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System represented by equation— y[n] = n[ x(n)]2 is—
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- linear and time-invariant
- non-linear and time-invariant
- linear and time-variant
- non-linear and time-variant
Correct Option: D
Given that y[n] = n[x(n)]2
First check for linearity
F[x1(n)] = n[x1(n)]2
and F[x2(n)] = n[x2(n)]2
Therefore,
F[x1(n)] + F[x2(n)] = n[{x2(n)}2 + {x2(n)}2]
F[x1(n) + x2(n)] = n[x1(n) + x2(n)]2
= n[{x2(n)}2 + {x2(n)}2 + 2x1(n) x2(n)]
∴ F[x1(n)] + F[x2(n)] ≠ F[x1(n) + x2(n)]
These for the system is non-linear. Now, check for time variance.
y[n] = n[x(n)]2
The response of delayed excitation is
F[x(n – n0)] = n[x(n – n0)2
and the delayed response is
y(n – n0)= (n – n0) [x(n – n0)]2
Since, y(n – n0) ≠ F[x(n – n0)]
Therefore, the system is time variant.
Hence, alternative (D) is the correct choice.
