Part 3Chapter 13

Eigenvalues

{contains corrections to the Texbook}

PowerPoints organized by Prof. Steve Chapra, Tufts UniversityAll images copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Chapter Objectives

• Understanding the mathematical definition of eigenvalues and eigenvectors.

• Understanding the physical interpretation of eigenvalues and eigenvectors within the context of engineering systems that vibrate or oscillate.

• Knowing how to implement the polynomial method.• Knowing how to implement the power method to

evaluate the largest and smallest eigenvalues and their respective eigenvectors.

• Knowing how to use and interpret MATLAB’s eig function.

Representing Linear Algebra

• Matrices provide a concise notation for representing and solving simultaneous linear equations:

a11x1 a12x2 a13x3 b1

a21x1 a22x2 a23x3 b2

a31x1 a32x2 a33x3 b3

a11 a12 a13

a21 a22 a23

a31 a32 a33

x1

x2

x3

b1

b2

b3

[A]{x} {b}

Dynamics of Three Coupled Bungee Jumpers in Time

Is there an underlying pattern???

Mathematics

[A] {x} = {b}

[A] {x} = 0

Up until now, heterogeneous systems:

What about homogeneous systems:

Trivial solution:

{x} = 0

Is there another way of formulating the systemso that the solution would be meaningful???

Mathematics

(a11 – l) x1 + a12 x2 + a13 x3 = 0 a21 x1 + (a22 – l) x2 + a23 x3 = 0 a31 x1 + a32 x2 + (a33 – l) x3 = 0

[A] – l [ I ] {x} = 0[ ]

homogeneous system:

or in matrix form

For this case, there could be a value of thatmakes the equations equal zero. This is calledan eigenvalue.

+ a11a22

+ × {correct textbook}

{correct textbook}

Graphical Depiction of Eigenvalues

Physical Background:Oscillations or Vibrations of Mass-

Spring Systems

m2

d 2x2

dt 2 = – k(x2 – x1) – kx2

m1

d 2x1

dt 2 = – k x1 + k(x2 – x1)

m2

d 2x2

dt 2– k (x1 – 2x2) = 0

m1

d 2x1

dt 2– k (– 2x1 + x2) = 0

Collect terms:

Model With Force Balances(AKA: F = ma)

xi = Xi sin ( w t) where w = 2pTp

Differentiate twice:

xi″ = – Xi w2 sin ( w t)

Substitute back into system and collect terms

Assume a Sinusoidal Solution

- w 2X1 – 2k

m1

X2 =0k

m1

- w 2X2 =02k

m2

X1 +k

m2

Given: m1 = m2 = 40 kg; k = 200 N/m

(10 – w 2) X1 – 5 X2 = 0

– 5 X1 + (10 – w 2) X2 = 0

This is now a homogeneous system where the eigenvalue represents the square of the

fundamental frequency.

Evaluate the determinant to yield a polynomial

Solution: The Polynomial Method

10 – w 2 -5 X1 0 – 5 10 – w 2 X2 0

=

= (w 2)2 -20w 2+7510 – w 2 -5 – 5 10 – w 2

The two roots of this "characteristic polynomial" are the system's eigenvalues (λ = w2)

w2 = w =3.873 rad/s2.36 rad/s

15 rad2/s2

5 rad2/s2

0.616 Hz0.356 Hz

f=

INTERPRETATIONw2 = 5 /s2

w = 2.236 /sTp = 2p/2.236 = 2.81 s

(10 – w2) X1 – 5 X2 = 0 – 5 X1 + (10 – w2) X2 = 0

(10 – 5) X1 – 5 X2 = 0 – 5 X1 + (10 – 5) X2 = 0

5 X1 – 5 X2 = 0– 5 X1 + 5 X2 = 0

X1 = X2

V =–0.7071–0.7071

(10 – 15) X1 – 5 X2 = 0 – 5 X1 + (10 – 15) X2 = 0

–5 X1 – 5 X2 = 0– 5 X1 – 5 X2 = 0

X1 = –X2

V =–0.70710.7071

w2 = 15 /s2

w = 3.873 /sTp = 2p/3.373 = 1.62 s

Principle Modes of Vibration

Tp = 1.62Tp = 1/f

Tp = 2.81Tp = 1/f

Iterative method to compute the largest eigenvalue and its associated eigenvector.

The Power Method

[[A] -l[I]]]{x} =0

[A]{x} =l{x}Simple Algorithm:

function [eval, evect] = powereig(A,es,maxit)n=length(A);evect=ones(n,1);eval=1;iter=0;ea=100; %initializewhile(1) evalold=eval; %save old eigenvalue value evect=A*evect; %determine eigenvector as [A]*{x) eval=max(abs(evect)); %determine new eigenvalue evect=evect./eval; %normalize eigenvector to eigenvalue iter=iter+1; if eval~=0, ea = abs((eval-evalold)/eval)*100; end if ea<=es | iter >= maxit,break,endend

Example

40 -20 0-20 40 -20 0 -20 40

111

=200

20

= 20101

First iteration:

40 -20 0-20 40 -20 0 -20 40

101

=40-4040

= 401-11

Second iteration:

|ea| = 40 -20

40 100% = 50%

Example: The Power MethodThird iteration:

|ea| = -80 -40

-80 100% = 150%

40 -20 0-20 40 -20 0 -20 40

= = -80-0.75

1-0.75

1-11

60-80 60

Fourth iteration:

|ea| = 70 -(-80)

70 100% = 214%

40 -20 0-20 40 -20 0 -20 40

= = 70-0.71429

1-0.71429

-50 70-50

-0.751

-0.75

Example: The Power MethodFifth iteration:

|ea | = 68.51714 - 70

70 100% = 2.08%

40 -20 0-20 40 -20 0 -20 40

= = 68.51714-0.708331

-0.70833

-48.51714 68.51714-48.51714

-0.714291

-0.71429

Note that the smallest eigenvalue and its associatedeigenvector can be determined by applying the

power method to the inverse of A

The process can be continued to determine the largesteigenvalue (= 68.284) with the associated eigenvector[-0.7071 1 -0.7071]

Determining Eigenvalues & Eigenvectors with MATLAB

>> A = [10 -5;-5 10]

A = 10 -5 -5 10

>> [v,lambda] = eig(A)

v = -0.7071 -0.7071 -0.7071 0.7071lambda = 5 0 0 15

Dynamics of Three Story Building

Principle Modes of Vibration