Correct Option: C
Expanding the given X(z) in terms of the positive powers of z.
x(z) = | 1 | |
(1 + z– 1) (1 – z– 1)2 |
Hence, x(z) = | 1 | |
(z + 1) (z – 1)2 |
Here, F(z) = | X(z) | = | z2 | |
z | (z + 1) (z – 1)2 |
= | A1 | + | A2 | + | A3 | |
(z + 1)2 | (z – 1) | (z – 1)2 |
A1 = (z + 1) F(z)|z = – 1 = | z2 | |z = – 1 = | 1 | |
(z – 1)2 | 4 |
A2 = | d |  | | z2 |  | | |
dz | z + 1 | z = 1 |
= | (z + 1) 2z – z2 | |z = 1 = | 3 | |
(z + 1)2 | 4 |
A3 = (z – 1)2 F(z)|z = 1 = | z2 | |z = 1 = | 1 | |
(z + 1) | 2 |
Therefore, F(z) = | 1 | 1 | + | 3 | 1 | + | 1 | 1 | |
4 | (z + 1) | 4 | (z – 1) | 2 | (z – 1)2 |
= | 1 | z | + | 3 | z | + | 1 | z | |
4 | (z + 1) | 4 | (z – 1) | 2 | (z – 1)2 |
Taking inverse z-transform of x(z), we obtain
x(n) = | 1 | (– 1)n u(n) + | 3 | u(n) + | 1 | nu(n) |
4 | 4 | 2 |
= |  | 1 | (– 1)n + | 3 | + | 1 | n |  | u(n) |
4 | 4 | 2 |
Hence alternative (C) is the correct choice.