of the measurement errors facilitate

such detection and identification.

As seen from Eq. (27), each estimated measurement error, ^

je = ( zj - ^

jz ) is a

Gaussian random variable with zero mean. A formula for the variance of ^

je

can be determined from Eq. (17) namely

(17) e)WHGHI(

e)WHG(He)XX(HeXHeXHzze

T1

T1

in two steps.

Step 1 Using Eq. (17) ^

e T^

e = (z -^

z ) (z -^

z )T = [ I - H G-1 HT R-1 ] e eT [ I - R-1 H G-1 HT ] (29) It is to be noted that 1. LHS of Eq. (29) is a square matrix.

2. The two matrices on the RHS contain only constant elements and are

transpose of one another.

3. As seen in Eq. (22) E [ e eT ] = R

Step 2 We take the expected value of both sides of Eq. (29)

E [^

e T^

e ] = [ I - H G-1 HT R-1 ] E [ e eT ] [ I - R-1 H G-1 HT ] = [ I - H G-1 HT R-1 ] R [ I - R-1 H G-1 HT ] = [ I - H G-1 HT R-1 ] [ R - H G-1 HT ] = [ I - H G-1 HT R-1 ] [ R - H G-1 HT R-1 R ] = [ I - H G-1 HT R-1 ] [ I - H G-1 HT R-1 ] R (30) It can be verified that the matrix [ I - H G-1 HT R-1 ] multiplied by itself

remains unaltered and so

E [^

e T^

e ] = [ I - H G-1 HT R-1 ] R = R - H G-1 HT (31)

The square matrix ^

e T^

e takes the form of Eq. (21) with typical entries given

by ^

j

^

i ee . Therefore, substituting for ^

je = ( zj - ^

jz ) in the diagonal entries,

we obtain

E [2^

je ] = E [ (zj - ^

jz )2 ] = R’j j (32)

where

R’j j is symbol for the jth diagonal element of the matrix R’ = R - H G-1 HT.

Eq. (32) is actually the formula for the variance of ^

je , which makes j j'R

the standard deviation. Dividing each side of Eq. (32) by the number R’j j

gives

E

jj'

2^

j

R

e = 1 (33)

E

jj'

2^

j

R

e = 1 (33)

(34)R

ethatinterpreatcanweabove,theFrom

jj'

^

j

is a STANDARD GAUSSIAN RANDAM VARIABLE with zero mean and

variance equals to 1.

Matrices R, given by R =

24

23

22

21

σ000

0σ00

00σ0

000σ

is the covariance matrix while

R’ = R - H G-1 HT is known as the modified covariance matrix.

Also from Eq. (15) namely eWHGxx T1^

we get

( x - ^

x ) ( x - ^

x )T = G-1 HT R-1 e eT R-1 H G-1. Therefore,

E [ ( x - ^

x ) ( x - ^

x )T ] = G-1 HT R-1 R R-1 H G-1

= G-1 HT R-1 H G-1 = G-1 = (HT R-1 H)-1 (35)

Now let us see how to detect the presence of bad data.

It is to be recalled that true measurement error ej, given by zj - zj true, is

never known in engineering applications. The best that can be done is to

calculate the estimated error ^

je , which then replaces ej in the objective

function. Accordingly, objective function value is obtained as

(36)σ

eewf

mm N

1j2

j

2^

j2N

1j

^

jj

^

where Nm is the number of measurements and the weighting factor w j is set

equal to 1/σj2. This weighted sum of squares of estimated errors,

^

f itself is

a random variable which has a well known probability distribution with

values (areas) already available in tabulated form. In order to use those

tables, we need to know the mean value of ^

f which can be determined as

discussed below.

(36)σ

eewf

mm N

1j2

j

2^

j2N

1j

^

jj

^

Multiplying the numerator and the denominator of each term in Eq. (36) by

the calculated variance R’j j we get

(37)Rσ

eRf

mN

1j

2

jj'2

j

^

jjj'^

Taking the expected value on both sides we get

E [ ]f^

= E

mN

1j jj'

2^

j

2j

jj'

R

e

σ

R (38)

Use of Eq. (33), namely E

jj'

2^

j

R

e = 1 in the above results in

E [ ]f^

=

mN

1j2

j

jj'

σ

R (39)

(The above result will be more clear by expanding Eq. (38), taking Nm = 4)

For the dc circuit considered in Example 1, it will be shown in Example 2,

that the right hand side of the above equation has the numerical value of 2

(= 4- 2) for the four measurements and two state variables. More generally,

the expected value of ^

f is always numerically equal to the number of

degrees of freedom; i.e.

E [ ]f^

= Nm - Ns (40)

where Nm is the number of measurements and Ns is the number of state

variables.

Thus the mean value of ^

f is an integer which is equal to Nm - Ns. This

number is also called as redundancy of the measurement scheme.

Statistical theory shows that the weighted sum of squares of estimated

errors ^

f has Chi-square distribution 2αk,χ where χ is a Greek letter called

Chi, k is the number of degrees of freedom and α relates to the area under

the 2αk,χ curve. The Chi-square distribution very closely matches the

standard Gaussian distribution when k is large (k > 30), which is often the

case in power system applications.

p( )χ2 for k degree of freedom

2χ

2χ k,α

Area (1 - α)

Area α

The figure below shows the probability density function of 2αk,χ for a

representative small value of k. As usual, the total area under the curve

equals to 1, but here it is not symmetrically distributed.

The area under the curve to the right of αk,

2χ in Fig. equals α. which is the

probability that ^

f exceeds αk,2χ . The remaining area under the curve is the

probability (1 - α) that the calculated value of the ^

f with k degree of

freedom, will take on a value less than αk,2χ i.e.

Pr (^

f < αk,2χ ) = (1 - α) (41)

Based on this, the critical value of ^

f can be determined using the tabulated

value of αk,2χ shown in the Table below.

Table 1: Critical values of 2αk,χ

α α k 0.05 0.025 0.01 0.005 k 0.05 0.025 0.01 0.005 1 3.84 5.02 6.64 7.88 11 19.68 21.92 24.73 26.76 2 5.99 7.38 9.21 10.60 12 21.03 23.34 26.22 28.30 3 7.82 9.35 11.35 12.84 13 22.36 24.74 27.69 29.82 4 9.49 11.14 13.28 14.86 14 23.69 26.12 29.14 31.32 5 11.07 12.83 15.09 16.75 15 25.00 27.49 30.58 32.80 6 12.59 14.45 16.81 18.55 16 26.30 28.85 32.00 34.27 7 14.07 16.01 18.48 20.28 17 27.59 30.19 33.41 35.72 8 15.51 17.54 20.09 21.96 18 28.87 31.53 34.81 37.16 9 16.92 19.02 21.67 23.59 19 30.14 32.85 36.19 38.58 10 18.31 20.48 23.21 25.19 20 31.41 34.17 37.57 40.00 For example, choosing α = 0.01 and k = (Nm - Ns) = 2, we can conclude that

the calculated value of ^

f is less than the critical value of 9.21 with a

probability of (1 - 0.01) or 99% confidence since 20.012,χ in Table.

The statistical properties of the measurement errors facilitate detection and

identification of bad measurement data.

The Chi-square distribution of

f provides a test for detection of bad

measurement. The procedure is as follows:

Use the raw measurements z from the system to determine the WLS

estimates x of the system state from

zWHGx T1

Substitute the estimates

x in equation

xHz .

Calculated the estimated errors

jjj zze

Evaluate the sum of weighted squared estimated errors from

mm N

1j2j

2

jN

1j

2

jj σ

eewf (29)

For the appropriate number of degrees of freedom sm NNk and

a specified probability α , find the critical Chi-square value 2αk,χ

from Table 1. Determine whether or not

f < 2αk,χ (30)

is satisfied. If it is, then the measured raw data and the estimated

values of the state variables are accepted as being accurate.

When the requirement of inequality (30) is not met, there is reason to suspect the presence of at least one bad measurement. In that case compute the diagonal elements of modified covariance matrix R’ given by

R]RHGHI[RRHGHRHGHRR 1T1-1T1T1' (31)

and designate them as 'jjR .

It is to be noted that matrix R’ is independent of measurements.

Omit the measurement corresponding to the largest standardized

error, (considering only the magnitudes) namely

'jj

j

R

e

(32)

and reevaluate the state estimates along with the sum of weighted

squared errors

f .

If the new value of

f satisfies the Chi-square test of inequality (30), then the omitted measurement has been successfully identified as the bad data.

Example 2 Suppose that the weighting factors 41 wtow in Example 1 are the corresponding reciprocals of error variances for the four meters of Fig,1.

Show that the expected value of

f is equal to the degree of freedom. Solution

Expected value of

f is given by

mN

1j2j

'jj

σ

R]f[E

First let us evaluate the diagonal elements of

T1' HGHRR To take advantage of previous calculations in Example 1, we write the

matrix 'R in the form

R]RHGHI[R 1T1'

Then from Example 1, taking value of 1T1 RHG we can compute

1T1 RHGH =

0.43860.22811.29820.0351

0.22810.43860.03511.2982

0.3750.125

0.1250.375

0.6250.125

0.1250.625

Only the diagonal elements resulting from this equations are required or

calculating ]f[E

.

1T1 RHGH =

0.1930***

*0.1930**

**0.8070*

***0.8070

24

23

22

21

'44

'33

'22

'11

σ...

.σ..

..σ.

...σ

0.1930***

*0.1930**

**0.8070*

***0.8070

1...

.1..

..1.

...1

R...

.R..

..R.

...R

31eq.From ,

The expected value of

f is

4

1j2j

'jj

σ

R]f[E

= 24

24

23

23

22

22

21

21

σ

σ)0.1931(

σ

σ)0.1931(

σ

σ)0.8071(

σ

σ)0.8071(

= ( 1 + 1 + 1 + 1 ) – ( 0.807 + 0.807 + 0.193 + 0.193 ) = 4 – 2 = 2 which is the number of degrees of freedom when the system of Example 1

has 4 measurements and 2 state variables.

Example 3

Using the Chi-square test of inequality

f < 2αk,χ check for the presence

of bad data in raw measurements of Example 1. Choose α = 0.01.

Solution

2Nand4N sm . Then k = 4 – 2 = 2

With k = 2 and α = 0.01, Chi-square critical value 20.012,χ as given in Table

1 is 9.21.

In Example 1,

0.00070

0.02596

0.00456

0.00875

e

e

e

e

4

3

2

1

Estimated sum of squares

f , is calculated as

4

1j2j

2

j

σ

ef =

2

4

2

3

2

2

2

1 e50e50e100e100

= 2222 )0.00070(50)0.02596(50)0.00456(100)0.00877(100 = 0.043507 This value is less than the Chi-square critical value of 9.21. Therefore, we conclude that the raw measurement set of Example 1 has no bad measurement.

Example 4 Suppose that the raw measurement set for the system of Fig.1 is given by

V4.40

V6.98

A3.02

A9.01

z

z

z

z

4

3

2

1

Check for the presence of bad data using Chi-square test for α = 0.01.

Eliminate any bad data detected and calculate the resultant state estimate

from the reduced data set.

Solution

For the given set of measurement we already calculated the estimated

errors as

0.49298-

0.05965

0.15439

0.06228

e

e

e

e

4

3

2

1

Therefore

f = 2

4

2

3

2

2

2

1 e50e50e100e100

= 2222 )0.49298(50)0.05965(50)0.15439(100)0.06228(100 = 0.3879 + 2.3836 + 0.1779 + 12.1515 = 15.1009

This value of

f exceeds 20.012,χ value of 9.21 and so we conclude that there

is at least one bad measurement. The standardized error estimates are next calculated using the diagonal elements '

jjR as follows:

Matrix R’ = [ I – H G-1 HT R-1 ] R Then taking value of 1T1 RHG from Example 1, we can compute

1T1 RHGH =

0.43860.22811.29820.0351

0.22810.43860.03511.2982

0.3750.125

0.1250.375

0.6250.125

0.1250.625

Only the diagonal elements resulting from this equations are required,

diag. elements of 1T1 RHGH =

0.1930***

*0.1930**

**0.8070*

***0.8070

From eq. 31, knowing R’ = [ I – H G-1 HT R-1] R

24

23

22

21

'44

'33

'22

'11

σ...

.σ..

..σ.

...σ

0.1930***

*0.1930**

**0.8070*

***0.8070

1...

.1..

..1.

...1

R...

.R..

..R.

...R

R’ =

50

0.193-1...

.50

0.1930-1..

..100

0.807-1.

...100

0.807 - 1

; 'R =

0.1270...

.0.1270..

..0.04393.

...0.04393

Standardized estimated errors are computed as

3.88040.127

0.49298

R

e

0.46970.127

0.05965

R

e

3.51450.04393

0.15439

R

e

1.41770.04393

0.06228

R

e

'44

4

'33

3

'22

2

'11

1

The magnitude of the largest standardized error corresponds to

measurement 4z in this case. Therefore we identify 4z as the bad

measurement and omit it, from the state estimation calculations. With the

three remaining measurements

ehenceandx are now computed.

H =

0.1250.375

0.6250.125

0.1250.625

H

50

100

100

0.1250.6250.125

0.3750.1250.625HRHG 1T

=

41.4062513.28125

13.2812547.65625

0.1250.375

0.6250.125

0.1250.625

6.2562.512.5

18.7512.562.5

zRHGx 1T1

=

0.0265220.007391

0.0073910.023043

6.2562.512.5

18.7512.562.5z

=

V8.0265

V16.0074

6.98

3.02

9.01

0.304351.565220.13044

0.4778260.173911.34783

This gives

A7.0061

A3.0157

A9.0013

8.0265

16.0074

0.1250.375

0.6250.125

0.1250.625

XHZ

Thus

0.0261

0.0043

0.0087

7.0061

3.0157

9.0013

6.98

3.02

9.01

ZZe

Sum of squares

f is calculated as

3

1j2j

2

j

σ

ef =

2

3

2

2

2

1 e50e100e100

= 0.0435 )0.0261(50)0.0043(100)0.0087(100 222

Corresponding to k = 1 and α = 0.01, Chi-square critical value is 6.64. Thus

f < 2αk,χ . This means that the calculated state of

V8.0265xV,16.0074x 21 is quite acceptable. The new state estimate

based on the three good measurements essentially match those of

Example 1. ( V8.0261xV,16.0072x 21 )

+

Fig. 1

V2 V1

1 Ω

1 Ω

1 Ω 1 Ω

1 Ω x1 x2 x3

+ +

+

-

- -

-

z1 z2

z3 z4

PROBLEM SET 1

1. Consider the circuit shown in Fig. 1, in which three loop current variables are

identified as x1, x2 and x3. Measurement weights are w1 = 100, w2 = 100, w3 = 50

and w4 = 50. Meter readings are z1 = 9.01 A, z2 = 3.02 A, z3 = 6.98 V and z4 = 5.01 V.

Determine the weighted least square estimate of the three loop currents.

Determine the source voltages V1 and V2 using (i) estimated loop currents (ii)

estimated measurements.

A A

V V

2 1

+

Fig. 1

V2 V1

1 Ω

1 Ω

1 Ω 1 Ω

1 Ω x1 x2 x3

+ +

+

-

- -

-

z1 z2

z3 z4

2. In the circuit shown in Fig. 1, consider the voltages at nodes 1 and 2 as state

variables. Measurements and measurement weights are same as in problem 1.

Determine the weighted least square estimates of the nodal voltages. Also

calculate the estimated measurements and estimated errors.

A A

V V

2 1

Fig. 2

1 Ω

3 Ω

3 Ω 1 Ω

3 Ω

-

+

+

+

+ +

-

- -

- I2 I1

3. Five ammeters numbered A1 to A5 are used in the dc circuit shown in Fig. 2 to

determine the two unknown sources I1 and I2. The standard deviations of meter

errors are 0.2 A for meters A2 and A5. and 0.1 A for the other three meters. The

readings of the five meters are: 0.12 A, 1.18 A, 3.7 A, 0.81 A and 7.1 A

respectively.

a) Determine the weighted least square estimate of the source currents I1 and I2.

b) Using the Chi-square test for α = 0.01, identify the bad measurement, if any.

c) Find the WLS estimates of the source currents using reduced data set and

check if the result is statistically acceptable.

A1

1 2 3 A2

A5 A3 A4

4. Redo Problem 3 when the unknowns to be determined are not the source

currents but, rather, the voltages at the nodes 1, 2 and 3 in Fig. 2.

5. Consider the circuit of Fig. 2 for which accuracy of ammeters and their

readings are the same as those specified in Problem 3. It is required to estimate

voltages at nodes 1, 2 and 3 treating those node voltages as state variables for

the following cases.

a) Suppose that meters A4 and A5 are found to be out of order, and therefore

only three measurements z1 = 0.12 A, z2 = 1.18 A and z3 = 3.7 A are available.

Determine WLS estimate of the nodal voltages and the estimated errors.

b) Suppose meters A2 and A5 are out of order and measurements z1 = 0.12 A, z3

= 3.7 A and z4 = 0.81 A, can the nodal voltages be estimated? Explain why by

examining the gain matrix.

6. Suppose that the two voltage sources of circuit shown in Fig. 1 are replaced

with new ones, and the meter readings are z1 = 2.9 A, z2 = 10.2 A, z3 = 5.1 V and z4

= 7.2 V. Measurement weights are: w1 = 100, w2 = 100, w3 = 50 and w4 = 50.

a) Taking the source voltages as state variables, determine their WLS estimates.

b) Using the Chi-square test for α = 0.005, detect bad data.

c) Eliminate the bad data and determine again the WLS estimates of the source

voltages.

d) Check the result in part c again using Chi-square test.

40 MW 60 MW

70 MW 50 MW Fig. 3

7. Five wattmeters are installed on the four-bus system of Fig. 3 to measure line

real power flows, where per-unit reactances of the lines are X12 = 0.05, X13 = 0.1,

X23 = 0.04, X24 = 0.0625 and X34 = 0.08. Suppose that the meter readings show that

z1 = P12 = 0.34 p.u., z2 = P13 = 0.26 p.u., z3 = P23 = 0.17 p.u., z4 = P24 = - 0.24 p.u. and

z5 = P34 = - 0.22 p.u. where the variances of measurement error of all the meters

are given by σi2 = 0.012.

z1

z5

z4 z3 z2

1

G

G

2

3 4

a) Apply the dc power flow method with bus 1 as reference and determine the

measurement Jacobian matrix H. Then compute WLS estimate of the phase

angles of the bus voltages in radians.

b) Using the Chi-square test for α = 0.01, identify two bad measurements. If both

bad measurements are eliminated simultaneously, would it be possible to

estimate the states of the system?

c) Eliminate one of the bad measurements identified in part b and determine the

WLS estimates of the phase angles of the bus voltages using the reduced data

set. Do the same for the other bad measurement.

ANSWERS

1.

^3

^2

^1

x

x

x

=

A2.0100

A3.0133

A9.0033

; V1 = 15.9966 V V2 = 8.0366 V

2.

2

1

V

V =

V5.0107

V7.0060; z^ =

5.0107

7.0060

3.0154

9.0013

; e^ =

0.0007-

0.0260-

0.0046

0.0087

3.

^

2

^1

I

I =

A8.0451

A3.7218; z4 is identified as bad measurement.

^

2

^1

I

I=

A8.2276

A3.7576

4.

3

2

1

V

V

V

=

V7.0375

V2.9351

V3.6595

; No bad data

5.

3

2

1

V

V

V

=

V6.88

V3.34

V3.70

; Node 3 is not represented and hence node voltages cannot be

estimated.

6.

2

1

V

V=

V17.6649

V8.0018; z2 is identified as the bad measurement.

2

1

V

V=

V16.5909

V7.9727; No bad data.

7. H =

12.512.50

16016

02525

0100

0020

;

4

3

2

δ

δ

δ

=

rad0.0040-

rad0.0240-

rad0.0175-

Note that the standardized errors for the 4th and 5th measurements are equally

bad. If both are discarded, bus 4 will be virtually be disconnected from the

system, making state estimation impossible, The elimination of both z4 and z5 is

equivalent to deleting the 4th and 5th rows from H. it is easy to check that the

resulting gain matrix cannot be inverted.

By deleting z4:

4

3

2

δ

δ

δ

=

rad0.0068-

rad0.0244-

rad0.0174-

By deleting z5:

4

3

2

δ

δ

δ

=

rad0.0024-

rad0.0244-

rad0.0174-