第7章3傅氏变换性质.

i
ℱ[H (t
t0 )]
ei t0
[1
i
( )]
ℱ[H (t
t0 )]
ei t0 [ 1 i
( )]
2
象函数的位移性质的应用1
如果 ℱ[ f (t)] F ( ) 则 ℱ[ ei0t f (t) ] F ( 0 )
ℱ[ ei0t f (t )] F ( 0 )
例1 ℱ[1] 2 ( ) ℱ[ei0 t] 2 ( 0 )
153页14.
已知 ℱ[ H (t)] 1 ( ) i
ℱ[et H (t )]
1
i
求下列函数的Fourier变换
(7) f (t ) et sin0t H (t )
解 f (t ) 1 [ ei0t e t H (t ) ei0t e t H (t )]
2i
ℱ[ f (t )] 1 { ℱ[ei0t et H (t )] ℱ[ei0tet H (t ) ]}
146页例7.9
i( 0 )
( 0 )
ℱ[H (t)cos0t]
153页14(6)
i(
2
2 0
)
2
[
(
0
)
(
0
)]
ℱ[H
(t
)
sin0
t
]
0 2 02
[
2i
(
0
)
(
04)]
153页14
已知 ℱ[ H (t)] 1 ( ) i
ℱ[et H (t )]
1
i
求下列函数 的Fourier变换
(4)
in
1
[ i
( )](n)
(3H) ℱ(t[)f(
1
t)]
0
t0
ℱ[tet H
t0
(t
)]iine[[(it1H10)nin(nt1)!]1(3e09(页nt )1(ti例t))70]2.2010
153页9(1). 求符号函数
1 sgn t
t0
的傅氏变换
1 t 0
1 t0 解 H(t)
t
dt
x at
t x
i x 1
f (x) e a d x
a
a
1
i xt
f ( xt ) e a d xt
a
1
F( )
1
F( )
a a |a| a
若 a 0 则也正确
7
翻转性质 如果 ℱ[f (t)] F ( )
则 ℱ[ f (t)] F ()
证明
因为 ℱ[ f (at)]
1
Leabharlann Baidu
F( )
|a| a
所以令 a 1 结论正确
例5. 计算 ℱ[H (t)]
解
ℱ[H (t]) 1 ( )
i
ℱ[H(t)] 1 ( ) i( )
1 ( )
i
8
6.象函数微分性质 如果 ℱ[ f (t ) ] F ( ) 则 ℱ[t n f (t )] i n [F ( )](n)
证明 F (ℱ)[tf(ft()t])e ii [tFd(t=ℱ)][f (t)]
2i
1 2i
[
1
i(
0 )
1
i(
0 ) ]
(
0 i )2 02
6
4. 相似性质 如果 ℱ[f (t)] F ( )
a0
153页11. 则 ℱ[ f (at)] 1 F ( )
|a| a
证明
ℱ[f (t)]
f
(t)
e i
t
dt
F ( )
若 a 0 则
ℱ[ f (at)]
f
(at )ei
(1)0
)
原函数的位移性质的应用
ℱ[ f (t t0 )] ei t0 ℱ[f (t )] ℱ[ f (t t0 )] ei t0 ℱ[f (t )]
例1 ℱ[ (t )] 1 ℱ[ (t t0 )] ei t0 ℱ[ (t )] ℱ[ (t t0 )] ei t0
例2
ℱ[H (t)] 1 ( )
0 t0
1 t 0 H(t)
0 t 0
sgn t
H (t) H (t)
ℱ[ei0 t ] 2 ( 0 )
ℱ[cos0
t]
cos0 [ (
t
1 ( ei0 t 2
0 ) (
ei0 t )
0 )]
例7.8
ℱ[sin0 t] i [ ( 0 ) ( 0 )]
ℱ[f
(t)
cos0
t]
1 2
[
ℱ[
f
(t)
ei0 t] ℱ[
f
(t)
ei0 t ]]
ℱ[f (t)sin0t ] 212i[[FF((00))F( 0 )]
§7.3 傅氏变换的性质
对 f (t) 进行傅氏变换 ℱ[f (t)] f (xt) ei xt dxt F ( )
1.线性性质 ℱ[ a f (t)] a ℱ[f (t)]
ℱ[f (t) g(t)] =ℱ[ f (t)] +ℱ[g(t)]
2.原函数的位移性质 ℱ[ f (t t0 )] ei t0 ℱ[ f (t ) ] ℱ[ f (t t0 )] ei t0 ℱ[ f (t )]
ℱ[ t n f (t )] i n F (n)( )
9
153页14.
已知 ℱ[ H (t)] 1 ( ) i
ℱ[et H (t )]
1
i
求下列函数的Fourier变换
t t0
(1)
f (t)
0 t0
解
ℱ[f
(t )] ℱ[t
H (t )]
i
[
1
i
( )]
1
2
i
( )
(2) ℱ[f (t)]ℱ[t n H (t)]
ℱ[
f
(t )]
ℱ[ ]e
i0t
H(t)
1
i( 0 )
(
0 )
(5) f (t ) e H i0t (t t0 )
解
ℱ][H(t t0 )
eit0 [ 1 ( )] i
ℱ[f (t )]ℱ[ei0t H(t t0 )]
e i( 0
[ )t0
1
i(
0
)
(
0
)]
147页例7.9(3)
5
3.象函数的位移性质 ℱ[ ei0t f (t )] F ( 0 ) ℱ[ ei0t f (t )] F ( 0 )
证明
ℱ[fe(it0t ft(0t))]
efi(t0tf t(0t)
eeii tt
dt
x t t0
t x t0
f
f(
x(t))
eeii(x
e0i)tt0ddtxe
iFt0(F
F (n)( )
ℱf[(tt)2(fe(ti)]t )(ni)
2
d
[tF
(
)]
ℱ[ft(8tf)((ti)]t )nei8i[tFd t( )](8)
(i)n
tn
f
(t)
e i
t
dt
(i)n ℱ[t n f (t)]
i n F (n)( ) i n (i)n ℱ[t n f (t )]
3
象函数的位移性质的应用2
若 ℱ[ f (t)] F ( ) 则 ℱ[ ei0t f (t)] F ( 0 )
例2
ℱ[H (t)]
1
ℱ[ e i0t ( )
f (t)] F ( 0 )
i
所以
ℱ[e H i0t (t ) ] 1 i( 0 )
( 0 )
ℱ[e H i0t (t )] 1