Inverse Z-transform: double pole at z=2

The inverse z-transform is found by the residue (inversion integral) method, since $X(z)$ has a double pole at $z = 2$ and a simple pole at $z = e^{-aT}$. Let $b = e^{-aT}$ for short.

$$x(k) = \sum \text{residues of } X(z)\,z^{k-1} = \sum \text{residues of } \frac{z^{k+1}}{(z-2)^2 (z-b)}$$

For $k \ge 0$ the factor $z^{k+1}$ adds no pole at $z = 0$, so only two poles count.

Residue at the simple pole $z = b$

$$R_1 = \left[\frac{z^{k+1}}{(z-2)^2}\right]_{z=b} = \frac{b^{k+1}}{(b-2)^2} = \frac{e^{-a(k+1)T}}{(2-e^{-aT})^2}$$

Residue at the double pole $z = 2$

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

Result

$$x(k) = \frac{1}{(2-e^{-aT})^2}\Big[(2-e^{-aT})\,k\,2^{k} - e^{-aT}\,2^{k} + e^{-a(k+1)T}\Big], \quad k = 0,1,2,\dots$$

Check

• $k = 0$: $\dfrac{-b + b}{(2-b)^2} = 0$. Since the numerator degree (2) is one less than the denominator degree (3), $X(z)$ starts with $z^{-1}$, so $x(0) = 0$ is correct.
• $k = 1$: $\dfrac{2(2-b) - 2b + b^2}{(2-b)^2} = \dfrac{(2-b)^2}{(2-b)^2} = 1$, which matches the first long-division term $z^{-1}$.
• Numerical check with $b = 0.5$: the formula gives $0, 1, 4.5, 14.25, 39.125$ for $k = 0$ to $4$, the same as long division of $X(z)$.

Answer: $x(k) = \dfrac{(2-e^{-aT})k\,2^k - e^{-aT}2^k + e^{-a(k+1)T}}{(2-e^{-aT})^2}$ for $k \ge 0$ (and $x(k) = 0$ for $k < 0$).