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The system equation is given as:

$$y(n)=(n-1)x(n-1)+(n+1)x(n+1)$$

I solved that the system is time variant: \begin{align} y(n-k)&=(n-k-1)x(n-k-1)+(n-k+1)x(n-k+1)\\ H[x(n-k)]&=(n-1)x(n-k-1)+(n+1)x(n-k+1) \end{align}

And for the linear/non-linear part, here is what I have so far: \begin{align} H\big[a_1x_1(n)+a_2x_2(n)\big]&=(n-1)\big[a_1x_1(n-1)+a_2x_2(n-1)\big]\\&+(n+1)\big[a_1x_1(n+1)+a_2x_2(n+2)\big] \end{align} It is ok?

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  • $\begingroup$ It is not time invariant. $\endgroup$ – IanJ Jan 13 at 23:02
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To be time invariant you would have to show that:

$$y(n-k) = (n-1)x(n-k-1) + (n+1)x(n-k+1)$$

(You shift the $x$ and you get a shift in $y$).

But you can't get rid of the $(n-k-1)$ and $(n-k+1)$.

In general any time you have the time term $n$ by itself it will be time variant unless it cancels out somehow.

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