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ECE 530 – Analysis Techniques for Large-Scale Electrical Systems. Lecture 1: Solution of Nonlinear Equations. Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign

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ece 530 analysis techniques for large scale electrical systems

ECE 530 – Analysis Techniques for Large-Scale Electrical Systems

Lecture 1: Solution of Nonlinear Equations

Prof. Tom Overbye

Dept. of Electrical and Computer Engineering

University of Illinois at Urbana-Champaign

Special Guest Lecture by (soon to be) Prof. Hao Zhu

course overview
Course Overview
  • Course presents the fundamental analytic, simulation and computation techniques for the analysis of large-scale electrical systems. The course stresses the importance of the structural characteristics of the systems, with an aim towards practical analysis.
  • Prof. Overbye will give a full introduction next lecture. Today Hao Zhu will jump into the analysis of nonlinear equations.
linear equations
Linear Equations
  • Course assumes that students are familiar with the solution of linear equations expressed as
  • Here we will use the style of bolding matrices and vectors; another common style is to underline them
  • 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
linear equations1
Linear Equations

A function fis linear if

f(a1m1+ a2m2) = a1f(m1) + a2f(m2)

That is

1) the output is proportional to the input

2) the principle of superposition holds

Linear Example:y = f(x) = Ax

y= A(x1+x2) = Ax1+ Ax2

Nonlinear Example: y = f(x) = c x2

y = c(x1+x2)2 ≠ (cx1)2 + (c x2)2

nonlinear equations
Nonlinear Equations
  • In this section 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!
nonlinear example of multiple solutions and no solution
Nonlinear Example of Multiple Solutions and No Solution

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

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

f(x) = x2 - 2

f(x) = x2 + 2

no solution f(x) = 0

two solutions where f(x) = 0

nonlinear equations1
Nonlinear Equations
  • The notation f(x) is short-hand for the vector functionso the problem is to solve n equations for n unknowns
nonlinear equations2
Nonlinear Equations
  • The nonlinear functions f(·) of interest include both algebraic and transcendental types
  • What we’ll find is the power flow problem becomes nonlinear when we consider constant power loads
  • We’ll first consider the problem in a single dimension, and then treat the more general case of n dimensions.
newton raphson method
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
newton raphson method scalar
Newton-Raphson Method (scalar)

Note, a priori we do NOT know x

sequential linear approximations
Sequential Linear Approximations

At each

iteration the

N-R method

uses a linear


to determine

the next value

for x

Function is f(x) = x2 - 2 = 0.

Solutions are points where

f(x) intersects f(x) = 0 axis

newton s method for a scalar equation


x (1)

x (4)


x (3)

x (2)

x (0)

Newton’s Method for a Scalar Equation

f (x)

example 2
Example 2
  • Find the positive root of

using Newton’s method starting

  • Computation must be done using radians!!!
example 2 iterations
Example 2 Iterations
  • We continue the iterations to obtain the following set of results
example 2 changed initial guess
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
newton raphson comments
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
normal convergence
Normal Convergence

f (x)

desired root

oscillatory convergence






Oscillatory Convergence

f (x)

Note that we actuallyovershoot the solution


f (x)





stopping criteria vector norms
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