傅里叶变换习题

f (t ) = a 0 +
∑ (a
n =1
∞
n
cos n ω 0 t + b n sin n ω 0 t )
其中an,bn分别用p89(3-3）,（3-4）表达 均方误差En为
E
n
N 1 N 2 2 2 = f 2 (t ) − a0 + ∑ (a1n + b1n ) − ∑ (a1n an + b1n bn ) 2 n =1 n =1
0
例三：p170.3 − 46
f (t)
E
−T − T − τ
2
τ T
2 2
T
t
2
解：单个梯形脉冲的傅立叶变换为（380.附录3）
8E (T + τ )ω (T − τ ) F1 (ω ) = 2 sin sin ω (T − τ ) 4 4 (T + τ ) E (τ + T )ω (T − τ ) = Sa[ Sa ] 2 4 4
− 2 n
2
2
1 N 2 2 + ∑(an + bn )] 2 n=1
证明:设 证明 设
1 2 其 : f (t) = ∫ f (t )dt 中 TT
2
sN (t) = a0 + ∑(a1n cos nω1t + b1n sin nω1t)
N
ε N (t) = f (t) − sN (t)
EN = ξn
*冲激抽样
∞
1 ∞ f (t ) ∑ δ (t − nTS ) ↔ ∑ F (ω − nω s ) Ts n = −∞ N = −∞ ∞ ωm f (t ) = ∑ f ( nT ) Sa[ω m (t − nT )] (内插公式) N = −∞ π
例五 : 对一最高频率为400 Hz的带限信号 f (t )抽样 要使抽样信号通过一理 想低通滤波器后，能 完全恢复 f (t ), 1.抽样间隔 T应满足何条件？ 2.若以 T = 1ms抽样，理想低通滤波器 的截止频率 f c应满足何条件？
E cos ω 0 t ↔ E π [δ (ω + ω 0 ) + δ (ω − ω 0 )]
E sin ω 0 t ↔ jE π [δ (ω + ω 0 ) − δ (ω − ω 0 )]
*抽样信号的傅立叶变换
1 ∞ f s ( t ) = s( t ).p( t ) ↔ ∑ p n F( ω − n ω s ) Ts n = −∞ 1 − jn ω s t pn = dt ∫T s p ( t ) e Ts
1 1 f = = = 50µs...then...v2 (t) = 0 −6 T 20 ×10
π
3 . if .τ = 15 µ s , T = 30 µ s
1 1 1 f = = = × 100 kHz −6 T 30 × 10 3
4 v3 (t ) = sin 3ω 0 t × 100 k Ω ≈ 42 . 4 sin 3ω 0 t 3π
1 = ∫ T ε T −2
T 2
2 N
( t ) dt
1 N 1 N 1 N 2 2 2 2 = f (t ) − a + ∑ (a1n − an ) + ∑ (b1n − bn ) − ∑ (an + bn ) 2 n =1 2 n =1 2 n=1
2 2 0
若: EN > 0.Then.a1n = an.b1n = bn
∑ (a
n =1
N
1n
cos n ω 0 t + b1n sin n ω 0 t )
ε N (t) = f (t) − sN (t)
ε N (t ) = [ f (t ) − a0 − (∑ a1n cos nω0t + b1n sin nω0t )]2
2 n =1 N
注意在要求均方误差En的过程中
f (t )e
jω 0 t
↔ F (ω − ω 0 )
j f (t ) sin ω 0 t ↔ [ F (ω + ω 0 ) − F (ω − ω 0 )] 2 1 f (t ) cosω 0t ↔ [ F (ω + ω 0 ) + F (ω − ω 0 )] 2
df (t ) ↔ ( jω ) F (ω ) dt
''
2π 2π T f (t ) = E cos t [ u ( t ) − u ( t − )] T T 2
'
2π T + E [ δ ( t ) − δ ( t − )] T 2
2π 2 2π T '' Q( ) f (t ) + f (t ) = E [δ ( t ) − δ ( t − )] 2 T T
1 f 1 ( t ). f 2 ( t ) ↔ F1 (ω ) ∗ F 2 (ω ) 2π
2π T f ( t ) = E sin t[u ( t ) − u ( t − )] T 2
2π 2 2π T f (t ) = − ( ) E sin t[u (t ) − u (t − )] T T 2
E
f (t)
E (T + τ ) 2
F( jω)
T τ − − 2 2
τ
2
T 2
t
ω
4π T +τ
f1(t)
−T E
f (t) ∗ f1(t)
T
π0 π − τ τ
F (ω)F1(ω)
ω
−T − T τ −
2
τ
2
2
T 2
T
t
π − τ
π τ
ω
例四： 例四：P164.3-13 i1 (t )
i1(t)
1 由题意， 解：.由题意， f ( t )的最高频率 f m = 400 Hz 1 1 则奈奎斯特抽样间隔为 = = 1.25 ms 2f c 2 × 400 所以抽样间隔应满足 T ≤ 1.25 ms
2. 已知抽样间隔T = 1ms,则抽样频率 1 fc = = 1KHz T 由抽样定理，fc 应满足 f m < fc < f s − f m 即 400Hz < fc < (1000 − 400)Hz = 600Hz
1 u (t) ↔ + πδ ( ω ) jω 2 sgn( t ) ↔ jω
G ( t ) = E [u ( t + ) − u ( t − )] ↔ EτSa ( ) 2 2 2
τ
τ
ωτ
Ee
Ee
−α t
E ↔ α + jω
(α 为正实数 )
jω 0 t
↔ 2π E δ (ω − ω 0 ) (ω0为正实数)
1当τ = 5µs T = 10µS .
bn = 4 nπ
0
n(even)
n(odd)
2π ω0 = T
1 1 5 f0 = = = 10 = 100kHz −6 T 10 ×10 4 v2 (t) = sin ω0t ×100kΩ ≅ 127.3sin ω0t
2.if .τ = 10µs, T = 20µs
dF (ω ) ( − jt ) f ( t ) ↔ dω
d F ( − jt ) f ( t ) ↔ dω
n
d f n ↔ ( jω ) F (ω ) dt n
n
−∞
∫
t
t
F (ω ) f (τ ) d τ ↔ + π F ( 0 )δ ( ω ) jω
−∞
∫
F (ω ) f (τ ) d τ ↔ ..( if , F ( 0 ) = 0 ) jω
三.傅立叶变换的定义 傅立叶变换的定义
F( jω) = ∫ f (t)e − jωt dt, F(0) = ∫ f (t)dt
1 1 jωt f (t) = ∫∞F( jω)e dω, f (0) = 2π −∫∞F( jω)dω 2π −
−∞ ∞ −∞
∞
∞
∞
典型信号的 傅立叶变换
δ (t ) ↔ 1 1 ↔ 2πδ (ω )
0
v2 (t )
τ
T
T
Q i1 ( t ) = − i1 ( − t )
bn 2 = T
∴ a0 = 0, a n = 0
∫i
0
1
( t ) sin n ω 0 tdt
T
τ 2 = [1 − cos( 2nπ )] nπ T
2 = ( ∫ sin n ω 0 tdt − T 0
τ
∫ τ
sin n ω 0 tdt )
付立叶级数(三角形式 二.付立叶级数 三角形式 付立叶级数 三角形式)
f (t) = a0 + ∑(an cos nω1t + bn sin nω1t)
n=1 ∞
1 a0 = ∫ f (t)dt T T
2 an = ∫ f (t) cos nω1tdt T T
2 bn = ∫ f (t) sin nω1tdt T T
2
n=1
1 (t ) = T
T 2
∫ξ
2 N
(t )dt
∂EN =0 ∂a n
∂EN =0 ∂bn
T − 2
∂E ∂a
N
0
∂E N ∂E = = .... = ∂a 1 ∂a
N n
= 0
设
∂E N ∂E N ∂E N = = .... = = 0 ∂ b1 ∂ b2 ∂ bn
S N (t ) = a 0 +
2π 2 2π 2 ( ) F (ω ) + ( j ω ) F (ω ) = E (1 + e T T