Transformada de Fourier
42 pág.

Transformada de Fourier


DisciplinaMatemática Aplicada5.531 materiais29.669 seguidores
Pré-visualização5 páginas
transformadas de
Fourier sa\u2dco dadas
(a) f\u2c6 (\u3c9) =
1
(2+ i\u3c9)(3+ i\u3c9)
.
(b) f\u2c6 (\u3c9) =
1
(1+ i\u3c9)2
.
(c) f\u2c6 (\u3c9) =
1
4\u2212\u3c92 + 2i\u3c9 .
3.3. Resolva a equac¸a\u2dco \u222b \u221e
\u2212\u221e
f (y)
(x\u2212 y)2 + 4dy =
1
x2 + 9
24
4 Aplicac¸o\u2dces a`s Equac¸o\u2dces Diferenciais Parciais
4.1 Equac¸a\u2dco do Calor em uma barra infinita
Vamos determinar a temperatura em func¸a\u2dco da posic¸a\u2dco e do tempo, u(x, t) em uma
barra infinita, sendo conhecida a distribuic¸a\u2dco de temperatura inicial, f (x), ou seja,
vamos resolver o problema de valor inicial\uf8f1\uf8f2
\uf8f3
\u2202u
\u2202t
= \u3b12
\u22022u
\u2202x2
u(x, 0) = f (x), x \u2208 R.
Vamos supor que existam a transformada de Fourier da soluc¸a\u2dco u(x, t) em relac¸a\u2dco
a varia´vel x e de suas derivadas
\u2202u
\u2202t
,
\u2202u
\u2202x
e
\u22022u
\u2202x2
. Ale´m disso vamos supor que
limx\u2192±\u221e |u(x, t)| = 0, limx\u2192±\u221e
\u2223\u2223\u2223\u2223\u2202u\u2202x
\u2223\u2223\u2223\u2223 = 0 e \u222b \u221e\u2212\u221e | f (x)|dx < \u221e. Enta\u2dco aplicando-se
a transformada de Fourier em relac¸a\u2dco a varia´vel x na equac¸a\u2dco diferencial obtemos
\u2202u\u2c6
\u2202t
(\u3c9, t) = \u2212\u3b12\u3c92u\u2c6(\u3c9, t).
Resolvendo esta equac¸a\u2dco diferencial obtemos que
u\u2c6(\u3c9, t) = c(\u3c9)e\u2212\u3b1
2\u3c92t.
Vamos supor que exista f\u2c6 (\u3c9). Neste caso, usando o fato de que u\u2c6(\u3c9, 0) = f\u2c6 (\u3c9) obte-
mos que
u\u2c6(\u3c9, t) = f\u2c6 (\u3c9)e\u2212\u3b1
2\u3c92t.
Seja k\u2c6(\u3c9, t) = e\u2212\u3b12\u3c92t. Enta\u2dco
k(x, t) =
1\u221a
2\u3b12t
e
\u2212 x2
4\u3b12t
e pelo Teorema da Convoluc¸a\u2dco (Teorema 13 na pa´gina 20) temos que
u(x, t) =
1\u221a
2pi
( f \u2217 k)(x, t) = 1
2
\u221a
pi\u3b12t
\u222b \u221e
\u2212\u221e
f (y)e
\u2212 (x\u2212y)2
4\u3b12t dy. (1)
Pode-se provar que se f e´ seccionalmente cont\u131´nua e limitada, enta\u2dco a expressa\u2dco
dada por (1) define uma func¸a\u2dco que satisfaz a equac¸a\u2dco do calor e
lim
t\u21920+
u(x, t) = f (x),
25
nos pontos em que f e´ cont\u131´nua.
Exemplo 14. Vamos resolver, usando a transformada de Fourier, o problema de valor
inicial \uf8f1\uf8f4\uf8f2
\uf8f4\uf8f3
\u2202u
\u2202t
=
\u22022u
\u2202x2
u(x, 0) = e\u2212
x2
4 , x \u2208 R.
Seja f (x) = e\u2212
x2
4 . Enta\u2dco f\u2c6 (\u3c9) =
\u221a
2e\u2212\u3c92 e
u\u2c6(\u3c9, t) = f\u2c6 (\u3c9)e\u2212\u3c9
2t =
\u221a
2e\u2212\u3c9
2(1+t).
Logo a soluc¸a\u2dco do problema de valor inicial e´
u(x, t) =
1
2
\u221a
1+ t
e
\u2212 x2
4(1+t) .
4.2 Equac¸a\u2dco da Onda em uma Dimensa\u2dco
Vamos resolver a equac¸a\u2dco diferencial da onda em uma dimensa\u2dco usando a trans-
formada de Fourier
\u22022u
\u2202t2
= a2
\u22022u
\u2202x2
, x \u2208 R.
Vamos supor que existam a transformada de Fourier da soluc¸a\u2dco u(x, t) em relac¸a\u2dco
a varia´vel x e de suas derivadas
\u2202u
\u2202t
,
\u2202u
\u2202x
,
\u22022u
\u2202x2
e
\u22022u
\u2202t2
. Ale´m disso vamos supor que
limx\u2192±\u221e |u(x, t)| = 0, limx\u2192±\u221e
\u2223\u2223\u2223\u2223\u2202u\u2202x
\u2223\u2223\u2223\u2223 = 0. Aplicando-se a transformada de Fourier em
relac¸a\u2dco a varia´vel x na equac¸a\u2dco diferencial obtemos
\u22022u\u2c6
\u2202t2
(\u3c9, t) = \u2212a2\u3c92u\u2c6(\u3c9, t).
Resolvendo esta equac¸a\u2dco diferencial obtemos que
u\u2c6(\u3c9, t) =
\uf8f1\uf8f2
\uf8f3
\u3c6\u2c61(\u3c9)e
\u2212ia\u3c9t + \u3c8\u2c61(\u3c9)e+ia\u3c9t, se \u3c9 > 0,
c1 + c2t, se \u3c9 = 0,
\u3c6\u2c62(\u3c9)e
\u2212ia\u3c9t + \u3c8\u2c62(\u3c9)e+ia\u3c9t, se \u3c9 < 0.
26
Definindo
\u3c6\u2c6(\u3c9) =
{
\u3c6\u2c61(\u3c9), se \u3c9 > 0,
\u3c6\u2c62(\u3c9), se \u3c9 < 0,
\u3c8\u2c6(\u3c9) =
{
\u3c8\u2c61(\u3c9), se \u3c9 > 0,
\u3c8\u2c62(\u3c9), se \u3c9 < 0,
temos que
u\u2c6(\u3c9, t) = \u3c6\u2c6(\u3c9)e\u2212ia\u3c9t + \u3c8\u2c6(\u3c9)e+ia\u3c9t. (2)
e pelo Teorema da Translac¸a\u2dco (Teorema 7 na pa´gina 12) temos que
u(x, t) = \u3c6(x\u2212 at) + \u3c8(x+ at),
que e´ a soluc¸a\u2dco de D\u2019Alembert para a equac¸a\u2dco corda ela´stica.
Vamos resolver o problema de valor inicial\uf8f1\uf8f4\uf8f4\uf8f2
\uf8f4\uf8f4\uf8f3
\u22022u
\u2202t2
= a2
\u22022u
\u2202x2
, x \u2208 R.
u(x, 0) = f (x),
\u2202u
\u2202t
(x, 0) = g(x), x \u2208 R.
Ale´m do que ja´ supomos anteriormente vamos supor tambe´m que f , g : R \u2192 R sejam
seccionalmente cont\u131´nuas, limitadas e tais que
\u222b \u221e
\u2212\u221e
| f (x)|dx < \u221e e
\u222b \u221e
\u2212\u221e
|g(x)|dx < \u221e.
Aplicando-se a transformada de Fourier nas condic¸o\u2dces iniciais em relac¸a\u2dco a varia´vel x
obtemos
u\u2c6(\u3c9, 0) = f\u2c6 (\u3c9),
\u2202u\u2c6
\u2202t
(\u3c9, 0) = g\u2c6(\u3c9).
Substituindo-se t = 0 em (2) obtemos
f\u2c6 (\u3c9) = u\u2c6(\u3c9, 0) = \u3c6\u2c6(\u3c9) + \u3c8\u2c6(\u3c9).
Derivando-se (2) e substituindo-se t = 0 obtemos
g\u2c6(\u3c9) = ia\u3c9(\u2212\u3c6\u2c6(\u3c9) + \u3c8\u2c6(\u3c9)).
Logo
\u3c8\u2c6(\u3c9) =
1
2
(
f\u2c6 (\u3c9) +
g\u2c6(\u3c9)
ia\u3c9
)
,
\u3c6\u2c6(\u3c9) =
1
2
(
f\u2c6 (\u3c9)\u2212 g\u2c6(\u3c9)
ia\u3c9
)
.
27
Substituindo-se em (2) obtemos
u\u2c6(\u3c9, t) =
1
2
(
f\u2c6 (\u3c9)\u2212 g\u2c6(\u3c9)
ia\u3c9
)
e\u2212ia\u3c9t +
1
2
(
f\u2c6 (\u3c9)\u2212 g\u2c6(\u3c9)
ia\u3c9
)
e+ia\u3c9t.
Aplicando-se a transformada de Fourier inversa obtemos
u(x, t) =
1
2
( f (x\u2212 at) + f (x+ at)) + 1
2a
\u222b x+at
x\u2212at
g(y)dy.
que e´ a soluc¸a\u2dco de d\u2019Alembert do problema de valor inicial.
4.3 Problema de Dirichlet no Semi-plano
Vamos considerar o problema de Dirichlet no semi-plano\uf8f1\uf8f4\uf8f2
\uf8f4\uf8f3
\u22022u
\u2202x2
+
\u22022u
\u2202y2
= 0, x \u2208 R, y > 0
u(x, 0) = f (x), x \u2208 R.
Vamos supor que existam a transformada de Fourier da soluc¸a\u2dco u(x, y) em relac¸a\u2dco
a varia´vel x e de suas derivadas
\u2202u
\u2202y
,
\u2202u
\u2202x
,
\u22022u
\u2202x2
e
\u22022u
\u2202y2
e
\u222b \u221e
\u2212\u221e | f (x)|dx < \u221e. Ale´m disso
vamos supor que limx\u2192±\u221e |u(x, y)| = 0, limx\u2192±\u221e
\u2223\u2223\u2223\u2223\u2202u\u2202x
\u2223\u2223\u2223\u2223 = 0. Enta\u2dco aplicando-se a
transformada de Fourier em relac¸a\u2dco a varia´vel x na equac¸a\u2dco diferencial obtemos
\u2212\u3c92u\u2c6(\u3c9, y) + \u2202
2u\u2c6
\u2202y2
(\u3c9, y) = 0.
Resolvendo esta equac¸a\u2dco diferencial obtemos que
u\u2c6(\u3c9, y) = c1(\u3c9)e
\u2212|\u3c9|y + c2(\u3c9)e|\u3c9|y.
Como lim\u3c9\u2192±\u221e u\u2c6(\u3c9, y) = 0, enta\u2dco c2(\u3c9) = 0. Vamos supor que exista f\u2c6 (\u3c9). Neste
caso, usando o fato de que u\u2c6(\u3c9, 0) = f\u2c6 (\u3c9) obtemos que
u\u2c6(\u3c9, y) = f\u2c6 (\u3c9)e\u2212|\u3c9|y.
Seja k\u2c6(\u3c9, y) = e\u2212|\u3c9|y. Enta\u2dco
k(x, y) =
2y\u221a
2pi
1
x2 + y2
28
e pelo Teorema da Convoluc¸a\u2dco (Teorema 13 na pa´gina 20) temos que
u(x, y) =
1\u221a
2pi
( f \u2217 k)(x, y) = y
pi
\u222b \u221e
\u2212\u221e
f (t)
(x\u2212 t)2 + y2 dt. (3)
Pode-se provar que se f e´ cont\u131´nua e limitada, enta\u2dco a expressa\u2dco dada por (3) define
uma func¸a\u2dco que satisfaz a equac¸a\u2dco de Laplace e
lim
y\u21920+
u(x, y) = f (x).
29
Exerc\u131´cios (respostas na pa´gina 40)
4.1. Resolva o problema de valor inicial\uf8f1\uf8f2
\uf8f3
\u2202u
\u2202t
+ 2
\u2202u
\u2202x
= g(x)
u(x, 0) = f (x), x \u2208 R.
4.2. Resolva o problema de valor inicial
\uf8f1\uf8f2
\uf8f3
\u2202u
\u2202t
= \u3b12
\u22022u
\u2202x2
\u2212 \u3b3u
u(x, 0) = f (x), x \u2208 R.
Aqui \u3b3 e´ uma constante positiva.
4.3. Determine a temperatura como func¸a\u2dco da posic¸a\u2dco e do tempo de uma barra infi-
nita com uma fonte externa de calor, ou seja, resolva o problema de valor inicial
\uf8f1\uf8f2
\uf8f3
\u2202u
\u2202t
= \u3b12
\u22022u
\u2202x2
+ g(x)
u(x, 0) = f (x), x \u2208 R.
4.4. Resolva a equac¸a\u2dco diferencial a seguir usando a transformada de Fourier
\u22022u
\u2202t2
= a2
\u22022u
\u2202x2
\u2212 2\u3b1 \u2202u
\u2202t
\u2212 \u3b12u, x \u2208 R.
Aqui \u3b1 e´ uma constante positiva.
30
5 Tabela de Transformadas de Fourier
Transformadas de Fourier Elementares
f (x) = F\u22121( f\u2c6 )(x) f\u2c6 (\u3c9) = F ( f )(\u3c9)
\u3c7[0,a](x) =
{
1, 0\u2264 x< a
0, caso contra´rio
1\u221a
2pi
1\u2212 e\u2212ia\u3c9
i\u3c9
e\u2212axu0(x) =
{
1, se x < 0
e\u2212ax, se x \u2265 0
1\u221a
2pi
1
a+ i\u3c9
, a > 0
1
x2 + a2
, para a > 0
\u221a
2pi
2a
e\u2212a|\u3c9|
e\u2212ax2 , para a > 0
1\u221a
2a
e\u2212
\u3c92
4a
f (ax), para a 6= 0 1|a| f\u2c6 (
\u3c9
a
)
x f (x) i
d f\u2c6
d\u3c9
(\u3c9)
f \u2032(x) i\u3c9 f\u2c6 (\u3c9)
\u222b x
0 f (y)dy
f\u2c6 (\u3c9)
i\u3c9
f (x\u2212 a) e\u2212ia\u3c9 f\u2c6 (\u3c9)
eiax f (x) f\u2c6 (\u3c9 \u2212 a)
f\u2c6 (x) f (\u2212\u3c9)
( f \u2217 g)(x) = \u222b \u221e\u2212\u221e f (y)g(x\u2212 y)dy \u221a2pi f\u2c6 (\u3c9).g\u2c6(\u3c9)
31
6 Relac¸a\u2dco com a Se´rie de Fourier e a Transformada de
Fourier Discreta
Usando fo´rmula de Euler podemos escrever a se´rie de Fourier de uma func¸a\u2dco
f : [\u2212L, L] \u2192 R seccionalmente cont\u131´nua com derivada tambe´m seccionalmente
cont\u131´nua como
f (x) =
a0
2
+
\u221e
\u2211
n=1
an cos
npix
L
+
\u221e
\u2211
n=1
bn sen
npix
L
=
a0
2
+
1
2
\u221e
\u2211
n=1
an
(
e
inpix
L + e\u2212
inpix
L
)
+
1
2i
\u221e
\u2211
n=1
bn
(
e
inpix
L \u2212 e\u2212 inpixL
)
=
a0
2
+
1
2
\u221e
\u2211
n=1
(an \u2212 ibn)e inpixL + 1
2
\u221e
\u2211
n=1
(an + ibn)e
\u2212 inpixL
=
a0
2
+
1
2
\u221e
\u2211
n=1
(an \u2212 ibn)e inpixL + 1
2
\u2212\u221e
\u2211
n=\u22121
(a\u2212n + ib\u2212n)e
inpix
L
=
\u221e
\u2211
n=\u2212\u221e
cne
inpix
L ,
em que
cn =
1
2L
\u222b L
\u2212L
f (x)e\u2212
inpix
L dx, para n = 0,±1,±2, . . .
pois
an =
1
L
\u222b L
\u2212L
f (x) cos
npix
L
dx para n = 0, 1, 2, . . .
bn =
1
L
\u222b L
\u2212L
f (x) sen
npix
L
dx, para n = 1, 2, . . .
Seja f : R\u2192 R uma func¸a\u2dco tal que f (x) = 0, para |x| > L. Enta\u2dco
f\u2c6
(npi
L
)
=
1\u221a
2pi
\u222b L
\u2212L
f (x)e\u2212
inpix
L dx =
2L\u221a
2pi
cn para n = 0,±1,±2, . . .
32
A transformada de Fourier discreta (DFT) de um vetor Y \u2208 Cn e´ definida por
X = FNY,
em que
FN =
1
N
\uf8ee
\uf8ef\uf8ef\uf8ef\uf8ef\uf8ef\uf8ef\uf8f0
1 1 1 . . . 1
1 e\u2212i2pi
1
N e\u2212i2pi
2
N . . . e\u2212i2pi
N\u22121
N
1 e\u2212i2pi
2
N e\u2212i4pi
4
N . . . e\u2212i2pi
2(N\u22121)
N
...
...
...
...
1 e\u2212i2pi
N\u22121
N e\u2212i2pi
2(N\u22121)
N . . . e\u2212i2pi
(N\u22121)(N\u22121)
N
\uf8f9
\uf8fa\uf8fa\uf8fa\uf8fa\uf8fa\uf8fa\uf8fb
(4)
Seja f : R\u2192 R uma func¸a\u2dco tal que f (x) = 0, para |x| > L. Enta\u2dco
f\u2c6
(npi
L
)
=
1\u221a
2pi
\u222b L
\u2212L
f (x)e\u2212ipi
nx
L dx, para n = 0,±1, . . . , N
2
.
Podemos, agora, aproximar a integral por uma soma de Riemann dividindo o intervalo
[0, 2L] em N subintervalos