信号与系统基础21 7新

7.5

离散冲激响应第七章第3讲1输入信号为离散冲激d(k)时离散系统的零状态响应，称为离散冲激响应h(k).它同连续系统中的冲激响应h(t)有相同的地位和作用。冲激响应为利用“卷积和”求解任意输入的零状态响应提供了极为有效的方法。通常使用冲激响应和阶跃响应来评价离散系统的时域性能。冲激响应的求法第七章第3讲2先考虑单个d(k)作用系统的情况:对于3阶后向差分方程a3

y(k

)

+

a2

y(k

-1)

+

a1

y(k

-

2)

+

a0

y(k

-

3)

=

d(k

)其响应记为h0(k)，则：a3h0

(k

)

+

a2

h0

(k

-1)

+

a1h0

(k

-

2)

+

a0

h0

(k

-

3)

=

d(k

)对于k

>0

时，由于d

(k

)=0，因而上式为a3h0

(k

)

+

a2

h0

(k

-1)

+

a1h0

(k

-

2)

+

a0

h0

(k

-

3)

=

0冲激响应h0

(k

)是一个特殊的零输入响应h

(k

)

=

C

lk

+

C

lk

+

C

lk0

1

1

2

2

3

3冲激响应的求法对于三阶后向差分方程的初始值令k=0

时，有a3h0

(0)

+

a2

h0

(-1)

+

a1h0

(-2)

+

a0

h0

(-3)

=

d(0)

=1由于是零状态响应，h0

(k

)=0,k

<0

。所以h0

(-1)

=

h0

(-2)

=

=

h0

(-n

+1)

=

0na0h

(0)

=

1n个常数Ci可由以上n个初始值确定。30a\

h

(0)

=

13第七章第3讲301故得等效初始值：

h0(-1)=h0(-2)=0,

h

(0)

=

a对于n阶后向差分方程的初始值冲激响应的求法对于多输入系统根据叠加定理再根据线性系统的时不变特性：线性非时变离散系统（零状态）d(k)第七章第3讲4h0(k)h0(k-n)Ah0(k-i)d(k-n)Ad(k-i)对于前向差分方程，可将其转换成后向差分方程再用上述方法求解冲激响应。例7.11已知系统的差分方程为试求系统的离散冲激响应h(k)。解：先只考虑d(k)的作用：6

6y(k

)

-

5

y(k

-1)

+

1

y(k

-

2)

=

d(k

)

-d

(k

-

2)6

600

0h

(k

)

-

5

h

(k

-1)

+

1

h

(k

-

2)

=

d

(k

)1+

=6

6特征方程：l2

-5

l2

31

20

解得：

l

=

1

,

l

=

1k第七章第3讲5k132+

C

(

)120

1\

h

(k

)

=

C

(例7.1113120k

ke(k

)(

-

2

(

)]\

h

(k

)

=[3系统的单位冲激响应为h(k

)

=

h0

(k

)

-

h0

(k

-

2)等效初始条件：h0

(0)=1,h0

(-1)=0将初始条件代入原方程，得：C1+C2=12C1+3C2=0解得：C1=3，C2=-2]e(k

-

2)31123112=[3(

-

2

(

)]k

-2k

-2k

ke(k

)

-[3(

-

2

(

)k第七章第3讲6k132+

C

(

)120

1\

h

(k

)

=

C

(序列表达式的化简]e(k

-

2)112

3321

1(

-

2

(

)k

-2k

-2k

ke(k

)

-[3由于e(k

-

2)

=

e(k

)

-d(k

)

-d(k

-1)h(k

)

=[3(

-

2

(

)]化简(][e(k

)

-d(k

)

-d(k

-1)]31-

2

(

)k

-2k

-212]e(k

-

2)

=[33112[3(

-

2

(

)k

-2

k

-2(k

)

+

6d(k

-1)]e d(k

)

+

03112=[3(

-

2

(

)k

-2k

-2d(k

)第七章第3讲7]e(k

)

-

6112

31312\

h(k

)

=[3(k(

+

2

(

)-

2

(

)

-

3k

-2k

-2k=

-6d

(k

)

+[-9(

1

)k

+16(

1

)k

]e(k

)2

3表达式为例7.12第七章第3讲8已知系统的差分方程为y(k

+

2)

-

5

y(k

+1)

+

6

y(k

)

=

f

(k

+

2)

-

3

f

(k

)试求系统的离散冲激响应h(k)。解:首先将前向差分方程转换成后向差分方程y(k

)

-

5

y(k

-1)

+

6

y(k

-

2)

=

d(k

)

-

3d(k

-

2)先只考虑d(k)的作用：h0

(k

)

-

5h0

(k

-1)

+

6h0

(k

-

2)

=

d(k

)特征根：l1

=

3,

l2

=

20

1

2k

ke(k

)\

h

(k

)

=[C

(3

+

C

(2

]例7.12第七章第3讲90

1

2kk2

]e(k

)\

h

(k

)

=[C

(3

+

C

(将初始条件代入原方程，得C1

+

C2

=

11

C1

+

1

C2

=

03

2=

-2解得，C1

=

3

C2h0

(k

)

=

[3(3

-

2(2

]e(k

)k

kh0

(-1)

=

0,

h0

(0)

=

1系统的初始条件为例7.12根据系统的非时变性质，系统的冲激响应为y(k

)

-

5

y(k

-1)

+

6

y(k

-

2)

=

d(k

)

-

3d(k

-

2)h(k

)

=

h0

(k

)

-

3h0

(k

-

2)=

[3(3)k

-

2(2)k

]e(k

)

-

3[3(3)k

-2

-

2(2)k-2

]e(k

-

2)第七章第3讲102

3=

[3(3

-

2(2

]

-[(

-

((k

-

2)k

k

k

ke(k

)

3

2

]e=[2(3)k

-

0.5(2)k

]e(k

)

-

0.5d(k

)d(k

)]e(k

)

-

0.52

3kk

k

k(3

-

2(2

]e(k

)

-[(3

-

(2h(k

)

=[3将上式化简,得阶跃响应与冲激响应的关系第七章第3讲11e(k

)

=

e(k

)

-

e(k

-1)g

(k

)

=

g

(k

)

-

g

(k

-1)

d(k

)

=\

h(k

)

=k\

g

(k

)

=

h(i)i=0

e(k

)

=

d(i)i=0k2第七章第3讲12已知阶跃响应为g

(k

)=6[1

-0.5(1

)k

]e(k

)，求离散冲激响应h(k)。解：离散冲激响应为h(k

)

=

g(k

)

-

g(k

-1)1212=

6

[1-

0.5(

)

](k

-1)k

-1ke(k

)

-

6

[1-

0.5(

)

]e=

6

[1-

0.5(

1

)k

]e(k

)

-

6

[1

-(

1

)k

]e(k

)2

22=

3(

1

)k

e(k

)例7.13例7.14，求阶跃响应。(3)

]1361ke(k

)-

0.5(2)k

+已知离散冲激响应

h(k

)

=[

d

(k

)解：阶跃响应为6

i

=0k

k

k

kg

(k

)

=

h(i)

=

1

d(k

)

-

1

(2)i

+

1

(3)ii

=0=

[

1

-

2k

+

1

3k

]e(k

)2

2e(k

)3

1

-

32

1

-

2e(k

)

-16e(k

)

+3

i

=01

1

-

3k

+12

i

=01

1

-

2k

+1=注：有限等比数列求和公式：式中：a1为首项，an为末项，q为比例系数。n第七章第3讲13a

-

qaS

=1-

q17.6

离散卷积卷积和的意义任意离散信号可分解为d(k)的线性组合：f

(k)=······+f(-1)d(k+1)+f

(0)d(k)+

f

(1)d(k-1)+······+

f

(i)d(k-i)+······¥f

(k

)第七章第3讲14kf

(i)d

(k

-

i)-1

0

1

2

3i¥=

f

(i)d(k

-

i)

=

f

(k

)

*d(k

)i

=-¥定义：f1

(k

)

*

f2

(k

)

=

f1

(i)

f2

(k

-

i)i

=-¥¥=

f2

(i)

f1

(k

-

i)i

=-¥称离散卷积或卷积和任意激励信号的零状态响应线性非时变离散系统（零状态）d(k)h

(k)d(k-n)h

(k-n)Ad(k-i)Ah

(k-i)第七章第3讲15任意信号：¥f

(k

)

=

f

(i)d(k

-

i)i

=-¥=

f

(k

)

*d(k

)¥零状态响应：yzs

(k

)=

f

(i)h(k

-

i)i=-¥=

f

(k

)

*h(k

)¥系统的零状态响应：yzs

(k

)

=

+

f

(-1)h(k

+1)

+

f

(0)h(k

)

+

f

(1)h(k

-1)

+

+

f

(i)h(k

-

i)

+=

f

(i)h(k

-

i)

=

f

(k

)

*h(k

)i

=-¥卷积和的性质交换律、分配律、结合律与卷积一样。kf1(k)、f2(k)均为因果序列，则

f1

(k

)

*

f2

(k

)

=

f1

(i)

f2

(k

-

i)i

=0¥kf1(k)为因果序列，f2(k)为一般序列，则f1

(k

)

*

f2

(k

)

=

f1

(i)

f2

(k

-

i)i

=0f1(k)为一般序列，f2(k)为因果序列，则f1

(k

)

*

f2

(k

)

=

f1

(i)

f2

(k

-

i)i

=-¥i0k第七章第3讲16卷积和的性质第七章第3讲17f(k)与d(k)的卷积和：f

(k

)*d(k

)

=

f

(k

),

f

(k

)*d(k

-

k0

)

=

f

(k

-

k0

)kf

(k

-

n)

*d

(k

-

m)

=

f

(k

-

n

-

m)f(k)与e(k)的卷积和：f

(k

)

*e(k

)

=

f

(i),i

=-¥k

-n

kf

(k

)

*e(k

-

n)

=

f

(i)

=

f

(i

-

n)i

=-¥

i

=-¥位移序列的卷积和：f1

(k-

n)

*

f2

(k

-

m)

=

f1

(k

-

m)*

f2

(k-

n)

=

f

(k

-

n

-

m)卷积和的计算图解法方法与连续系统的卷积类似例：求y(k)=f1(k)*

f2(k)f1

(k

)k-

2

-1

0

1

21f2(k)k10

1

2

332f1

(-i)i-

2

-1

0

1

21i01

23f2

(i)321-

2

-10k

<

-23k

=

-25k

=

-1y(k

)

=6k

=

0,

1,

23k

=

31k

=

40k

>

4k

=0时第七章第3讲18卷积和的计算第七章第3讲19不进位乘法法例：求y(k)=f1(k)*

f2(k)对于两个有限序列，可以利用一种“不进位乘法”较快地求出卷积结果。

2,

k

=

0,1,

2

0,

k

为其它f

(k

)

=1

0,k

=1,

2,

3k

为其它

k

2

,f2

(k

)

=

888+

222210282618f1

(k

)

:222f2

(k

)

:·149181818将两序列样值以各自的最高值按右端对齐，进行不进位乘法。对位排列如下：其中，两序列样值的最低值之和为卷积和序列的最低值，即起点为0+1=1。卷积和为y(k

)

={0,

2,

10,

28,

26,

18,

0}›卷积和的计算第七章第3讲20解析法方法与连续系统的卷积类似，用求和公式得解析式。例7.19(a)：求y(k)=e(k)*

e(k)k

k解：y(k

)=e(i)e(k

-i)=1

=(1

+k

)e(k

)i

=0

i

=0例7.191

-

0.5i

=01求y(k)=(0.5)k

e(k)*

[e(k)-e(k-5)]kke(k

)

=[2

-(0.5)

]e(k

)ki=

(0.5)

=1

-(0.5)k

+1ik(0.5)

e(i)e(k

-

i)e(k

)

=(k

)

*解：

y

(k

)

=

(0.5)

ei

=0]e(k

-

5)ky2

(k

)

=

(0.5)

e(k

)

*e(k

-

5)1=

y

(k

-

5)=[2

-(0.5)k

-5\

y(k

)

=

y1

(k

)

-

y2

(k

)=[2

-(0.5)k

]e(k

)

-[2

-(0.5)k

-5

]e(k

-

5)n第七章第3讲21a

-

qaa

=n

ii

=111

-

q例7.19第七章第3讲222求信号的卷积和：f

(k

)=(1

)k

e(-k

)*(3)k

[e(k

)-e(k

-2)][e(k

)

-

e(k

-

2)]

=

d(k

)

+

d(k

-1)2解：\

f

(k

)

=

(

1

)k

e(-k

)

*[d

(k

)

+

3d

(k

-1)]2=

(

1

)k

e(-k

)

*d

(k

)

+

(

1

)ke(-k

)

*3d

(k

-1)=

(=

(121221212(-k

)(-k

)(-k

+1)(-k

)kk

-1k

-1k[e

+d(k

-1)])

e

+

3(

)

e)

e

+

3(

)1212(-k

)

+

3(-k

)

+

3(

)k

k

-1d

(k

-1)=

(

)

e

e2=

7(

1

)k

e(-k

)

+

3d

(k

-1)例7.19i

=-¥

i

=k求信号的卷积和：f

(k

)=(2)k

e(-k

)*(3)k

e(-k

)¥

0解：f