第十二章傅立葉級數2009／03／11Fourier法國數學家、物理學家1768.3.21-1830.3.16

1817年當選為科學院院士1807年，《熱的傳播》熱傳導方程指，

出任一周期函數都可以展成三角函數的無窮級數§12.1

周期函數的Fourier級數一、問題的提出簡諧波：x(t

)=a

sin(w

t+j

)wj－－初相；a－－振幅.T

=2p

－－周期；w

－－頻率；x1

(t

)

=

sin

t,

x2

(t

)

=

sin

3t，

x1

(t

)

+

x2

(t

)?問題￥f

(t

)

=

A0

+

An

sin(nw

t

+

jn

)?n=1otu--1-

p

￡

t

<

0

1, 0

￡

t

<

p非正弦周期函數:矩形波u(t

)=-1,不同頻率正弦波的疊加4

sin

t,

4 1

sin

3t,

4 1

sin

5t,

4 1

sin

7t,

p p

3

p

5

p

74

sin

tp34

(sin

t+

1

sin

3t

)p3

54

(sin

t+

1

sin

3t+

1

sin

5t

)p3

5

74

(sin

t

+

1

sin

3t

+

1

sin

5t

+

1

sin

7t

)pp

3

5

7u(t

)

=

4

(sin

t

+

1

sin

3t

+

1

sin

5t

+

1

sin

7t

+

)(-p

<

t

<

p,

t

?

0)3

5

7

94

(sin

t

+

1

sin

3t

+

1

sin

5t

+

1

sin

7t

+

1

sin

9t

)p二、三角級數三角函數系的正交性￥1.三角級數￥n=1f

(t

)

=

A0

+

An

sin(nw

t

+

jn

)n=1=

A0

+

(

An

sin

jn

cos

nw

t

+

An

cosjn

sin

nw

t

)￥2n=10

+

(an

cos

nx

+

bn

sin

nx)a200a=

A

,令an

=

An

sin

jn

,

bn

=

An

cosjn

,

w

t

=

x,三角級數p-pcos

nxdx

=

0,p-psin

nxdx

=

0,2.三角函數系的正交性三角函數系1,cos

x,sin

x,cos

2

x,sin

2

x,cos

nx,sin

nx,正交:任意兩個不同函數的乘積在[-p

,p

]上的積分等于零.(n

=

1,2,3,),0,

m

?

n1,

m

=

nmnd

=

0,

m

?

n=

,mnp

,

m

=

npsin

mx

sin

nxdx

=

pd-p0,

m

?

n=

,mnp

,

m

=

npcos

mx

cos

nxdx

=pd-pp-p

sin

mx

cos

nxdx

=

0.(其中m,n

=1,2,)2sin

mx

sin

nx

=

1[cos(m

-

n)

x

-

cos(m

+

n)

x]2cos

mx

cos

nx

=

1[cos(m

-

n)

x

+

cos(m

+

n)

x]2cos

mx

sin

nx

=

1[sin(m

+

n)

x

-

sin(m

-

n)

x]三、函數的傅里葉級數問題:1.若能展開,an

,bn是什么?2.展開的條件是什么? (下節課介紹)1.傅里葉系數￥2

k

=1k

k(a

cos

kx

+

b

sin

kx)若有f

(x)=a0

+(1)求a0

.kk[2(a

cos

kx

+

b

sin

kx)]dxa0

dx

+f

(

x)dx

=p-p￥k

=1p-pp-p22p,=

a0p-pa

=f

(

x)dx1p0kkb

sin

kxdx2a0￥￥a

cos

kxdx

+=

dx

+p-p

k

=1p-p

k

=1p-p(2)求an

.p-pp-pacos

nxdx2f

(

x)cos

nxdx

=0p-pp-p￥+sin

kx

cos

nxdx]cos

kx

cos

nxdx

+

b[akk

=1kp-p=

an2cos

nxdxn=

a

p,p-pa

=nf

(

x)cos

nxdx1p(n

=

1,2,3,)p-pb

=nf

(

x)sin

nxdx1p(n

=

1,2,3,)p-p(3)求bn

.p-pasin

nxdx2f

(

x)sin

nxdx

=0p-pp-p￥+[acos

kx

sin

nxdx

+

bkk

=1ksin

kx

sin

nxdx]

=

bnp,p-pp-p=

1p=

1p(n

=

1,2,)f

(

x)sin

nxdx,b(n=

0,1,2,)f

(

x)cos

nxdx

,ann00=

1p=

1p(n

=

1,2,)2pf

(

x)sin

nxdx,b(n=

0,1,2,)2pf

(

x)cos

nxdx

,ann或傅里葉系數￥2n=1nn+

b

sin

nx)(a

cos

nxa0

+2.

傅里葉級數若

f

是以

2p為周期的可積或絕對可積函數，?

?￥+2n=1(an

cos

nx

+

bn

sin

nx)a0f

(

x)

~解例1

以2p

為周期的矩形脈沖的波形-

p

￡

t

<

0

1, 0

￡

t

<

pu(t

)

=

-

1,寫出它的Fourier級數？tu-

o-

11anpu(t

)

cos

ntdt-p=

1p=p-ppp00111cos

ntdt(-1)cos

ntdt

+=

0 (n

=

0,1,2,)看奇偶得有無正弦余弦bnpu(t

)

sin

ntdt-p=

1p=p-ppp00111sin

ntdt(-1)sin

ntdt

+np=

2

(1

-

cos

np

)

=

2

[1

-

(-1)n

]np0,4,

n

=

2k

-

1,

k

=

1,2,n

=

2k,

k

=

1,2,=

(2k

-

1)p￥4n=1sin(2n

-

1)t\

u(t

)~(2n

-

1)p作業習題12.12、3、4.思考題若函數j

(-x)=y

(x)，問：j

(x)與y

(x)(n

=

0,1,2,)的傅里葉系數an

、bn

與a

n

、bn之間有何關系？思考題解答-ppanpj

(

x)cos

nxdx=

1=-pp1j

(-t

)cos(-nt

)d

(-t

)=p-pj

(-

x)cos

nxdxp1=p-ppy

(

x)cos

nxdxp1n=

a(n

=

0,1,2,)-