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# Fourier Series - PowerPoint PPT Presentation

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.

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### Fourier Series

Dr. K.W. Chow

Mechanical Engineering

• 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?

• 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.

• 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)?)

• 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.

• 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.

• 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 .

• Consider:

f(x)

f(x)

n = 3

f(x)

n = 3

n = 7

f(x)

n = 3

n = 7

n = 15

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

Even function

Odd function

• 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.)

Even extension

Odd extension

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

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

f(x)

f(x)

n = 3

f(x)

n = 3

n = 7

f(x)

n = 3

n = 7

n = 50

The Fourier series converges to this value at the discontinuity.

Approximation using n = 1 series

Approximation using n = 2 series

Approximation using n = 3 series

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

• Consider the square wave again:

5 terms

25 terms

25 terms

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

• 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)

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

• (Integration = a global operator).

• Configuration:

Density of conductor:

Specific heat capacity: c

Heat conduction coefficient: k

Cross-sectional area: A

Temperature:

Temperature:

L

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).

Assumptions:

• Heat flow in the xdirection only

• No external heat source

• No heat loss

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

Heat out

Heat in

• 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

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

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

• When , the above equation becomes exact:

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

Solution procedure

• Separation of variables:

• Substitute back into the heat equation:

• Assume both ends are kept at :

• For non-trivial solution, choose:

• F satisfies the differential equation:

• For non-trivial solutions:

• The temporal part:

• Overall solution:

• Using superposition principle, we obtain general solution:

• is the Fourier sine coefficient:

• is called the eigen-value

• is called the

corresponding eigen-function

• Consider:

t = 0.5

t = 0.5

t = 1.2

t = 0.5

t = 1.2

t = 5

t = 0.5

t = 1.2

t = 5

t = 15

• 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.

• 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.

• Consider:

t = 0.5

t = 0.5

t = 1.2

t = 0.5

t = 1.2

t = 5

t = 0.5

t = 1.2

t = 5

t = 12

• For heat conduction in higher dimensions,

where is the Laplacian.

• 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:

3 types of boundary conditions:

• Dirichlet boundary condition

• Neumann boundary condition

• Robin boundary condition

• Dirichlet boundary condition

• Neumann boundary condition

• Robin boundary condition

Consider (Dirichlet b.c.)

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

• Solution is obtained using separation of variables:

• Consider

x

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

• Configuration:

• Assumptions:

- CONSTANT TENSION and density,

- the slope of the vibration is small,

- gravity much smaller than tension.

L

• Vertical force at the left end:

Vertical force at the right end:

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

• Net vertical upward force on the element:

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

• When , the equation becomes exact.

• If no external force is present:

• 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).

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

• i.e. initial and boundary conditions:

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)

• Solution obtained by separation of variables:

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

First mode

Second mode

Third mode

• The mode shapes will oscillate with time

• 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.

• Consider:

Forward cycle

Forward cycle

t = L/6c

Forward cycle

t = L/6c

t = L/3c

Forward cycle

t = L/6c

t = L/3c

t = L/2c

Forward cycle

t = L/6c

t = L/3c

t = L/2c

t = 2L/3c

Forward cycle

t = L/6c

t = L/3c

t = L/2c

t = 2L/3c

t = 5L/6c

t = L/c

t = 7L/6c

t = L/c

t = 4L/3c

t = 7L/6c

t = L/c

t = 3L/2c

t = 4L/3c

t = 7L/6c

t = L/c

t = 5L/3c

t = 3L/2c

t = 4L/3c

t = 7L/6c

t = L/c

t = 11L/6c

t = 5L/3c

t = 3L/2c

t = 4L/3c

t = 7L/6c

t = L/c

t = 2L/c

t = 11L/6c

t = 5L/3c

t = 3L/2c

t = 4L/3c

t = 7L/6c

t = L/c

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

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

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

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

• 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

• 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.

• 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.

• 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

• 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.

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

• Use integral transforms:

- Fourier transform

- Laplace transform

- …

• Fourier integrals are analogous to Fourier series:

• For functions defined in semi-infinite domain,

even extension Fourier cosine integral

odd extension Fourier sine integral

• Fourier transform pair:

• 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

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.

• 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.

Common techniques:

• Integration by parts

• Exchange order of integrations

• Contour integrals

• Gaussian integrals

• Example : heat equation

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

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

t = 0.1

t = 0.1

t = 0.5

t = 0.1

t = 0.5

t = 5

t = 0.1

t = 0.5

t = 5

t = 50