Fundamental Principles & Methods

NATIONAL ELECTRIFICATION ADMINISTRATIONU. P. NATIONAL ENGINEERING CENTER

Certificate in

Power System Modeling and Analysis

Competency Training and Certification Program in Electric Power Distribution System Engineering

U. P. NATIONAL ENGINEERING CENTERU. P. NATIONAL ENGINEERING CENTER

Competency Training and Certification Program in Electric Power Distribution System Engineering

Fundamental Principles and Methods in Power System Analysis

Training Course in

Competency Training & Certification Program in Electric Power Distribution System Engineering

U. P. National Engineering CenterNational Electrification Administration

Training Course in Fundamental Principles and Methods in Power System Analysis

Course Outline

1. Circuit Conventions and Notations

2. Power System Representation

3. Per Unit Quantities

4. Symmetrical Components

5. Network Equations and Methods of Solution

� Voltage and Current Directions

� Double Subscript Notation

� Voltage, Current and Phasor Notation

� Complex Impedance and Phasor Notation

Circuit Conventions & Notations

Polarity Marking of Voltage Source Terminals:

Plus sign (+) for the terminal where positive current comes out

Specification of Load Terminals: Plus sign (+) for the terminal where positive current enters

Specification of Current Direction:Arrows for the positive current (i.e., from the source towards the load)

Voltage and Current Directions

The letter subscripts on a voltage indicate the nodes of the circuit between which the voltage exists. The first subscript denotes the voltage of that node with respect to the node identified by the second subscript.

VS = Vao Vb = VbnI = Iab Zb

Double Subscript Notation

The current direction is from first subscript to the second subscript .

Vao - IabZA - Vbn = 0

Iab =Vao - Vbn

Voltage, Current and Phasor Notation

tjmVv ωε= tsinjtcostj ωωε ω +=

voltsV

V m o02

-4 -2 0 2 4ππ π π

1j −=

θθθθ

( )θωε −= tjmIi

amperesI

I m oθ−∠=2

Complex Impedance and Phasor Notation

VLi(t)R (Resistance)

L (Inductance)tj

mS VV ωε=

The first order linear differential equation has a particular solution of the form . tjK)t(i ωε=

)t(diL)t(Ri ωε=+Applying Kirchoff’s voltage law,

tjtj VLKjRK ωωω εεωε =+Hence,

Solving for the current

Dividing voltage by current to get the impedance,

V)t(i ωε

tjm ω

Therefore, the impedance Z is a complex quantity with real part R and an imaginary (j) part ωωωωL

ωωωωL

θθθθ

VLi(t)R (Resistance)

C (Capacitance)tj

mS VV ωε=

For Capacitive Circuit, .

ω−=

Z= |Z|ejφφφφ or Z = |Z|(cosφφφφ + jsinφφφφ) or Z = |Z|∠φ∠φ∠φ∠φ

φφφφR

ωωωωC

v = 141.4 cos(ωωωωt + 30°) voltsi = 7.07 cos(ωωωωt) amperesVmax = 141.4 |V| = 100 V = 100∠∠∠∠30

Imax = 7.07 |I| = 5 I = 5∠∠∠∠0

Z= |Z|ejφφφφ or Z = |Z|(cosφφφφ + jsinφφφφ) or Z = |Z|∠φ∠φ∠φ∠φ

30°203020

30100Z ∠=

∠∠

1032.17)30sin30(cos20 jjZ +=+=

� Electrical Symbols

� Three-Line and Single-Line Diagram

� Equivalent Circuit of Power System Components

� Impedance and Admittance Diagram

� Bus Admittance Matrix

Power System Representation

Generator

TransformerCircuit Breaker

Transmission or Distribution Line

G Switch

Electrical Symbols

Power System representation

3-phase wye neutral grounded

3-phase delta connection

Ammeter

Voltmeter

3-phase wye neutral ungrounded

Protective Relay

Current Transformer

Potential Transformer

Power System representation

Three Line DiagramThe three-line diagram is used to represent each phase of a three-phase power system.

Relays

CTs Distribution Lines

Transformer

The three-line diagram becomes rather cluttered for large power systems. A shorthand version of the three-line diagram is referred to as the Single Line Diagram.

Single Line Diagram

TransformerDistribution Line

CT and Relay

Recloser

Equivalent Circuit of Power System Components: Generator

sa jXR +

Three-Phase Equivalent Single-Phase Equivalent

Equivalent Circuit of Power System Components: Transformer

Core Loss

Primary Secondary

6x6 Admittance Matrix

Y66Y65Y64Y63Y62Y61

Y56Y55Y54Y53Y52Y51

Y46Y45Y44Y43Y42Y41

Y36Y35Y34Y33Y32Y31

Y26Y25Y24Y23Y22Y21

Y16Y15Y14Y13Y12Y11

Three-Phase Equivalent

HT RaRR +=

HT XaXX +=

Equivalent Circuit of Power System Components: Transformer

Single-Phase Equivalent

XX XjaRa 22 +HH jXR +

Equivalent Circuit of Power System Components: Transmission & Distribution Lines

T&D Lines can be represented by an infinite series of resistance and inductance and shunt capacitance.

Equivalent Circuit of Power System Components: Distribution Lines

ZccZcbZca

ZbcZbbZba

ZacZabZaa

Equivalent ππππ-Network

YccYcbYca

YbcYbbYba

YacYabYaa

1/2YccYcbYca

YbcYbbYba

YacYabYaa

Capture Unbalanced

Characteristics Three-Phase Equivalent

Equivalent Circuit of Power System Components: Long Transmission Lines

• •

••+

Length = Longer than 240 km. (150 mi.)lsinhZ'Z c γ=

zZ C =

Characteristic Impedance zy=γPropagation Constant

Equivalent Circuit of Power System Components: Medium-Length

Transmission LinesLength = 80 – 240 km. (50 - 150 mi.)

• •

••+

( )ljxrZ L+=

Equivalent Circuit of Power System Components: Short Transmission Lines

Length = up to 80 km. (50 mi.)

• •

••+

Is = IR

( )ljxrZ L+=

Single Phase Equivalent of Balanced Three-Phase System

Eao = |E|∠∠∠∠0° V

Ebo= |E| ∠∠∠∠240° V

Eco = |E| ∠∠∠∠120° V

Eao = |E|∠∠∠∠ 0° V

Ebo= |E| ∠∠∠∠240° V b

Eco = |E| ∠∠∠∠120° V c

−∠=∠∠

)240(IZ

θ−∠=

∠∠

)120(IZ

θ−∠=

∠∠

Note: Currents are Balanced

Single Phase Representation of a Balanced Three-Phase System

Eao = |E|∠∠∠∠ 0° V

Eao = |E|∠∠∠∠0° V

Ebo= |E| ∠∠∠∠240° V

Eco = |E| ∠∠∠∠120° V

Impedance and Admittance Diagrams

1 - 32 - 31 - 42 - 43 - 4

Single Line Diagram

Impedance Diagram

Ea Ec Eb

Generator

ILZpIs

The two sources will be equivalent if VLand IL are the same for both circuits.

Eg = ISZPZg = Zp

1 - 32 - 31 - 42 - 43 - 4

Single Line Diagram

Impedance Diagram

Ea Ec Eb

Generator

Equivalent Sources

ILZpIs

The two sources will be equivalent if VLand IL are the same for both circuits.

Eg = ISZPZg = Zp

Admittance Diagram

y03y01

y14 y24

I1 I3 I2

I1 = Ea/zay01 = 1/za

I2 = Eb/zby02 = 1/zb

I3 = Ec/zcy03 = 1/zc

y13 = 1/zd

y23 = 1/ze

y14 = 1/zf

y24 = 1/zg

y34 = 1/zh

at node 1:

( ) ( ) 144113310111 yVVyVVyVI −+−+=

at node 4:

( ) ( ) ( ) 343424241414 yVVyVVyVV0 −+−+−=

at node 2:( ) ( ) 244223320222 yVVyVVyVI −+−+=

at node 3:( ) ( ) ( ) 1313344323230333 yVVyVVyVVYVI −+−+−+=

Applying Kirchoff’s Current Law

Nodal Voltage Equations

Rearranging the equations,( ) 14413314130111 yVyVyyyVI −−++=

( ) 3432421413424144 yVyVyVyyyV0 −−−++=

( ) 1313442321334230333 yVyVyVyyyyVI −−−+++=

In matrix form,

⎥⎥⎥⎥

⎢⎢⎢⎢

⎥⎥⎥⎥

⎢⎢⎢⎢

++−−−

⎥⎥⎥⎥

⎢⎢⎢⎢

342414342414

34133423032313

2423242302

1413141301

yyyyyy

y-yyyyy-y-

y-y-yyy0

y-y-0yyy

( )24423324230222 yVyVyyyVI −−++=

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

nn3n2n1n

n3333231

n2232221

n1131211

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

[ I ] = [Ybus][V]

Yii = self-admittance, the sum of all admittances terminating on the node (diagonal elements)

Yij = mutual admittance, the negative of the admittances connected directly between the nodes identifed by the double subscripts

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

nn3n2n1n

n3333231

n2232221

n1131211

[YBUS] =

Ybus is also called Bus Admittance Matrix

� The Per Unit System

� Per Unit Impedance

� Changing Per Unit Values

� Consistent Per Unit Quantities of Power System

� Advantages of Per Unit Quantities

Per Unit Quantities

Base Value

Actual ValuePer Unit Value =

Per-unit Value is a dimensionless quantity

Per-unit value is expressed as decimal

Actual ValuePercent =

Per Unit Value

The Per Unit System

Base Power

Actual Value of PowerPer Unit Power =

Base Voltage

Actual Value of VoltagePer Unit Voltage =

Base Current

Actual Value of CurrentPer Unit Current =

Base Impedance

Actual Value of ImpedancePer Unit Impedance =

Per Unit Value

The Per Unit System

PU VoltagePU Current =

PU Impedance

PU Power = PU Voltage x PU Current

Per Unit Calculations

The Per Unit System

Zline = 1.4 ∠∠∠∠75° ΩΩΩΩ

Zload = 20 ∠∠∠∠30° ΩΩΩΩ 2540 ∠∠∠∠0° V

Example:

Vs = ?

Determine Vs

Per Unit Calculations

The Per Unit System

Choose: Base Impedance = 20 ohms (single phase)Base Voltage = 2540 volts (single phase)

PU Impedance of the load = 20∠∠∠∠30°/20 = ______ p.u.PU Impedance of the line = 1.4∠∠∠∠75°/20 = ______ p.u.PU Voltage at the load = 2540∠∠∠∠0° /2540 = ______ p.u.

Line Current in PU = PU voltage / PU impedance of the load= ______ / ______ = ______ p.u.

PU Voltage at the Substation = Vload(pu) + IpuZLine(pu)= ________ + _______ x _______ = _______ p.u.

The Per Unit SystemPer Unit Calculations

The magnitude of the voltage at the substation is

1.05 p.u. x 2540 Volts = _______ Volts

1.0 ∠∠∠∠-30° p.u.

1.05∠∠∠∠2.70°

0.07 ∠∠∠∠75° p.u.

1.0∠∠∠∠30° p.u. 1.0∠∠∠∠0° p.u.

The Per Unit SystemPer Unit Calculations

1. Base values must satisfy fundamental electrical laws (Ohm’s Law and Kirchoff’s Laws)

2. Choose any two electrical parameters

• Normally, Base Power and Base Voltage are chosen

3. Calculate the other parameters

• Base Impedance and Base Current

The Per Unit SystemEstablishing Base Values

Base PowerBase Current =

Base Voltage

Base Voltage (Base Voltage)2

Base Impedance = = Base Current Base Power

For a Given Base Power and Base Voltage,

For Single Phase System,

Pbase(1φφφφ)------------Vbase(1φφφφ)

Ibase =

Vbase(1φφφφ)------------Ibase(1φφφφ)

Zbase =

[Vbase(1φφφφ)]²= ------------Pbase(1φφφφ)

For Three Phase System,

Pbase(3φφφφ)------------√√√√3Vbase(LL)

Ibase =

Vbase(LN)------------Ibase(L)

Zbase =

[Vbase(LL)]²= ------------Pbase(3φφφφ)

kV Base kV Base

MVABase3

1MVABase

• Base MVA is the same base value for Apparent, Active and Reactive Power

• Base Z is the same base value for Impedance, Resistance and Reactance

• Base Values can be established from Single Phase or Three Phase Quantities

Base kVA1φφφφ = 10,000 kVA= 10 MVA

Base kVLN = 69.282 kV

Base Z = (69.282)2/10= 480 ohms

Base kVA3φφφφ = 30,000 kVA= 30 MVA

Base kVLL = 120 kV

Base Z = (120)2/30= 480 ohms

Amps144.34

)120(3

1000x30CurrentBase

Amps144.34 282.69

1000x10CurrentBase

Example:

� Manufacturers provide the following impedance in per unit:1. Armature Resistance, Ra

2. Direct-axis Reactances, Xd”, Xd’ and Xd

3. Quadrature-axis Reactances, Xq”, Xq’ and Xq

4. Negative Sequence Reactance, X2

5. Zero Sequence Reactance, X0

� The Base Values used by manufacturers are:1. Rated Capacity (MVA, KVA or VA)

2. Rated Voltage (kV or V)

} Positive Sequence Impedances

Per Unit Impedance Generators

)(L)pu(L Z

XX ΩΩΩΩ====

)()pu( Z

RR ΩΩΩΩ====

)pu(C Z

XX Ω=

Per Unit ImpedanceTransmission and Distribution Lines

� The ohmic values of resistance and leakage reactance of a transformer depends on whether they are measured on the high- or low-tension side of the transformer

� The impedance of the transformer is in percent or per unit with the Rated Capacity and Rated Voltages taken as base Power and Base Voltages, respectively

� The per unit impedance of the transformer is the same regardless of whether it is referred to the high-voltage or low-voltage side

� The per unit impedance of the three-phase transformer is the same regardless of the connection

Per Unit ImpedanceTransformers

ExampleA single-phase transformer is rated 110/440 V, 2.5 kVA.The impedance of the transformer measured from the low-voltage side is 0.06 ohms. Determine the impedance in per unit (a) when referred to low-voltage side and (b) when referred to high-voltage side

Solution

Low-voltage Zbase = = ______ ohms 1000/5.2

110.0 2

PU Impedance, Zpu = = ______ p.u.

Per Unit Impedance

High Voltage, Zbase = = _______ ohms

PU Impedance, Zpu = = _______ p.u.

If impedance had been measured on the high-voltage side, the ohmic value would be

ohmsZ _______110

44006.0

=⎟⎠⎞

⎜⎝⎛=

Note: PU value of impedance referred to any side of the transformer is the same

Per Unit Impedance

Example:Consider a three-phase transformer rated 20 MVA, 67 kV/13.2 kV voltage ratio and a reactance of 7%. The resistance is negligible.

a) What is the equivalent reactance in ohms referred to the high voltage side?

b) What is the equivalent reactance in ohms referred to the low voltage side?

c) Calculate the per unit values both in the high voltage and low voltage side at 100 MVA.

Changing Per-Unit Values

SOLUTION:

a) Pbase = 20 MVAVbase = 67 kV (high voltage)

( kV)² = ________ ohms

( MVA)Zbase =

Xhigh = Xp.u. x Zbase = _______ x _______= _______ ohms

b) Pbase = 20 MVAVbase = 13.2 kV (low voltage)

= ________ ohmsZbase =

Xlow = Xp.u. x Zbase = _______ x _______= _______ ohms

( kV)²

( MVA)

SOLUTION:

c) Pbase = 100 MVAVbase,H = 67 kV

(67)² = ________ ohms

100Zbase,H =

ohms = ______ p.u.Xp.u.,H =

Vbase,L = 13.2 kV

(13.2)² = _______ ohms

100Zbase,L =

= ______ p.u.Xp.u.,L =

Note that the per unit quantities are the same regardless of the voltage level.

Three parts of an electric system are designated A, B and C and are connected to each other through transformers,as shown in the figure. The transformer are rated as follows:

A-B 10 MVA, 3φφφφ, 13.8/138 kV, leakage reactance 10%B-C 10 MVA, 3φφφφ, 138/69 kV, leakage reactance 8%

Determine the voltage regulation if the voltage at the load is 66 kV.

SOURCE A B C

300 ΩΩΩΩ/ φφφφPF=100 %A-B B-C

A-B B-C13.8/138 kV 138/69 kV

XAB=10% XBC=8%

Solve using actual quantities

300 ΩΩΩΩ/ φφφφPF=100%

SOLUTION USING PER UNIT METHOD:

Pbase = 10 MVAVA,base = 13.8 kVVB,base = 138 kVVC,base = 69 kV

(69)² -------- = _____ ohms

10ZC,base =

---------- = ______ + j _____ p.u.ZLOAD,p.u. =

-------------- = _______ p.u.VC,p.u. =

---------------- = _______ p.u.Ip.u. =

VA = _______ + ( ________ ) x ( ________ + ________ )= ________ + j ________ p.u.= _________ p.u.

VNL - VL---------------- x 100%

V.R. =

------------------------ x 100%V.R. =

Consider the previous example, What if transformer A-B is 20 MVA instead of 10 MVA. The transformer nameplate impedances are specified in percent or per-unit using a base values equal to the transformer nameplate rating.

The PU impedance of the 20 MVA transformer cannot be added to the PU impedance of the 10 MVA transformer because they have different base values

The per unit impedance of the 20 MVA can be referred to 10 MVA base power

Convert per unit value of 20 MVA transformer,

Pbase = 20 MVA (Power Rating)Vbase,H = 138 kV (Voltage Rating)

(138)² ---------- = _______ ohms

20Zbase,H =

0.10 p.u. x _______ ohms = _______ ohmsXactual,H =

At Pbase = 10 MVA (new base)(138)²

---------- = 1904.4 ohms10

Zbase,H =

95.22 ---------- = 0.05 p.u. 1904.4

Xp.u.(new) =

The per unit impedance of the 20MVA and 10 MVA transformer can now be added.

Zactual = Zpu1 • Zbase1 Zactual = Zpu2 • Zbase2

2base2pu1base1pu ZZZZ ⋅=⋅

1base1pu2pu Z

ZZZ ⋅=

Note that the transformer can have different per unit impedance for different base values (i.e., the actual ohmicimpedances of the equipment is independent of the selected base values), then

Recall:( )

Powerbase

voltagebaseZ

base =

( )2,3

baseLL

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

1base,3

2base,3

2base,LL

1base,LL1pu2pu MVA

A three-phase transformer is rated 400 MVA, 220Y/22 Δ kV. The impedance measured on the low-voltage side of the transformer is 0.121 ohms (approx. equal to the leakage reactance). Determine the per-unit reactance of the transformer for 100 MVA, 230 kV base values at the high voltage side of the transformer.

Example

)given(base

)new(base

)given(base

)given.(u.p)new.(u.p P

]V[xZZ =

Solution

On its own base the transformer reactance is

On the chosen base the reactance becomes

= ________ puX =( )

X = ( ) x x = ________ pu( )2

( )2( )

Consistent Per Unit Quantities of Power System

Procedure:a) Establish Base Power and Base Voltages

• Declare Base Power for the whole Power System

• Declare Base Voltage for any one of the Power System components

• Compute the Base Voltages for the rest of the Power System Components using the voltage ratio of the transformers

Note: Define each subsystem with unique Base Voltage based on separation due to magnetic coupling

b) Compute Base Impedance and Base Current

• Using the Declared Base Power and Base Voltages, compute the Base Impedances and Base Currents for each Subsystem

c) Compute Per Unit Impedance

• Using the declared and computed Base Values, compute the Per Unit values of the impedance by:� Dividing Actual Values by Base Values

� Changing Per Unit Impedance with change in Base Values

Generator 1 (G1): 300 MVA; 20 kV; 3φ; Xd” = 20 %

Transmission Line(L1): 64 km; XL = 0.5 Ω / km

Transformer 1 (T1): 3φ; 350 MVA; 230 / 20 kV; XT1 = 10 %

Transformer 2 (T2): 3-1φ; 100 MVA; 127 / 13.2 kV; XT2 = 10 %

Generator 2 (G2): 200 MVA; 13.8kV, Xd” = 20 %

Generator 3 (G3): 100 MVA; 13.8kV, Xd” = 20 %

Use Base Power = 100 MVA

T1Transmission Line

G3XLINE XT2

XG1 XG2 XG3

a) & b) Establish Base Power, Base Voltages, Base Impedance, and Base Current

Ibase (Amp)Zbase (Ohm)Vbase (kV)Sub-System

Base Power:

c) Compute Per Unit Impedance

Advantages of Per-Unit Quantities

The computation for electric systems in per-unit simplifies the work greatly. The advantages of Per Unit Quantities are:

1. Manufacturers usually specify the impedances of equipments in percent or per-unit on the base of the nameplate rating.

2. The per-unit impedances of machines of the same type and widely different rating usually lie within a narrow range. When the impedance is not known definitely, it is generally possible to select from tabulated average values.

3. When working in the per-unit system, base voltages can be selected such that the per-unit turns ratio of most transformers in the system is equal to 1:1.

4. The way in which transformers are connected in three-phase circuits does not affect the per-unit impedances of the equivalent circuit, although the transformer connection does determine the relation between the voltage bases on the two sides of the transformer.

5. Per unit representation yields more meaningful and easily correlated data.

6. Network calculations are done in a much more handier fashion with less chance of mix-up • between phase and line voltages• between single-phase and three-phase powers, and• between primary and secondary voltages.

� Sequence Components of Unbalanced Phasor

� Sequence Impedance of Power System Components

� Practical Implications of Sequence Components of Electric Currents

Symmetrical Components

Sequence Components of Unbalanced Phasor

In a balanced Power System,� Generator Voltages are three-phase balanced

� Line and transformer impedances are balanced

� Loads are three-phased balanced

Single-Phase Representation and Analysis can be used for the Balanced Three-Phase Power System

In a practical Power Systems,� Lines are not transposed.� Single-phase transformers used to form three-phase

banks are not identical.� Loads are not balanced.� Presence of vee-phase and single phase lines.� Faults

Single-phase Representation and Analysis cannot be use for an unbalanced three-phase power system.

Any unbalanced three-phase system of phasorsmay be resolved into three balanced systems of phasors which are referred to as the symmetrical components of the original unbalanced phasors, namely:

a) POSITIVE-SEQUENCE PHASOR

b) NEGATIVE-SEQUENCE PHASOR

c) ZERO-SEQUENCE PHASOR

Phase a

Phase b

Phase c

120°120°

REFERENCE PHASE SEQUENCE: abc

� Zero Sequence Phasors are single-phase, equal in magnitude and in the same direction.

� Positive Sequence Phasors are three-phase, balanced and have the phase sequence as the original set of unbalanced phasors.

� Negative Sequence Phasors are three-phase, balanced but with a phase sequence opposite to that original set of unbalncedphasors.

Va = Va1 + Va2 + Va0

Vb = Vb1 + Vb2 + Vb0

Vc = Vc1 + Vc2 + Vc0

Each of the original unbalanced phasor is the sum of it’s sequence components. Thus,

Where,Va1 – Positive Sequence component of Voltage VaVa2 – Negative Sequence component of Voltage VaVa0 – Zero Sequence component of Voltage Va

a = 1 ∠∠∠∠ 120°

OPERATOR “a”

An operator “a” causes a rotation of 120° in the counter clockwise direction of any phasor.

Operating V by a

a² = 1 ∠∠∠∠ 240°

a³ = 1 ∠∠∠∠ 0°

Vb in terms of Va

Vb = a² VaVb1 = a² Va1

Vc in terms of Va

Vc = a VaVc1 = a Va1

Vc = aVa

120°120°

Vb = a2Va

The original Phasor and Positive Sequence components in terms of phase a

Vb in terms of Va

Vb2 = aVa2

Vc in terms of Va

Vc2 = a²Va2

Vb2 = aVa2

120°120°

Vc2 = a2Va2

The Negative Sequence components in terms of phase a

Vb in terms of Va

Vb0 = Va0

Vc in terms of Va

Vc0 = Va0

Va0= Vb0 = Vc0

The Zero Sequence components in terms of phase a

Writing again the phasors in terms of phasor Vaand operator “a”,

Va = Va0 + Va1 + Va2 Vb = Va0 + a²Va1 + aVa2 Vc = Va0 + aVa1 + a²Va2

Computing for Va0, Va1 & Va2

[ ] [ ]c2

ba1acba0a VaaVV3

1V VVV

1V ++=++=

[ ]cb2

a2a aVVaV3

1V ++=

EXAMPLE:

Determine the symmetrical components of the followingunbalanced voltages.

Vc = 8 ∠∠∠∠143.1

Vb = 3 ∠∠∠∠-90

Va = 4 ∠∠∠∠0

18.384.9

143.1)240)(8(1 90)-120)(3(1 0 43

)Va aV V(3

∠∠+∠∠+∠=

For Phasor VFor Phasor Vaa: :

143.051

143.1) 8 90- 3 0 4(3

)V V V(3

1V cba0a

∠+∠+∠=

For Phasor VFor Phasor Vaa: :

86.082.15

143.1)120)(8(1 90)-240)(3(1 0 43

)aV Va V(3

−∠=

∠∠+∠∠+∠=

Components of Vb can be obtained by operating the sequence components of phasor Va.

33.922.15

86.08)-120)(2.15(1

101.62-4.9

258.384.9

18.38)240)(4.9(1

143.05 1 143.05 1

∠∠=

∠=∠=

Similarly, components of phasor Vc can be obtained byoperating Va.

3.92512.15

86.08)-0)(2.1542(1

8.38314.9

18.38)0)(4.921(1

143.05 1 143.05 1

∠∠=

∠=∠=

Positive Sequence Components Negative Sequence Components

Zero Sequence Components

Va0 = Vb0 = Vc0Vc1

Vb1Va2

Vb2Vc2

°18.38

°143.05

°86.08

Components of Vc

Components of Va

Components of Vb

Add Sequence Components Graphically

Vc = 8 ∠∠∠∠143.1

Vb = 3 ∠∠∠∠-90

Va = 4 ∠∠∠∠0

The results can be checked either mathematically or graphically.

143.18

153.922.15138.384.905.143 1

V V V V

33.922.15101.62-4.9143.05 1

VV V V

86.08-2.1518.384.905.143 1

V V V V

c2c1c0c

b2b1b0b

a2a1a0a

∠+∠+∠=

Ia1 + Ia2 + Ia0

Ib1 + Ib2 + Ib0

Ic1 + Ic2 + Ic0

Sequence Impedance of Power System Components

Positive Sequence Negative Sequence Zero Sequence

2a2a2ZI -V ====

1a1fa1ZI – V V ====

oaoaoZI -V ====

Sequence Networks

Sequence Impedance of Power System Components

� In general,

Z1 ≠ Z2 ≠ Z0 for generators

Z1 = Z2 = Z0 for transformers

Z1 = Z2 ≠ Z0 for lines

Practical Implications of Sequence Components of Electric Currents

ZERO-SEQUENCE CURRENTS:

The neutral return (ground) carries the in-phase zero-sequence currents.

ZERO-SEQUENCE CURRENTS:

The neutral return (ground) carries the in-phase zero-sequence currents.

In-phase zero-sequence currents circulates in the delta-connected transformer windings.

There is “balancing ampere-turns” for the zero-sequence currents.

I0 = 0

NEGATIVE-SEQUENCE CURRENTS:

� A three-phase unbalanced load produces a reaction field which rotates synchronously with the rotor-field system of generators.

� Any unbalanced condition will have negative sequence components.This negative sequence currents rotates counter to the synchronously revolving field of the generator.

� The flux produced by sequence currents cuts the rotor field at twice the rotational velocity, thereby inducing double frequency currents in the field system and in the rotor body.

� The resulting eddy-currents are very large and cause severe heating of the rotor.

� Network Equations

� Matrix Representation of System of Equations

� Type of Matrices

� Matrix Operations

� Direct Solutions of System of Equations

� Iterative Solutions of System of Equations

Network Equations and Methods of Solution

The standard form of n independent equations:

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

nn3n2n1n

n3333231

n2232221

n1131211

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

[ I ] = [Ybus][V]

Network Equations

Ypp = self-admittance, the sum of all admittances terminating on the node (diagonal elements)

Ypq = mutual admittance, the negative of the admittances connected directly between the nodes identifed by the double subscripts

1 - 32 - 31 - 42 - 43 - 4

Single Line Diagram

Network Equations

Impedance Diagram

Ea Ec Eb

Generator

Network Equations

Admittance Diagram

y03y01

y14 y24

I1 I3 I2

I1 = Ea/zay01 = 1/za

I2 = Eb/zby02 = 1/zb

I3 = Ec/zcy03 = 1/zc

y13 = 1/zd

y23 = 1/ze

y14 = 1/zf

y24 = 1/zg

y34 = 1/zh

Network Equations

( ) 14413314130111 yVyVyyyVI −−++=

( ) 3432421413424144 yVyVyVyyyV0 −−−++=

( ) 1313442321334230333 yVyVyVyyyyVI −−−+++=

In matrix form,

⎥⎥⎥⎥

⎢⎢⎢⎢

⎥⎥⎥⎥

⎢⎢⎢⎢

++−−−

⎥⎥⎥⎥

⎢⎢⎢⎢

342414342414

34133423032313

2423242302

1413141301

yyyyyy

y-yyyyy-y-

y-y-yyy0

y-y-0yyy

( )24423324230222 yVyVyyyVI −−++=

Network Equations

Matrix Representations of System of Equations

� System of n Linear Equations

In the following system of equations:

x1, x 2, and x3, are unknown variables, a11, a12,… …, a33are the coefficient of these variables and y1, y2, and y3are known parameters.

3333232131

2323222121

1313212111

yxaxaxa

y xaxaxa

⎥⎥⎥

⎢⎢⎢

333231

232221

131211

which is called the Coefficient Matrix of the system of equations.

The coefficients form an array

Similarly, the variables and parameters can be written in matrix form as.

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

The system of equations in matrix notation is

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

333231

232221

131211

AX = Y

� System of Equations in Matrix Form

Va = Va0 + Va1 + Va2 Vb = Va0 + a²Va1 + aVa2 Vc = Va0 + aVa1 + a²Va2

Rearranging and writing in matrix form

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

[ ] [ ]c2

ba1acba0a VaaVV3

1V VVV

1V ++=++=

[ ]cb2

a2a aVVaV3

1V ++=

In matrix form

Definition of a MATRIX

A matrix consists of a rectangular array of elements represented by a single symbol.

[A] is a shorthand notation for the matrix and aijdesignates an individual element of the matrix.

A horizontal set of elements is called a row and a vertical set is called a column.

The first subscript i always designates the number of the row in which the element lies. The second subscript j designates the column. For example, element a23 is in row 2 and column 3.

Definition of a MATRIX

[ ] [ ]

⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢

mnm3m2m1

3n333231

2n232221

1n131211

The matrix has m rows and n columns and is said to have a dimension of m by n (or m x n).

[aij]mxn

Definition of a Vector

A vector X is defined as an ordered set of elements. The components x1, X2…, Xn may be real or complex numbers or functions of some dependent variable.

⎥⎥⎥⎥

⎢⎢⎢⎢

“n” defines the dimensionality or size of the vector.

Matrices with only one row (n = 1) are called RowVectors while those with one column (m=1) are called Column Vectors. The elements of a vectors are denoted by single subscripts as the following:

[ ]n21 rrr R L=

Thus, R is a row vector of dimension n while C is a column vector of dimension m.

⎥⎥⎥⎥

⎢⎢⎢⎢

� Square Matrix� Upper Triangular Matrix� Lower Triangular Matrix� Diagonal Matrix� Identity or Unit Matrix� Null Matrix� Symmetric Matrix� Skew-symmetric Matrix

Type of Matrices

⎥⎥⎥

⎢⎢⎢

333231

232221

131211

A square matrix is a matrix in which m = n.

For a square, the main or principal diagonal consists of the elements of the form aii; e.g., for the 3 x 3 matrix shown

the elements a11, a22, and a33 constitute the principal diagonal.

⎥⎥⎥

⎢⎢⎢

131211

An upper triangular matrix is one where all the elements below the main diagonal are zero.

Type of Matrices

⎥⎥⎥

⎢⎢⎢

333231

A lower triangular matrix is one where all elements above the main diagonal are zero.

⎥⎥⎥

⎢⎢⎢

A diagonal matrix is a square matrix where all elements off the diagonal are equal to zero.Note that where large blocks of elements are zero, they are left blank.

Type of Matrices

⎥⎥⎥

⎢⎢⎢

An identity or unit matrix is a diagonal matrix where all elements on the main diagonal are equal to one.

Type of Matrices

The null matrix is matrix whose elements are

equal to zero.

⎥⎥⎥

⎢⎢⎢

Type of Matrices

⎥⎥⎥

⎢⎢⎢

A symmetric matrix is one where aij = aji for all i’s and j’s.

Type of Matrices

A skew-symmetric matrix is a matrix which has the property aij = -aji for all i and j; this implies aii = 0

⎥⎥⎥

⎢⎢⎢

−−

� Addition of Matrices� Product of a Matrix with a Scalar� Multiplication of Matrices� Transpose of a Matrix� Kron Reduction Method� Determinant of a Matrix� Minors and Cofactors of a Matrix� Inverse of a Matrix

Matrix Operations

Addition of Matrices

Two matrices A = [aij] and B = [bij] can be added together if they are of the same order (mxn). The sum C = A + B is obtained by adding the corresponding elements.

C = [cij] = [aij + bij]

⎥⎦

⎤⎢⎣

⎡=⎥

⎤⎢⎣

Example:

⎥⎦

⎤⎢⎣

⎡=⎥

⎤⎢⎣

⎡+++

1)(31)(70)(2

6)(02)(45)(1 B A

⎥⎦

⎤⎢⎣

⎡ −−=⎥

⎤⎢⎣

⎡−−−

−−−=−

1)(31)(70)(2

6)(02)(45)(1 B A

⎥⎥⎥

⎢⎢⎢

+−−

⎥⎥⎥

⎢⎢⎢

−−+

5j64j55j7

4j56j41j2

5j71j22j3

9j81j13j6

1j13j51j4

3j61j42j1

Example:

( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )⎥

⎥⎥

⎢⎢⎢

+++−+−−++

+++++++++

++−−+−+++

5j69j84j51j15j73j6

4j51j16j43j51j21j4

5j73j61j21j42j32j1

⎥⎥⎥

⎢⎢⎢

+−−

14j145j62j13

5j69j92j6

2j132j64j4

⎥⎥⎥

⎢⎢⎢

++−+−

−−−+

−−++−

4j23j48j1

3j43j10j2

8j10j20j2

Product of a Matrix with a Scalar

A matrix is multiplied by a scalar k by multiplying all elements mn by k , that is,

⎥⎥⎥⎥

⎢⎢⎢⎢

mn2m1m

n22221

n11211

kakaka

Example:

3 k and

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

3 Ak B

Example:

3 k and

4j1j36

5j2j2-5

6j3j14

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

12j39j18

15j66j15

18j93j12

⎥⎥⎥

⎢⎢⎢

j4-1j36

j52j2-5

j6-3j14

3 Ak B

Multiplication of Matrices

Two matrices A = [aij] and B = [bij] can be multiplied in the order AB if and only if the number of columns of A is equal to the number of rows of B .

That is, if A is of order of (m x l), then B should be of order (l x n).

If the product matrix is denoted by C = A B, then C is of order (m x n).

[ ] [ ] [ ] n x mn x ll x m CBA =

An easy way to check whether two matrices can be multiplied.

Interior dimensions are equal multiplication is possible

Exterior dimensions definethe dimensions of the result

If the product matrix is denoted by C = A B, then C is of order (m x n). The elements cij are given by

1kkjikij bac for all i and j.

2x33231

⎥⎥⎥

⎢⎢⎢

2x22221

bbB ⎥

⎤⎢⎣

⎡=and

⎥⎦

⎤⎢⎣

⎥⎥⎥

⎢⎢⎢

C = A B =

Example:

2112111111 babac +=∑=

1kkjikij bac

⎥⎥⎥

⎢⎢⎢

)baba()bab(a

)baba()baba(

)bab(a)bab(a

2232123121321131

2222122121221121

2212121121121111

1kkjikij bac

⎥⎥⎥

⎢⎢⎢

87B ⎥

⎤⎢⎣

⎡=and

⎥⎦

⎤⎢⎣

⎥⎥⎥

⎢⎢⎢

Example:

232475

⎥⎥⎥

⎢⎢⎢

23)0683()9673(

)0582()9572(

)0481()9471(

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

+−−

⎥⎥⎥

⎢⎢⎢

−−+

5j64j55j7

4j56j41j2

5j71j22j3

9j81j13j6

1j13j51j4

3j61j42j1

( )( ) ( )( ) ( )( ) 41j355j73j61j21j42j32j1c11 −=−−++−+++=

( )( ) ( )( ) ( )( ) 42j725j63j64j51j45j72j1c13 +=+−++−+++=

Example:

( )( ) ( )( ) ( )( ) 16j444j53j66j41j41j22j1c12 −=−−++−+−+=

( )( ) ( )( ) ( )( ) 24j295j71j11j23j52j31j4c21 +=−+++++++=

( )( ) ( )( ) ( )( ) 73j375j61j14j53j55j71j4c 23 +=++++++++=( )( ) ( )( ) ( )( ) 41j204j51j16j43j51j21j4c22 +=−+++++−+=( )( ) ( )( ) ( )( ) 24j295j71j11j23j52j31j4c21 +=−+++++++=

( )( ) ( )( ) ( )( ) 144j395j69j84j51j15j73j6c33 +=++++−+++=( )( ) ( )( ) ( )( ) 15j1014j59j86j41j11j23j6c32 +=−+++−+−+=( )( ) ( )( ) ( )( ) 43j1165j79j81j21j12j33j6c31 +=−+++−+++=

[ ] [ ]⎥⎥⎥

⎢⎢⎢

+−−

144j3915j10143j116

73j3741j2024j29

42j7216j4441j35

Transpose of a Matrix

2x33231

⎥⎥⎥

⎢⎢⎢

If the rows and columns of an m x n matrix are interchanged, the resultant n x m matrix is the transpose of the matrix and is designated by AT.

For the matrix

The transpose is

3x2322212

312111

aaaA ⎥

⎤⎢⎣

Example:

⎥⎥⎥

⎢⎢⎢

⎥⎦

⎤⎢⎣

Example:

⎥⎦

⎤⎢⎣

⎡−+−

+−+=

3j61j45j2

2j56j34j1A

⎥⎥⎥

⎢⎢⎢

3j62j5

1j46j3

5j24j1

Determinant of a Matrix

Two simultaneous equations:

Determinant of a 2 x 2 Matrix

)2(yxaxa

)1(yxaxa

2222121

1212111

In Matrix Form

⎥⎦

⎤⎢⎣

⎡=⎥

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

The solutions of two simultaneous equations can be obtained by eliminating the variables one at a time. Solving for x2 in terms of x1 from the second equation and substituting this expression for x2 in the first equation, the following is obtained:

yx −=

1212222 xayxa −=

21122211

2121221

212122121122211

1221211221212211

212111

yayax )aaaa(

yaxaayaxaa

y(a xa

−−

−=−

substituting x2 and solving for x1

The expression (a11a22 – a12a21) is the value of the determinant of the coefficient matrix A, denoted by |A|.

aa |A| =

Then, substituting x1 in either equation (1) or (2), x2 is obtained

21122211

1212112 aaaa

−−

[ ]⎥⎥⎥

⎢⎢⎢

−−=

The determinant of A is can be determined by reducing the size of the original matrix by eliminating rows.

−−=

For example:

Eliminate row 1 by striking out the row1 and jth column

−−=

The determinant of A is

212)1(

111)1(

121)1(|A| 312111

−−−+

−−+

−−= +++

[ ] [ ][ ]

[ ] [ ] [ ]44

124)2(62)1(44)1(

)6)(2()4)(1()2)(1(

)6)(1()2)(1()1)(1()4)(1()2)(2()1)(1(

−−++−−−−=

−−−−++

−−−−+−−+=

The determinant obtained by striking out the ith

row and jth column is called the minor element aij.

Example:

333231

232221

131211

Determinant of a MatrixMinors and Cofactors of a Matrix

)aaaa(a of minorThe 1332331221 −=

The cofactor of an element aij designated by Aij is

ij a of minor)1(A +−=

( )2121

a ofminor the 1- A

)a min((-1)

)a of min()1(A

−= +

oforthe

ortheExample:

)aaaa(a of minortheSince 1332231221 −=

)aaaa(1A of cofactorthe 1332331221 −−=∴

4)]6)(1()2)(1[(1

)1(A 21

12 −=−−−−=

−−−= +

Example:

⎥⎥⎥

⎢⎢⎢

8)]4)(1()2)(2[(1

)1(A 11

11 −=−−=

−−−= +

14)]6)(2()2)(1[(1

)1(A 22

22 =−−=

−−−= +

16)]6)(2()4)(1[(1

)1(A 31

13−=−−−−=

−−−= +

6)]2)(4()2)(1[(1

)1(A 12

21 =−−=

−−−= +

10)]6)(1()4)(1[(1

)1(A 32

23−=−−−=

−−−= +

5)]2)(2()1)(1[(1

)1(A 13

31 =−−=

−−−= +

3)]2)(1()1)(1[(1

)1(A 23

32 −=−−−=

−−−= +

1A 10A 16A

3A 14A 4A

5A 6A 8A

332313

322212

312111

−=−=−=

−==−=

==−=

Therefore the cofactors of matrix A are:

1)]1)(1()2)(1[(1

)1(A 33

33−=−−−=

−−−= +

⎥⎥⎥

⎢⎢⎢

333231

232221

131211

A of Cofactors

⎥⎥⎥

⎢⎢⎢

−−

−−−

A of Cofactors

and in matrix form:

Inverse of a MatrixDivision does not exist in matrix algebra except in the case of division of a matrix by a scalar. However, for a given set of equations.

or in matrix form [AX] = [Y]. It is desirable to express x1, x2, and x3 a function of y1, y2, and y3, i.e.. [X] = [BY], where B is the inverse of A designated by A-1.

3333232131

2323222121

1313212111

yxaxaxa

y xaxaxa

If the determinant of A is not zero, the equations can be solved for x ’s as follows;

1 y|A|

Ax ++=

Ax ++=2

3 y|A|

Ax ++=

Inverse of a Matrix

where A11, A12, …, A33 are cofactors of a11, a12,,a33 and |A| is the determinant of A.

⎥⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢⎢

332313

322212

312111

A+ is called the adjoint of A. It should be noted that the elements of adjoint A+ are the cofactors of the elements of A, but are placed in transposed position.

Inverse of a Matrix

⎥⎥⎥

⎢⎢⎢

−−−

−−

⎥⎥⎥

⎢⎢⎢

332313

322212

312111

Example: Get the inverse of A

⎥⎥⎥

⎢⎢⎢

the Adjoint of A is

Inverse of a Matrix

⎥⎥⎥

⎢⎢⎢

−−−

−−

Hence, the inverse of matrix A is

⎥⎥⎥

⎢⎢⎢

−−

−−−

−−−−

⎥⎥⎥

⎢⎢⎢

−−−

−−

−=−

Kron Reduction Method

⎥⎥⎥⎥

⎢⎢⎢⎢

⎥⎥⎥⎥

⎢⎢⎢⎢

⎥⎥⎥⎥

⎢⎢⎢⎢

44434241

34333231

24232221

14131211

⎥⎥⎥⎥

⎢⎢⎢⎢

⎥⎥⎥⎥

⎢⎢⎢⎢

⎥⎥⎥⎥

⎢⎢⎢⎢

44434241

34333231

24232221

14131211

The four equations can be reduced to three equations by Kron Reduction Method since the independent variable of the fourth equation is zero.

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=⎥⎦

⎤⎢⎣

[ ]⎥⎥⎥

⎢⎢⎢

[ ] [ ]42 xX =[ ] [ ]4342413 aaaA = [ ] [ ]444 aA =

[ ]⎥⎥⎥

⎢⎢⎢

333231

232221

131211

A [ ]⎥⎥⎥

⎢⎢⎢

[ ] [ ]02 =Y

[ ]⎥⎥⎥

⎢⎢⎢

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=⎥⎦

⎤⎢⎣

[ ] [ ][ ] [ ][ ]22111 XAXAY +=

[ ] [ ][ ] [ ][ ]24130 XAXA +=

[ ][ ] [ ][ ]1324 XAXA −=

[ ] [ ] [ ][ ]131

42 XAAX −−=

From (2)

Substitute [X2] to equation (1)

[ ] [ ][ ] [ ] [ ] [ ][ ]( )131

42111 XAAAXAY −−+=

[ ] [ ] [ ][ ] [ ]{ }[ ]131

4211 XAAAAY −−=

⎥⎥⎥⎥

⎢⎢⎢⎢

⎥⎥⎥⎥

⎢⎢⎢⎢

⎥⎥⎥⎥

⎢⎢⎢⎢

yExample:

[ ] [ ]⎥⎥⎥

⎢⎢⎢

⎟⎟⎟

⎜⎜⎜

⎥⎥⎥

⎢⎢⎢

⎡−

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

⎡−

[ ][ ]⎥⎥⎥

⎢⎢⎢

⎟⎟⎟

⎜⎜⎜

⎥⎥⎥

⎢⎢⎢

⎡−

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

742125.0

[ ]⎥⎥⎥

⎢⎢⎢

⎟⎟⎟

⎜⎜⎜

⎥⎥⎥

⎢⎢⎢

⎡−

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

⎟⎟⎟

⎜⎜⎜

⎥⎥⎥

⎢⎢⎢

⎡−

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

25.535.1

375.45.225.1

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

25.00.150.1

625.15.475.6

50.000

The 4x4 matrix was reduced to a 3x3matrix.

⎥⎥⎥⎥

⎢⎢⎢⎢

⎥⎥⎥⎥

⎢⎢⎢⎢

⎥⎥⎥⎥

⎢⎢⎢⎢

� Cramer’s Rule of Determinants

� Matrix Inversion Method

� Gaussian Elimination Method

� Gauss-Jordan Method

Direct Solutions of System of Equations

Solutions of System of Equations by Cramer’ s Rule

3333232131

2323222121

1313212111

yxaxaxa

y xaxaxa

The system of three linear equations in three unknowns x1, x2, x3:

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

333231

232221

131211

or AX = Y

written in matrix form as :

can be solved by Cramer’s Rule of determinants. The determinant of coefficient matrix A is

333231

232221

131211

x 33323

x1 can be obtained by :

Note that values in the numerator are the values of the determinant of A with the first column were replaced by the Y vector elements.

Similarly, x2 and x3 can be obtained by:

x 33331

x 33231

142x4x6x-

7x2xx-

3 2xxx

=++Example:

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

Solutions of System of Equations by Cramer’s Rule

⎥⎥⎥

⎢⎢⎢

|A| −=

x1 −=−

x 33323

x2 −=−

x3 =−−

=⎥⎥⎥

⎢⎢⎢

−−

3 x-1 x2- x 321 ===Therefore,

x 33331

x 33231

Solutions of System ofEquations by Matrix Inversion

The system of equations in matrix form can be manipulated as follows:

YA AXA

Hence, the solution X can be obtained by multiplying

The inverse of the coefficient matrix by the constant

matrix Y.

142x4x6x-

7x2xx-

3 2xxx

=++Example:

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

−−−

−−

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

−−−

−−

1YA X 1-

⎥⎥⎥

⎢⎢⎢

−+−+−

−++−

)14(1 )7(10 )3(16

(14)3 )7(14 )3(4

)14(5 )7(6 )3(8

1YA X 1-

⎥⎥⎥

⎢⎢⎢

⎡−

⎥⎥⎥

⎢⎢⎢

1YA X 1-

=Therefore:

Gaussian Elimination Method

The following are the rules in matrix manipulation:

(1) Interchange rows

(2) Multiply row by constant

(3) Add rows

Example:

⎥⎥⎥

⎢⎢⎢

−−

⎥⎥⎥

⎢⎢⎢

⎡−

3214100

Add row 1 to row 2 to get row 2.Add 6 times row 1 to row 3 to get row 3.

142x4x6x-

7x2x x-

3 2xx x

⎥⎥⎥

⎢⎢⎢

⎡−

3214100

⎥⎥⎥

⎢⎢⎢

⎡−

1324400

Multiply row 2 by 10 then add to row 3 to obtained row 3.

By Back Substitution:

3)3(2)1(x

10)3(3xx0

132x44x00x

=+−+

-2x 1- x 3 x 123 ===

Therefore:

Forwardelimination

Backsubstitution

The two phases ofGauss Elimination: forward elimination& back substitution.The primes indicate the number of timesthat the coefficients and constants havebeen modified.

1131321211

1131211

3333231

2232221

1131211

a/)xaxac(x

a/)xac(x

−−=

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

MGaussian Elimination Method

⎥⎥⎥

⎢⎢⎢

⎡−−

1324400

Multiply row 2 by -1.

Gauss-Jordan Method

⎥⎥⎥

⎢⎢⎢

⎡−

1324400

From Gauss Elimination Method

⎥⎥⎥

⎢⎢⎢

⎡−−

Divide row 3 by 44.

Gauss-Jordan Method

⎥⎥⎥

⎢⎢⎢

⎡−−

Multiply row 2 by -1 then add to row 1 to get row 1.

Gauss-Jordan Method

Multiply row 3 by -5 then add to row 1 to get row 1.

Multiply row 3 by 3 then add to row 2 to get row 2.

⎥⎥⎥

⎢⎢⎢

⎡−

Gauss-Jordan Method

Therefore:

⎥⎥⎥

⎢⎢⎢

⎡−

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

−=Then,

Gauss-Jordan Method

� The Gauss-Jordan method is a variation of Gauss Elimination. The major differences is that when an unknown is eliminated in the GJM, it is eliminated from all other equations rather than just the subsequent ones.

� In addition, all rows are normalized by dividing them by their pivot elements. Thus, the elimination steps results in an identity matrix rather than a triangular matrix.

3333231

2232221

1131211

⎥⎥⎥

⎢⎢⎢

⎥⎥⎥

⎢⎢⎢

Graphical depiction of theGauss-Jordan Method.The superscript (n) meansthat the elements of the right-hand-side vector have been modified ntimes (for this case, n=3).

Gauss-Jordan Method

� Gauss Iterative Method

� Gauss-Seidel Method

� Newton-Raphson Method

Iterative Solutions of System of Equations

An iterative method is a repetitive process for obtaining the solution of an equation or a system of equation. It is applicable to system of equations where the main-diagonal elements of the coefficient matrix are larger in magnitude in comparison to the off-diagonal elements.

The Gauss and Gauss-Seidel iterative techniques are for solving linear algebraic solutions and the Newton-Raphson method applied to the solution of non-linear equations.

The solutions starts from an arbitrarily chosen initial estimates of the unknown variables from which a new set of estimates is determined. Convergence is achieved when the absolute mismatch between the current and previous estimates is less than some pre-specified precision index for all the variables.

Example:

Assume a convergence index of ε = 0.001 and the following initial estimates:

5 3x x x

6 x 4x x

4 x x 4x

0.5 x x x b)

0.0 x x x a)0

Gauss Iterative Method

Solution:

a) The system of equation must be expressed in standard form.

)x- x 4 (4

) x -x - 5(3

) x - x - 6 ( 4

Iteration 1 (k = 0):

1.6667 xmax

6667.106667.1x

5.105.1x

1.6667 ) 0 -0 - 5(3

1.5 ) 0 - 0 - 6 ( 4

1.0) 0 - 0 4 (4

0 x x x witha) 0

0.83334 xmax

83334.06667.1833333.0x

66667.05.1833333.0x

041667.01958325.0x

0.833333 ) 1.5 -1.0 - 5(3

0.833333 ) 1.6667 - 1.0 - 6 ( 4

0.958333) 1.6667 - 1.5 4 (4

−=−=

0.23617 xmax

23617.08333.00695.1x

21877.0833325.00521.1x

041667.0958325.01x

1.0695 ) 0.8333 -0.9583 - 5(3

1.0521 ) 0.8333 - 0.9583 - 6 ( 4

1.0 ) 0.8333 - 0.83334 (4

0.0869 xmax

0869.00695.19826.0x

0695.00521.19826.0x

0044.019956.0x

0.9826 ) 1.0521 -1.0 - 5(3

0.9826 ) 1.0695 - 1.0 - 6 ( 4

0.9956) 1.0695 - 1.05214 (4

−=−=

0.0247 xmax

0247.09826.00073.1x

0228.09826.00054.1x

0044.09956.01x

1.00730.9826) -0.9956 - 5(3

1.0054 ) 0.9826 - 0.9956 - 6 ( 4

1.0 ) 0.9826 - 0.98264 (4

−=−=

0.0091 xmax

0091.00073.19982.ox

0072.00054.19982.0x

0005.019995.0x

0.9982 ) 1.0054 -1.0 - 5(3

0.9982 ) 1.0071 - 1.0 - 6 ( 4

0.9995) 1.0073 - 1.00544 (4

−=−=

0.0026 xmax

0026.09982.00008.1x

0024.09982.00006.1x

0005.09995.01x

1.00080.9982) -0.9995 - 5(3

1.0006 ) 0.9982 - 0.9995 - 6 ( 4

1.0 ) 0.9982 - 0.99824 (4

0.0010xmax

0010.00008.19998.0x

0008.00006.19998.0x

0005.019995.0x

0.9998 ) 1.0008 -1.0 - 5(3

0.9998 ) 1.0008 - 1.0 - 6 ( 4

0.9995) 1.0008 - 1.00064 (4

−=−=

The Gauss iterative method has converged at iteration 7. The method yields the following solution.

0.9998x

0.9995x

0.5 x x x withb) 0

1 ===Gauss Iterative Method

Note: Number of iterations to achieve convergence is also dependent on initial estimates

Given the system of algebraic equations,

In the above equation, the x’s are unknown.

3nnn232131

2n2n222121

1n1n212111

yxaxaxa

y xaxaxa

↓↓↓↓

From the first equation,

)xaxay(a

1x n1n2121

1 −−= K

n1n2121111 xaxayxa −−= K

Similarly, x2, x3…xn of the 2nd to the nth equations can be obtained.

↓↓↓↓

−−−=

)xaxaxa(ba

1x n2n3231212

)xaxaxab(a

1x 1-n1-nn,2n21n1n

n −−−−= K

In general, the jth equation may be written

)xab(a

1i jij

∑≠=−=

n2,1,j K=

equation “a”

In general, the Gauss iterative estimates are:

where k is the iteration count

yx −−−−=+

21k2 −−−−=+

1-nn,k

yx −−−−=+

From an initial estimate of the unknowns (x10,

x20,…xn

0), updated values of the unknown variables are computed using equation “a”. This completes one iteration. The new estimates replace the original estimates. Mathematically, at the kth iteration,

)xab(a

1i jij

∑≠=

+ −=

n2,1,j K=

equation “b”

A convergence check is conducted after each iteration. The latest values are compared with their values respectively.

k xxx −= +Δn2,1,j K=

equation “c”

The iteration process is terminated when

t)(convergen |x |max k

j εΔ <

)convergent-(non itermax k =

Solution:

a) The system of equation must be expressed in standard form.

) x -x - 5(3

) x - x - 6 ( 4

)x- x 4 (4

Gauss-Seidel Method

Example: Solve the system of equations using the Gauss-Seidel method. Used a convergence index of ε = 0.001

5 3x x x

6 x 4x x

4 x x 4x

0.5 x x x 0

Gauss-Seidel Method

Iteration 1 (k =0):

0.625 | x |max

4583.050.09583.0x

625.050.0125.1x

50.05.01x

0.9583 ) 1.125 -1.0 - 5(3

1.125 ) 0.5 - 1.0 - 6 ( 4

1.0) 0.5 - 0.54 (4

0.5 x x x with 0

Gauss-Seidel Method

0.125 | x |max

0323.09583.09861.0x

125.0125.11x

0417.010417.1x

0.9861 ) 1.0 -1.0417 - 5(3

1.0 ) 0.9583 - 1.0417 - 6 ( 4

1.0417) 0.9583 - 1.1254 (4

−=−=

Gauss-Seidel Method

0.0119 | x |max

0119.09861.09980.0

0026.010026.1

0382.00417.10035.1

0.9980 ) 1.0026 -1.0035 - 5(3

1.0026 ) 0.9861 - 1.0035 - 6 ( 4

1.0035) 0.9861 - 1.0 4 (4

=−=Δ

−=−=Δ

Gauss-Seidel Method

0.0024 | x |max

0015.09980.09995.0x

0024.00026.10002.1x

0023.00035.10012.1x

0.9995 1.0002) -1.0012 -1.0 - 5(3

1.0002 0.9980) - 1.0012 - 6 ( 4

1.0012)0.9980 -1.00264 (4

−=−=

Gauss-Seidel Method

0.001 | x|max

0004.09995.09999.0x

0001.00002.10001.1x

001.00012.10002.1x

0.9999 1.0001) -1.0002 - 5(3

1.00010.9995) - 1.0002 - 6 ( 4

1.0002)0.9995 - 1.00024 (4

−=−=

Gauss-Seidel Method

The Gauss-Seidel Method has converged after 4 iterations only with the following solutions:

0.9999x

1.0001x

1.0002x

Gauss-Seidel Method

The Gauss-Seidel method is an improvement over the Gauss iterative method. As presented in the previous section, the standard form of the jth equation may be written as follows.

)xab(a

1i jij

∑≠=

−= n2,1,j K=

Gauss-Seidel Method

From an initial estimates (x10, x2

0,…xn0), an updated

value is computed for x1 using the above equation with j set to 1.This new value replaces x1

0 and is then used together with the remaining initial estimates to compute a new value for x2. The process is repeated until a new estimate is obtained for xn. This completes one iteration.

Note that within an iteration, the latest computed values are used in computing for the remaining unknowns. In general, at iteration k,

)xab(a

1i jij

α∑≠=

+ −=

n2,1,j K=

Gauss-Seidel Method

j i if 1k

jiif kwhere

After each iteration, a convergence check is conducted. The convergence criterion applied is the same with Gauss Iterative Method.

An improvement to the Gauss Iterative Method

Gauss-Seidel Method

1−−−=

2−−−= ++

in1k1i

1ii,1k1-i

1-ii,1ki

+++ −−−−−=+

1-nn,1k1

++ −−−=+

Solve the non-linear equations

Newton-Raphson Method

The Newton-Raphson method is applied when the system of equations is non-linear.

Consider a set of 2 non-linear equations in 2 unknowns.

( )( )2122

00 )()()( 20

1011 xx

fxfy Δ

∂∂

+Δ∂∂

)()()( 20

02022 xx

fxfy Δ

∂∂

+Δ∂∂

The system of non-linear equations can be linearizedusing the first order Taylor’s Series

Where:x0 = (x1

0, x20) are set of initial estimates

fi(x0) = the function fi (x1,x2) evaluated using the set of initial estimates.

= the partial derivatives of the function fi(x1,x2)evaluated using the set of original estimates.j

∂∂

⎥⎦

⎤⎢⎣

⎥⎥⎦

⎢⎢⎣

⎡=⎥

⎤⎢⎣

∂∂

The equation may be written in matrix form as follows:

The matrix of partial derivatives is known as the Jacobian. The linearized system of equations may be solved for ∆x’s which are then used to update the initial estimates.

At the kth iteration:

≤−

Convergence is achieved when

Where ε is pre-set precision indices.

Solve the non-linear equation

00 −==

x x :use

2211 xxf 4−=

Solution: First, form the Jacobian

∂∂ 4−=

∂∂

212 xxf −= 2 1−=∂∂

2 2=∂∂

⎥⎦

⎤⎢⎣

⎥⎥⎥⎥

⎢⎢⎢⎢

∂∂

=⎥⎦

⎤⎢⎣

⎡−

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

=⎥⎦

⎤⎢⎣

⎡−−

−−

)xx2(2

)x4x(4

In Matrix form

2(1)2x

4y ,5)1(41)x(f

−=∂∂

==∂∂

==−−=

2y ,3)1()1(2)x(f

−=∂∂

=∂∂

==−−=

Newton-Raphson MethodIteration 0:

The equations are:

x)1(x)2(32

x)4(x)2(54

−+=−

−+=−⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−=⎥

⎤⎢⎣

⎡−

In matrix form:

5.0)5.0(1x1

−=+−=

=−+=

Solving,

Repeating the process with the new estimates,Iteration 1:

0.1)5.0(2x

4y ,25.4)1(4)5.0()x(f

−=∂

==−−=

2y ,2)1()5.0(2)x(f

−=∂

==−−=

The equations are: In matrix form:

Solving,

x4x25.44

−=−

−=−⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−=⎥

⎤⎢⎣

⎡−1

07143.0x

03571.0x1

92857.007143.01x

53571.003571.05.0x2

−=+−=

Repeating the process with the new estimates,Iteration 2:

07142.1)53571.0(2x

4y ,001265.4)92857.0(4)53571.0()x(f

−=∂

==−−=

299999.1)92857.0()53571.0(2)x(f

−=∂

≅=−−=

The equations are:

In matrix form:

Solving,

xx20.22

x4x07142.1001265.44

−=−

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−=⎥

⎤⎢⎣

⎡−2

407142.1

001265.0

00036.0x

00018.0x2

92893.000035.092857.0x

53553.000018.053571.0x3

−=−−=

4y ,2512400.4)92893.0(4)53553.0()x(f 1

11 ==−−=

2y ,99928.1)92893.0()53553.0(2)x(f 2

12 ==−−=

00072.0fy

0025.0fy

−=−

92893.0x

53553.0x

Substituting to the original equation:

Therefore,

Note the rapid convergence of the Newton-RaphsonMethod.

The Newton-Raphson method is applied when the system of equations is non-linear.

Consider a set of n non-linear equations in n unknowns.

)x,,x,(xfy

)x,,x,x(fy

11 x)x(x

f)x(fy ΔΔΔ

∂∂

++∂∂

+∂∂

22 x)x(x

f)x(fy ΔΔΔ

∂∂

++∂∂

+∂∂

nn x)x(x

f)x(fy ΔΔΔ

∂∂

++∂∂

+∂∂

M MM M

The system of non-linear equations can be linearizedusing Taylor’s Series

Where:X0 = (x1

0,x20, …, xn

= set of initial estimates

fi(x0) = the function fi (x1,x2, …, xn)

evaluated using the set of initial estimates.

= the partial derivatives of the function fi(x1,x2,…,xn) evaluated using the set of original estimates.

∂∂

⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢

⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢

⎥⎥⎥⎥⎥

⎢⎢⎢⎢⎢

∂∂

The equation may be written in matrix form as follows:

The matrix of partial derivatives is known as the Jacobian. The linearized system of equations may be solved for ∆x’s which are then used to update the initial estimates.

n..., 2, 1, j xxx k

1kj =+== Δ

At the kth iteration:

n..., 2, 1, j )x(fy 1

jj =≤− ε

n ..., 2, 1, j x 2

j =≤ εΔ

Convergence is achieved when

Where ε1 and ε2 are pre-set precision indices.

Fundamental Principles & Methods

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