CEE203 NUMERICAL METHODS IN ELECTRICAL ENGINEERING Assist. Prof. Dr. Çağatay ULUIŞIK...

CEE203NUMERICAL METHODS

IN ELECTRICAL ENGINEERING

Assist. Prof. Dr. Çağatay ULUIŞIK

[email protected]

2

OUTLINE

Introducing MatLab

Fundamental Terms, Basic Definitions

Solving Nonlinear Equations – Root Finding

Solving Systems of Linear Equations

Taylor Series

Numerical Differentiation

Numerical Integration

Curve Fitting and Interpolation

Ordinary Differential Equations

3

INTRODUCING MATLAB

Vectors, Arrays and Matrices

Assign a value to a variable (observe the effect of the semicolon):

Creating a row vector of size (14) Creating a column vector of size (41)

Creating an array of size (19) using the colon sign (:)

4

INTRODUCING MATLAB

Creating an array of size (19) using the for loop :

Creating a matrix of size (33) : Creating a matrix of size (33) using 2 for loops :

5

BASIC MATRIX OPERATIONS IN MATLAB

6

GRAPHIC UTILITIES OF MATLAB

7

GRAPHIC UTILITIES OF MATLAB

8

COMPLEX NUMBERS IN MATLAB

Addition : Subtraction : Multiplication : Division :

Conjugate : Magnitude : Angle in Radians : Angle in Degrees :

Defining a Complex Number :

9

LOADING DATA FROM A FILE AND SAVING DATA TO A FILE

10

STATISTICS IN MATLAB

Mean Value : Variance : Standard Deviation :

Generation of Uniformly Distributed Random Numbers :

Generation of Uniformly Distributed Random Integers :

11

INTRODUCING MATLAB

432)( 23 xxxxP

Roots of a Polynom :

Symbolic Integration : Numeric Integration :

clc : To clear the command windowclear all : To clear all the variables in memory

12

FORMAT OF DATA

roundRound towards nearest integer

floorRound towards minus infinity

ceilRound towards plus infinity

13

EXAMPLES OF MATLAB SCRIPTSExample 1 : Calculating of the sum of integers from 1 to a user-specifed value n

a) using a formula b)using iterations

Example 2 : Creating and Calling a function

14

EXAMPLES OF MATLAB SCRIPTS

Example 3 : Error of the first n terms in the Taylor Series of the function ex

a) Calculating the error for a given n value

b) Calculating the minimum required number of terms n for a given error.

a) b)

15

Accuracy: Closeness of agreement between a measured / computed value and a true value.

Measurement accuracy: the ability of an instrument to measure the true value to within some stated error specifications.

Numerical accuracy: the degree to which the numerical solution to the approximate physical problem approximates the exact solution to the approximate physical problem.

FUNDAMENTAL TERMS

16

Comparison of a measured value with the real value

Meas. ValueM

eas.

freq

uenc

yMean

Value interval

Accuracy = (%)True val. – Meas. val

True value

True value ?

MEASUREMENT ACCURACY

17

It is the difference between a measured or calculated value of a quantity and its exact value.

Systematic error plagues experiments or calculations caused by negative factors. For example, a DC voltage component, which unintentionally is present, e.g., because of a failure on the blockage capacitor, is a systematic error. These can be removed once understood/discovered via controls and calibration.

Random error is an error which is always present, but varies unpredictably in size and direction. They are related to the scatter in the data obtained under fixed conditions which determine the repeatability (precision) of the measurement and follow well-behaved statistical rules.

ERROR

18

Absolute error: It is an error that is expressed in physical units. It is the absolute value of the difference between the measured value and the true value (or the average value if the true value is not known) of a quantity.

Relative error: An error expressed as a fraction of the absolute error to the true (or average) value of a quantity. It is always given as a percentage.

ERROR

19

A number is represented with a finite (fixed) number of digits called word length.

Precision of a measurement or the accuracy of a computation bounds the number of significant digits. All non-zero digits beyond the number of significant digits at the right of a number are removed in one of two ways; truncation or round-off.

For example, 53.0534 has 6 significant digits. If it is going to be represented only by 4 significant digits both truncation and round-off processes yield 53.05. On the other hand, if the number of significant digits will be 3, then truncation and round-off processes yield, 53.0 and 53.1, respectively.

ERROR

20

A measurement/calculation result should be given as (a a), which means the value may be anything between (a - a) and (a + a).

For example, if the measured speed is 98 3 km/hr, then the real value may be anything between 95 and 101 km/hr.

Total error is the sum of individual errors for the arithmetic combination of two measurements.

Total relative error is the sum of individual relative errors for the mulitlicative combination of two measurements.

ERROR CALCULATIONS

21

A parameter (say “A”) is going to be estimated / calculated from two measurements (say “B” and “C”), with the measurement errors of “b” and “c”, respectively.

If A = B + C then total error will be

If A = B C then total error will be

cba

C

c

B

bAa

ERROR CALCULATIONS

22

In general, the total (propagated) error is obtained from these two properties as

,...),,( 321 xxxfy

B

bn

A

am

C

cBAC nm

...33

22

11

xx

fx

x

fx

x

fy

For a multi-variable function, the total error is calculated from

ERROR CALCULATIONS

23

Uncertainty is a range that is likely to contain the true value of a quantity being measured or calculated. Uncertainty can be expressed in absolute or relative terms.

Modeling uncertainty is defined as the potential deficiency due to a lack of information.

Modeling error is the recognizable deficiency not due to a lack of information but due to the approximations and simplifications made there.

Measurement error is the difference between the measured and true values, while measurement uncertainty is an estimate of the error in a measurement.

UNCERTAINTY

24

Modeling and simulation uncertainties occur during the phase of

- Conceptual modeling of the physical system- Mathematical modeling of the conceptual model- Discretization - Computer modeling of the mathematical model- Computer modeling of the discrete model- Numerical solution

Numerical uncertainties occur during computations due to round-off, turncation, non-convergence, artificial dissipations, etc.

UNCERTAINTY

25

Digital Voltmeter Accuracy:

(0.25 % Reading + 2 digit)

Meaning; multiply the value you read on the multimeter by 0.25 % (i.e., by 0.0025) . This is called scaling error.

Add 2 times the value of the least significant digit to the result.This is called quantization error.

Example: You read 15.00 V from a 20 V scale.• Measurement error : 0.0025 15.00

= 0.0375 V• Quantization error : 2 0.01 = 0.02

V• Total error : 0.0375 +

0.02 = 0.0575 V

ERROR / UNCERTAINTY

26

Truncation error is also defined for mathematical modeling. For example, Taylor’s expansion or a Fourier-series representation are used to replace a function in terms of an infinite summation.

Taking only a given number of low-order terms, and neglecting the rest of the higher-order terms, introduces a truncation error.

For example, the Taylor expansion of the exponent function:

yields (for n=3)

2 3 4

1 .... ...1! 2! 3! 4! !

nx x x x x x

en

4 !

n

n

xAbsolute error

n

MODELING ERROR

27

Derivative

Mathematical definition

x

xfxxfLimxf

x x

)()()(

0

Finite difference approximation

x

xfxxfxf

x

)()(

)(

x

xxfxxfxf

x

2

)()()(

+ O(∆x)

+ O(∆x2)

2Pt

x

f(x)

FDTD MODELING ERROR

3Pt

28

SOLVING NONLINEAR EQUATIONS

Fixed Point Iteration Method

Bisection Method

Secant Method

Newton-Raphson Method

MatLab Built-in Functions

29

FIXED POINT ITERATION METHOD

The equation f(x)=0 is rewritten in the form :

The intersection point of the graphs of the functions y=x and y=g(x) is the

solution and is called the fixed point .

The numerical solution is obtained by an iterative process.

First a xi value near the fixed point is chosen and substituted into g(x).

The solution is substituted back into g(x). So the iteration formula is given by:

)x(gx

)x(gx ii 1

30


x1x2=

g(x1)

x

g(x1)

g(x2)

x3=

g(x2)

g(x3)

x4=

g(x3)

g(x4)

x5=

g(x4)

g(x5)

x6=

g(x5)

x7=

g(x6)

g(x6)

g(x7)

g(x)

y=x

31


0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 60

1

2

3

4

5

6

1

6

1

6

1

i

ii x)x(gx

x)x(g

13

2

12

62

1

6

2

2

)()(g

)x()x(g

i xi

1 6.0000

2 0.8571

3 3.2308

4 1.4182

5 2.4812

6 1.7235

7 2.2030

8 1.8732

9 2.0882

10 1.9429

11 2.0388

g(x)=6/(x+1)

y=x

Convergence Test: Example:

)x/(x

)x(x

xx

16

61

062

32


Choosing the appropriate iteration function g(x)

The fixed point iteration method converges if in the neighborhood of

the fixed point the derivative of g(x) has an absolute value smaller

than 1. (also called Lipschitz continious) 1 )x(g

052150 x.ex x.Example:

0 0.5 1 1.5 2 2.5 3-5

0

5

10

15

x

f(x)=xe0.5x+1.2x-5

The plot of the function shows that the

equation has a solution between 1 and 2.

The actual solution is at x=1.5050 .

33

FIXED POINT ITERATION METHODCase A Case B Case C

Divergent Convergent Divergent

21

550 .e

x)x(gx.

x.e

x.x)x(g

50

215

21

5 50

.

exx)x(g

x.

21

50150

.

)x.(e)x(g

x.

250

50

212

5

).e(

e)x(g

x.

x.

x.e

x..)x(g

50

6073

0609221

5011

50.

.

).(e)(g

.

5305421

112 .

.

)(e)(g

50790212

51

250

50.

).e(

e)(g

.

.

44260212

52

2.

).e(

e)(g

880216073

150

.e

..)(g

.

9197026073

2 .e

..)(g

i xi

1 1.00002 2.79273 -5.23674 4.48495 -31.02626 4.16677 -23.71958 4.16689 -23.7223

10 4.166811 -23.7223

i xi

1 1.00002 1.75523 1.38694 1.56225 1.47766 1.51827 1.49878 1.50809 1.503510 1.505711 1.5046

i xi

1 1.00002 2.30483 0.70574 2.91835 0.34826 3.85007 0.05548 4.79869 -0.068810 5.260611 -0.0946

34

BISECTION METHOD

STEP 1: Choose two points a and b such that a solution exists between them :

f(a) f(b) < 0

STEP 2: Find c=(a+b)/2

STEP 3: Determine whether the true solution is between a and c or between c and b.

Then select the subinterval that contains the true solution

If f(a) f(c) < 0 , solution is between a and c. anew=a, bnew=c

If f(b) f(c) < 0 , solution is between c and b. anew=c, bnew=b

STEP 4: Repeat steps 2 and 3 until (b-a) is less than a specified tolerance

35

BISECTION METHOD

5 6 7 8 9 10 11 12 13-20

0

20

40

60

80

a bc

f(a)

f(b)

5 6 7 8 9 10 11 12 13-20

0

20

40

60

80

a bc

f(b)

f(a)

5 6 7 8 9 10 11 12 13-20

0

20

40

60

80

ab

f(b)

f(a)

5 6 7 8 9 10 11 12 13-20

0

20

40

60

80

a bc

f(a)

f(b)

f(a) f(b) < 0

c=(a+b) /2

True Root

36

BISECTION METHOD

a b c f(c) abs(b-a)

5.0000 12.0000 8.5000 18.7500 7.0000

5.0000 8.5000 6.7500 -2.6875 3.5000

6.7500 8.5000 7.6250 7.2656 1.7500

6.7500 7.6250 7.1875 2.0977 0.8750

6.7500 7.1875 6.9688 -0.3428 0.4375

6.9688 7.1875 7.0781 0.8655 0.2188

6.9688 7.0781 7.0234 0.2584 0.1094

6.9688 7.0234 6.9961 -0.0430 0.0547

6.9961 7.0234 7.0098 0.1075 0.0273

6.9961 7.0098 7.0029 0.0322 0.0137

6.9961 7.0029 6.9995 -0.0054 0.0068

6.9995 7.0029 7.0012 0.0134 0.0034

6.9995 7.0012 7.0004 0.0040 0.0017

2832 xx)x(f

37

NEWTON RAPSON METHOD

)x(f

)x(fxx

i

iii 1

x1

f(x1)

f(x2)

f(x)

x2x3x4x5

x

Slope: f’(x1)

Slope: f’(x2)Slope:

f’(x3)Slope: f’(x4)

Exact Root

38

NEWTON RAPSON METHOD

)x(f

)x(fxx

i

iii 1

xi f(xi) xi -xi-1

6.00000000 31.00000000

4.60238918 11.14583002 1.3976

3.27847524 3.85164919 1.3239

2.13314198 1.19335902 1.1453

1.34820293 0.27297398 0.7849

1.03883431 0.02728345 0.3094

1.00051801 0.00035912 0.0383

1.00000009 0.00000006 0.0005

1250 x.)x(f

39

SECANT METHOD

)x(f)x(f)x(f

xxxx i

ii

iiii

1

11

x1

f(x1)

f(x2)

f(x)

x2x3x4x5

x

Exact Root

x6

f(x3)

f(x4)f(x5)

40

SECANT METHOD

xi f(xi) xi -xi-1

6.00000000 31.00000000

5.00000000 15.00000000 1.0000

4.06250000 7.35419026 0.9375

3.16075727 3.47149501 0.9017

2.35451439 1.55711029 0.8062

1.69873759 0.62308391 0.6558

1.26127246 0.19853535 0.4375

1.05669682 0.04008167 0.2046

1.00494836 0.00343583 0.0517

1.00009654 0.00006692 0.0049

1.00000017 0.00000011 0.0001

1250 x.)x(f

)x(f)x(f)x(f

xxxx i

ii

iiii

1

11

41

MATLAB BUILT-IN FUNCTIONS

The roots(p) command finds the roots of a polynomial, whose coefficients

are defined in the row vector p. The following MatLab script find the roots of

the polynom P(x)=x3-3x2-13x+15

>> coef=[1 -3 -13 15];

>> xi=roots(coef)xi =

5.0000-3.00001.0000

>> x=fzero('x^3-3*x^2-13*x+15',7)x = 5

>> x=fzero('x^3-3*x^2-13*x+15',3)x = 1

The fzero(‘function’,x0) command finds the roots of the ‘function’ near x0.

The following MatLab script find the roots of the function f(x)=x3-3x2-13x+15

42

SYSTEMS OF LINEAR EQUATIONS

Gauss Elimination Method

Gauss-Jordan Elimination Method

LU Decomposition Method

43

GAUSS ELIMINATION METHOD

1525

22524

42

321

321

321

xxx

xxx

xxx

15125

225-24

4112

15125

225-24

250501 ..

15125

307-40

250501 ..

2553540

30740

250501

..

..

2553540

7.51.7510

250501

..

..

8.754.37500

7.51.7510

250501 ..

2100

7.51.7510

250501 ..

2

4

1

3

2

1

x

x

x

12250450

4572751

57751

2

11

22

32

3

x)(..x

x.)(.x

.x.x

x

Backward Substitution :

44

LU DECOMPOSITION METHOD

bLy

yxU

bxLU

LUA

bxA

• A is the coefficients matrix.

• U is an upper triangular matrix and is obtained at the end of the Gauss

elimination procedure.

• L is a lower triangular matrix which has all 1’s on the diagonal and the

elements below the diagonal are the multipliers mij that multiply the pivot

equation when it is used to eliminate the elements below the pivot

coefficient at the Gauss elimination procedure.

• L is a lower triangular matrix and y is obtained from Ly=b using forward

substitution.

• U is an upper triangular matrix and x is obtained from Ux=y using

backward substitution.

45

LU DECOMPOSITION METHOD

1525

22524

42

321

321

321

xxx

xxx

xxx

A

125

5-24

112

125

7-40

112

53540

740

112

..

4.37500

740

112

U

2

4

1

3

2

1

x

x

x

m21 =2 m31 =2.5 m32 =1.125

11.1252.5

012

001

1mm

01m

001

3231

21L

byL

y

y

y

15

22

4

11.1252.5

012

001

3

2

1

75815301.125452

151.12552

302242

222

4

33

321

22

21

1

.yy)(.

yyy.

yy)(

yy

y

yxU

.x

x

x

758

30

4

4.37500

740

112

3

2

1

14242

42

430)2(-74

3074

27584.375

11

321

22

32

33

xx

xxx

xx

xx

x.x

LUA

Forward Substitution : Backward Substitution :

46

GAUSS-JORDAN ELIMINATION METHOD

1525

22524

42

321

321

321

xxx

xxx

xxx

15125

225-24

4112

15125

225-24

250501 ..

15125

307-40

250501 ..

2553540

30740

250501

..

..

2553540

7.51.7510

250501

..

..

8.754.37500

7.51.7510

250501 ..

2100

7.51.7510

250501 ..

2100

4010

250501 ..

2100

4010

10501 .

2100

4010

1001

2

4

1

3

2

1

x

x

x

47

GAUSS-JORDAN ELIMINATION METHOD

221-43-2-

39-57-43

31-462

163-21-1

221-43-2-

39-57-43

29-5080

163-21-1

221-43-2-

87-1413-70

29-5080

163-21-1

547-85-0

87-1413-70

29-5080

163-21-1

547-85-0

87-1413-70

3.625-0.625010

163-21-1

547-85-0

61.625-9.62513-00

3.625-0.625010

163-21-1

35.8753.875-800

61.625-9.62513-00

3.625-0.625010

163-21-1

35.8753.875-800

4.74040.74038-100

3.625-0.625010

163-21-1

2.0481-2.0481000

4.74040.74038-100

3.625-0.625010

163-21-1

1-1000

4.74040.74038-100

3.625-0.625010

163-21-1

1-1000

40100

3.625-0.625010

163-21-1

1-1000

40100

3-0010

163-21-1

1-1000

40100

3-0010

13021-1

1-1000

40100

3-0010

5001-1

1-1000

40100

3-0010

20001

22432

395743

3462

1632

4321

4321

4321

4321

xxxx

xxxx

xxxx

xxxx

-1

4

-3

2

4

3

2

1

x

x

x

x

48

GAUSS-JORDAN ELIMINATION METHODMATLAB SCRIPTA=[2 -1 1 -4; 4 2 -5 22; 5 2 -1 15];[n,col]=size(A);%-------------------------------------------------------------------for i=1:n A(i,:)=A(i,:)/A(i,i); disp('A ='); disp(A); pause; for k=i+1:n dd=A(k,i); A(k,:)=A(k,:)-dd*A(i,:); A pause; endend%-------------------------------------------------------------------for i=n:-1:2 for k=i-1:-1:1 dd=A(k,i); A(k,:)=A(k,:)-dd*A(i,:); A pause; end endx=A(:,col);disp(' The unknowns are :'); disp(x);rref(A) % MatLab built-in function

49

TAYLOR SERIES

40

04

30

020

0000 !4!3!2

)xx()x(f

)xx()x(f

)xx()x(f

)xx)(x(f)x(f)x(f)(

44

32

!4

0

!3

0

!2

000 x

)(fx

)(fx

)(fx)(f)(f)x(f

)(

Taylor Series :

Maclaurin Series :

Example :

)xcos()x(f

)xsin()x(f

)xcos()x(f

)xsin()x(f

)xcos()x(f

)xsin()x(f

)xcos()x(f

)xsin()x(f

)xcos()x(f

)(

)(

)(

)(

)(

8

7

6

5

4

1

00

10

00

10

00

10

00

10

8

7

6

5

4

)x(f

)(f

)(f

)(f

)(f

)(f

)(f

)(f

)(f

)(

)(

)(

)(

)(

12108642

!12

1

!10

1

!8

1

!6

1

!4

1

2

11 xxxxxx)xcos(

50

TAYLOR SERIES

-8 -6 -4 -2 0 2 4 6 8-2

-1.5

-1

-0.5

0

0.5

1

1.5

2y=cos(x)2 Terms3 Terms4 Terms5 Terms6 Terms7 Terms8 Terms

51

NUMERICAL DIFFERENTIATION

Two-Point Forward Difference Method

Two-Point Backward Difference Method

Two-Point Central Difference Method

Three-Point Forward Difference Method

Three-Point Backward Difference Method

52


xixi-1

f(xi)

h

f(xi-1)

True derivative

Approximated derivative

f(x)

xi xi+1

f(xi)

h

f(xi+1)

True derivative


f(x)

True derivative

xi+1xi-1

f(xi+1)

2h

f(xi-1)


f(x)

Forward Difference Backward Difference

Central Difference

ii

ii

xx xx

)x(f)x(f

dx

df

i

1

1

1

1

ii

ii

xx xx

)x(f)x(f

dx

df

i

11

11

ii

ii

xx xx

)x(f)x(f

dx

df

i

Forward Difference:

Backward Difference:

Central Difference:

53


h

)x(f)x(f

dx

df ii

xx i

1

h

)x(f)x(f

dx

df ii

xx i2

11

2 Point Forward Difference:

2 Point Backward Difference:

2 Point Central Difference:

h

)x(f)x(f

dx

df ii

xx i

1



h

)x(f)x(f)x(f

dx

df iii

xx i2

43 21

h

)x(f)x(f)x(f

dx

df iii

xx i2

34 12

hxxxx iiii 11

)h(O

)h(O

)h(O 2

)h(O 2

)h(O 2

Truncation Error:

54


Example : For f(x)=ln(x) estimate the value of f’(0.2) taking h=0.01.

xi-2 0.18

xi-1 0.19

xi 0.2

xi+1 0.21

xi+2 0.22

f(xi-2) ln(0.18)

f(xi-1) ln(0.19)

f(xi) ln(0.2)

f(xi+1) ln(0.21)

f(xi+2) ln(0.22)

12935010

1902020 1 .

.

).ln().ln(

h

)x(f)x(f).(f ii

00425020

190210

220 11 .

.

).ln().ln(

h

)x(f)x(f).(f ii

8794010

2021020 1 .

.

).ln().ln(

h

)x(f)x(f).(f ii

99254020

2202104203

2

4320 21 .

.

).ln().ln().ln(

h

)x(f)x(f)x(f).(f iii

99064020

2031904180

2

3420 12 .

.

).ln().ln().ln(

h

)x(f)x(f)x(f).(f iii

520

1

).(f

x)x(f

Method Used f’(0.2)Abs.

Error

2 Point Forward Difference 4.879 0.121

2 Point Backward Difference

5.1293 0.1293

2 Point Central Difference 5.0042 0.0042

3 Point Forward Difference 4.9925 0.0075

3 Point Backward Difference

4.9906 0.0094

55

NUMERICAL DIFFERENTIATIONh f’(0.8) Abs. Error

0.7500 0.8819 0.36810.5000 0.9710 0.27900.4000 1.0137 0.23630.3000 1.0615 0.18850.2000 1.1157 0.13430.1000 1.1778 0.07220.0500 1.2125 0.03750.0100 1.2423 0.00770.0050 1.2461 0.00390.0010 1.2492 0.00080.0005 1.2496 0.00040.0001 1.2499 0.0001

h f’(0.8) Abs. Error0.7500 3.6968 2.44680.5000 1.9617 0.71170.4000 1.7329 0.48290.3000 1.5667 0.31670.2000 1.4384 0.18840.1000 1.3353 0.08530.0500 1.2908 0.04080.0100 1.2579 0.00790.0050 1.2539 0.00390.0010 1.2508 0.00080.0005 1.2504 0.00040.0001 1.2501 0.0001

h f’(0.8) Abs. Error0.7500 2.2893 1.03930.5000 1.4663 0.21630.4000 1.3733 0.12330.3000 1.3141 0.06410.2000 1.2771 0.02710.1000 1.2566 0.00660.0500 1.2516 0.00160.0100 1.2501 6.51e-0050.0050 1.2500 1.63e-0050.0010 1.2500 6.51e-0070.0005 1.2500 1.63e-0070.0001 1.2500 6.51e-009

25180

1

.).(f

x/)x(f

)xln()x(f

h f’(0.8) Abs. Error0.7500 1.0597 0.19030.5000 1.1311 0.11890.4000 1.1609 0.08910.3000 1.1903 0.05970.2000 1.2178 0.03220.1000 1.2399 0.01010.0500 1.2472 0.00280.0100 1.2499 0.00010.0050 1.2499679 3.21e-0050.0010 1.2499987 1.298e-0060.0005 1.24999967 3.25e-0070.0001 1.249999987 1.30e-008

2 Point Forward

Difference:


2 Point Central

Difference:


56


1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 2.1 2.2 2.3 2.4 2.5 2.6 2.71

1.2

1.4

1.6

1.8

2

2.2

2.4

2.6

2.8

3

151

42

242

).(fSlope

x)x(f

xx)x(f

57


h f’(1.5) Abs. Error

1.0000 0 1.0000

0.8000 0.2000 0.8000

0.5000 0.5000 0.5000

0.2000 0.8000 0.2000

0.1000 0.9000 0.1000

0.0500 0.9500 0.0500

0.0100 0.9900 0.0100

0.0050 0.9950 0.0050

0.0010 0.9990 0.0010

151

42

242

).(f

x)x(f

xx)x(f

58

NUMERICAL INTEGRATION

Rectangle Methods

Midpoint Method

Trapezoidal Method

Simpsons MethodsSimpsons 1/3 Method

Simpsons 3/8 Method

59


a

f(a)

b

f(b)f(x)

a

f(a)

b

f(b)f(x)

a

f(a)

b

f(b)f(x)

a

f(a)

b

f(b)f(x)

1. Rectangle Method 2. Rectangle Method

Midpoint Method Trapezoidal Method

)a(f)ab(dx)x(fb

a

b

a)

ba(f)ab(dx)x(f

2

b

a)b(f)ab(dx)x(f

b

a

)b(f)a(f)ab(dx)x(f

2

60


N

ii

b

a

N

ii )x(fhIdx)x(f

11

x1

ax2 xi xi+1 xN-1 xN xN+1

bx3

f(x1)

I1 I2 I3 Ii Ii+1 IN-1 IN

h h h h h h h

f(x2)

f(x)1. Rectangle Method

N

abh

61


x1


bx3

f(x1)


h h h h h h h

f(x2)

f(x)

1

21

N

ii

b

a

N

ii )x(fhIdx)x(f

2. Rectangle Method

62


x1


bx3


h h h h h h h

f(x)

N

i

iib

a

N

ii

xxfhIdxxf

1

1

1 2)(

221 xx

f

232 xx

f

(x1+x2)2

(x2+x3)2

Midpoint Method

63


x1


bx3

f(x1)


h h h h h h h

f(x2)

f(x)

N

iii

b

a

N

ii )x(f)x(f

hIdx)x(f

11

12

Trapezoidal Method

64

Simpson 1/3 MethodA second order polynomial g(x) is used to approximate the integrand f(x)


x1

ax2 x3

b

f(x1)

h/2=k

f(x)

h/2=k

)()

2(4)(

3)( bf

bafaf

kdxxf

b

a

g(x)

65


x1

ax2 x4

b

f(x1)

h/3=k

x3

h/3=k h/3=k

Simpson 3/8 MethodA third order polynomial g(x) is used to approximate the integrand f(x)

f(x)

g(x)

)()(3)(3)(

8

3)( 32 bfxfxfaf

kdxxf

b

a

66


0 1 2 3 4 5 6 7 8 9 10 11 120

50

100

150

200

0 1 2 3 4 5 6 7 8 9 10 11 120

50

100

150

200

0 1 2 3 4 5 6 7 8 9 10 11 120

50

100

150

200

0 1 2 3 4 5 6 7 8 9 10 11 120

50

100

150

200

96x18x-x 23 )x(f

52.h 2h

1h 50.h

67


1440dx96x18x-x11

123

1. Rectangular Method

h Num Int.

Abs. Error

Rel. Error %

5 1115 325 22.5692.5 1277.5 162.5 11.2852 1310 130 9.02781 1375 65 4.51390.5 1407.5 32.5 2.25690.1 1433.5 6.5 0.451390.05 1436.8 3.25 0.225690.01 1439.3 0.65 0.0451390.005 1439.7 0.325 0.0225690.001 1439.9 0.065 0.00451390.0005 1440 0.0325 0.00225690.0001 1440 0.0065 0.00045139

2. Rectangular Method

h Num Int.

Abs. Error Rel. Error %

5 1765 325 22.5692.5 1602.5 162.5 11.2852 1570 130 9.02781 1505 65 4.51390.5 1472.5 32.5 2.25690.1 1446.5 6.5 0.451390.05 1443.3 3.25 0.225690.01 1440.6 0.65 0.0451390.005 1440.3 0.325 0.0225690.001 1440.1 0.065 0.00451390.0005 1440 0.0325 0.00225690.0001 1440 0.0065 0.00045139

68


389.333

1168dx40-x7316x-x

10

223

1. Rect. Method 2. Rect. Method

h Num Int.

Abs. Error

Num Int.

Abs. Error

8 400 10.667 720 330.674 352 37.333 512 122.672 360 29.333 440 50.6671 372 17.333 412 22.6670.5 380 9.3333 400 10.6670.1 387.36 1.9733 391.36 2.02670.05 388.34 0.99333 390.34 1.00670.01 389.13 0.19973 389.53 0.200270.005 389.23 0.099933 389.43 0.100070.001 389.31 0.019997 389.35 0.0200030.0005 389.32 0.009999 389.34 0.0100010.0001 389.33 0.002 389.34 0.002

Midpoint Method

h Num Int.

Abs. Error

8 304 85.3334 368 21.3332 384 5.33331 388 1.33330.5 389 0.333330.1 389.32 0.0133330.05 389.33 0.00333330.01 389.33 0.000133330.005 389.33 3.3333e-0050.001 389.33 1.3333e-0060.0005 389.33 3.3333e-0070.0001 389.33 1.3332e-008

Trapezoidal Method

h Num Int.

Abs. Error

8 560 170.674 432 42.6672 400 10.6671 392 2.66670.5 390 0.666670.1 389.36 0.0266670.05 389.34 0.00666670.01 389.33 0.000266670.005 389.33 6.6667e-0050.001 389.33 2.6667e-0060.0005 389.33 6.6667e-0070.0001 389.33 2.6663e-008

69


(h=0.1)Numerical

Integration

Absolute

Error

1. Rectangle Method 387.36 1.9733

2. Rectangle Method 391.36 2.0267

Midpoint Method 389.32 0.013333

Trapezoidal Method 389.36 0.026667

Simpsons 1/3 Method 389.33 5.6843e-014

Simpsons 3/8 Method 389.33 5.6843e-014

389.333

1168dx40-x7316x-x

10

223

70

CURVE FITTING WITH A LINEAR FUNCTION

Curve Fitting with a Linear Function

x1 x2 xi xN

y1

y2

yi

yN

r1 r2

ri

rN

(x1 ,y1)

(x2 ,y2)

(xi ,yi)

(xN ,yN)

f(x)=a1x+a0

A linear function f(x)=a1x+a0 is used to best fit the given data points (xi ,yi).

A residual ri at point (xi , yi) is the difference between the value yi of the data point

and the value of the function f(xi) used to approximate the data points : ri= yi -f(xi)

71

The overall error E is defined as the sum of the squares of the residuals :

N

iii

N

ii axayrE

1

201

1

2

Linear least-squares regression :The coefficients a1 and a0 of the linear function f(x)=a1x+a0 are determined such that the

error E has the smallest possible value. E is minimum if:

00

a

E0

1

a

E

The coefficients a1 and a0 are found as:

N

iix xS

1

N

iiy yS

1i

N

iixy yxS

1

N

iixx xS

1

2

21xxx

yxxy

SnS

SSnSa

20xxx

xxyyxx

SnS

SSSSa


72

x 1 2 3 4 5 6 7 8 9 10

y 6 9.2 13 14.7 19.7 21.8 22.8 29.1 30.2 32.2

Example : Use linear least-squares regression to determine the coefficients a1 and a0 in the function f(x)=a1x+a0 that best fits the data given below.

55109876543211

N

iix xS

198.732.2 30.2 29.1 22.8 21.8 19.7 14.7 13 9.2 61

N

iiy yS

1337.732.21030.2929.18 22.8721.8619.7514.741339.22611

i

N

iixy yxS

38510987654321 2222222222

1

2

N

iixx xS

2.9679

5538510

7198557133710221

..

SnS

SSnSa

xxx

yxxy

3.5467

5538510

55713377198385220

..

SnS

SSSSa

xxx

xxyyxx


73

0 1 2 3 4 5 6 7 8 9 10 110

5

10

15

20

25

30

35

40

f(x)=2.9679x+3.5467

yi f(xi) ri= yi -f(xi) (ri)2

6.0000 6.5145 -0.5145 0.2648

9.2000 9.4824 -0.2824 0.0798

13.0000 12.4503 0.5497 0.3022

14.7000 15.4182 -0.7182 0.5158

19.7000 18.3861 1.3139 1.7264

21.8000 21.3539 0.4461 0.1990

22.8000 24.3218 -1.5218 2.3159

29.1000 27.2897 1.8103 3.2772

30.2000 30.2576 -0.0576 0.0033

32.2000 33.2255 -1.0255 1.0516

E= 9.736

+


74

x 2 5 6 8 9 13 15

y 7 8 10 11 12 14 15

Example : Use linear least-squares regression to determine the coefficients a1 and a0 in the function f(x)=a1x+a0 that best fits the data given below.

581513 986521

N

iix xS

7715 14 12 11 10 8 71

N

iiy yS

17715151413 12911810685721

i

N

iixy yxS

6041513 98652 2222222

1

2

N

iixx xS

0.64

586047

77587177221

xxx

yxxy

SnS

SSnSa

5.6968

586047

5871777604220

xxx

xxyyxx

SnS

SSSSa


75

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 164

6

8

10

12

14

16

yi f(xi) ri= yi -f(xi) (ri)2

7.0000 6.9769 0.0231 0.00058.0000 8.8970 -0.8970 0.8046

10.0000 9.5370 0.4630 0.214311.0000 10.8171 0.1829 0.033412.0000 11.4572 0.5428 0.294714.0000 14.0174 -0.0174 0.000315.0000 15.2975 -0.2975 0.0885

E= 1.4363+

f(x)=0.64x+5.6968


76

x -8 -7 -5 -1 0 2 4 5 6

y m 15 12 5 2 0 n -5 -9

Example : Use linear least-squares regression to determine the coefficients a1 and a0 in the function f(x)=a1x+a0 that best fits the data given below. Use f(x) to determine the interpolated value n for x=4 and the extrapolated value m for x= -8.

06 5 2 0 1- 5- -71

N

iix xS

029- 5- 0 2 5 12 151

N

iiy yS

-249(-9)6 (-5)5 02 20 5(-1)12(-5)15(-7)1

i

N

iixy yxS

1406 5 2 0 (-1)(-5)(-7) 2222222

1

2

N

iixx xS

1.7786-

01407

200(-249)7221

xxx

yxxy

SnS

SSnSa

2.8571

01407

0(-249)20140220

xxx

xxyyxx

SnS

SSSSa


77

-10 -8 -6 -4 -2 0 2 4 6 8-10

-5

0

5

10

15

20

f(x)=1.7786x+ 2.8571

Interpolation : Using the data points for estimating the expexted values between the known points.

Extrapolation : Using the data points for estimating the expexted values beyond the known points.

17.08572.85718)(778618 .)(fm

4.25712.8571)4(778614 .)(fn

m

n


78

LAGRANGE POLYNOMIALS

)x(Ly)x(Ly)x(Ly)x(Ly)x(Ly)x(P nnkk

n

kkkn

1100

0

)xx).....(xx)(xx).....(xx)(xx(

)xx).....(xx)(xx).....(xx)(xx()x(L

nkkkkkkk

nkkk

1110

1110

ki

ki,)x(L ik 0

1

iin y)x(P

For n+1 data points (x0 , y0), (x1 , y1), (x2 , y2), ….. (xk , yk), ….. (xn , yn) a unique polynomial of

order n can be found, which passes through these n data points. A Lagrange interpolating

polynomial Pn(x) that passes through n points is defined as :

79

)x)(.x())(.(

)x)(.x(

)xx)(xx(

)xx)(xx()x(L 452

42522

452

2010

210

3

522

52424

522

1202

102

).x)(x(

).)((

).x)(x(

)xx)(xx(

)xx)(xx()x(L

1514250050

55412

186

3

61105650

3

522250

43

424045250

2

222

221100

2

02

.x.x.

)x.x()xx(.

)x.x(.

).x)(x(.

/

)x)(x(.)x)(.x(.

)x(Ly)x(Ly)x(Ly)x(Ly)x(Pk

kk


x 2 2.5 4

y 0.5 0.4 0.25

Example : Find the Lagrange interpolation polynomial that passes through the points :

2504

4052

502

22

11

00

.y,x

.y,.x

.y,x

43

42

452252

42

2101

201 /

)x)(x(

).)(.(

)x)(x(

)xx)(xx(

)xx)(xx()x(L

MATLAB SCRIPT

x=[2 2.5 4]; y=[0.5 0.4 0.25];polyfit(x,y,2)

80

1514250050 22 .x.x.)x(P


x 2 2.5 4

y 0.5 0.4 0.25The Lagrange interpolation polynomial that passes through the points :

0 1 2 3 4 5 60

0.2

0.4

0.6

0.8

1

1.2

81

)x)(x(.))((

)x)(x(

)xx)(xx(

)xx)(xx()x(L 2150

2010

21

2010

210

)x(x.))((

)x)(x(

)xx)(xx(

)xx)(xx()x(L 150

1202

10

1202

102

12

5353422350

15072221501

2

222

221100

2

02

xx

x.x.xx)xx(.

)x(x.)x(x)x)(x(.

)x(Ly)x(Ly)x(Ly)x(Ly)x(Pk

kk


x 0 1 2

y -1 2 7


72

21

10

22

11

00

y,x

y,x

y,x

)x(x))((

)x)(x(

)xx)(xx(

)xx)(xx()x(L 2

2101

20

2101

201

82


The Lagrange interpolation polynomial that passes through the points : x 0 1 2

y -1 2 7

1222 xx)x(P

-2 -1 0 1 2 3 4-5

0

5

10

15

20

25

83

)x)(x)(x())()((

)x)(x)(x(

)xx)(xx)(xx(

)xx)(xx)(xx()x(L 431

12

1

403010

431

302010

3210

)x(L2

3311313

4434

115413

4

1

314

5431

4

1

232

2

332021

0100

3

03

xxx)x)(x)(x()x)(x(

)x)(x()x(x)x)(x()x(

)x)(x(x)x)(x)(x(

)x(Ly)x(Ly)x(Ly)x(Ly)x(Ly)x(Pk

kk


x 0 1 3 4

y 3 0 0 15


154

03

01

30

33

22

11

00

y,x

y,x

y,x

y,x

)x(L1

)x)(x(x))()((

)x)(x)(x(

)xx)(xx)(xx(

)xx)(xx)(xx()x(L 31

12

1

341404

310

231303

2103

84


The Lagrange interpolation polynomial that passes through the points : x 0 1 3 4

y 3 0 0 15

33 233 xxx)x(P

-2 -1 0 1 2 3 4 5-15

-10

-5

0

5

10

15

20

25

30

85

)x)(x)(x())()((

)x)(x)(x(

)xx)(xx)(xx(

)xx)(xx)(xx()x(L 543

60

1

504030

543

302010

3210

)x(L2

825120307515

15

1

3241892720090101209424215

1

43275410543215

1

4310

1854

6

4543

60

8

2323

232323

332021100

3

03

xxxxxx

xxxxxxxxx

)x)(x(x)x)(x(x)x)(x)(x(

)x)(x(x)x)(x(x)x)(x)(x(

)x(Ly)x(Ly)x(Ly)x(Ly)x(Ly)x(Pk

kk


x 0 3 4 5

y 8 -4 0 18


18,5

0,4

4,3

8,0

33

22

11

00

yx

yx

yx

yx

)x)(x(x))()((

)x)(x)(x(

)xx)(xx)(xx(

)xx)(xx)(xx()x(L 43

10

1

453505

430

231303

2103

)x)(x(x))()((

)x)(x)(x(

)xx)(xx)(xx(

)xx)(xx)(xx()x(L 54

6

1

534303

540

312101

3201

86


The Lagrange interpolation polynomial that passes through the points : x 0 3 4 5

y 8 -4 0 18

825 233 xxx)x(P

-1 0 1 2 3 4 5 6-5

0

5

10

15

20

25

30

35

87

ORDINARY DIFFERENTIAL EQUATIONS

A differential equation is an equation that contains derivatives of an unknown

function.

A differential equation that has one independent variable is called an

ordinary differential equation (ODE).

If an ODE involves only first derivatives of the dependent variable (y) with

respect to the independent variable (x), it is a first – order ordinary

differential equation.

A first –order ODE is linear, if it is a linear function of y and dy/dx.

)nonlinear(ybayxdx

dy

)linear(byaxdx

dy

0

02

88

EULER’S METHOD

xixi+1

yi

yi+1

h

Slope =f(xi, yi)

y(x)

Numerical Solution

Exact Solution

)y,x(fhyy

hxx

)y,x(fdx

dySlope

)y,x(fdx

dy

iiii

ii

iixx i

1

1

89

EULER’S METHOD Example : Use Euler’s method to solve the Ordinary Differential Equation below from x=0 to x=1

with the initial conditions x=0 and y=2. (Take h=0.2) Compute the errors in each step for the exact solution y=3ex-x-1

2010 )(yxyxdx

dy

)y,x(fhyy

hxx

yxyx)y,x(f

iiii

ii

1

1

00 20

422020220202

20200

0001

01

..),(f.)y,x(fhyy

..hxx

9224220204242202042

402020

1112

12

.....).,.(f..)y,x(fhyy

...hxx

584392240209229224020922

602040

2223

23

.....).,.(f..)y,x(fhyy

...hxx

42084584360205843584360205843

802060

3334

34

.....).,.(f..)y,x(fhyy

...hxx

46554208480204208442084802042084

12080

4445

45

.....).,.(f..)y,x(fhyy

..hxx

90

EULER’S METHOD

xi yi

(Euler)

yi

(Exact)

Abs.Error

0 2.0000 2.0000 00.2500 2.5000 2.6021 0.10210.5000 3.1875 3.4462 0.25870.7500 4.1094 4.6010 0.49161.0000 5.3242 6.1548 0.8306

xi yi

(Euler)

yi

(Exact)

Abs.Error

0 2.0000 2.0000 00.2000 2.4000 2.4642 0.06420.4000 2.9200 3.0755 0.15550.6000 3.5840 3.8664 0.28240.8000 4.4208 4.8766 0.45581.0000 5.4650 6.1548 0.6899

xi yi

(Euler)

yi

(Exact)

Abs.Error

0 2.0000 2.0000 00.1000 2.2000 2.2155 0.01550.2000 2.4300 2.4642 0.03420.3000 2.6930 2.7496 0.05660.4000 2.9923 3.0755 0.08320.5000 3.3315 3.4462 0.11460.6000 3.7147 3.8664 0.15170.7000 4.1462 4.3413 0.19510.8000 4.6308 4.8766 0.24590.9000 5.1738 5.4788 0.30501.0000 5.7812 6.1548 0.3736

xi yi

(Euler)

yi

(Exact)

Abs.Error

0 2.0000 2.0000 00.0500 2.1000 2.1038 0.00380.1000 2.2075 2.2155 0.00800.1500 2.3229 2.3355 0.01260.2000 2.4465 2.4642 0.01770.2500 2.5788 2.6021 0.02320.3000 2.7203 2.7496 0.02930.3500 2.8713 2.9072 0.03590.4000 3.0324 3.0755 0.04310.4500 3.2040 3.2549 0.05100.5000 3.3867 3.4462 0.05950.5500 3.5810 3.6498 0.06870.6000 3.7876 3.8664 0.07880.6500 4.0069 4.0966 0.08970.7000 4.2398 4.3413 0.10150.7500 4.4868 4.6010 0.11420.8000 4.7486 4.8766 0.12800.8500 5.0261 5.1689 0.14290.9000 5.3199 5.4788 0.15900.9500 5.6309 5.8071 0.17631.0000 5.9599 6.1548 0.1950

h=0.25

h=0.2

h=0.1 h=0.05

91

EULER’S METHOD

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 12

2.5

3

3.5

4

4.5

5

5.5

6

6.5

92

EULER’S METHOD Example : Use Euler’s method to solve the Ordinary Differential Equation below from x=0 to x=1.8 with the initial conditions x=0 and y=1. (Take h=0.6) Compute the errors in each step for the exact solution.

21081025023 x.

exact exy)(y.xxyxdx

dy

)y,x(fhyy

hxx

yxxyx)y,x(f

iiii

ii

1

1

003 10

100160110601

60600

30001

01

.),(f.)y,x(fhyy

..hxx

1.230460601601160601

216060

31112

12

...),.(f.)y,x(fhyy

...hxx

1.0795212123041602304123041216023041

816021

32223

23

.....).,.(f..)y,x(fhyy

...hxx

93

EULER’S METHOD

xi yi

(Euler)

yi

(Exact)

Abs.Error

0 1.0000 1.0000 00.6000 1.0000 1.1628 0.16281.2000 1.2304 1.3856 0.15521.8000 1.0795 0.1869 0.8926

xi yi

(Euler)

yi

(Exact)

Abs.Error

0 1.0000 1.0000 00.3000 1.0000 1.0440 0.04400.6000 1.0819 1.1628 0.08090.9000 1.2118 1.3107 0.09891.2000 1.3203 1.3856 0.06521.5000 1.2773 1.1698 0.10751.8000 0.8395 0.1869 0.6526

xi yi

(Euler)

yi

(Exact)

Abs.Error

0 1.0000 1.0000 00.1000 1.0000 1.0050 0.00500.2000 1.0099 1.0198 0.00990.3000 1.0293 1.0440 0.01470.4000 1.0575 1.0767 0.01920.5000 1.0934 1.1169 0.02350.6000 1.1355 1.1628 0.02720.7000 1.1821 1.2124 0.03030.8000 1.2305 1.2629 0.03230.9000 1.2778 1.3107 0.03291.0000 1.3199 1.3513 0.03141.1000 1.3518 1.3787 0.02691.2000 1.3675 1.3856 0.01811.3000 1.3587 1.3620 0.00331.4000 1.3157 1.2955 0.02011.5000 1.2255 1.1698 0.05571.6000 1.0718 0.9634 0.10841.7000 0.8337 0.6481 0.18551.8000 0.4841 0.1869 0.2972

h=0.6

h=0.3

h=0.1

94

EULER’S METHOD

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.80

0.2

0.4

0.6

0.8

1

1.2

1.4

Date post:	26-Dec-2015
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CEE203 NUMERICAL METHODS IN ELECTRICAL ENGINEERING Assist. Prof. Dr. Çağatay ULUIŞIK...

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