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The system represented by equation—
y[n] = a[x(n)]2 + bx[n]
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- linear and time-invariant
- linear and time-variant
- non-linear and time-invariant
- non-linear and time-variant
Correct Option: C
Given y(n) = a[x(n)]2 + bx(n)
First check for linearity
Since here,
F[x1(n)] = a[x1(n)]2 + bx1(n)
and F[x2(n)] = a[x2(n)]2 + bx2(n)
∴ F[x1(n)] = F[x2(n)]
= a[{x2(n)}2 + {x2(n)2}] + b[x1(n) + x2(n)]
Also, F[x1(n) + x2(n)]
= a[x1(n) + x2(n)]2 + b[x1(n) + x2(n)]
= a[{x1(n)}2 + {x2(n)}2 + 2x1(n) x2(n)] + bx1(n) + bx2(n)
Since here,
F[x1(n) + x2(n)] ≠ F[x1(n)] + F[x2(n)]
Therefore, the system is non-linear.
Check now for the time-in-variance.
y(n) = a[x(n)]2 + bx(n)
The response of delayed excitation is
F[x(n – n0)] = a[x(n – n0)]2 + b[x(n – n0)]
and the delayed response is
y[(n – n0)] = a[x(n – n0)]2 + b[x(n – n0)]
Since, y[(n – n0)] = F[x(n – n0)]
Therefore, the system is time-invariant.
Hence, alternative (C) is the correct choice.
