PDF《第8章习题及解答》

0 )

0 ( = = x x (2)

0 ) (

2 )

1 (

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2 ( = + + ? + k x k x k x 初始条件:

1 )

1 ( ,

1 )

0 ( = = x x 题解: (1) ) ( ) (

2 )

1 (

3 )

2 ( k u k x k x k x = + + ? + , ) (

1 ) ( k k u = ,

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2 )

1 (

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2 ( [ k u Z k x k x k x Z = + + ? +

1 ) ( )

2 3 (

2 ? = + ? z z z X z z 得到输出离散序列的 z 变换为

1 )

2 3 (

1 ) (

2 ? + ? = z z z z z X )

2 ( )

1 (

2 ? ? = z z z 作z反变换得到差分方程的解为

96 nT nT T nT X

2 1 ) (

1 ) ( + ? = 迭代法:写出迭代式为 ) ( ) (

2 )

1 (

3 )

2 ( k u k x k x k x + ? + = + 迭代解为 k

0 1

2 3

4 5

6 ... x(k)

0 0

1 4

11 26

57 ... x(k+1)

0 1

4 11

26 57

120 ... x(k+2)

1 4

11 26

57 120

247 ... u(k)

1 1

1 1

1 1

1 ... 差分方程的解序列为 ... )

3 (

4 )

2 (

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1 (

0 ) (

0 ) ( + ? ? + ? ? + ? ? + ? = t t t t k x δ δ δ δ ... )

6 (

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5 (

20 )

4 (

9 ... + ? ? + ? ? + ? ? + t t t δ δ δ (2)

0 ) (

2 )

1 (

3 )

2 ( = + + ? + k x k x k x ,

1 )

1 ( ,

1 )

0 ( = = x x z 变换法:方程两边作 z 变换,代入初值有

0 )] (

2 )

1 (

3 )

2 ( [ = + + ? + k x k x k x Z

0 ) (

2 )]

0 ( ) ( [

3 )]

1 ( )

0 ( ) ( [

2 2 = + ? ? ? ? z X zx z zX zx x z z X z z z z X z z

2 ) ( )

2 3 (

2 2 ? = + ?

1 2

3 2 ) (

2 2 ? = + ? ? = z z z z z z z X 作........