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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