Inverse Z-transform by inversion integral (cubic)

In the inversion integral method, $x(k)$ is the sum of the residues of $X(z)z^{k-1}$ at its poles:

$$x(k) = \frac{1}{2\pi j}\oint_C X(z)z^{k-1}\,dz = \sum_i\text{Res}\left[X(z)z^{k-1}\right]_{z=z_i}$$

Step 1: Rewrite $X(z)$ in powers of $z$

$1 + 1.6z^{-1} + 0.64z^{-2} = (1 + 0.8z^{-1})^2$. Multiply numerator and denominator by $z^3$:

$$X(z) = \frac{z^{-1}}{(1 - z^{-1})(1 + 0.8z^{-1})^2} = \frac{z^2}{(z-1)(z+0.8)^2}$$ $$X(z)z^{k-1} = \frac{z^{k+1}}{(z-1)(z+0.8)^2}$$

For $k \ge 0$ there is no pole at the origin. The poles are a simple pole at $z = 1$ and a double pole at $z = -0.8$.

Step 2: Residue at $z = 1$

$$R_1 = \left[\frac{z^{k+1}}{(z+0.8)^2}\right]_{z=1} = \frac{1}{(1.8)^2} = \frac{1}{3.24} = 0.3086$$

Step 3: Residue at the double pole $z = -0.8$

$$\begin{aligned} R_2 &= \frac{d}{dz}\left[\frac{z^{k+1}}{z-1}\right]_{z=-0.8} = \left[\frac{(k+1)z^{k}(z-1) - z^{k+1}}{(z-1)^2}\right]_{z=-0.8} \\ &= \frac{(-0.8)^k\left[(k+1)(-1.8) - (-0.8)\right]}{(-1.8)^2} = \frac{(-0.8)^k\left(-1.8k - 1\right)}{3.24} \end{aligned}$$

Step 4: Result

$$x(k) = R_1 + R_2 = \frac{1}{3.24}\left[1 - (1 + 1.8k)(-0.8)^k\right], \quad k = 0, 1, 2, \dots$$

Equivalently $x(k) = 0.3086 - (0.3086 + 0.5556k)(-0.8)^k$.

Step 5: Check

Expanding the denominator gives $(1 - z^{-1})(1 + 1.6z^{-1} + 0.64z^{-2}) = 1 + 0.6z^{-1} - 0.96z^{-2} - 0.64z^{-3}$. Direct division of $\dfrac{z^{-1}}{1 + 0.6z^{-1} - 0.96z^{-2} - 0.64z^{-3}}$ gives $z^{-1} - 0.6z^{-2} + 1.32z^{-3} - 0.728z^{-4} + \dots$, which matches.

The final value theorem gives $\lim_{z\to1}(1 - z^{-1})X(z) = \dfrac{1}{(1.8)^2} = 0.3086$, which also matches.

Answer: $x(k) = \dfrac{1}{3.24}\left[1 - (1 + 1.8k)(-0.8)^k\right]$ for $k \ge 0$.