# Determine if system is linear time variant

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?

• It is not time invariant. – IanJ Jan 13 at 23:02

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