# Is this discrete-time system time-invariant?

For a discrete-time system, the input signal is $x(n)$, the output signal is

$y[n] = x[n+1] - x[1-n]$

I think it's time-invariant, but the solution's manual says it's not.

My procedure is: $T\{x[n-k]\} = x[n - k + 1] - x[1 - (n - k)] = x[n - k + 1] - x[1 - n + k]$

$y[n-k] = x[n-k+1]-x[1-n+k]$

Am I wrong?

• What is $T{}{}$? – Ian Oct 13 '17 at 10:56
• Are you sure the book does not rather ask about $y[n] = x[n+1] - x[n-1]$? – Did Mar 3 '18 at 11:49

The system you have given is $$y[n]=x[n+1]-x[1-n]=x[n+1]-x[-n+1]$$ You can see that the system is subtracting a shifted and reflected version of the input $x[n]$ from a shifted version of the same input. In particular, the second term shifts $x[n]$ one unit to the right and then reflects it about the y-axis.
In this case, for the second term, the system further shifts $x[n-k]$ to the right so it becomes $x[n-k+1]$. It then flips it so it becomes $x[-n-k+1]$. This is not equivalent to the term you obtain when the output is directly shifted by $k$, given by $x[-(n-k)+1]=x[-n+k+1]$. The system is not time-invariant.