Fourier series
Download
1 / 131

Fourier Series - PowerPoint PPT Presentation


  • 252 Views
  • Uploaded on

Fourier Series. Dr. K.W. Chow Mechanical Engineering. Introduction. Conceptual question: While one can readily see that two vectors can be ‘perpendicular’ or ‘orthogonal’, how can we extend this concept to a sequence of functions?. Introduction.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'Fourier Series' - tad


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
Fourier series

Fourier Series

Dr. K.W. Chow

Mechanical Engineering


Introduction
Introduction

  • Conceptual question: While one can readily see that two vectors can be ‘perpendicular’ or ‘orthogonal’, how can we extend this concept to a sequence of functions?


Introduction1
Introduction

  • A general formulation: For a sequence of functions {φn} and

    f(x) = Σcn φn

    What is cn?

    IF ∫ φm φn dx = 0 for m, n different, then

    cn can be found fromthis ‘orthogonal’ property.


Introduction2
Introduction

  • A general theory has been developed for linear, second order differential equations regarding these issues:

    (a) Orthogonal ‘eigenfunctions’?

    (b) Completeness in terms of expansion? (i.e. is it possible for any arbitrary function f(x) to be represented as a sum of φn(x)?)


Introduction3
Introduction

  • Sin(x) and Cos(x) are the solutions of the simplest second order ordinary differential equations (ODEs)

  • d2y/dx2 + y = 0,

  • subject to certain boundary conditions.


Introduction4
Introduction

  • Fourier series is an infinite series of sine and cosine

  • Dirichlet’s theorem (Sufficient, but not necessary):

    If a function is 2L-periodic and piecewise continuous, then its Fourier series converges

  • At a point of discontinuity, the series converges to the mean value.


Fourier series1
Fourier series

  • an and bn are calculated by the orthogonal properties of sines and cosines.

  • If one uses a0 as the constant term, two schemes for defining an, n = 0and n > 0.

  • If one uses a0/2, then one definition for all an .


Introduction5
Introduction

  • Consider:


Fourier series

f(x)


Fourier series

f(x)

n = 3


Fourier series

f(x)

n = 3

n = 7


Fourier series

f(x)

n = 3

n = 7

n = 15


Introduction6
Introduction

  • The n = 15 series gives a very good approximation to the original function


Introduction7
Introduction

Even function

Odd function


Introduction8
Introduction

  • Even extension of a function results in Fourier cosine series

  • Odd extension of a function results in Fourier sine series

    (Assuming function is given only in half the interval.)


Introduction9
Introduction

Even extension

Odd extension


Introduction10
Introduction

Even extension of f : Fourier series of f1will have cosine terms only

Odd extension of f : Fourier series of f2will have sine terms only


Fourier series


Fourier series

f(x)


Fourier series

f(x)

n = 3


Fourier series

f(x)

n = 3

n = 7


Fourier series

f(x)

n = 3

n = 7

n = 50


Introduction11
Introduction

The Fourier series converges to this value at the discontinuity.



Introduction13
Introduction

Approximation using n = 1 series


Introduction14
Introduction

Approximation using n = 2 series


Introduction15
Introduction

Approximation using n = 3 series


Introduction16
Introduction

  • Gibbs phenomenon – large oscillations of the series near a discontinuity.


Introduction17
Introduction

  • Consider the square wave again:

5 terms


Introduction18
Introduction

25 terms


Introduction19
Introduction

25 terms


Introduction20
Introduction

  • Animation to show the Gibbs phenomenon using different partial sums.


Differentiation of series
Differentiation of series

  • Fourier coefficients are determined uniquely using orthogonality of trigonometric functions.

    only when the series on the right is uniformly convergent. (d/dx = a local operator)


Integration of series
Integration of series

  • Always permitted, but the resulting series is not a Fourier series, unless the constant term a0 is zero.

  • (Integration = a global operator).


1d heat conduction in finite domain
1D heat conduction in finite domain

  • Configuration:

Density of conductor:

Specific heat capacity: c

Heat conduction coefficient: k

Cross-sectional area: A

Temperature:

Temperature:

L


1d heat conduction in finite domain1
1D heat conduction in finite domain

Thermal conductivity defined by

  • Heat flux = - k A∂u/∂x

    where A = cross sectional area,

    u = temperature

    (i.e. k is heat flux per unit area per unit temperature gradient).


1d heat conduction in finite domain2
1D heat conduction in finite domain

Assumptions:

  • Heat flow in the xdirection only

  • No external heat source

  • No heat loss


1d heat conduction in finite domain3
1D heat conduction in finite domain

  • Consider heat conduction across an infinitesimal element of the conductor:

Heat out

Heat in


1d heat conduction in finite domain4
1D heat conduction in finite domain

  • Heat flux at the left surface : - k A∂u(x,t)/∂x

  • Heat flux at the right surface:

    - k A ∂u(x,t)/∂x - ∂[k A ∂u(x,t)/∂x]/∂x dx + …

  • Net heat flux INTO the element:

  • ∂[k A ∂u(x,t)/∂x]/∂x dx


1d heat conduction in finite domain5
1D heat conduction in finite domain

  • Net heat must be used to heat up the element (c = specific heat capacity):


1d heat conduction in finite domain6
1D heat conduction in finite domain

  • Therefore (if k and A not functions of x):


1d heat conduction in finite domain7
1D heat conduction in finite domain

  • When , the above equation becomes exact:

    This is the 1D heat conduction equation in finite domain.


1d heat conduction in finite domain8
1D heat conduction in finite domain

Solution procedure

  • Separation of variables:

  • Substitute back into the heat equation:



1d heat conduction in finite domain10
1D heat conduction in finite domain

  • Assume both ends are kept at :

  • For non-trivial solution, choose:


1d heat conduction in finite domain11
1D heat conduction in finite domain

  • F satisfies the differential equation:

  • For non-trivial solutions:

  • The temporal part:


1d heat conduction in finite domain12
1D heat conduction in finite domain

  • Overall solution:

  • Using superposition principle, we obtain general solution:


1d heat conduction in finite domain13
1D heat conduction in finite domain

  • is the Fourier sine coefficient:


1d heat conduction in finite domain14
1D heat conduction in finite domain

  • is called the eigen-value

  • is called the

    corresponding eigen-function




Fourier series

t = 0

t = 0.5


Fourier series

t = 0

t = 0.5

t = 1.2


Fourier series

t = 0

t = 0.5

t = 1.2

t = 5


Fourier series

t = 0

t = 0.5

t = 1.2

t = 5

t = 15


1d heat conduction in finite domain16
1D heat conduction in finite domain

  • The above procedure cannot be applied directly when the end points are not at

  • First find steady-state temperature distribution v:

Note that the steady-state temperature depends on x only.


1d heat conduction in finite domain17
1D heat conduction in finite domain

  • Introduce a new function (transient):

  • w satisfies the heat equation with homogeneous boundary conditions.

  • Solve for w with separation of variables and hence u can be found.




Fourier series

t = 0

t = 0.5


Fourier series

t = 0

t = 0.5

t = 1.2


Fourier series

t = 0

t = 0.5

t = 1.2

t = 5


Fourier series

t = 0

t = 0.5

t = 1.2

t = 5

t = 12


Laplace s equation
Laplace’s equation

  • For heat conduction in higher dimensions,

    where is the Laplacian.


Laplace s equation1
Laplace’s equation

  • The steady-state solution in 2D satisfies:

    which is the Laplace’s equation

  • In the presence of heat sources, u satisfies the Poisson’s equation:


Laplace s equation2
Laplace’s equation

3 types of boundary conditions:

  • Dirichlet boundary condition

  • Neumann boundary condition

  • Robin boundary condition


Laplace s equation3
Laplace’s equation

  • Dirichlet boundary condition


Laplace s equation4
Laplace’s equation

  • Neumann boundary condition


Laplace s equation5
Laplace’s equation

  • Robin boundary condition


Laplace s equation6
Laplace’s equation

Consider (Dirichlet b.c.)


Laplace s equation7
Laplace’s equation

u(x, y) = F(x) G(y)

F’’/F = – G’’/G = constant

F(0) = F(a) = 0 and thus the constant is

– n2π2 / L2 .

Hence

F ~ sin (n πx/L)

G ~ cosh


Laplace s equation8
Laplace’s equation

  • Solution is obtained using separation of variables:



Fourier series

y

x

Isotherms (curves joining points with the same temperature) of the problem


1d wave equation in finite domain
1D wave equation in finite domain

  • Configuration:

  • Assumptions:

    - CONSTANT TENSION and density,

    - the slope of the vibration is small,

    - gravity much smaller than tension.

L



1d wave equation in finite domain2
1D wave equation in finite domain

  • Vertical force at the left end:

    Vertical force at the right end:

  • u = u(x, t) = displacement of the string


1d wave equation in finite domain3
1D wave equation in finite domain

  • Net vertical upward force on the element:


1d wave equation in finite domain4
1D wave equation in finite domain

  • Newton’s second law (ρ = linear density):


1d wave equation in finite domain5
1D wave equation in finite domain

  • When , the equation becomes exact.

  • If no external force is present:


1d wave equation in finite domain6
1D wave equation in finite domain

  • c = speed of the wave

  • Check the dimensions: Square root of (force/mass per unit length).

  • Mathematically, signs of two second derivatives same (contrast with Laplace equation).


1d wave equation in finite domain7
1D wave equation in finite domain

  • Consider a vibrating string with two ends fixed and initial position and velocity given;

  • i.e. initial and boundary conditions:


Separation of variables
Separation of Variables

u(x, y) = F(x) G(t)

F’’/F = G’’/(c2 G) = constant

F(0) = F(a) = 0 and thus the constant is

– n2π2 / L2 .

Hence

F ~ sin (n πx/L)

G ~ C1sin (c n πt/L)+C2 cos (c n πt/L)


1d wave equation in finite domain8
1D wave equation in finite domain

  • Solution obtained by separation of variables:


1d wave equation in finite domain9
1D wave equation in finite domain

  • Modes of vibration are the profiles of the envelopes corresponding to different n

First mode

Second mode

Third mode


1d wave equation in finite domain10
1D wave equation in finite domain

  • The mode shapes will oscillate with time


1d wave equation in finite domain11
1D wave equation in finite domain

  • Another example – Vibration of a stretched string with a triangular initial profile.

  • Qualitatively, the shapes of the string at subsequent times are similar to the n = 1 mode previously. However, the shapes remain piecewise linear, and some ‘corners’ persist.




Fourier series

t = 0

Forward cycle


Fourier series

t = 0

Forward cycle

t = L/6c


Fourier series

t = 0

Forward cycle

t = L/6c

t = L/3c


Fourier series

t = 0

Forward cycle

t = L/6c

t = L/3c

t = L/2c


Fourier series

t = 0

Forward cycle

t = L/6c

t = L/3c

t = L/2c

t = 2L/3c


Fourier series

t = 0

Forward cycle

t = L/6c

t = L/3c

t = L/2c

t = 2L/3c

t = 5L/6c



Fourier series

Backward cycle

t = 7L/6c

t = L/c


Fourier series

Backward cycle

t = 4L/3c

t = 7L/6c

t = L/c


Fourier series

Backward cycle

t = 3L/2c

t = 4L/3c

t = 7L/6c

t = L/c


Fourier series

Backward cycle

t = 5L/3c

t = 3L/2c

t = 4L/3c

t = 7L/6c

t = L/c


Fourier series

Backward cycle

t = 11L/6c

t = 5L/3c

t = 3L/2c

t = 4L/3c

t = 7L/6c

t = L/c


Fourier series

Backward cycle

t = 2L/c

t = 11L/6c

t = 5L/3c

t = 3L/2c

t = 4L/3c

t = 7L/6c

t = L/c


1d wave equation in finite domain14
1D wave equation in finite domain

  • General solution (d’Alembert) to the wave equation:

  • Using trigonometric identities, the Fourier series solution can also be rewritten in the above form.


1d wave equation in finite domain15
1D wave equation in finite domain

  • x- ctrepresents a wave traveling to the right with speed c


1d wave equation in finite domain16
1D wave equation in finite domain

  • x+ ct represents a wave traveling to the left with speed c


1d wave equation in finite domain17
1D wave equation in finite domain

  • The lines are known as characteristic lines (or just characteristics).

  • The forms of functions Fand G depend on the initial conditions.

  • The initial profile splits into two waves of same amplitude traveling in opposite directions



1d wave equation in finite domain19
1D wave equation in finite domain

  • Method of characteristics is of tremendous theoretical significance but less practical interest.

  • Usually we just use separation of variables for finite domains and integral transform for infinite ones.


Pdes in infinite domain
PDEs in infinite domain

  • Fourier series gives information in the interval

  • Fourier series will give the periodic extension outside this domain.

  • Fourier ‘fails’ if the given function is already defined along the whole real-axis.


From fourier series to fourier integrals
From Fourier series to Fourier integrals

  • To represent a function from minus infinity to plus infinity, we use a Fourier series over the interval (- L, L) and let L go to infinity.

  • Result (Fourier integral):

  • f = integral (integral f dξ) dx


From fourier series to fourier integrals1
From Fourier series to Fourier integrals

  • f(x) = sum over An cos (nπx/L)

  • An = 2/L integral f(ξ) cos (nπξ/L)dξ

  • Hence

  • f(x) = sum over n [ 2/L (integral of

    f(ξ) cos (nπ (x – ξ)/L) dξ)]

    Now convert ‘sum over n and (1/L)’ into another integral.


Pdes in infinite domain1
PDEs in infinite domain

  • Separation of variables is usually not feasible or will fail.

  • Use integral transforms:

    - Fourier transform

    - Laplace transform

    - …


Pdes in infinite domain2
PDEs in infinite domain

  • Fourier integrals are analogous to Fourier series:


Pdes in infinite domain3
PDEs in infinite domain

  • For functions defined in semi-infinite domain,

    even extension Fourier cosine integral

    odd extension Fourier sine integral


Pdes in infinite domain4
PDEs in infinite domain

  • Fourier transform pair:


Pdes in infinite domain5
PDEs in infinite domain

  • In some alternative versions, the + and - signs in the exponentials are interchanged.

  • There is no universally accepted format.

  • The constants in front of the integral signs are arbitrary, as long as their product is


Applications of fourier transform
Applications of Fourier transform

Idea:

  • A PDE defined in an infinite domain is given.

  • Apply transform on each term in the equation, with respect to a certain independent variable, e.g. x

  • Derivatives in x become algebraic in ω.


Applications of fourier transform1
Applications of Fourier transform

  • The transformed equation becomes an ODE in t (ω is a parameter, no derivatives in ω), rather than PDE in x and t.

  • Solve for the transformed function.

  • Apply inverse transform to obtain solution in the original coordinates.


Applications of fourier transform2
Applications of Fourier transform

Common techniques:

  • Integration by parts

  • Exchange order of integrations

  • Contour integrals

  • Gaussian integrals


Applications of fourier transform3
Applications of Fourier transform

  • Example : heat equation

    Spatial conditions: u(x, t) decaying in far field.

    This represents an initially concentrated source of unit intensity at the origin




Fourier series

t = 0.05

t = 0.1


Fourier series

t = 0.05

t = 0.1

t = 0.5


Fourier series

t = 0.05

t = 0.1

t = 0.5

t = 5


Fourier series

t = 0.05

t = 0.1

t = 0.5

t = 5

t = 50