Inverse Z-transform of z⁻¹(1−z⁻²)/(1+z⁻¹)²

In the inversion integral method, $x(k)$ is the sum of the residues of $X(z)z^{k-1}$ at all its poles inside the contour (which encloses every pole):

$$x(k) = \frac{1}{2\pi j}\oint_C X(z)z^{k-1}\,dz = \sum \text{residues of } X(z)z^{k-1}$$

Step 1: Simplify $X(z)$

$1 - z^{-2} = (1 - z^{-1})(1 + z^{-1})$, so

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

The pole at $z = 0$ exists only for $k = 0$ and $k = 1$, so those cases are handled separately.

Case $k = 0$: $\dfrac{z-1}{z^2(z+1)}$

• Double pole at $z = 0$: $\dfrac{d}{dz}\left[\dfrac{z-1}{z+1}\right]_{z=0} = \left[\dfrac{2}{(z+1)^2}\right]_{z=0} = 2$
• Pole at $z = -1$: $\left[\dfrac{z-1}{z^2}\right]_{z=-1} = -2$

$$x(0) = 2 - 2 = 0$$

Case $k = 1$: $\dfrac{z-1}{z(z+1)}$

• Pole at $z = 0$: $\left[\dfrac{z-1}{z+1}\right]_{z=0} = -1$
• Pole at $z = -1$: $\left[\dfrac{z-1}{z}\right]_{z=-1} = 2$

$$x(1) = -1 + 2 = 1$$

Case $k \ge 2$: only the pole at $z = -1$

$$x(k) = \left[(z-1)z^{k-2}\right]_{z=-1} = (-2)(-1)^{k-2} = -2(-1)^k$$

Result

$$x(k) = \begin{cases} 0, & k = 0 \\ 1, & k = 1 \\ -2(-1)^k, & k \ge 2 \end{cases}$$

Check by long division

$\dfrac{z^{-1}(1-z^{-1})}{1+z^{-1}} = z^{-1}(1 - 2z^{-1} + 2z^{-2} - 2z^{-3} + \dots) = z^{-1} - 2z^{-2} + 2z^{-3} - \dots$, which matches.

The pole at $z = -1$ lies on the unit circle, so the sequence keeps oscillating between $\pm2$.

Answer: $x(0) = 0$, $x(1) = 1$, $x(k) = -2(-1)^k$ for $k \ge 2$.