Correct Answer : Option (B) : y(n) = x(n) – x(n-1)
Explanation : We know that, for any system y (n) = k x (n), to be a time invariant system, it must satisfy the relation, y (n-n1) = k x (n-n1) [where k is a constant or a function of n].
For y (n) = n x (n), y (n-n1) = (n-n1) x (n-n1)
This does not satisfy the criteria as stated above. Hence not time invariant.
For y (n) = x (-n), y (n-n1) = x (-n+n1)
This also does not satisfy the criteria as stated above. Hence not time invariant.
For y (n) = x (n) cos 2nf, y (n-n1) = x (n-n1) cos 2(n-n1) f
This also does not satisfy the criteria as stated above. Hence not time invariant.
For y (n) = x (n) – x (n-1), y (n-n1) = x (n-n1) – x (n-n1-1)
This satisfies the above criteria. Hence given system is time invariant.
