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Fourier Series

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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

- Consider
using odd extension

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

t = 0

t = 0.5

t = 0

t = 0.5

t = 1.2

t = 0

t = 0.5

t = 1.2

t = 5

t = 0

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

t = 0

t = 0.5

t = 0

t = 0.5

t = 1.2

t = 0

t = 0.5

t = 1.2

t = 5

t = 0

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

y

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:

t = 0

Forward cycle

t = 0

Forward cycle

t = L/6c

t = 0

Forward cycle

t = L/6c

t = L/3c

t = 0

Forward cycle

t = L/6c

t = L/3c

t = L/2c

t = 0

Forward cycle

t = L/6c

t = L/3c

t = L/2c

t = 2L/3c

t = 0

Forward cycle

t = L/6c

t = L/3c

t = L/2c

t = 2L/3c

t = 5L/6c

Backward cycle

t = L/c

Backward cycle

t = 7L/6c

t = L/c

Backward cycle

t = 4L/3c

t = 7L/6c

t = L/c

Backward cycle

t = 3L/2c

t = 4L/3c

t = 7L/6c

t = L/c

Backward cycle

t = 5L/3c

t = 3L/2c

t = 4L/3c

t = 7L/6c

t = L/c

Backward cycle

t = 11L/6c

t = 5L/3c

t = 3L/2c

t = 4L/3c

t = 7L/6c

t = L/c

Backward cycle

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 = 0.05

t = 0.05

t = 0.1

t = 0.05

t = 0.1

t = 0.5

t = 0.05

t = 0.1

t = 0.5

t = 5

t = 0.05

t = 0.1

t = 0.5

t = 5

t = 50

- This represents a uniform initial temperature distribution