ECE 530 – Analysis Techniques for Large-Scale Electrical Systems

Prof. Hao ZhuDept. of Electrical and Computer Engineering

University of Illinois at Urbana-Champaign

haozhu@illinois.edu

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Lecture 4: Newton-Raphson Method

Power Flow Analysis

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i1

In the Newton-Raphson power flow we use Newton's

method to determine the voltage magnitude and angle

at each bus in the power system.

We need to solve the power balance equations

P ( cosn

i k ik ikk

V V G

i1

sin )

Q ( sin cos )

ik ik Gi Di

n

i k ik ik ik ik Gi Dik

B P P

V V G B Q Q

• When analyzing power systems, we know the complex power being consumed by the load, and the power being injected by the generators plus their voltage magnitudes• Want to find voltage magnitude and angle at every bus

Solving Nonlinear Equations

• Most common technique for solving the nonlinear power flow is to use the Newton-Raphson method•

• Key idea behind Newton-Raphson is to use sequential linearization• More tractable for linear systems

General form of problem: Find an x such that

( ) 0ˆf x

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Linear Equations

• Linear system of equations

• Here we will use the style of bolding matrices and vectors• Later in course we’ll consider solution methods for

sparse linear equations, which are quite common in electric power systems• Linear equations are conceptually easy to solve,

provided A is nonsingular; then there is a single solution

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Ax = b

Linear Power System Elements

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Resistors, inductors, capacitors, independent

voltage sources and current sources are linear

circuit elements

1V = R I V = V =

Such systems may be analyzed by superposition

j L I Ij C

Nonlinear Equations

• Motivated by power flow analysis, we’ll consider the solution of nonlinear equations of the form:

• Problem may be restated as finding a root x of f where both x and f(x) are n-vectors• A key challenge with nonlinear equations is there may

be one, none or multiple solutions!

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f(x) = 0

Nonlinear Example of Multiple Solutions and No Solution

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f(x) = x2 - 2 f(x) = x2 + 2

two solutions where f(x) = 0 no solution f(x) = 0

Example 1: x2 - 2 = 0 has solutions x = 1.414…

Example 2: x2 + 2 = 0 has no real solution

Nonlinear Equations

• The notation f(x) is short-hand for the vector function

so the problem is to solve n equations for n unknowns

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1 1 2

2 1 2

1 2

, , ,

, , ,

, , ,

n

n

n n

f x x x

f x x x

f x x x

Newton-Raphson Method

• Newton developed his method for solving for the roots of nonlinear equations in 1671, but it wasn’t published until 1736• Raphson developed a similar method in 1690;

Raphson’s approach was actually simpler than Newton’s, and is what is used today • General form of scalar problem is to find an x such that

f(x) = 0• Key idea behind the Newton-Raphson method is to use

sequential linearization

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Newton-Raphson Method (scalar)

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( )

( ) ( )

( )( ) ( )

2 ( ) 2( )2

1. For each guess of , , define

-

2. Represent ( ) by a Taylor series about ( )

( )( ) ( )

1 ( )higher order terms

2

v

v v

vv v

vv

x x

x x x

f x f x

df xf x f x x

dx

d f xx

dx

Note, a priori we do NOT know x

Newton-Raphson Method, cont’d

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( )( ) ( )

( )

1( )( ) ( )

3. Approximate ( ) by neglecting all terms

except the first two

( )( ) 0 ( )

4. Use this linear approximation to solve for

( )( )

5. Solve for a new estimat

vv v

v

vv v

f x

df xf x f x x

dx

x

df xx f x

dx

( 1) ( ) ( )

e of xv v vx x x

Newton-Raphson Example

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2

1( )( ) ( )

( ) ( ) 2( )

( 1) ( ) ( )

( 1) ( ) ( ) 2( )

Use Newton-Raphson to solve ( ) - 2 0

The equation we must iteratively solve is

( )( )

1(( ) - 2)

2

1(( ) - 2)

2

vv v

v vv

v v v

v v vv

f x x

df xx f x

dx

x xx

x x x

x x xx

Newton-Raphson Example, cont’d

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( 1) ( ) ( ) 2( )

(0)

( ) ( ) ( )

3 3

6

1(( ) - 2)

2

Guess x 1. Iteratively solving we get

v ( )

0 1 1 0.5

1 1.5 0.25 0.08333

2 1.41667 6.953 10 2.454 10

3 1.41422 6.024 10

v v vv

v v v

x x xx

x f x x

Sequential Linear Approximations

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Function is f(x) = x2 - 2 = 0.Solutions are points wheref(x) intersects f(x) = 0 axis

At each iteration theN-R methoduses a linearapproximationto determine the next valuefor x

Newton’s Method for a Scalar Equation

x (3) x (2) x (0)

root x*

x (4) x (1) x

f (x)

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Example 2

• Find the positive root of

using Newton’s method starting

• Computation must be done using radians!!!

. .. .

.

0

1 0

0

1 0 0 785391 57079 2 00001

0 0 5

f x x x

f x

0.5 0sin x x

x (0) 2

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Example 2 Graphical View

2

x

2

xf x sin x

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Example 2 Iterations

• We continue the iterations to obtain the following set of results

iteration number v

0 1.57079

1 2.00001

2 1.90100

3 1.89551

4 1.89549

vx

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Example 2, Changed Initial Guess

• It is interesting to note that we get to the value of 1.89549 also if we start at 3.14159

iteration number v

0 3.14159

1 2.09440

2 1.91322

3 1.89567

4 1.89549

vx

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Newton-Raphson Comments

• When close to the solution the error decreases quite quickly -- method has quadratic convergence• f(x(v)) is known as the mismatch, which we would like

to drive to zero• Stopping criteria is when f(x(v)) < • Results are dependent upon the initial guess. What if

we had guessed x(0) = 0, or x (0) = -1?• A solution’s region of attraction (ROA) is the set of

initial guesses that converge to the particular solution. The ROA is often hard to determine

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Normal Convergence

desired root

( )x 0( )1x ( )2x x

f (x)

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Oscillatory Convergence

x (3) x

(1) x (2) x

(0)x

(4)

f (x)Note that we actuallyovershoot the solution

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Convergence to an Unwanted Root

x

desired root

undesired root

x (1)

x (0)

f (x)

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Divergence

x (1) x

(0)

x (2) x

f (x)

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Multi-Variable Newton-Raphson

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1 1

2 2

Next we generalize to the case where is an n-

dimension vector, and ( ) is an n-dimension function

( )

( )( )

( )

Define the solution so ( ) 0 andˆ ˆ

ˆ

n n

x f

x f

x f

x

f x

x

xx f x

x

x f x

x x x

Multi-Variable Case, cont’d

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i

1 11 1 1 2

1 2

1

n nn n 1 2

1 2

n

The Taylor series expansion is written for each f ( )

f ( ) f ( )f ( ) f ( )ˆ

f ( )higher order terms

f ( ) f ( )f ( ) f ( )ˆ

f ( )higher order terms

nn

nn

x xx x

xx

x xx x

xx

x

x xx x

x

x xx x

x

Multi-Variable Case, cont’d

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1 1 1

1 21 1

2 2 22 2

1 2

1 2

This can be written more compactly in matrix form

( ) ( ) ( )

( )( ) ( ) ( )

( )( )ˆ

( )( ) ( ) ( )

n

n

nn n n

n

f f fx x x

f xf f f

f xx x x

ff f fx x x

x x x

xx x x

xf x

xx x x

higher order terms

nx

Jacobian Matrix

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1 1 1

1 2

2 2 2

1 2

1 2

The n by n matrix of partial derivatives is known

as the Jacobian matrix, ( )

( ) ( ) ( )

( ) ( ) ( )

( )

( ) ( ) ( )

n

n

n n n

n

f f fx x x

f f fx x x

f f fx x x

J x

x x x

x x x

J x

x x x

Multi-Variable N-R Procedure

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1

( 1) ( ) ( )

( 1) ( ) ( ) 1 ( )

( )

Derivation of N-R method is similar to the scalar case

( ) ( ) ( ) higher order termsˆ

( ) 0 ( ) ( )ˆ

( ) ( )

( ) ( )

Iterate until ( )

v v v

v v v v

v

f x f x J x x

f x f x J x x

x J x f x

x x x

x x J x f x

f x

Multi-Variable Example

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1

2

2 21 1 2

2 22 1 2 1 2

1 1

1 2

2 2

1 2

xSolve for = such that ( ) 0 where

x

f ( ) 2 8 0

f ( ) 4 0

First symbolically determine the Jacobian

f ( ) f ( )

( ) =f ( ) f ( )

x x

x x x x

x x

x x

x f x

x

x

x x

J xx x

Multi-variable Example, cont’d

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1 2

1 2 1 2

11 1 2 1

2 1 2 1 2 2

(0)

1(1)

4 2( ) =

2 2

Then

4 2 ( )

2 2 ( )

1Arbitrarily guess

1

1 4 2 5 2.1

1 3 1 3 1.3

x x

x x x x

x x x f

x x x x x f

J x

x

x

x

x

Multi-variable Example, cont’d

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1(2)

(2)

2.1 8.40 2.60 2.51 1.8284

1.3 5.50 0.50 1.45 1.2122

Each iteration we check ( ) to see if it is below our

specified tolerance

0.1556( )

0.0900

If = 0.2 then we wou

x

f x

f x

ld be done. Otherwise we'd

continue iterating.

Stopping Criteria: Vector Norms

• When x is a vector the stopping criteria is determined by calculating the vector norm. Any norm could be used, but the most common norm used is the infinity norm, , where

• Other common norms are the one norm, which is the sum of the element absolute values and the Euclidean (or two norm) defined as

max iixx

=

x

1

22

2 21

n

ii

x @

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=

=

Newton-Raphson Power Flow

i1

In the Newton-Raphson power flow we use Newton's

method to determine the voltage magnitude and angle

at each bus in the power system.

We need to solve the power balance equations

P ( cosn

i k ik ikk

V V G

i1

sin )

Q ( sin cos )

ik ik Gi Di

n

i k ik ik ik ik Gi Dik

B P P

V V G B Q Q

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Power Flow Variables

2 2 2

n

2

Assume the slack bus is the first bus (with a fixed

voltage angle/magnitude). We then need to determine

the voltage angle/magnitude at the other buses.

( )

( )

G

n

P P

V

V

x

x f x

2

2 2 2

( )

( )

( )

D

n Gn Dn

G D

n Gn Dn

P

P P P

Q Q Q

Q Q Q

x

x

x

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N-R Power Flow Solution

( )

( )

( 1) ( ) ( ) 1 ( )

The power flow is solved using the same procedure

discussed with the general Newton-Raphson:

Set 0; make an initial guess of ,

While ( ) Do

( ) ( )

1

End While

v

v

v v v v

v

v v

x x

f x

x x J x f x

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Power Flow Jacobian Matrix

1 1 1

1 2

2 2 2

1 2

1 2

The most difficult part of the algorithm is determining

and inverting the n by n Jacobian matrix, ( )

( ) ( ) ( )

( ) ( ) ( )

( )

( ) ( ) ( )

n

n

n n n

n

f f fx x x

f f fx x x

f f fx x x

J x

x x x

x x x

J x

x x x

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Power Flow Jacobian Matrix, cont’d

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i1

Jacobian elements are calculated by differentiating

each function, f ( ), with respect to each variable.

For example, if f ( ) is the bus i real power equation

f ( ) ( cos sin )n

i k ik ik ik ik Gik

x V V G B P P

x

x

i

1

i

f ( )( sin cos )

f ( )( sin cos ) ( )

Di

n

i k ik ik ik iki k

k i

i j ik ik ik ikj

xV V G B

xV V G B j i

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