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?

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

Yes, you are wrong.

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.

When you are evaluating what output you would get had you introduced a shifted input x[n-k] to the system, you must understand that your input is already shifted. The system then does additional operations on this input.

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.

  • $\begingroup$ Thanks very much! I got it! $\endgroup$ – Johnny Ji Oct 14 '17 at 1:28
  • $\begingroup$ @JohnnyJi You're welcome $\endgroup$ – Undertherainbow Oct 14 '17 at 8:34

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