Correct Option: C

(i) Linearity: The given system
y(n)= 3y²(n – 1) – nx(n) + 4x(n – 1) – 2x(n + 1)
will be linear if
F[ax(n)] = aF[x(n)]
F[ax(n)] = ay(n) = 3a²y²(n – 1) – anx(n) + 4x(n – 1) – 2ax(n + 1)
aF[x(n)] = a[y(n)] = 3ay²(n – 1) – an x(n) + 4x(n – 1) – 2ax(n + 1)
Since here,
F[ax(n)] ≠ aF[x(n)]
Therefore, the system is non-linear.
(ii) Shift invariance: The given system equation will be time invariant if
y[n – n₀] = F[x(n – n₀)]
The response of delayed excitation is
y[x(n – n₀)] = 3y²[(n – n₀) y(n – n₀)]
= F[x(n – n₀ – n)]
= 3ay²(n – 1) – nx(n – n₀) + 4x(n – n₀ – 1) – 2x(n – n₀ + 1)
and the delayed response is
y(n – n₀)= 3y²[(n –n₀ – 1) – (n – n₀) x(n – n₀) + 4x(n – n₀ – 1) – 2x(n – n₀ + 1)]
Here, y(n – n₀) ≠ F[x(n – n₀)]
Hence, system is not time-invariant. (iii) Causality: In the given equation
y(n)= 3y²[(n – 1) – 1] – nx(n) x(n – n₀) + 4x(n – 1) – 2x(n + 1)
output y(n) is depends on a future input sample value x(n + 1).
Hence, the system is non-causal. Hence, alternative (C) is the correct choice.