Transformada de Fourier
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Transformada de Fourier


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segue o resultado.
Exemplo 8. Seja f (x) = e\u2212ax2 . Derivando obtemos
f \u2032(x) = \u22122ax f (x).
Aplicando-se a transformada de Fourier a ambos os membros obtemos
i\u3c9 f\u2c6 (\u3c9) = \u22122ai f\u2c6 \u2032(\u3c9).
Resolvendo esta equac¸a\u2dco diferencial obtemos
f\u2c6 (\u3c9) = f\u2c6 (0)e\u2212
\u3c92
4a .
12
Mas,
f\u2c6 (0) =
1\u221a
2pi
\u222b \u221e
\u2212\u221e
e\u2212ax
2
dx =
1\u221a
2pi
(\u222b \u221e
\u2212\u221e
\u222b \u221e
\u2212\u221e
e\u2212a(x
2+y2)dxdy
)1/2
=
1\u221a
2pi
(\u222b 2pi
0
\u222b \u221e
0
e\u2212ar
2
rdrd\u3b8
)1/2
= \u2212 1\u221a
2a
\u221a
2pi
(\u222b 2pi
0
e\u2212ar
2
\u2223\u2223\u2223\u221e
0
d\u3b8
)1/2
=
=
1\u221a
2a
.
Logo
F (e\u2212ax2)(\u3c9) = 1\u221a
2a
e\u2212
\u3c92
4a .
Em particular
F (e\u2212 x
2
2 )(\u3c9) = e\u2212
\u3c92
2 .
Teorema 7 (Translac¸a\u2dco). Seja a uma constante. Se a transformada de Fourier da func¸a\u2dco
f : R\u2192 R e´ f\u2c6 (\u3c9), enta\u2dco
(a) F ( f (x\u2212 a))(\u3c9) = e\u2212ia\u3c9 f\u2c6 (\u3c9), para \u3c9 \u2208 R. e
(b) F (eiax f (x))(\u3c9) = f\u2c6 (\u3c9 \u2212 a).
Demonstrac¸a\u2dco. (a)
F ( f (x\u2212 a))(\u3c9) = 1\u221a
2pi
\u222b \u221e
\u2212\u221e
e\u2212i\u3c9x f (x\u2212 a)dx
=
1\u221a
2pi
\u222b \u221e
\u2212\u221e
e\u2212i\u3c9(x
\u2032+a) f (x\u2032)dx\u2032 = e\u2212ia\u3c9 f\u2c6 (\u3c9).
(b)
F (eiax f (x))(\u3c9) = 1\u221a
2pi
\u222b \u221e
\u2212\u221e
e\u2212i\u3c9xeiax f (x)dx
=
1\u221a
2pi
\u222b \u221e
\u2212\u221e
e\u2212i(\u3c9\u2212a)x f (x)dx = f\u2c6 (\u3c9 \u2212 a).
13
f (x)
f (x\u2212 a)
f\u2c6 (\u3c9)
e\u2212ia\u3c9 f\u2c6 (\u3c9)
F
Figura 6: Teorema da Translac¸a\u2dco (a)
f (x)
eiax f (x)
f\u2c6 (\u3c9)
f\u2c6 (\u3c9 \u2212 a)
F
Figura 7: Teorema da Translac¸a\u2dco (b)
14
Exemplo 9. Seja f : R\u2192 R dada por
f (x) =
{
cos ax se \u2212 b < x < b
0 caso contra´rio
Como
f (x) = (cos ax)\u3c7[\u2212b,b](x) =
(
eiax + e\u2212iax
2
)
\u3c7[\u2212b,b](x),
e pela linearidade da transformada de Fourier e pelo Teorema da Dilatac¸a\u2dco (Teorema
1 na pa´gina 4), para \u3c9 6= 0 temos que
F (\u3c7[\u2212b,b])(\u3c9) = F
(
\u3c7[0,b](\u2212x) + \u3c7[0,b](x)
)
(\u3c9)
=
1\u221a
2pi
(
eib\u3c9 \u2212 1
i\u3c9
+
1\u2212 e\u2212ib\u3c9
i\u3c9
)
=
2\u221a
2pi
sen(b\u3c9)
\u3c9
, para \u3c9 6= 0,
F (\u3c7[\u2212b,b])(0) =
2b\u221a
2pi
enta\u2dco, pelo Teorema da Translac¸a\u2dco (Teorema 7 (b) na pa´gina 12) e pela linearidade da
transformada de Fourier, temos que
f\u2c6 (\u3c9) =
1
2
(
F (\u3c7[\u2212b,b])(\u3c9 \u2212 a) +F (\u3c7[\u2212b,b])(\u3c9 + a)
)
=
1\u221a
2pi
(
sen b(\u3c9 \u2212 a)
\u3c9 \u2212 a +
sen b(\u3c9 + a)
\u3c9 + a
)
, para \u3c9 6= ±a
f\u2c6 (\u2212a) = f\u2c6 (a) = 1\u221a
2pi
(
2b+
sen 2ab
2a
)
.
15
Exerc\u131´cios (respostas na pa´gina 35)
1.1. Determine a transformada de Fourier das seguintes func¸o\u2dces f : R\u2192 R
(a) f (x) = (1\u2212 |x|/a)\u3c7[\u2212a,a](x) =
{
1\u2212 |x|/a, se \u2212 a < x < a,
0, caso contra´rio.
(b) f (x) = sen(ax)\u3c7[\u2212b,b](x) =
{
sen(ax), se \u2212 b < x < b
0, caso contra´rio.
(c) f (x) = xe\u2212x2 .
(d) f (x) = x2e\u2212x2 .
(e) f (x) = e\u2212(a+ib)xu0(x) =
{
e\u2212(a+ib)x, se x > 0
0, caso contra´rio,
para a > 0 e b \u2208 R.
(f) f (x) = e(a+ib)xu0(\u2212x) =
{
e(a+ib)x, se x < 0
0, caso contra´rio,
para a > 0 e b \u2208 R.
16
2 Inversa\u2dco
Teorema 8. Se f : R\u2192 R e´ seccionalmente cont\u131´nua e tal que \u222b \u221e\u2212\u221e | f (x)|dx < \u221e, enta\u2dco
lim
\u3c9\u2192±\u221e f\u2c6 (\u3c9) = 0.
Demonstrac¸a\u2dco. Pelo Lema de Riemann-Lesbegue, temos que
lim
\u3c9\u2192±\u221e
\u222b M
\u2212M
e\u2212i\u3c9x f (x)dx = lim
\u3c9\u2192±\u221e
\u222b M
\u2212M
f (x) cos\u3c9xdx+ i lim
\u3c9\u2192±\u221e
\u222b M
\u2212M
f (x) sen\u3c9xdx = 0.
Para todo e > 0, existe M > 0 tal que
\u222b
|x|>M | f (x)|dx < e. Logo
\u221a
2pi lim
\u3c9\u2192±\u221e | f\u2c6 (\u3c9)| = lim\u3c9\u2192±\u221e
\u2223\u2223\u2223\u2223
\u222b \u221e
\u2212\u221e
e\u2212i\u3c9x f (x)dx
\u2223\u2223\u2223\u2223
\u2264 lim
\u3c9\u2192±\u221e
\u2223\u2223\u2223\u2223
\u222b M
\u2212M
e\u2212i\u3c9x f (x)dx
\u2223\u2223\u2223\u2223+
\u222b
|x|>M
| f (x)|dx \u2264 e.
Lema 9. Se g : R \u2192 R e´ seccionalmente cont\u131´nua tal que \u222b \u221e\u2212\u221e |g(x)|dx < \u221e, g(0) = 0 e
g\u2032(0) existe, enta\u2dco \u222b \u221e
\u2212\u221e
g\u2c6(\u3c9)d\u3c9 = 0.
Demonstrac¸a\u2dco. Seja
h(x) =
\uf8f1\uf8f2
\uf8f3
g(x)
x
, se x 6= 0,
g\u2032(0), se x = 0.
Enta\u2dco g(x) = xh(x) e
\u222b \u221e
\u2212\u221e |h(x)|dx < \u221e. Logo\u222b \u221e
\u2212\u221e
g\u2c6(\u3c9)d\u3c9 = i
\u222b \u221e
\u2212\u221e
h\u2c6\u2032(\u3c9)d\u3c9 = ih\u2c6(\u3c9)
\u2223\u2223\u2223\u221e
\u2212\u221e
= 0,
pelo Teorema 8.
17
Teorema 10. Se f : R\u2192 R e´ seccionalmente cont\u131´nua tal que \u222b \u221e\u2212\u221e | f (x)|dx < \u221e, enta\u2dco
f (x) =
1\u221a
2pi
\u222b \u221e
\u2212\u221e
eix\u3c9 f\u2c6 (\u3c9)d\u3c9,
para todo x \u2208 R em que f e´ cont\u131´nua.
Demonstrac¸a\u2dco. Vamos demonstrar para o caso em que f \u2032(x) existe. Seja g : R \u2192 R
definida por
g(x\u2032) = f (x+ x\u2032)\u2212 f (x)e\u2212 x
\u20322
2 .
Como g(0) = 0, pelo Lema 9 temos que
0 =
\u222b \u221e
\u2212\u221e
g\u2c6(\u3c9)d\u3c9 =
\u222b \u221e
\u2212\u221e
eix\u3c9 f\u2c6 (\u3c9)d\u3c9 \u2212 f (x)
\u222b \u221e
\u2212\u221e
e\u2212
\u3c92
2 d\u3c9
=
\u222b \u221e
\u2212\u221e
eix\u3c9 f\u2c6 (\u3c9)d\u3c9 \u2212 f (x)
\u221a
2pi.
Corola´rio 11. Se f : R\u2192 R e´ cont\u131´nua tal que \u222b \u221e\u2212\u221e | f (x)|dx < \u221e, enta\u2dco
F ( f\u2c6 )(\u3c9) = f (\u2212\u3c9).
Demonstrac¸a\u2dco. Pelo Teorema 10 temos que
f (\u2212\u3c9) = 1\u221a
2pi
\u222b \u221e
\u2212\u221e
e\u2212i\u3c9
\u2032\u3c9 f\u2c6 (\u3c9\u2032)d\u3c9\u2032 = F ( f\u2c6 )(\u3c9)
18
Exemplo 10. Seja a um nu´mero real positivo. Seja f : R\u2192 R dada por
f (x) =
1
x2 + a2
Como F (e\u2212a|x|)(\u3c9) = 1\u221a
2pi
2a
\u3c92 + a2
, enta\u2dco
f (\u3c9) = g\u2c6(\u3c9), em que g(x) =
\u221a
2pi
2a
e\u2212a|x|.
Logo
F ( f )(\u3c9) = F (g\u2c6)(\u3c9) = g(\u2212\u3c9) =
\u221a
2pi
2a
e\u2212a|\u3c9|.
Corola´rio 12 (Injetividade). Dadas duas func¸o\u2dces f (x) e g(x) seccionalmente cont\u131´nuas tais
que
\u222b \u221e
\u2212\u221e | f (x)|dx < \u221e e
\u222b \u221e
\u2212\u221e |g(x)|dx < \u221e, se
F ( f )(\u3c9) = F (g)(\u3c9), para todo \u3c9 \u2208 R,
enta\u2dco f (x) = g(x), exceto possivelmente nos pontos de descontinuidade.
Demonstrac¸a\u2dco. Pela linearidade da transformada de Fourier, basta provarmos que se
F ( f )(\u3c9) = 0, enta\u2dco f (x) = 0 nos pontos em que f e´ cont\u131´nua. Mas isto e´ decorre\u2c6ncia
imediata do Teorema 10.
Exemplo 11. Vamos determinar a func¸a\u2dco f : R \u2192 R cuja transformada de Fourier e´
f\u2c6 (\u3c9) =
1
a+ ib+ i\u3c9
, para a > 0 e b \u2208 R.
f\u2c6 (\u3c9) =
1
a+ ib+ i\u3c9
=
1
a+ i(b+ \u3c9)
f (x) = e\u2212ibx
\u221a
2pie\u2212axu0(x) =
\u221a
2pie\u2212(a+ib)xu0(x).
19
Exerc\u131´cios (respostas na pa´gina 36)
2.1. Determine as func¸o\u2dces f : R\u2192 C cujas 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) =
i\u3c9
1+ \u3c92
.
(d) f\u2c6 (\u3c9) =
1
\u3c92 + \u3c9 + 1
.
(e) f\u2c6 (\u3c9) =
1
a+ ib\u2212 i\u3c9 , para a > 0 e b \u2208 R.
(f) f\u2c6 (\u3c9) =
1
4\u2212\u3c92 + 2i\u3c9 .
Calcule a transformada de Fourier das func¸o\u2dces f : R\u2192 R:
(a) f (x) =
x
1+ x2
.
(b) f (x) =
x
(1+ x2)2
.
20
3 Convoluc¸a\u2dco
A convoluc¸a\u2dco de duas func¸o\u2dces f : R \u2192 R e g : R \u2192 R 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, e´ definida por
( f \u2217 g)(x) =
\u222b \u221e
\u2212\u221e
f (y)g(x\u2212 y)dy, para x \u2208 R.
Exemplo 12. Seja f : R\u2192 R definida por f (x) = \u3c7[0,1](x) =
{
1, se 0 \u2264 x \u2264 1,
0, caso contra´rio.
( f \u2217 f )(x) =
\u222b \u221e
\u2212\u221e
\u3c7[0,1](y)\u3c7[0,1](x\u2212 y)dy =
\u222b 1
0
\u3c7[0,1](x\u2212 y)dy
=
\u222b 1
0
\u3c7[\u22121,0](y\u2212 x)dy =
\u222b 1
0
\u3c7[\u22121+x,x](y)dy =
\uf8f1\uf8f4\uf8f4\uf8f2
\uf8f4\uf8f4\uf8f3
0, se x < 0,
x, se 0 \u2264 x < 1,
2\u2212 x, se 1 \u2264 x < 2,
0, se x \u2265 2.
Teorema 13 (Convoluc¸a\u2dco). Sejam f : R \u2192 R e g : R \u2192 R 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. Enta\u2dco
F ( f \u2217 g)(\u3c9) =
\u221a
2pi f\u2c6 (\u3c9).g\u2c6(\u3c9)
Demonstrac¸a\u2dco. Pelas definic¸o\u2dces temos que
F ( f \u2217 g)(\u3c9) = 1\u221a
2pi
\u222b \u221e
\u2212\u221e
e\u2212i\u3c9x
[\u222b \u221e
\u2212\u221e
f (y)g(x\u2212 y)dy
]
dx.
Sob as hipo´teses consideradas pode-se mostrar que podemos trocar a ordem de
integrac¸a\u2dco para obtermos
F ( f \u2217 g)(\u3c9) = 1\u221a
2pi
\u222b \u221e
\u2212\u221e
f (y)
[\u222b \u221e
\u2212\u221e
e\u2212i\u3c9xg(x\u2212 y)dx
]
dy.
21
Fazendo-se a mudanc¸a de varia´veis x\u2212 y = z obtemos
F ( f \u2217 g)(\u3c9) = 1\u221a
2pi
\u222b \u221e
\u2212\u221e
f (y)
[\u222b \u221e
\u2212\u221e
e\u2212i\u3c9(z+y)g(z)dz
]
dy
=
1\u221a
2pi
\u222b \u221e
\u2212\u221e
e\u2212i\u3c9y f (y)
[\u222b \u221e
\u2212\u221e
e\u2212i\u3c9zg(z)dz
]
dy
=
\u221a
2pi f\u2c6 (\u3c9).g\u2c6(\u3c9).
Exemplo 13. Seja f : R\u2192 R dada por
f (x) =
\uf8f1\uf8f4\uf8f4\uf8f2
\uf8f4\uf8f4\uf8f3
0, se x < 0,
x, se 0 \u2264 x < 1,
2\u2212 x, se 1 \u2264 x < 2,
0, se x \u2265 2.
Como, pelo Exemplo 12, f = \u3c7[0,1] \u2217 \u3c7[0,1], enta\u2dco
f\u2c6 (\u3c9) =
\u221a
2pi (\u3c7\u302[0,1](\u3c9))
2 =
\u221a
2pi
(
1\u221a
2pi
1\u2212 e\u2212ia\u3c9
i\u3c9
)2
= \u2212 1\u221a
2pi
(1\u2212 e\u2212ia\u3c9)2
\u3c92
.
Teorema 14. A convoluc¸a\u2dco satisfaz as seguintes propriedades:
(a) f \u2217 g = g \u2217 f
(b) f \u2217 (g1 + g2) = f \u2217 g1 + f \u2217 g2
(c) ( f \u2217 g) \u2217 h = f \u2217 (g \u2217 h)
(d) f \u2217 0 = 0 \u2217 f = 0
Demonstrac¸a\u2dco.
(a)
( f \u2217 g)(x) =
\u222b \u221e
\u2212\u221e
f (y)g(x\u2212 y)dy =
\u222b \u2212\u221e
\u221e
f (x\u2212 y\u2032)g(y\u2032)(\u2212dy\u2032) =
=
\u222b \u221e
\u2212\u221e
f (x\u2212 y\u2032)g(y\u2032)dy\u2032 = (g \u2217 f )(x).
22
(b)
( f \u2217 (g1 + g2))(x) =
\u222b \u221e
\u2212\u221e
f (y)(g1(x\u2212 y) + g2(x\u2212 y))dy =
=
\u222b \u221e
\u2212\u221e
f (x\u2212 y)g1(x\u2212 y)dy+
\u222b \u221e
\u2212\u221e
f (x\u2212 y)g2(x\u2212 y)dy =
= ( f \u2217 g1)(x) + ( f \u2217 g2)(x).
(c)
(( f \u2217 g) \u2217 h)(x) =
\u222b \u221e
\u2212\u221e
( f \u2217 g)(x\u2212 y)h(y)dy =
=
\u222b \u221e
\u2212\u221e
[\u222b \u221e
\u2212\u221e
f (y\u2032)g(x\u2212 y\u2212 y\u2032)dy\u2032
]
h(y)dy =
=
\u222b \u221e
\u2212\u221e
\u222b \u221e
\u2212\u221e
f (y\u2032)g(x\u2212 y\u2212 y\u2032)h(y)dydy\u2032 =
=
\u222b \u221e
\u2212\u221e
f (y\u2032)
[\u222b \u221e
\u2212\u221e
g(x\u2212 y\u2212 y\u2032)h(y)dy
]
dy\u2032 =
=
\u222b \u221e
\u2212\u221e
f (y\u2032)(g \u2217 h)(x\u2212 y\u2032)dy\u2032 = ( f \u2217 (g \u2217 h))(x).
(d) ( f \u2217 0)(x) = \u222b \u221e\u2212\u221e f (y\u2212 x) · 0 dy = 0 = (0 \u2217 f )(x).
23
Exerc\u131´cios (respostas na pa´gina 38)
3.1. Calcule a convoluc¸a\u2dco f \u2217 g para f , g : R\u2192 R dadas por
(a) f (x) = e\u2212xu0(x) =
{
e\u2212x, se x > 0
0, caso contra´rio,
,
g(x) = e\u22122xu0(x) =
{
e\u22122x, se x > 0
0, caso contra´rio,
.
(b) f (x) = \u3c7[\u22121,1](x) =
{
1, se \u2212 1 < x < 1
0, caso contra´rio,
,
g(x) = e\u2212xu0(x) =
{
e\u2212x, se x > 0
0, caso contra´rio,
.
3.2. Determine, usando convoluc¸a\u2dco, as func¸o\u2dces f : R \u2192 C cujas