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

2

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

3






� Voltage and Current Directions

� Double Subscript Notation

� Voltage, Current and Phasor Notation

� Complex Impedance and Phasor Notation

Circuit Conventions & Notations

4






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)

Vs ZB

+

-

++

-

-

VA

ZA

VB

a b

o n

I

II

I

Voltage and Current Directions


5






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.

+

-

+ -VA

ZAa b

o n

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

ZA

6






Voltage, Current and Phasor Notation


tjmVv ωε= tsinjtcostj ωωε ω +=

voltsV

V m o02

∠=

-1

-0.5

0

0.5

1

-4 -2 0 2 4ππ π π

v

i

1j −=

V

I

θθθθ

( )θωε −= tjmIi

amperesI

I m oθ−∠=2

7






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=

tjmV

dt

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

tjm

tjtj VLKjRK ωωω εεωε =+Hence,


8







Solving for the current

Dividing voltage by current to get the impedance,

tjm

LjR

V)t(i ωε

ω+=

LjR

LjR

VV

)t(i

)t(vZ

tjm

tjm ω

εω

ε

ω

ω

+=

+

==

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

ωωωωL

R

θθθθ

Z


9






+

-

+

-

VLi(t)R (Resistance)

C (Capacitance)tj

mS VV ωε=

For Capacitive Circuit, .

)C

1(jRZ

ω−=


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

φφφφR

Z

1

ωωωωC


10






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

10

17.32

30°203020

05

30100Z ∠=

∠∠

=

1032.17)30sin30(cos20 jjZ +=+=


11






� Electrical Symbols

� Three-Line and Single-Line Diagram

� Equivalent Circuit of Power System Components

� Impedance and Admittance Diagram

� Bus Admittance Matrix

Power System Representation

12






Generator

TransformerCircuit Breaker

Transmission or Distribution Line

Bus

or

G Switch

Node

Fuse

Electrical Symbols

Power System representation

13






3-phase wye neutral grounded

3-phase delta connection

Ammeter

Voltmeter

3-phase wye neutral ungrounded

Protective Relay

V

A

R

Current Transformer

Potential Transformer

Power System representation

14






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

Relays

Cir

cuit

Bre

aker

Mai

n B

us

R

R

R

R

CTs Distribution Lines

Transformer

A B C


Cir

cuit

Rec

lose

r

15






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

R

BusCB

TransformerDistribution Line

CT and Relay

Recloser


16






Equivalent Circuit of Power System Components: Generator

Ec

Eb

Ea

Ic

Ib

Ia

b

a

c

sa jXR +

aI +

-

aV

+

-gE

Za

Zb Zc

Three-Phase Equivalent Single-Phase Equivalent


17






Equivalent Circuit of Power System Components: Transformer

A

B

C

a

b

c

Core Loss

Primary Secondary

6x6 Admittance Matrix

Y66Y65Y64Y63Y62Y61

Y56Y55Y54Y53Y52Y51

Y46Y45Y44Y43Y42Y41

Y36Y35Y34Y33Y32Y31

Y26Y25Y24Y23Y22Y21

Y16Y15Y14Y13Y12Y11

Three-Phase Equivalent


18






X2

HT RaRR +=

X2

HT XaXX +=

Equivalent Circuit of Power System Components: Transformer

+

-

+

-

TZ

Single-Phase Equivalent


HIr+

-HVr

+

-XVar

mjXcR

exIv

XX XjaRa 22 +HH jXR +

19






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.

Δl


20






Equivalent Circuit of Power System Components: Distribution Lines

ZccZcbZca

ZbcZbbZba

ZacZabZaa

Equivalent ππππ-Network

A

BC

a

bc

YccYcbYca

YbcYbbYba

YacYabYaa

1/2YccYcbYca

YbcYbbYba

YacYabYaa

1/2

Capture Unbalanced

Characteristics Three-Phase Equivalent


21






Equivalent Circuit of Power System Components: Long Transmission Lines

• •

••+

-

VsVR

+

-

IS IR

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

y

zZ C =

Characteristic Impedance zy=γPropagation Constant

2

ltanh

Z

1

2

'Y

c

γ=

2

'Y



22






Equivalent Circuit of Power System Components: Medium-Length

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

• •

••+

-

+

-

Vs

IS IR

( )ljxrZ L+=

2

c/1

2

Y ω=

2

Y VR



23






Equivalent Circuit of Power System Components: Short Transmission Lines

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

• •

••+

-

Vs VR

+

-

Is = IR


( )ljxrZ L+=


24






Single Phase Equivalent of Balanced Three-Phase System

ZR

ZR

ZR

n

Eao = |E|∠∠∠∠0° V

Ebo= |E| ∠∠∠∠240° V

Eco = |E| ∠∠∠∠120° V

a

b

c

o

Ic

b

c

Iaa

Ib


25






ZR

n

Eao = |E|∠∠∠∠ 0° V

ao

ZRZR

n

Ebo= |E| ∠∠∠∠240° V b

o

ZR

ZR

n

Eco = |E| ∠∠∠∠120° V c

o

c

b

a

Ic

Ib

Ia

θθ

−∠=∠∠

= IZ

0EI

Ra

)240(IZ

240EI

Rb θ

θ−∠=

∠∠

=

)120(IZ

120EI

Rc θ

θ−∠=

∠∠

=

Note: Currents are Balanced


26






Single Phase Representation of a Balanced Three-Phase System

Eao = |E|∠∠∠∠ 0° V

n

a

o

a

Ia

ZR

ZR

ZR

n

Eao = |E|∠∠∠∠0° V

Ebo= |E| ∠∠∠∠240° V

Eco = |E| ∠∠∠∠120° V

a

b

c

o

Ic

b

c

Iaa

Ia

ZR


27







Impedance and Admittance Diagrams

21 3

4

bca

Bus

1234

Gen

abc

Line

1 - 32 - 31 - 42 - 43 - 4

Single Line Diagram

28








21 3

4

Impedance Diagram

0

0

0

1

31

Ea za

z13

zd

zcza

ze

zf zg

zb

Ea Ec Eb

Generator

Line

zh

29








VL

ILZpIs

Zg

+

-Eg

VL

IL

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

Eg = ISZPZg = Zp

30






21 3

4

bca

Bus

1234

Gen

abc

Line

1 - 32 - 31 - 42 - 43 - 4

Single Line Diagram


31






21 3

4

Impedance Diagram

0

0

0

1

31

Ea za

z13

zd

zcza

ze

zf zg

zb

Ea Ec Eb

Generator

Line

zh


32






Equivalent Sources

VL

ILZpIs

Zg

+

-Eg

VL

IL

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

Eg = ISZPZg = Zp


33






21 3

4

Admittance Diagram

0

y13

y03y01

y23

y14 y24

y02

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


34






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

35






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

( ) 3432421413424144 yVyVyVyyyV0 −−−++=

( ) 1313442321334230333 yVyVyVyyyyVI −−−+++=

In matrix form,

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

++−−−

+++

++

++

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

4

3

2

1

342414342414

34133423032313

2423242302

1413141301

3

2

1

V

V

V

V

yyyyyy

y-yyyyy-y-

y-y-yyy0

y-y-0yyy

0

I

I

I

( )24423324230222 yVyVyyyVI −−++=


36






====

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

n

3

2

1

I

I

I

I

M

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

nn3n2n1n

n3333231

n2232221

n1131211

YYYY

YYYY

YYYY

YYYY

L

MMMM

L

L

L

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

n

3

2

1

V

V

V

V

M

[ 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

37






⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

nn3n2n1n

n3333231

n2232221

n1131211

YYYY

YYYY

YYYY

YYYY

L

MMMM

L

L

L

[YBUS] =


Ybus is also called Bus Admittance Matrix

38






� 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

39






Base Value

Actual ValuePer Unit Value =

Per-unit Value is a dimensionless quantity

Per-unit value is expressed as decimal

100

Actual ValuePercent =

Per Unit Value

The Per Unit System

40






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

41






PU VoltagePU Current =

PU Impedance

PU Power = PU Voltage x PU Current

Per Unit Calculations

The Per Unit System

42






I

Zline = 1.4 ∠∠∠∠75° ΩΩΩΩ

Zload = 20 ∠∠∠∠30° ΩΩΩΩ 2540 ∠∠∠∠0° V

+

-

+

-

Example:

Vs = ?

Determine Vs

Per Unit Calculations

The Per Unit System

43






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

44






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

45






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

46






Base PowerBase Current =

Base Voltage

Base Voltage (Base Voltage)2

Base Impedance = = Base Current Base Power

For a Given Base Power and Base Voltage,


47






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)


48






3

kV Base kV Base

MVABase3

1MVABase

LL1

31

=

=

φ

φφ

• 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


49






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:


50






� 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

51






base

)(L)pu(L Z

XX ΩΩΩΩ====

base

)()pu( Z

RR ΩΩΩΩ====

base

)(C

)pu(C Z

XX Ω=

Per Unit ImpedanceTransmission and Distribution Lines

52






� 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

53






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

54






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

2

=⎟⎠⎞

⎜⎝⎛=

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

Per Unit Impedance

55






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

56






SOLUTION:

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

( kV)² = ________ ohms

( MVA)Zbase =

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


57






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

= ________ ohmsZbase =

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

( kV)²

( MVA)

SOLUTION:


58






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.

ohms

ohms

ohms


59






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


LOAD

60






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

XAB=10% XBC=8%

VcVA

Solve using actual quantities

300 ΩΩΩΩ/ φφφφPF=100%


61






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. =


62






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

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

VNL

V.R. =

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

=


63






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


64






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.


65






Zactual = Zpu1 • Zbase1 Zactual = Zpu2 • Zbase2

2base2pu1base1pu ZZZZ ⋅=⋅

2base

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


66






Recall:( )

Powerbase

voltagebaseZ

2

base =

( )

( )2,3

22,

1,3

21,

12

base

baseLL

base

baseLL

pupu

MVA

kV

MVA

kV

ZZ

φ

φ=

Then,

or,

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

φ

φ

1base,3

2base,3

2

2base,LL

1base,LL1pu2pu MVA

MVA

kV

kVZZ


67






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

2

)new(base

2

)given(base

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

Px

]V[

]V[xZZ =

68






Solution

On its own base the transformer reactance is

On the chosen base the reactance becomes

= ________ puX =( )

( )2

( )

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

( )2( )


69






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

70






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


71






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 %

T1

L1

T2

G1G2

G3


Use Base Power = 100 MVA

72






E1

XT1

E2 E3

T1Transmission Line

T2

G1G2

G3XLINE XT2

XG1 XG2 XG3


1 2 3

4

73






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


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

Base Power:

74






c) Compute Per Unit Impedance


G2:

T1:

G1:

G3:

L1:

T2:

75






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.

76






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.


77






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.


78






� Sequence Components of Unbalanced Phasor

� Sequence Impedance of Power System Components

� Practical Implications of Sequence Components of Electric Currents

Symmetrical Components

79






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

80






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.


81






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


82






Phase a

Phase b

Phase c

120°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.

83






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


84






a = 1 ∠∠∠∠ 120°

OPERATOR “a”

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


V

120°

aV

Operating V by a

a² = 1 ∠∠∠∠ 240°

a³ = 1 ∠∠∠∠ 0°

85






Vb in terms of Va

Vb = a² VaVb1 = a² Va1

Vc in terms of Va

Vc = a VaVc1 = a Va1


Va

Vc = aVa

120°120°

120°

Vb = a2Va

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

86






Vb in terms of Va

Vb2 = aVa2

Vc in terms of Va

Vc2 = a²Va2


Va2

Vb2 = aVa2

120°120°

120°

Vc2 = a2Va2

The Negative Sequence components in terms of phase a

87






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

88






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

3

1V ++=++=

[ ]cb2

a2a aVVaV3

1V ++=

89






EXAMPLE:

Determine the symmetrical components of the followingunbalanced voltages.

Vc = 8 ∠∠∠∠143.1

Vb = 3 ∠∠∠∠-90

Va = 4 ∠∠∠∠0


90






[ ]

18.384.9

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

1

)Va aV V(3

1V c

2ba1a

∠=

∠∠+∠∠+∠=

++=

For Phasor VFor Phasor Vaa: :

143.051

143.1) 8 90- 3 0 4(3

1

)V V V(3

1V cba0a

∠=

∠+∠+∠=

++=


91






For Phasor VFor Phasor Vaa: :

[ ]

86.082.15

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

1

)aV Va V(3

1V cb

2a2a

−∠=

∠∠+∠∠+∠=

++=


92






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

33.922.15

86.08)-120)(2.15(1

aV V

101.62-4.9

258.384.9

18.38)240)(4.9(1

Va V

143.05 1 143.05 1

VV

a2b2

1a2

b1

a0b0

∠=

∠∠=

=

∠=

∠=

∠∠=

=

∠=∠=

=


93






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

3.92512.15

86.08)-0)(2.1542(1

Va V

8.38314.9

18.38)0)(4.921(1

Va V

143.05 1 143.05 1

VV

a22

c2

1ac1

a0c0

∠=

∠∠=

=

∠=

∠∠=

=

∠=∠=

=


94






Positive Sequence Components Negative Sequence Components

Zero Sequence Components

Va0 = Vb0 = Vc0Vc1

Va1

Vb1Va2

Vb2Vc2


°18.38

°143.05

°86.08

95






Components of Vc

Components of Va

Components of Vb

Add Sequence Components Graphically

Va0

Vb0

Vc0

Vc1

Va1

Vb1

Va2

Vb2

Vc2

Vc = 8 ∠∠∠∠143.1

Vb = 3 ∠∠∠∠-90

Va = 4 ∠∠∠∠0


96






The results can be checked either mathematically or graphically.


143.18

153.922.15138.384.905.143 1

V V V V

90- 3

33.922.15101.62-4.9143.05 1

VV V V

0 4

86.08-2.1518.384.905.143 1

V V V V

c2c1c0c

b2b1b0b

a2a1a0a

∠=

∠+∠+∠=

++=

∠=

∠+∠+∠=

++=

∠=

∠+∠+∠=

++=

97







Ia

b

a

c

b

a

c

Ib

Ic

Ia1 + Ia2 + Ia0

Ib1 + Ib2 + Ib0

Ic1 + Ic2 + Ic0

98






Sequence Impedance of Power System Components

Positive Sequence Negative Sequence Zero Sequence

+

-

Z1

Ia1

Vf+

Va1

+

-

Ia2

Va2Z2

+

-

Ia0

Va0Z0

2a2a2ZI -V ====

1a1fa1ZI – V V ====

oaoaoZI -V ====

Sequence Networks

99






Sequence Impedance of Power System Components

� In general,

Z1 ≠ Z2 ≠ Z0 for generators

Z1 = Z2 = Z0 for transformers

Z1 = Z2 ≠ Z0 for lines

100






Practical Implications of Sequence Components of Electric Currents

ZERO-SEQUENCE CURRENTS:

Ia0

3Io

Ic0

Ib0

b

a

c

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

IA0

3I0

IC0

IB0

B

A

C

101







ZERO-SEQUENCE CURRENTS:

Ia0

3Io

Ic0

Ib0

b

a

c

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.

IA0

IC0

IB0

BA

C

I0 = 0

I0 = 0

I0 = 0

102







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.

103






� 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

104






The standard form of n independent equations:

====

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

n

3

2

1

I

I

I

I

M

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

nn3n2n1n

n3333231

n2232221

n1131211

YYYY

YYYY

YYYY

YYYY

L

MMMM

L

L

L

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦⎦⎦⎦

⎤⎤⎤⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣⎣⎣⎣

⎡⎡⎡⎡

n

3

2

1

V

V

V

V

M

[ 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

105






21 3

4

bca

Bus

1234

Gen

abc

Line

1 - 32 - 31 - 42 - 43 - 4

Single Line Diagram

Network Equations

106






21 3

4

Impedance Diagram

0

0

0

1

31

Ea za

z13

zd

zcza

ze

zf zg

zb

Ea Ec Eb

Generator

Line

zh

Network Equations

107






21 3

4

Admittance Diagram

0

y13

y03y01

y23

y14 y24

y02

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

108






( ) 14413314130111 yVyVyyyVI −−++=

( ) 3432421413424144 yVyVyVyyyV0 −−−++=

( ) 1313442321334230333 yVyVyVyyyyVI −−−+++=

In matrix form,

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

++−−−

+++

++

++

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

4

3

2

1

342414342414

34133423032313

2423242302

1413141301

3

2

1

V

V

V

V

yyyyyy

y-yyyyy-y-

y-y-yyy0

y-y-0yyy

0

I

I

I

( )24423324230222 yVyVyyyVI −−++=

Network Equations

109






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

yxaxaxa

y xaxaxa

=++

=++

=++

110






⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

333231

232221

131211

aaa

aaa

aaa

A

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


The coefficients form an array

111






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

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

3

2

1

x

x

x

X and

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

3

2

1

y

y

y

Y


112






The system of equations in matrix notation is

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

3

2

1

3

2

1

333231

232221

131211

y

y

y

x

x

x

aaa

aaa

aaa

AX = Y


� System of Equations in Matrix Form

113






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

Rearranging and writing in matrix form


⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

2a

1a

0a

2

2

c

b

a

V

V

V

aa1

aa1

111

V

V

V


114







⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

c

b

a

2

2

2a

1a

0a

V

V

V

aa1

aa1

111

3

1

V

V

V

[ ] [ ]c2

ba1acba0a VaaVV3

1V VVV

3

1V ++=++=

[ ]cb2

a2a aVVaV3

1V ++=

In matrix form

115






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.

116






Definition of a MATRIX

[ ] [ ]

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

==

mnm3m2m1

3n333231

2n232221

1n131211

ij

aaaa

aaaa

aaaa

aaaa

aA

K

MMM

K

K

K

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

117






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

2

1

x

x

x

XM

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

118






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.

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

m

2

1

c

c

c

CM

119






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

Type of Matrices

120






Type of Matrices

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

333231

232221

131211

aaa

aaa

aaa

A

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.

121






⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

33

2322

131211

u00

uu0

uuu

U

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

Type of Matrices

122






Type of Matrices

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

333231

2221

11

lll

0ll

00l

L

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

123






⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

33

22

11

d00

0d0

00d

D

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

124






⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

100

010

001

I

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

Type of Matrices

125






Type of Matrices

The null matrix is matrix whose elements are

equal to zero.

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

000

000

000

N

126






Type of Matrices

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

872

731

215

S

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

7aa

2aa

1aa

3223

3113

2112

==

==

==

127






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

063

605

350

K

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−

=5a

5a

21

12

+=

−=

128






� 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

129






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]

130






⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡=

110

625B

372

041A

Example:

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡+++

+++=+

482

666

1)(31)(70)(2

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

then,

⎥⎦

⎤⎢⎣

⎡ −−=⎥

⎦

⎤⎢⎣

⎡−−−

−−−=−

262

624

1)(31)(70)(2

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


131






⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+−−

+++

+−+

=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+−+

+++

−−+

=

5j64j55j7

4j56j41j2

5j71j22j3

B

9j81j13j6

1j13j51j4

3j61j42j1

A

Example:

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

⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+++−+−−++

+++++++++

++−−+−+++

=+

5j69j84j51j15j73j6

4j51j16j43j51j21j4

5j73j61j21j42j32j1

B A

then,


132






⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+−−

+++

+−+

=+

14j145j62j13

5j69j92j6

2j132j64j4

B A


⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

++−+−

−−−+

−−++−

=−

4j23j48j1

3j43j10j2

8j10j20j2

B A

133






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

kakaka

kakaka

AkkA

L

MMMM

L

L

134






Example:

3 k and

16

25

34

A =

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

318

615

912

B


⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡==

16

25

34

3 Ak B

135






Example:

3 k and

4j1j36

5j2j2-5

6j3j14

A =

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−+

+

−+

=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−+

+−

−+

=

12j39j18

15j66j15

18j93j12

B


⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+

+

+

==

j4-1j36

j52j2-5

j6-3j14

3 Ak B

136






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).

137






[ ] [ ] [ ] 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

138







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

∑=

=l

1kkjikij bac for all i and j.

139







2x33231

2221

1211

aa

aa

aa

A

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

2x22221

1211

bb

bbB ⎥

⎦

⎤⎢⎣

⎡=and

⎥⎦

⎤⎢⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

2221

1211

3231

2221

1211

bb

bb

aa

aa

aa

C = A B =

then

Example:

2112111111 babac +=∑=

=2

1kkjikij bac

140






⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

++

++

++

==

)baba()bab(a

)baba()baba(

)bab(a)bab(a

AB C

2232123121321131

2222122121221121

2212121121121111


∑=

=2

1kkjikij bac

141







2363

52

41

x

A

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

2x209

87B ⎥

⎦

⎤⎢⎣

⎡=and

⎥⎦

⎤⎢⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡==

09

87

63

52

41

ABC

then

Example:

142







232475

1659

843

x

C

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

23)0683()9673(

)0582()9572(

)0481()9471(

xxxxx

xxxx

xxxx

ABC

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

++

++

++

==

143







⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+−−

+++

+−+

=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+−+

+++

−−+

=

5j64j55j7

4j56j41j2

5j71j22j3

B

9j81j13j6

1j13j51j4

3j61j42j1

A

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

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

Example:

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

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

144







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

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

[ ] [ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+++

+++

+−−

=

144j3915j10143j116

73j3741j2024j29

42j7216j4441j35

Bx A

145






Transpose of a Matrix

2x33231

2221

1211

T

aa

aa

aa

A

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

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

aaa

aaaA ⎥

⎦

⎤⎢⎣

⎡=

146






Example:

2x3

T

3x2

65

43

21

A

642

531 A

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎦

⎤⎢⎣

⎡=

then,


147






Example:

then,


⎥⎦

⎤⎢⎣

⎡−+−

+−+=

3j61j45j2

2j56j34j1A

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−+

+−

−+

=

3j62j5

1j46j3

5j24j1

AT

148






Determinant of a Matrix

Two simultaneous equations:

Determinant of a 2 x 2 Matrix

)2(yxaxa

)1(yxaxa

2222121

1212111

=+

=+

In Matrix Form

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

2

1

2

1

2221

1211

y

y

x

x

aa

aa

149







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:

1

22

21

22

22 x

a

a

a

yx −=

1212222 xayxa −=

150







21122211

2121221

212122121122211

1221211221212211

11

22

21

22

212111

aaaa

yayax

yayax )aaaa(

yaxaayaxaa

y)xa

a

a

y(a xa

−−

=

−=−

=−+

=−+

substituting x2 and solving for x1

151






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

2221

1211

aa

aa |A| =


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

21122211

1212112 aaaa

yayax

−−

=

152






[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−−=

246

121

211

A

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


246

121

211

|A|

−

−−=

For example:

Eliminate row 1 by striking out the row1 and jth column

153






246

121

211

|A|

−

−−=

The determinant of A is

46

212)1(

26

111)1(

24

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(

−=

−−++−−−−=

−−−−++

−−−−+−−+=

|A|

|A|

154






The determinant obtained by striking out the ith

row and jth column is called the minor element aij.

Example:

3332

1312

333231

232221

131211

aa

aa

aaa

aaa

aaa

=

Determinant of a MatrixMinors and Cofactors of a Matrix

)aaaa(a of minorThe 1332331221 −=

155






The cofactor of an element aij designated by Aij is

( )ij

ji

ij a of minor)1(A +−=

( )2121

213

2112

21

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 −−=∴

156






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

246

121

211

)1(A 21

12 −=−−−−=

−

−−−= +

Example:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

246-

12-1-

211

A

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

246

121

211

)1(A 11

11 −=−−=

−

−−−= +


157







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

246

121

211

)1(A 22

22 =−−=

−

−−−= +

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

246

121

211

)1(A 31

13−=−−−−=

−

−−−= +

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

246

121

211

)1(A 12

21 =−−=

−

−−−= +

158






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

246

121

211

)1(A 32

23−=−−−=

−

−−−= +

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

246

121

211

)1(A 13

31 =−−=

−

−−−= +

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

246

121

211

)1(A 23

32 −=−−−=

−

−−−= +


159






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

246

121

211

)1(A 33

33−=−−−=

−

−−−= +


160







⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

333231

232221

131211

AAA

AAA

AAA

A of Cofactors

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−

−−−

=

135

10146

1648

A of Cofactors

and in matrix form:

161






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

yxaxaxa

y xaxaxa

=++

=++

=++

162






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

331

221

111

1 y|A|

Ay

|A|

Ay

|A|

Ax ++=

332

222

112 y

|A|

Ay

|A|

Ay

|A|

Ax ++=2

333

223

113

3 y|A|

Ay

|A|

Ay

|A|

Ax ++=

Inverse of a Matrix

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

163






Thus,

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

==

|A|

A

|A|

A

|A|

A|A|

A

|A|

A

|A|

A|A|

A

|A|

A

|A|

A

AB

332313

322212

312111

1-

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

|A|

AA 1-

+

=or

164






Inverse of a Matrix

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

−−

−

=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=+

11016

3144

568

AAA

AAA

AAA

A

332313

322212

312111

Example: Get the inverse of A

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

246-

12-1-

211

A

the Adjoint of A is

|A|

AA 1-

+

=

165






Inverse of a Matrix

44

11016

3144

568

A

AA 1

−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

−−

−

==+

−

Hence, the inverse of matrix A is

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

−−

−−

−−

−−

−−−

−−−−

441

4410

4416

443

4414

444

445

446

448

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

−−

−

−=−

11016

3144

568

44

1A 1

166






Kron Reduction Method

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

4

3

2

1

44434241

34333231

24232221

14131211

3

2

1

x

x

x

x

aaaa

aaaa

aaaa

aaaa

0

y

y

y

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

4

3

2

1

44434241

34333231

24232221

14131211

3

2

1

x

x

x

x

aaaa

aaaa

aaaa

aaaa

0

y

y

y

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

167







⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=⎥⎦

⎤⎢⎣

⎡

2

1

43

211

X

X

AA

AA

0

Y

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

3

2

1

1

x

x

x

X

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

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

333231

232221

131211

1

aaa

aaa

aaa

A [ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

34

24

14

2

a

a

a

A

[ ] [ ]02 =Y

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

3

2

1

1

y

y

y

Y

168







⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=⎥⎦

⎤⎢⎣

⎡

2

1

43

211

X

X

AA

AA

0

Y

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

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

[ ][ ] [ ][ ]1324 XAXA −=

[ ] [ ] [ ][ ]131

42 XAAX −−=

(1)

(2)

From (2)

169







Substitute [X2] to equation (1)

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

42111 XAAAXAY −−+=

Thus,

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

4211 XAAAAY −−=

170







⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

4

3

2

1

3

2

1

x

x

x

x

8742

6543

5678

4321

0

y

y

yExample:

[ ] [ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

3

2

11

3

2

1

7428

6

5

4

543

678

321

x

x

x

y

y

y

171







[ ][ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

3

2

1

3

2

1

742125.0

6

5

4

543

678

321

x

x

x

y

y

y

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

3

2

1

3

2

1

742

75.0

625.0

5.0

543

678

321

x

x

x

y

y

y

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

3

2

1

3

2

1

25.535.1

375.45.225.1

5.321

543

678

321

x

x

x

y

y

y

172







⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−

=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

3

2

1

3

2

1

25.00.150.1

625.15.475.6

50.000

x

x

x

y

y

y

The 4x4 matrix was reduced to a 3x3matrix.

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

4

3

2

1

3

2

1

x

x

x

x

8742

6543

5678

4321

0

y

y

y

173






� Cramer’s Rule of Determinants

� Matrix Inversion Method

� Gaussian Elimination Method

� Gauss-Jordan Method

Direct Solutions of System of Equations

174






Solutions of System of Equations by Cramer’ s Rule

3333232131

2323222121

1313212111

yxaxaxa

yxaxaxa

y xaxaxa

=++

=++

=++

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

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

3

2

1

3

2

1

333231

232221

131211

y

y

y

x

x

x

aaa

aaa

aaa

or AX = Y

written in matrix form as :

175






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

333231

232221

131211

aaa

aaa

aaa

|A| =

|A|

aay

aay

aay

x 33323

23222

13121

1 =

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.

176






Similarly, x2 and x3 can be obtained by:

and

|A|

aya

aya

aya

x 33331

23221

13111

2 =

|A|

yaa

yaa

yaa

x 33231

22221

11211

3 =

177






142x4x6x-

7x2xx-

3 2xxx

321

321

321

=++

=+−

=++Example:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

14

7

3

x

x

x

246-

12-1-

211

3

2

1

Solutions of System of Equations by Cramer’s Rule

178






⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

246-

12-1-

211

A 44

246

121-

211

|A| −=

−

−=

244

88

44-

2414

127

213

x1 −=−

=

−

=


|A|

aay

aay

aay

x 33323

23222

13121

1 =

179






144

44

44-

2146

171

231

x2 −=−

=−

−

=

344

132

44-

1446

721

311

x3 =−−

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−−

=

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


|A|

aya

aya

aya

x 33331

23221

13111

2 =

|A|

yaa

yaa

yaa

x 33231

22221

11211

3 =

180






Solutions of System ofEquations by Matrix Inversion

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

YAX

YA IX

YA AXA

YAX

1-

1-

1-1-

=

=

=

=

Hence, the solution X can be obtained by multiplying

The inverse of the coefficient matrix by the constant

matrix Y.

181






142x4x6x-

7x2xx-

3 2xxx

321

321

321

=++

=+−

=++Example:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

246-

12-1-

211

A


182






⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

−−

−

−=

11016

3144

568

44

1A 1-

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

−−

−

−==

14

7

3

11016

3144

568

44

1YA X 1-

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−+−+−

−++−

++−

−==

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

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

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

44

1YA X 1-


183






⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

−

=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−==

3

1

2

132

44

88

44

1YA X 1-

3 x

1- x

2- x

3

2

1

=

=

=Therefore:


184






Gaussian Elimination Method

The following are the rules in matrix manipulation:

(1) Interchange rows

(2) Multiply row by constant

(3) Add rows

185






Example:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−−

14246

7121

3211

M

M

M

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

3214100

10310

3211

M

M

M

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

321

321

321

=++

=+−

=++

186






⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

3214100

10310

3211

M

M

M

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

1324400

10310

3211

M

M

M

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


187






By Back Substitution:

3)3(2)1(x

10)3(3xx0

132x44x00x

1

21

321

=+−+

=+−

=++

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

Therefore:


188






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

'

223

'

23

'

22

"

33

"

33

"

3

"

33

'

2

'

23

'

22

1131211

3333231

2232221

1131211

a/)xaxac(x

a/)xac(x

a/cx

ca

caa

caaa

caaa

caaa

caaa

−−=

−=

=

⇓

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⇓

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

M

M

M

M

M

MGaussian Elimination Method

189






⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−−

1324400

10310

3211

M

M

M

Multiply row 2 by -1.

Gauss-Jordan Method

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

1324400

10310

3211

M

M

M

From Gauss Elimination Method

190






⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−−

3100

10310

3211

M

M

M

Divide row 3 by 44.

Gauss-Jordan Method

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−−

3100

10310

13501

M

M

M

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

191






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.

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

−

3100

1010

2001

M

M

M

192






Gauss-Jordan Method

Therefore:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

−

=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

3

1

2

x

x

x

100

010

001

3

2

1

3x

1x

2x

3

2

1

=

−=

−=Then,

193






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.

194






(n)

3 3

(n)

22

(n)

11

(n)

3

(n)

2

(n)

1

3333231

2232221

1131211

c x

c x

c x

c100

c010

c001

caaa

caaa

caaa

=

=

=

↓

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

↓

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

M

M

M

M

M

M

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

195






� Gauss Iterative Method

� Gauss-Seidel Method

� Newton-Raphson Method

Iterative Solutions of System of Equations

196







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.

197







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.

198






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

3 21

32 1

321

=++

=++

=+−

0.5 x x x b)

0.0 x x x a)0

3

0

2

0

1

0

3

0

2

0

1

===

===

Gauss Iterative Method

199






Solution:

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

)x- x 4 (4

1x k

3

k

2

1k

1 +=+


) x -x - 5(3

1x k

2

k

1

1k

3 =+

) x - x - 6 ( 4

1x k

3

k

1

1k

2 =+

200






Iteration 1 (k = 0):

1.6667 xmax

6667.106667.1x

5.105.1x

101x

1.6667 ) 0 -0 - 5(3

1x

1.5 ) 0 - 0 - 6 ( 4

1x

1.0) 0 - 0 4 (4

1x

0

3

0

3

0

2

0

1

1

3

1

2

1

1

=

=−=

=−=

=−=

==

==

=+=

Δ

Δ

Δ

Δ

0 x x x witha) 0

3

0

2

0

1 ===


201






0.83334 xmax

83334.06667.1833333.0x

66667.05.1833333.0x

041667.01958325.0x

0.833333 ) 1.5 -1.0 - 5(3

1x

0.833333 ) 1.6667 - 1.0 - 6 ( 4

1x

0.958333) 1.6667 - 1.5 4 (4

1x

1

3

1

3

1

2

1

1

2

3

2

2

2

1

=

−=−=

=−=

=−=

==

==

=+=

Δ

Δ

Δ

Δ



202







0.23617 xmax

23617.08333.00695.1x

21877.0833325.00521.1x

041667.0958325.01x

1.0695 ) 0.8333 -0.9583 - 5(3

1x

1.0521 ) 0.8333 - 0.9583 - 6 ( 4

1x

1.0 ) 0.8333 - 0.83334 (4

1x

2

3

2

3

2

2

2

1

3

3

3

2

3

1

=

=−=

=−=

=−=

==

==

=+=

Δ

Δ

Δ

Δ


203






0.0869 xmax

0869.00695.19826.0x

0695.00521.19826.0x

0044.019956.0x

0.9826 ) 1.0521 -1.0 - 5(3

1x

0.9826 ) 1.0695 - 1.0 - 6 ( 4

1x

0.9956) 1.0695 - 1.05214 (4

1x

3

3

3

3

3

2

3

1

4

3

4

2

4

1

=

−=−=

−=−=

−=−=

==

==

=+=

Δ

Δ

Δ

Δ



204







0.0247 xmax

0247.09826.00073.1x

0228.09826.00054.1x

0044.09956.01x

1.00730.9826) -0.9956 - 5(3

1x

1.0054 ) 0.9826 - 0.9956 - 6 ( 4

1x

1.0 ) 0.9826 - 0.98264 (4

1x

4

3

4

3

4

2

4

1

5

3

5

2

5

1

=

=−=

−=−=

=−=

==

==

=+=

Δ

Δ

Δ

Δ


205






0.0091 xmax

0091.00073.19982.ox

0072.00054.19982.0x

0005.019995.0x

0.9982 ) 1.0054 -1.0 - 5(3

1x

0.9982 ) 1.0071 - 1.0 - 6 ( 4

1x

0.9995) 1.0073 - 1.00544 (4

1x

5

3

5

3

5

2

5

1

6

3

6

2

6

1

=

−=−=

−=−=

−=−=

==

==

=+=

Δ

Δ

Δ

Δ



206







0.0026 xmax

0026.09982.00008.1x

0024.09982.00006.1x

0005.09995.01x

1.00080.9982) -0.9995 - 5(3

1x

1.0006 ) 0.9982 - 0.9995 - 6 ( 4

1x

1.0 ) 0.9982 - 0.99824 (4

1x

6

3

6

3

6

2

6

1

7

3

7

2

7

1

=

=−=

=−=

=−=

==

==

=+=

Δ

Δ

Δ

Δ


207






0.0010xmax

0010.00008.19998.0x

0008.00006.19998.0x

0005.019995.0x

0.9998 ) 1.0008 -1.0 - 5(3

1x

0.9998 ) 1.0008 - 1.0 - 6 ( 4

1x

0.9995) 1.0008 - 1.00064 (4

1x

7

3

7

3

7

2

7

1

8

3

8

2

8

1

=

−=−=

−=−=

−=−=

==

==

=+=

Δ

Δ

Δ

Δ



208






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

0.9998x

0.9998x

0.9995x

3

2

1

=

=

=


209







=

=

=

=

=

=

=

xmax

x

x

x

x

x

x

0

3

0

2

0

1

1

3

1

2

1

1

Δ

Δ

Δ

Δ

0.5 x x x withb) 0

3

0

2

0

1 ===Gauss Iterative Method

210






xmax

x

x

x

x

x

x

1

1

3

1

2

1

1

2

3

2

2

2

1

=

=

=

=

=

=

=

Δ

Δ

Δ

Δ



211








xmax

x

x

x

x

x

x

2

2

3

2

2

2

1

3

3

3

2

3

1

=

=

=

=

=

=

=

Δ

Δ

Δ

Δ

212








xmax

x

x

x

x

x

x

3

3

3

3

2

3

1

4

3

4

2

4

1

=

=

=

=

=

=

=

Δ

Δ

Δ

Δ

213








xmax

x

x

x

x

x

x

4

4

3

4

2

4

1

5

3

5

2

5

1

=

=

=

=

=

=

=

Δ

Δ

Δ

Δ

214








xmax

x

x

x

x

x

x

5

5

3

5

2

5

1

6

3

6

2

6

1

=

=

=

=

=

=

=

Δ

Δ

Δ

Δ

215








xmax

x

x

x

x

x

x

6

6

3

6

2

6

1

7

3

7

2

7

1

=

=

=

=

=

=

=

Δ

Δ

Δ

Δ

216








xmax

x

x

x

x

x

x

7

7

3

7

2

7

1

8

3

8

2

8

1

=

=

=

=

=

=

=

Δ

Δ

Δ

Δ

217






x

x

x

3

2

1

=

=

=


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

218







Given the system of algebraic equations,

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

3nnn232131

2n2n222121

1n1n212111

yxaxaxa

yxaxaxa

y xaxaxa

=+++

↓↓↓↓

=+++

=+++

L

L

L

219






From the first equation,

)xaxay(a

1x n1n2121

11

1 −−= K

n1n2121111 xaxayxa −−= K

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

↓↓↓↓

−−−=

)xaxaxa(ba

1x n2n3231212

22

2 K

)xaxaxab(a

1x 1-n1-nn,2n21n1n

nn

n −−−−= K


220






In general, the jth equation may be written

as

)xab(a

1x i

n

1i jij

jj

jji

∑≠=−=

n2,1,j K=

equation “a”


221






In general, the Gauss iterative estimates are:

where k is the iteration count


k

n

11

1nk

3

11

13k

2

11

12

11

11k

1 xa

a...x

a

ax

a

a

a

yx −−−−=+

xa

a...x

a

ax

a

a

a

yx k

n22

2nk3

22

23k1

22

21

22

21k2 −−−−=+

k

1-n

nn

1-nn,k

2

nn

n2k

1

nn

n1

nn

n1k

n xa

a...x

a

ax

a

a

a

yx −−−−=+

222






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

1x kn

1i jij

jj

1k

j iji

∑≠=

+ −=

n2,1,j K=

equation “b”


223






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

k

j

1k

j

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 =


225






Solution:

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

) x -x - 5(3

1x

) x - x - 6 ( 4

1x

)x- x 4 (4

1x

1k

2

1k

1

1k

3

k

3

1k

1

1k

2

k

3

k

2

1k

1

+++

++

+

=

=

+=

Gauss-Seidel Method

224






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

3 21

32 1

321

=++

=++

=+−

0.5 x x x 0

3

0

2

0

1 ===

Gauss-Seidel Method

226






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

1x

1.125 ) 0.5 - 1.0 - 6 ( 4

1x

1.0) 0.5 - 0.54 (4

1x

0

2

0

3

0

2

0

1

1

3

1

2

1

1

=

=−=

=−=

=−=

==

==

=+=

Δ

Δ

Δ

Δ

0.5 x x x with 0

3

0

2

0

1 ===

Gauss-Seidel Method

227






0.125 | x |max

0323.09583.09861.0x

125.0125.11x

0417.010417.1x

0.9861 ) 1.0 -1.0417 - 5(3

1x

1.0 ) 0.9583 - 1.0417 - 6 ( 4

1x

1.0417) 0.9583 - 1.1254 (4

1x

1

2

1

3

1

2

1

1

2

3

2

2

2

1

=

=−=

−=−=

=−=

==

==

=+=

Δ

Δ

Δ

Δ


Gauss-Seidel Method

228







0.0119 | x |max

0119.09861.09980.0

0026.010026.1

0382.00417.10035.1

0.9980 ) 1.0026 -1.0035 - 5(3

1x

1.0026 ) 0.9861 - 1.0035 - 6 ( 4

1x

1.0035) 0.9861 - 1.0 4 (4

1x

23

23

22

21

3 3

3 2

3 1

=Δ

=−=Δ

=−=Δ

−=−=Δ

==

==

=+=

x

x

x

Gauss-Seidel Method

229






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

1x

1.0002 0.9980) - 1.0012 - 6 ( 4

1x

1.0012)0.9980 -1.00264 (4

1x

3

2

3

3

3

2

3

1

4

3

4

2

4

1

=

=−=

−=−=

=−=

==

==

=+=

Δ

Δ

Δ

Δ


Gauss-Seidel Method

230







εΔ

Δ

Δ

Δ

0.001 | x|max

0004.09995.09999.0x

0001.00002.10001.1x

001.00012.10002.1x

0.9999 1.0001) -1.0002 - 5(3

1x

1.00010.9995) - 1.0002 - 6 ( 4

1x

1.0002)0.9995 - 1.00024 (4

1x

4

4

3

4

2

4

1

5

3

5

2

5

1

<=

=−=

−=−=

−=−=

==

==

=+=

Gauss-Seidel Method

231






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

0.9999x

1.0001x

1.0002x

3

2

1

=

=

=

Gauss-Seidel Method

232






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

1x i

n

1i jij

jj

jji

∑≠=

−= 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.

233






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

)xab(a

1x i

n

1i jij

jj

1k

jij

α∑≠=

+ −=

n2,1,j K=

Gauss-Seidel Method

j i if 1k

jiif kwhere

<+=

>=α

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

234






An improvement to the Gauss Iterative Method

Gauss-Seidel Method

xa

a...x

a

a

a

yx k

n11

1nk2

11

12

11

11k

1−−−=

+

xa

a...x

a

a

a

yx k

n22

2n1k1

22

21

22

21k

2−−−= ++

1kn

ii

in1k1i

ii

1ii,1k1-i

ii

1-ii,1ki

ii

ij

ii

i xa

ax

a

ax

a

a...x

a

a

a

yx

1k

i

+++

+++ −−−−−=+

xa

a...x

a

a

a

yx 1k

1-nnn

1-nn,1k1

nn

n1

nn

n1k

n

++ −−−=+

235






Solve the non-linear equations

22

44

=−

=−

21

221

xx

xx

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

2111

xxfy

xxfy

,

,

=

=

236






00 )()()( 20

2

11

0

1

1011 xx

x

fxx

x

fxfy Δ

∂∂

+Δ∂∂

+=

00

1

)()()( 20

2

21

02022 xx

x

fxx

x

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

0

i

x

)x(f

∂∂

237






⎥⎦

⎤⎢⎣

⎡

Δ

Δ

⎥⎥⎦

⎤

⎢⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡

−

−

∂∂

∂∂

∂∂

∂∂

0

0

)()(

))(

0

1

0

)(

(

2

1

xxf

xxf

x(xf

xxf

022

011

x

x

xfy

)xfy

2

22

2

01

1

o1

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.


238






k2

k2

k2

k1

k1

k

xxx

xxx

Δ+=

Δ+=+

+

1

11

At the kth iteration:

ε

ε

≤−

≤−

)xfy

)xfy

k22

k11

(

(

Convergence is achieved when

Where ε is pre-set precision indices.


239






Solve the non-linear equation

1,1

22

44

00 −==

=−

=−

21

21

221

x x :use

xx

xx

2211 xxf 4−=

Solution: First, form the Jacobian


11

1 xx

f2=

∂∂ 4−=

∂∂

2

1

x

f

212 xxf −= 2 1−=∂∂

2

2

x

fx

f

1

2 2=∂∂

240






⎥⎦

⎤⎢⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

∂∂

∂∂

∂∂

∂∂

=⎥⎦

⎤⎢⎣

⎡−

−

2

1

2

0

2

1

0

2

2

0

1

1

0

1

022

0

11

x

x

x

)x(f

x

)x(fx

)x(f

x

)x(f

)x(fy

)x(fy

ΔΔ

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

=⎥⎦

⎤⎢⎣

⎡−−

−−

2

11

21

2

2

1

x

x

12

4-x2

)xx2(2

)x4x(4

Δ

Δ


In Matrix form

241






4x

f

2(1)2x

f

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

2

1

1

1

1

20

1

−=∂∂

==∂∂

==−−=

1x

f

2x

f

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

2

2

1

2

2

0

2

−=∂∂

=∂∂

==−−=

Newton-Raphson MethodIteration 0:

242






The equations are:

02

01

02

01

x)1(x)2(32

x)4(x)2(54

ΔΔ

ΔΔ

−+=−

−+=−⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−=⎥

⎦

⎤⎢⎣

⎡−

−0

2

01

x

x

12

42

1

1

Δ

Δ

In matrix form:

0x

5.0x0

2

0

1

=

−=

Δ

Δ

101x

5.0)5.0(1x1

2

11

−=+−=

=−+=

Solving,

Thus,


243






Repeating the process with the new estimates,Iteration 1:

4x

)x(f

0.1)5.0(2x

)x(f

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

2

11

1

11

121

1

−=∂

∂

==∂

∂

==−−=

1x

)x(f

2x

)x(f

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

2

12

1

12

21

2

−=∂

∂

=∂

∂

==−−=


244






The equations are: In matrix form:

Solving,

Thus,


1

2

1

1

1

2

1

1

xx222

x4x25.44

ΔΔ

ΔΔ

−=−

−=−⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−=⎥

⎦

⎤⎢⎣

⎡−1

2

1

1

x

x

12

41

0

25.0

Δ

Δ

07143.0x

03571.0x1

2

1

1

=

=

Δ

Δ

92857.007143.01x

53571.003571.05.0x2

2

2

1

−=+−=

=+=

245






Repeating the process with the new estimates,Iteration 2:

4x

)x(f

07142.1)53571.0(2x

)x(f

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

2

1

1

1

2

1

1

22

1

−=∂

∂

==∂

∂

==−−=

1x

)x(f

2y 2x

)x(f

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

2

2

2

2

1

2

2

2

2

−=∂

∂

==∂

∂

≅=−−=


246






The equations are:

In matrix form:

Solving,

Thus,


2

2

2

1

2

2

2

1

xx20.22

x4x07142.1001265.44

ΔΔ

ΔΔ

−=−

−=−

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−=⎥

⎦

⎤⎢⎣

⎡−2

2

2

1

x

x

12

407142.1

0

001265.0

Δ

Δ

00036.0x

00018.0x2

2

2

1

=

−=

Δ

Δ

92893.000035.092857.0x

53553.000018.053571.0x3

2

3

1

−=−−=

=−=

247






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

23

11 ==−−=

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

3

12 ==−−=

00072.0fy

0025.0fy

22

11

=−

−=−

92893.0x

53553.0x

2

1

−=

=

Substituting to the original equation:

Therefore,

Note the rapid convergence of the Newton-RaphsonMethod.


248






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

)x,,x,x(fy

n21nn

n2122

n2111

K

M

K

K

=

=

=

249






0

n

0

n

10

2

0

2

10

1

0

1

10

11 x)x(x

fx)x(

x

fx)x(

x

f)x(fy ΔΔΔ

∂∂

++∂∂

+∂∂

+= K

0

n

0

n

20

2

0

2

20

1

0

1

20

22 x)x(x

fx)x(

x

fx)x(

x

f)x(fy ΔΔΔ

∂∂

++∂∂

+∂∂

+= K

0

n

0

n

10

2

0

2

10

1

0

1

n0

nn x)x(x

fx)x(

x

fx)x(

x

f)x(fy ΔΔΔ

∂∂

++∂∂

+∂∂

+= K

M MM M


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

250






Where:X0 = (x1

0,x20, …, xn

0)

= 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.

j

0

i

x

)x(f

∂∂


251






⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−

−

−

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

0n

02

01

x)x(f

x)x(f

x)x(f

x)x(f

x)x(f

x)x(f

x)(xf

x)(xf

x)x(f

0nn

022

011

x

x

x

)x(fy

)x(fy

)x(fy

n

0n

2

0n

1

0n

n

02

2

02

1

02

n

01

2

01

1

o1

Δ

Δ

Δ

M

K

MMM

K

K

M

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.


252






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

j

k

j

1kj =+== Δ

At the kth iteration:

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

k

jj =≤− ε

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

k

j =≤ εΔ

Convergence is achieved when

or

Where ε1 and ε2 are pre-set precision indices.


253






Fundamental Principles & Methods

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