# 一、判断系统是否 " 非时变 "

## 1、案例二

x

(

n

)

=

{

0

,

1

,

2

,

3

,

4

,

5

,

0

}

x(n) = { 0, 1 , 2, 3, 4, 5 , 0 }

x(n)={0,1,2,3,4,5,0} ,

n

n

n 取值

1

-1

1 ~

5

5

5

y

(

n

)

=

x

(

n

2

)

y(n) = x(n^2)

y(n)=x(n2) 的 " 变换 " 操作是否是 " 时不变 " 的 ;

y

(

n

)

=

x

(

n

2

)

y(n) = x(n^2)

y(n)=x(n2) 变换操作 :

y

(

n

)

y(n)

y(n) 只有在

n

=

1

,

0

,

1

,

2

n = -1 , 0 , 1 , 2

n=1,0,1,2 取值时 , 才有值 ,

n

=

3

n = 3

n=3 ,

n

2

=

9

n^2 = 9

n2=9 ,

x

(

9

)

x(9)

x(9) 没有值 ;

n

=

4

n = 4

n=4 ,

n

2

=

16

n^2 = 16

n2=16 ,

x

(

16

)

x(16)

x(16) 没有值 ;

n

=

5

n = 5

n=5 ,

n

2

=

25

n^2 = 25

n2=25 ,

x

(

10

)

x(10)

x(10) 没有值 ;

y

(

n

)

y(n)

y(n) 的取值是

n

=

0

,

1

,

2

n = 0 , 1 , 2

n=0,1,2 时的取值 ,

n

=

1

n = -1

n=1 时 ,

y

(

n

)

=

x

(

n

2

)

=

x

(

(

1

)

2

)

=

x

(

1

)

=

2

y(n) = x(n^2) = x((-1)^2) = x(1) = 2

y(n)=x(n2)=x((1)2)=x(1)=2 ;

n

=

0

n = 0

n=0 时 ,

y

(

n

)

=

x

(

n

2

)

=

x

(

0

2

)

=

x

(

0

)

=

1

y(n) = x(n^2) = x(0^2) = x(0) = 1

y(n)=x(n2)=x(02)=x(0)=1 ;

n

=

1

n = 1

n=1 时 ,

y

(

n

)

=

x

(

n

2

)

=

x

(

1

2

)

=

x

(

1

)

=

2

y(n) = x(n^2) = x(1^2) = x(1) = 2

y(n)=x(n2)=x(12)=x(1)=2 ;

n

=

2

n = 2

n=2 时 ,

y

(

n

)

=

x

(

n

2

)

=

x

(

2

2

)

=

x

(

4

)

=

5

y(n) = x(n^2) = x(2^2) = x(4) = 5

y(n)=x(n2)=x(22)=x(4)=5 ;

1

-1

1

1

1

1 的平方都为

1

1

1 , 合并成一个 ;

x

(

n

)

x(n)

x(n) 正常变换后的取值为 :

y

(

n

)

=

{

1

,

2

,

5

}

y(n) = { 1, 2, 5 }

y(n)={1,2,5}

### ① 时不变系统概念

y

(

n

m

)

=

T

[

x

(

n

m

)

]

y(n - m) = T[x(n-m)]

y(nm)=T[x(nm)]

### ② 先变换后移位

" 输出序列 " 进行移位 , 先 " 变换 "" 移位 " ;

x

(

n

)

x(n)

x(n) 进行 变换 操作 , 得到 输出序列

x

(

n

2

)

x(n^2)

x(n2) ,

x

(

n

2

)

x(n^2)

x(n2) 输出序列 进行移位

n

n

0

n - n_0

nn0 得到

x

(

(

n

n

0

)

2

)

x((n-n_0)^2)

x((nn0)2) ,

y

(

n

n

0

)

=

x

(

(

n

n

0

)

2

)

y(n - n_0) = x((n-n_0)^2)

y(nn0)=x((nn0)2)

y

(

n

)

=

{

1

,

2

,

5

}

y(n) = { 1, 2, 5 }

y(n)={1,2,5}

y

(

n

1

)

=

{

0

,

1

,

2

,

5

}

y(n-1) = { 0, 1, 2, 5 }

y(n1)={0,1,2,5}

### ③ 先移位后变换

" 输入序列 " 进行移位 , 先进行移位 , 将 " 输入序列

x

(

n

)

x(n)

x(n) " 先进行 " 移位 " 操作 , 得到 新的 " 输入序列 " 为

x

(

n

n

0

)

x(n-n_0)

x(nn0) , 然后 对新的输入序列进行 " 变换 " 操作 , 得到 " 输出序列 " ;

T

[

x

(

n

n

0

)

]

=

x

(

n

2

n

0

)

T[x(n - n_0)] = x(n^2 - n_0)

T[x(nn0)]=x(n2n0) , 变换时 , 只是将

n

n

n 值变为

n

2

n^2

n2 ,

n

0

n_0

n0 值不动 ;

x

(

n

n

0

)

x(n-n_0)

x(nn0) 变换时 , 只将

n

n

n 乘以

2

2

2 ,

n

0

n_0

n0 不变 , 变换结果如为

x

(

2

n

n

0

)

x(2n - n_0)

x(2nn0) ;

T

[

x

(

n

n

0

)

]

=

x

(

n

2

n

0

)

T[x(n - n_0)] = x(n^2 - n_0)

T[x(nn0)]=x(n2n0)

x

(

n

)

=

{

0

,

1

,

2

,

3

,

4

,

5

,

0

}

x(n) = { 0, 1 , 2, 3, 4, 5 , 0 }

x(n)={0,1,2,3,4,5,0} ,

n

n

n 取值

1

-1

1 ~

5

5

5 , 向右移位 , 移位后的序列 :

x

(

n

)

=

{

0

,

1

,

2

,

3

,

4

,

5

}

x(n) = { 0, 1 , 2, 3, 4, 5 }

x(n)={0,1,2,3,4,5}

n

n

n 取值

0

0

0 ~

6

6

6 , 移位后的序列图式如下 :

n

n

n 取值 由原来的

1

-1

1 ~

5

5

5 变为了

0

0

0 ~

6

6

6 ,

y

(

n

)

y(n)

y(n) 只有在

n

=

0

,

1

,

2

n = 0 , 1 , 2

n=0,1,2 取值时 , 才有值 ,

n

=

3

n = 3

n=3 ,

n

2

=

9

n^2 = 9

n2=9 ,

x

(

9

)

x(9)

x(9) 没有值 ;

n

=

4

n = 4

n=4 ,

n

2

=

16

n^2 = 16

n2=16 ,

x

(

16

)

x(16)

x(16) 没有值 ;

n

=

5

n = 5

n=5 ,

n

2

=

25

n^2 = 25

n2=25 ,

x

(

10

)

x(10)

x(10) 没有值 ;

y

(

n

)

y(n)

y(n) 的取值是

n

=

0

,

1

,

2

n = 0 , 1 , 2

n=0,1,2 时的取值 ,

n

=

0

n = 0

n=0 时 ,

y

(

n

)

=

x

(

n

2

)

=

x

(

0

2

)

=

x

(

0

)

=

0

y(n) = x(n^2) = x(0^2) = x(0) = 0

y(n)=x(n2)=x(02)=x(0)=0 ;

n

=

1

n = 1

n=1 时 ,

y

(

n

)

=

x

(

n

2

)

=

x

(

1

2

)

=

x

(

1

)

=

1

y(n) = x(n^2) = x(1^2) = x(1) = 1

y(n)=x(n2)=x(12)=x(1)=1 ;

n

=

2

n = 2

n=2 时 ,

y

(

n

)

=

x

(

n

2

)

=

x

(

2

2

)

=

x

(

4

)

=

4

y(n) = x(n^2) = x(2^2) = x(4) = 4

y(n)=x(n2)=x(22)=x(4)=4 ;

x

(

n

1

)

x(n - 1)

x(n1) 正常变换后的取值为 :

T

(

x

(

n

1

)

)

=

{

0

,

1

,

4

}

T(x(n -1 )) = { 0, 1, 4 }

T(x(n1))={0,1,4}

### ④ 结论

x

(

(

n

n

0

)

2

)

x((n-n_0)^2)

x((nn0)2) , 输出序列 为

y

(

n

1

)

=

{

0

,

1

,

2

,

5

}

y(n-1) = { 0, 1, 2, 5 }

y(n1)={0,1,2,5}

x

(

n

2

n

0

)

x(n^2 - n_0)

x(n2n0) , 输出序列为

T

(

x

(

n

1

)

)

=

{

0

,

1

,

4

}

T(x(n -1 )) = { 0, 1, 4 }

T(x(n1))={0,1,4}

THE END