The correct option is (d) System is Linear and stable
For explanation: y1 (t) = v (\(\frac{t}{2}\))
And y2 (t) = k v (\(\frac{t}{2}\)) = k y1 (t)
If x3 = v (t) + w (t),
Then, y3 (t) = v (\(\frac{t}{2}\)) + w (\(\frac{t}{2}\))
= y1 (t) + y2 (t)
Since, it satisfies both law of homogeneity and additivity, it is also linear.
Again, y1 (t) = v (\(\frac{t}{2}\)), y2 (\(\frac{t}{2}\)-t0) ≠ y (t-t0) = v (\(\frac{t-t_0}{2})\)
∴ The system is time variant.
If x (t) is bounded then y (t) is bounded, so a stable system.
At time t=-2, y (-2) = x (-1), therefore, the response at time t=-2 depends on the excitation at a later time, t=-1, so a Non-causal system.