信息处理 - is F(x,n)=x(−n+2)F(x,n)=x(−n+2) time invariant? - 吾爱随笔录

信息处理 discrete-signals system-identification

2022-02-15 16:38:15

My book with answers to proakis and manolakis book says that $F(x,n)=x(-n + 2)$

First we shift output signal:

F (x, n - k) = x (- (n - k) + 2) = x (- n + k + 2)

$F(x, n - k) = x(-(n-k) + 2) = x(-n+k+2)$

now lets put $x'(n) = x(n-k)$

F (x^{'}, n) = x^{'} (- n + 2) = x (- n + 2 - k) = x (- n - k + 2)

$F(x', n) = x'(-n+2)=x(-n+2-k)=x(-n-k+2)$

since $F(x, n - k) \neq F(x', n)$

1个回答

Your reasoning is correct and we arrive at the same conclusion.

The system's output is given by

\begin{matrix} (1) & y [n] = x [- n + 2] \end{matrix}

$y[n]=x[-n+2]\tag{1}$

Let

\begin{matrix} (2) & x_{2} [n] = x [n - k] \end{matrix}

$x_2[n]=x[n-k]\tag{2}$

The corresponding output is

\begin{matrix} (3) & y_{2} [n] = x_{2} [- n + 2] = x [- n - k + 2] \end{matrix}

$y_2[n]=x_2[-n+2]=x[-n-k+2]\tag{3}$

From $(1)$ the shifted output is

\begin{matrix} (4) & y [n - k] = x [- (n - k) + 2] = x [- n + k + 2] \end{matrix}

$y[n-k]=x[-(n-k)+2]=x[-n+k+2]\tag{4}$

Since $(3)$ and $(4)$ are not identical, the system is time-varying.

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