Is y(t) = x(t−2) + x(2−t) LTI

Given $y(t) = x(t-2) + x(2-t)$. Check linearity and time invariance separately.

Linearity

Let $x_1(t) \to y_1(t) = x_1(t-2) + x_1(2-t)$ and $x_2(t) \to y_2(t) = x_2(t-2) + x_2(2-t)$.

For $x_3(t) = a x_1(t) + b x_2(t)$:

$$\begin{aligned} y_3(t) &= x_3(t-2) + x_3(2-t) \\ &= a x_1(t-2) + b x_2(t-2) + a x_1(2-t) + b x_2(2-t) \\ &= a\big[x_1(t-2) + x_1(2-t)\big] + b\big[x_2(t-2) + x_2(2-t)\big] \\ &= a y_1(t) + b y_2(t) \end{aligned}$$

Superposition holds, so the system is linear.

Time invariance

Apply a delayed input $x_1(t) = x(t - t_0)$. The output is obtained by replacing $x(\cdot)$ with $x(\cdot - t_0)$:

$$y_1(t) = x_1(t-2) + x_1(2-t) = x(t - 2 - t_0) + x(2 - t - t_0)$$

Now delay the original output by $t_0$ (replace $t$ by $t - t_0$):

$$y(t - t_0) = x(t - t_0 - 2) + x(2 - t + t_0)$$

The first terms match, but the second terms differ: $x(2 - t - t_0) \ne x(2 - t + t_0)$. So $y_1(t) \ne y(t - t_0)$ and the system is time-variant.

Counter-example: let $x(t) = \delta(t)$. Then $y(t) = \delta(t-2) + \delta(2-t) = 2\delta(t-2)$. For $x(t) = \delta(t-1)$: $y(t) = \delta(t-3) + \delta(1-t) = \delta(t-3) + \delta(t-1)$, which is not $y(t-1) = 2\delta(t-3)$.

The cause is the time-reversal term $x(2 - t)$: a delay of the input moves this part of the output earlier, not later.

Answer: the system is linear but time-variant, so it is not an LTI system. (It is also non-causal: $y(0) = x(-2) + x(2)$ needs a future input.)