傅里叶变换的基本性质

傅里叶变的基本性质际信号分线性一f(t)(j1

)f(F(j2

(()j21

(j

(3-55)ab均为常例

F(j

解

f(t)U()

t)式(3-55)得Fj二、对性

11211sgn(t22jjf(t)F(j

)F(jt)(

(3-56)证

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f()

2

F(j

j

(t

F(j

jd(

Fj

2

f(

F(jx)

jxt

t(

Fjx)

j

xt替2

F(jt

F(jt)2(

f(t)

ff(t)

f(f(

(3-56)成为F(jt)2(

(3-57)中的f(t))F(j

F(jt)(

(例若信

f(t)

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F(j

22

ft)

解

F(j

F(j

的实函数F(jt)()0(

22(

F(t)2f(A(

f(

换成，则得f(t)

f()Sa)3-20所F(j)

f(t)A

0

t图320三、折叠fF(j

)精选资料，欢迎下载

F(jf((jj

ftft

四、尺变换性看动画fF(j

)fat)证a

1F(j)aa

)（3-59)

at

f()

j

，dx，代入前()e

dx1(j)aa

证

f(at)f(

(时间尺度a

Fj

F(j

)

(或)a续例已知)

E0

44,

F(j

解

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tt(t)

E

2F(jE

)

f(t

f(t)

F(j

F(j)()3-21所

f(t

ft)

t

tE

(j

E

Fj

0

0

图-21五、时移fF(j

)f(t)j

)e

j

(3-60)

ft)

0

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jj

例

)

E0

tt

F(j

解:根Fj

E

e

/2六、频性fF(jf(t)

j

(3-61)证f(t)e

j

t

f()e

j

e

f(t)e

j(

)t

)0

ft)

e

t

e

t

f(t)

cos

t

t

ft)

t

f()七、时微分性

t精选资料，欢迎下载

fF(j

)

()dtn

(j

nF(j

证

f()

2

F(j

j

t导数，得()2

j

j

j()

(j

F(j

nf(t)dt

(j

Fj

证例求

f(t

()

(t)

F(j

解

(t)1F(j

j

例图

f(t

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ttttttf(t))

F(j

解

f(t

f

''

21())

f''()j

2

2f(t)(j

)

sin(j(2)

2)2

)2

/

-1/

f

t

(1/

f(1/(a)(b)图-八、频域分性fF(j

)()j

dF(j精选资料，欢迎下载

tt

()(j)n

F(j

(3-63)例

ftU

F(j

)

解:因U(t)

1jtU()j

djdj

1九、时域分性f(Fj

)

t

(t)dt

F(jj

(0)

(3-64)例根

(t)

ft)(t)

解

(t)1Ut)

(x)U(t)

1j

例求3-23

f(t)

F(j

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tt解:

f(t

1f''()2)f

''

t)

2j/2j)f(t

t

f()

sin(

)

sin(

)(

)f(t

t

f(

j

1)(0)(j

(1/

/2

t

/2

/2

t

/2

/2

t

(-1/

)(a)(b)(c)图-十、频积分性fF(j(0)(t)jtj

F()

(3-65)例

ft)

sin()t

F(j)

解

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t)t)t)

12(jtjt)2jj

jdx(tj十一、时卷积定理

f(t)(j1

f(t)F(j2

f(t)(t)F(j1

F(j

(3-66)证F(t)f(t)12

f1

)(t2

)dt(

f(t2

dtd

fje12

d(j2

fj1

djF(j21

证毕例图tf(t)G(t3

F(j

)

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f(t)1

(t)

t

/

/

t(a)(b)图3-解()

sin(2

)

2

)2f()G(t)()

Fj

(

例一

ft)

f(t)ft

的f(t)f(t)

1

1

f(

d解sgn()

2j2jt

2精选资料，欢迎下载

f(t)f()

sgn((jFj

jsgn((j

十二、频卷积定理f(t)(j1

)f(t)(j2

f(t)f(t)12

1

(j(j12

(3-67)ft)f(t)(j2(j2)1例

f(t)(t

Fj

)

解

(t)j

jt2

tj2

U(t)

1j精选资料，欢迎下载

j2(jj2(j22,f)tU()F(j

1j

j)

'Fj

j

十三、帕瓦尔定理f(t)(j1

)f(F(j2

f(t)f(t)1

12

(j1

j2

d

f()dt1

1

(j1

ft)1

f(t)dt1

1

1

(j

d

ftf(t)2

f(t)f(dt1

12

(j1

F(j2

d

例求

2(d

解精选资料，欢迎下载

2

Sa2Sa)(t

(d

()tdt2十四、奇性

ft)F(jF(ej

(

当

f(t)

F(j(

R((X(

ft)

f(tf)

j

X0

(实函)

ft)

f()()

FjR0

(虚奇函)当

f(t

f(tt)

F(F(

RX(X

3-3所示。精选资料，欢迎下载

名称

域af(t)t)12F(jt)

域(jbFj12(

f(

(

)

f()

Fj)a

f()0

F(j

)

j

e

f(t

d

n

ft)

(j

n

F(j

n

tf(t)

(j)

n

d

n

F(jd

t

f(