[digital signal processing] linear time invariant system LTI (judge whether a system is a "non time variant" system | case 3)

One 、 Judge whether the system is " Non time varying "

---

1、 Case 2

Given Input sequence $x(n)=\{0,1,2,3,4,5,0\}$ , $n$ Value $-1$ ~ $5$

Determine the output sequence $y(n)=x(n^2)$ Of " Transformation " Operation Yes No " Time does not change " Of ;

$y(n)=x(n^2)$ Change operation :

$y(n)$ Only in $n=-1,0,1,2$ When the value is taken , It's worth it ,

If $n=3$ , $n^2=9$ , $x(9)$ No value ;
If $n=4$ , $n^2=16$ , $x(16)$ No value ;
If $n=5$ , $n^2=25$ , $x(10)$ No value ;

therefore , After normal transformation , $y(n)$ The value is $n=0,1,2$ The value of time ,

When $n=-1$ when , $y(n)=x(n^2)=x((-1{)}^2)=x(1)=2$ ;
When $n=0$ when , $y(n)=x(n^2)=x(0^2)=x(0)=1$ ;
When $n=1$ when , $y(n)=x(n^2)=x(1^2)=x(1)=2$ ;
When $n=2$ when , $y(n)=x(n^2)=x(2^2)=x(4)=5$ ;

among $-1$ and $1$ The square of is $1$ , Merge into one ;

$x(n)$ The value after normal transformation is :

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

① Time invariant system concept

Time invariant system ( time-invariant ) : System features , Does not change with time ;

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

After input delay , The output is also delayed ;

And " Time does not change " The system corresponds to " time varying " System ;

② First transform and then shift

take " Output sequence " Shift , First " Transformation " after " displacement " ;

First the " Input sequence " Conduct " Transformation " operation , obtain " Output sequence " , Then on Output sequence Conduct " displacement " operation ;

among " Transformation " refer to , Discrete time systems , take " Input sequence " Transformation by " Output sequence " , Input sequence To Output sequence Operation between , yes " Transformation " ;

Change operation : First the Input sequence $x(n)$ Conduct Transformation operation , obtain Output sequence $x(n^2)$ ,

Shift operation : then Yes $x(n^2)$ Output sequence Shift $n-n_0$ obtain $x((n-n_0{)}^2)$ ,

The complete operation process is as follows :

$$y(n-n_0)=x((n-n_0{)}^2)$$

First change , After transformation, the output is :
$$y(n)=\{1,2,5\}$$

The value of post shift is : Move one bit to the right ;

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

③ Shift first and then transform

take " Input sequence " Shift , Shift first , take " Input sequence $x(n)$ " to " displacement " operation , obtain new " Input sequence " by $x(n-n_0)$ , then The new input sequence is " Transformation " operation , obtain " Output sequence " ;

The transformation process is $T[x(n-n_0)]=x(n^2-n_0)$ , Transformation time , Just to $n$ Value to $n^2$ , $n_0$ Fixed value ;

$x(n-n_0)$ Transformation time , Only will $n$ multiply $2$ , $n_0$ unchanged , The transformation result is as follows $x(2n-n_0)$ ;

The whole process is as follows :

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

First the $x(n)=\{0,1,2,3,4,5,0\}$ , $n$ Value $-1$ ~ $5$ , Shift right , The shifted sequence :
$$x(n)=\{0,1,2,3,4,5\}$$$n$ Value $0$ ~ $6$ , The shifted sequence schema is as follows :

Shift right 1 after , $n$ Value From the original $-1$ ~ $5$ Change into $0$ ~ $6$ ,

$y(n)$ Only in $n=0,1,2$ When the value is taken , It's worth it ,

If $n=3$ , $n^2=9$ , $x(9)$ No value ;
If $n=4$ , $n^2=16$ , $x(16)$ No value ;
If $n=5$ , $n^2=25$ , $x(10)$ No value ;

therefore , After normal transformation , $y(n)$ The value is $n=0,1,2$ The value of time ,

When $n=0$ when , $y(n)=x(n^2)=x(0^2)=x(0)=0$ ;
When $n=1$ when , $y(n)=x(n^2)=x(1^2)=x(1)=1$ ;
When $n=2$ when , $y(n)=x(n^2)=x(2^2)=x(4)=4$ ;

$x(n-1)$ The value after normal transformation is :

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

④ Conclusion

First " Transformation " after " displacement " , The result is $x((n-n_0{)}^2)$ , Output sequence by $y(n-1)=\{0,1,2,5\}$

First " displacement " after " Transformation " , The result is $x(n^2-n_0)$ , The output sequence is $T(x(n-1))=\{0,1,4\}$

The system is " Time varying system " ;