1 Fourier Series

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A (generalized) function f(t) of period 2L has a Fourier series of the form ... The Fourier series it convergent at every point t for which both the right limit f(t+) and ...
MA 18.03, R05

FOURIER SERIES, GENERALIZED FUNCTIONS, LAPLACE TRANSFORM

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Fourier Series

1.1

General facts

1. A (generalized) function f (t) of period 2L has a Fourier series of the form f (t) ∼

a0 πt 2πt πt 2πt + a1 cos + a2 cos + · · · + b1 sin + b2 sin + ··· 2 L L L L

or



f (t) ∼



a0 X nπt X nπt + an cos + bn sin 2 L L n=1 n=1

2. The coefficients an and bn can be found by the following formulas. an

bn

=

=

1 L

Z

1 L

Z

L

f (t) cos

nπt dt L

f (t) sin

nπt dt L

−L

L

−L

3. The Fourier series it convergent at every point t for which both the right limit f (t+) and left limit f (t−) exist and are finite. The sum of the Fourier series at t is the average of these two limits. 4. The terms in the Fourier series of a function f (t) must have the same symmetries as f (t) itself. For instance, • an odd function will only have a sine functions in its Fourier series (no constant); • an even function will only have a cosine functions in its Fourier series and the constant term; • the Fourier series of a function that is odd about L/2 will only have cos (2n+1)πt and sin 2nπt L L ; (2n+1)πt • the Fourier series of a function that is even about L/2 will only have cos 2nπt . L and sin L

It might be useful to recall that • even · even = even • even · odd = odd • odd · odd = even

5. We differentiate Fourier series term by term. So given f (t) ∼ we get f 0 (t) ∼

∞ X n=1

∞ ∞ a0 X nπt X nπt + an cos + bn sin , 2 n=1 L L n=1



bn

nπ nπt X nπ nπt cos + −an sin L L L L n=1

6. Integration is also done term by term, but the result is a Fourier series only of the series we integrated does not have a non-zero constant term. 1

MA 18.03, R05

1.2

How do we compute Fourier series?

Directly from the definition using the formulas for an and bn . Reducing to a known Fourier series (or to a Fourier series given on the exam) by differentiating, or integrating, or translating, or taking a dilation, etc. . . and maybe employing some trig identities. Trig identities are especially useful for computing Fourier series of, say, sin2 t. Try to remember that sin2 A =

1 − cos 2A 2

sin(A ± B) = sin A cos B ± cos A sin B;

sin A sin B =

cos(A − B) − cos(A + B) ; 2 sin A cos B =

cos2 A =

1 + cos 2A 2

cos(A ± B) = cos A cos B ∓ sin A sin B;

cos A cos B =

cos(A − B) + cos(A + B) ; 2

sin(A + B) − sin(A − B) ; 2

and the special values sin

(2n + 1)π = (−1)n ; 2

π π 1 sin = cos = √ ; 4 4 2

1.3

cos nπ = (−1)n ;

π 1 π sin = cos = ; 6 3 2

√ 3 π π sin = cos = . 3 6 2

Use in ODE

Consider the linear ODE with constant coefficients cn y (n) + . . . + c1 y 0 + c0 y = f (t),

cn 6= 0,

f (t) periodic with period 2L.

We can make use of Fourier series to find a particular solution yp of this ODE, as follows. (But the general solution is still y = yp + yh .) 1. First write down the Fourier series of f (t). 2. Replace f (t) by its Fourier series in the RHS of the ODE. It becomes something of the form ∞

cn y (n) + . . . + c1 y 0 + c0 y =



a0 X nπt X nπt + + . an cos bn sin 2 L L n=1 n=1

3. For each term that appears in the Fourier series find a particular solution of the corresponding equation (use ERF or ERF’ or ERF” etc. . . ). a0 cn y (n) + . . . + c1 y 0 + c0 y = ; get y0 2 cn y (n) + . . . + c1 y 0 + c0 y

=

an cos

nπt L

;

get

y1,n

cn y (n) + . . . + c1 y 0 + c0 y

=

bn sin

nπt L

;

get

y2,n

2

MA 18.03, R05

4. A particular solution to the original ODE is the sum of all these pieces yp = y0 +

∞ X n=1

2

y1,n +

∞ X

y2,n .

n=1

Step and Delta

2.1

Definitions ( 1 The step function u(t) = 0

( 1 ua (t) = u(t − a) = 0

t>0 ta t 0)

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1 s−a

MA 18.03, R05

5.2

Use in ODE

Finding solutions of ODEs: take he Laplace transforms of both sides of an ODE, one obtains an (algeraic) equation for the Laplace transform. Solve that and take the inverse Laplace transform. That gives you a solution of the original ODE.

Finding the weight function of an operator p(D):

L(w(t)) =

1 p(s)

Pole diagram: the rightmost pole(s) of the Laplace transform F (s) of a function f (t) tells us how f (t) behaves for very large t (as t → ∞). • If the right most pole is at s = a, then, for large t, f (t) ∼ eat • If the right most poles are at s = a ± ib, then, for large t, f (t) ∼ eat cos bt • In particular, f (t) → 0 as t → ∞ if and only if the poles of F (s) have real part negative. More precisely, if F (s) has simple poles at a, and α ± iβ, i.e. if F (s) =

∗ , (s − a)((s − α)2 + β 2 )

then F (s) will be written as a sum of simple fractions F (s) =

A B C A B0s + C 0 . + + = + s − a s − α − iβ s − α + iβ s − a (s − α)2 + β 2

Therefore, f (t) = Aeat + Be(α+iβ)t + Ce(α−iβ)t or, equivalently, f (t) = A00 eat + B 00 eαt cos βt + C 00 eαt sin βt. The dominant term as t gets very large will be the one with the largest coefficient on the exponential, which is to say the one corresponding to the rightmost pole.

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Transfer function of p(D) W (s) =

1 = L(w(t)) p(s)

It has the property that a particular solution to p(D)x = ert is given by xp = W (r)ert (if p(r) 6= 0.) Note that this is an exponential solution.

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