例7-4-1n

=

0

y(0)=

3

y(-1)+1

=

1n

=

1

y(1)=

3

y(0)+

1

=

4n

=

2

y(2)=

3

y(1)+

1

=

13n

=

3

y(3)=

3

y(2)+

1

=

40

由遞推關(guān)系,可得輸出值：

?4,

13,

40,

n=0y(n)=

1,已知y(n)=3

y(n

-1)+u(n)，且y(-1)=0,求解方程n

=

0

y(0)=

C1

+

C2=

2n

=

1

y(1)=

2C1

+

3C2

=

1求解二階差分方程

y(n)-

5

y(n

-1)+

6

y(n

-

2)=

0已知y(0)=

2,

y(1)=

1。(r

-

2)(r

-

3)=

0r

2

-

5r

+

6

=

0r1

=

2,

r2

=

3y(n)=

C1

(2)

+

C

(3)n

n2解出C1

=

5,

C2

=

-3\

y(n)=

5(2)n

-

3(3)n例7-4-2特征方程特征根齊次解定C1,C2求方程y(n)+6

y(n

-1)+12

y(n

-2)+8

y(n

-3)=0的解特征方程(r

+

2)3

=

0r

3

+

6r

2

+

12r

+

8

=

0\r

=-2

三重根y(n)=

C1

(-

2)

+

C2n(-

2)

+

C3n

(-

2)n

n

2

n給定邊界條件即可求出常數(shù)C1

,

C2

,

C3例7-4-3例6-4-4設(shè)jjr=

Me

-

jjr

=

Me21(

)

(

)

(

)nn1

1

2

2+

C

ry

n

=

C

r=

C1

(Me

)

+

C

(Me

)jj

n

-

jj

n2=

C1

M

(cos

nj

+

j

sin

nj

)+

C

M

(cos

nj

-

j

sin

nj

)n

n2=

PM

n

cos

nj

+

QMn

sin

njP,Q為待定系數(shù)M

=

1M

>

1M

<

1y(n)為等幅正弦序列

y(n)為增幅正弦序列

y(n)為減幅正弦序列例7-4-5

求全解且y(-1)=1y(n)+

2

y(n

-

1)=

5u(n)

r

+

2

=

0

r

=

-2n

?0

時(shí)全為5（常數(shù)）

x(n)=

5u(n)\

y

p

(n)=

CC

+

2C

=

5 (n

?

0)3\

C

=

5代入原方程求特解(

)

(

)

(

)

(

)531+nphn

+

y

n

=

C

-

2y

n

=

yyh

(n)=

C1

(-

2)n齊次解特解全解形式由y(-1)=1迭代出n

=

0

y(0)

=

5

-

2

y(-1)

=

3(

)

(

)351+，得n代入 解

y

n

=

C

-

231y(0)=

3

=

C

+

531=

4\

C\

y(n)=

4

(-

2)n

+

5

n

?

03

3由邊界條件定系數(shù)例7-4-6LTIS

的差分方程y(n)+3

y(n

-1)+2

y(n

-2)=x(n)-x(n

-1)已知x(n)=

(-

2)n

u(n)

y(0)=

y(1)=

0求系統(tǒng)的零輸入響應(yīng)。零輸入響應(yīng)yzi(n),即當(dāng)x(n)=0時(shí)的解y(n)+

3

y(n

-1)+

2

y(n

-

2)=

0r

2

+

3r

+

2

=

0

r1

=

-2,

r2

=

-1yzi

(n)=

C1

(-

2)

+

C2

(-1)n

n求初始狀態(tài)（0-狀態(tài)）題目中

y(0)=

y(1)=

0,是激勵(lì)加上以后的,不能說明狀態(tài)為0,需迭代求出

y(-1),

y(-

2)

。n

=

1

y(1)+

3

y(0)+

2

y(-1)=

(-

2)u(1)+

(-

2)0

u(0)0

+

0

+

2

y(-1)=

(-

2)+

1

=

-1\

y(-1)=

-

12n

=

0

y(0)+

3

y(-1)+

2

y(-

2)=

(-

2)0

u(0)+

(-

2)-1

u(-1)0

+

3

y(-1)+

2

y(-

2)=

14\

y(-

2)=

5以y(-1),y(-2)代入方程:(

)

(

)(

)(

)

(

)

(

)

=5412-

1

=

--22-21-12-11+

C

-

1-

2

=

C

-

2y+

C-

1

=

C

-

2yzizi

1

C2

=

2

C

=

-3\

yzi

(n)=

-3(-

2)

+

2