第二章 (4)卷积积分的性质

实际上利用
推广4 推广
若f (t ) = f1(t ) f2(t ),则 f1(t t1 ) f2(t t2 ) = f1(t t2 ) f2 (t t1 ) = f (t t1 t2 )
∵ε (t ) ε (t ) = tε (t )
∴ε (t + 3) ε (t 5) = (t 2)ε (t 2)
f 2 (t t 2 )
f (t t1 t 2 )
0
t1
t
0
t2
t
0
t1+ t2
t
f1 (t t 2 )
f 2 (t t1 )
f (t t1 t 2 )
0
t2
t
0 t1
t
0
t1+ t2
t
图 2.4-6
计算下列卷积积分: 例2.4-2 计算下列卷积积分:
(1) ε(t + 3) ε(t 5)
0 α(t τ )
ε (t τ )dτ
(1 eαt )ε (t )
dτ =
1
α
分配律的应用
f1(t ) [ f2 (t ) + f3 (t )] = f1(t ) f2 (t ) + f1(t ) f3 (t )
f (t )
h1(t )
+ +
∑
y f (t ) f (t )
∵ f (t ) h (t ) + f (t ) h2(t ) = f (t ) [h (t ) + h2(t )] 1 1
(2) e ε(t + 3) ε(t 5) 2t e ε (t + 3) ε (t 5) ∞ 2τ = ∫ e ε (τ + 3) ε (t τ 5)dτ ∞
2t
=∫
t 5
3
e
2τ
1 2(t 5) 6 e = e 2 6 1 2( t 2) = e 1 e 2
[
1 2τ dτ = e 2
f 1 (t )
f 2 (t )
2
1
0
2
0 1
1
2 3
t
1
3
t
解法一: 解法一:图示法
f 1 (τ
t <1 ,
)
f (t ) = 0
2
0
1
2 3
τ
f 2 (t τ
t2
)
1
t 0
1
τ
解法一: 解法一:图示法
f 1 (τ
t <1 ,
)
t
f (t ) = 0
1< t < 2 ,
f (t ) = ∫ 2dτ = 2(t 1)
t
图 2.4-4函数与冲积函数的卷积 函数与冲积函数的卷积
f (t t1 )
δ (t t 2 )
f (t t1 t 2 )
0
t1
t
0
t2
t
0
t1+ t2
t
f (t t 2 )
δ (t t1 )
f (t t1 t 2 )
0
t2
t
0
t1
t
0
t1+ t2
t
图 2.4-5
f1 (t t1 )
若f (t ) = f1(t ) f2(t ),则 f1(t t1 ) f2(t t2 ) = f1(t t2 ) f2 (t t1 ) = f (t t1 t2 )
推广4 推广
推广4 推广
若f (t ) = f1(t ) f2(t ),则 f1(t t1 ) f2(t t2 ) = f1(t t2 ) f2 (t t1 ) = f (t t1 t2
积分: 积分:
f
( 1)
(t ) = f
( 1) 1
(t ) f2 (t ) = f1(t ) f
( 1) 2
( 1) 2
(t )
推论: f 推论:(t ) = f (t ) f
(1) 1
(t ) = f (t )
( 1) 1
(t ) f (t )
(1) 2
f (t ) = f
(i )
( j) 1
结论:并联系统的冲激响应, 结论:并联系统的冲激响应,等于组成并联系 统的各个子系统冲激响应之和. 统的各个子系统冲激响应之和.
h2(t )
h (t ) + h2 (t ) 1
h(t ) =
y f (t )
结合律及其应用: 结合律及其应用:
[ f1(t ) f2 (t )] f3 (t ) = f1(t ) [ f2 (t ) f3 (t )] f (t ) y f (t ) h (t ) h (t ) 2 1
= ∫ ε (τ ) e =∫ e
0 α(t τ )
ε (t τ )dτ
(1 eαt )ε (t )
dτ =
1
α
2 例 .4 1 设 1(t ) = eαtε (t ),f2 (t ) = ε (t ),分别求: : f f1(t ) f2 (t )和f2 (t ) f1(t ).
解: f1(t ) f2(t ) = ∫
∞ ∞ ∞ ∞ t
f1(τ ) f2(t τ )dτ
= ∫ eατε (τ ) ε (t τ )dτ =∫ e
0 ατ
dτ =
1
f2(t ) f1(t ) = ∫
∞ ∞
α
(1 eαt )ε (t )
∞ ∞ t
f2(τ ) f1(t τ )dτ
α(t τ )
= ∫ ε (τ ) e =∫ e
解: f1(t ) f2(t ) = ∫
∞ ∞ ∞wenku.baidu.com∞ t
f1(τ ) f2(t τ )dτ
= ∫ eατε (τ ) ε (t τ )dτ =∫ e
0 ατ
dτ =
1
f2(t ) f1(t ) = ∫
∞ ∞
α
(1 eαt )ε (t )
∞ ∞ t
f2(τ ) f1(t τ )dτ
α(t τ )
' ∞ ∞
上式称为杜阿密尔积分. 上式称为杜阿密尔积分. 杜阿密尔积分 其物理含义为: 其物理含义为:LTI系统的零状态响应等于激励的 系统的零状态响应等于激励的
f ' (t )与系统的阶跃响应 g(t )的卷积积分. 的卷积积分. 导数
例2.4-4 求图示函数 f1(t ) 与 f2 (t ) 的卷积 f (t ) .
(t ) f
(i j ) 2
LTI系统的零状态响应等于激励与系统冲激响应的卷积 系统的零状态响应等于激励与系统冲激响应的卷积 积分,利用上面的结论可得: 积分,利用上面的结论可得:
y f (t ) = f (t ) h(t ) = f (1) (t ) h(1) (t ) = f (1) (t ) g(t ) = ∫ f (τ )g(t τ )dτ
推广1 推广
f (t ) δ (t t1 ) = δ (t t1 ) f (t ) = f (t t1 ) 若 f (t ) = δ (t t1 ) 得:
推广2 推广 推广3 推广
δ ( t t1 ) δ ( t t 2 ) = δ ( t t1 t 2 )
f (t t1 )δ (t t2 ) = f (t t2 )δ (t t1 ) = f (t t1 t2 )
f (t )
h (t ) 2
h (t ) 1
y f (t )
f (t )
h(t ) = h (t ) h2 (t ) 1
y f (t )
∵[ f (t ) h (t )] h2(t ) = f (t ) [h (t ) h2(t )] 1 1 [ f (t ) h2(t )] h (t ) = f (t ) [h2(t ) h (t )] 1 1 = f (t ) [h (t ) h2(t )] 1
δT (t )表示
δT (t ) =
m=∞
∑δ (t mT )
∞
式中 m为整数 .函数 fo (t ) 如图(b)所示,试求 : 如图( 所示 所示, 为整数
δ T (t )
f (t ) = fo (t ) δT (t )
f 0 (t )
-2T -T 0 T 2T
t
(a)
(b)
0
t
解:
δ T (t )
∴若f (t ) = f1(t ) f2(t ),则 f1(t t1 ) f2(t t2 ) = f1(t t2 ) f2(t t1 ) = f (t t1 t2 )
f (t )
δ (t )
f (t )
0
t
0
t
0
t
f (t )
δ (t t1 )
f (t t1 )
0
t
0
t1
t
0
t1
∞ ∞
f (t ) δ (t t1 ) = δ (t t1 ) f (t ) = f (t t1 )
f (t ) δ (t t1 ) = δ (t t1 ) f (t ) = ∫ δ (τ t1 ) f (t τ )dτ = f (t t1 )
∞ ∞
f (t ) δ (t ) = δ (t ) f (t ) = f (t )
复习
1,卷积的图解法求解 2,卷积的解析法求解
第四节 卷积积分的性质 一,卷积的代数运算
交换律 f1(t ) f2 (t ) = f2 (t ) f1(t )
下面我们来看一道例题. 下面我们来看一道例题.
2 例 .4 1 设 1(t ) = eαtε (t ),f2 (t ) = ε (t ),分别求: : f f1(t ) f2 (t )和f2 (t ) f1(t ).
t 5 3
]
[
]
上式适用于
t 5 ≥ 3 t ≥ 2
∴e ε (t + 3) ε (t 5) 6 1 2( t 2) = e 1 e ε (t 2) 2
2t
[
]
(2) e2tε(t + 3) ε(t 5)
1 2t ∵e ε(t ) ε(t ) = 1 e ε(t ) 2
2t
(
)
∴e ε (t + 3) ε (t 5)
解:1) ε (t + 3) ε (t 5) = (
(2) e ε(t + 3) ε(t 5)
2t
∫ ε(τ + 3) ε(t τ 5)dτ = ∫ dτ = t 2
∞ ∞ t 5 3
上式适用于
t 5 ≥ 3 t ≥ 2
∴ε(t + 3) ε(t 5) = (t 2)ε(t 2)
2t
=e e
6
6 2( t +3)
ε (t + 3) ε (t 5)
1 2( t 2) = e 1 e ε (t 2) 2
[
]
下页图( 画出了周期为 画出了周期为T的周期性单位冲激 例2.4-3 下页图(a)画出了周期为 的周期性单位冲激 函数序列,可称为梳状函数, 函数序列,可称为梳状函数,它可用符号 它可写为: 它可写为:
结论:串联系统的冲激响应, 结论:串联系统的冲激响应,等于组成串联系统 的各个子系统的冲激响应的卷积. 的各个子系统的冲激响应的卷积.
二,函数与冲激函数的卷积
f (t ) δ (t ) = δ (t ) f (t ) = f (t )
f (t ) δ (t ) = δ (t ) f (t ) = ∫ δ (τ ) f (t τ )dτ = f (t )
f (t ) = fo (t ) δT (t ) ∞ = fo (t ) ∑δ (t mT ) m=∞
-2T -T 0 T 2T
t
(a)
f 0 (t )
=
m=∞
∑[ f (t ) δ (t mT )]
o
∞
=
m=∞
∑ f (t mT )
o
∞
(b) f 0 (t ) δ T (t )
T 0 T
0
t
t
(c)
图 2.4-15 δT (t )与 fo (t )的卷积 与 的卷积
三,卷积的微分与积分
f (t ) ,用符号 f (1) (t ) 表示其一阶导数, 对于任一函数 表示其一阶导数, f (1)表示一次积分,即 (t ) 表示一次积分, 用符号
df (t ) f (t ) = dt
(1)
Def
f
( 1)
(t ) =
Def
∫
t
∞
f ( x)dx
其中: 其中: f
( 1)
( ∞) = 0
若
f (t ) = f1(t ) f2 (t ) = f2 (t ) f1(t )
f (t ) = f (t ) f2 (t ) = f1(t ) f (t )
(1) (1) 1 (1) 2
则其导数: 则其导数:
1
2
f 2 (t τ
)
0
1
2 3
τ
1
t2 0
t
τ
1
解法一: 解法一:图示法
f 1 (τ
)
τ
t <1 , f (t ) = 0 1< t < 2 ,
f (t ) = ∫ 2dτ = 2(t 1)
t 1
2
2< t < 3
f (t ) = ∫
t-1 1
0
1
2
3
f 2 (t τ
)
t2
(- 2) dτ + ∫t2dτ 1
证明: 证明: f1(t t1 ) f2(t t2 ) =
令τ t1 = x则
∞
∫∞ f1(τ t1) f2(t τ t2 )dτ
∞
= f (t t1 t2 )
同理: 同理:
原式 = ∫ f1( x) f2 (t x t1 t2 )dx
∞
f1(t t2 ) f2(t t1) = f (t t1 t2 )
t
1
0 1
t
τ
= 2(t 2) + 2 = 2(t 3)
解法一: 解法一:图示法
f 1 (τ
)
2
t < 1 , f (t ) = 0 1< t < 2 , t f (t ) = ∫ 2dτ = 2(t 1)
3
4 τ