Inverse Z-transform: double pole at z=1

Let $p = e^{-aT}$ (with $a > 0$, so $p \ne 1$). By the inversion integral method,

$$x(k) = \sum \text{residues of } X(z)z^{k-1} = \sum \text{Res}\left[\frac{z^{k+1}}{(z-1)^2(z-p)}\right]$$

For $k \ge 0$ there is no pole at the origin. Poles: a double pole at $z=1$ and a simple pole at $z=p$.

Residue at $z = p$ (simple pole)

$$K_1 = \left.\frac{z^{k+1}}{(z-1)^2}\right|_{z=p} = \frac{p^{k+1}}{(1-p)^2} = \frac{e^{-a(k+1)T}}{(1-e^{-aT})^2}$$

Residue at $z = 1$ (double pole)

$$\begin{aligned} K_2 &= \lim_{z\to1}\frac{d}{dz}\left[\frac{z^{k+1}}{z-p}\right] \\ &= \lim_{z\to1}\frac{(k+1)z^k(z-p) - z^{k+1}}{(z-p)^2} \\ &= \frac{(k+1)(1-p) - 1}{(1-p)^2} = \frac{k+1}{1-p} - \frac{1}{(1-p)^2} \end{aligned}$$

Result

$$\begin{aligned} x(k) &= K_1 + K_2 = \frac{k+1}{1-e^{-aT}} - \frac{1 - e^{-a(k+1)T}}{(1-e^{-aT})^2} \\ &= \frac{k - (k+1)e^{-aT} + e^{-a(k+1)T}}{(1-e^{-aT})^2} \end{aligned}$$

Check

• $k=0$: numerator $= 0 - p + p = 0$, so $x(0) = 0$. Initial value theorem: $\lim_{z\to\infty}X(z) = 0$. ✔
• $k=1$: numerator $= 1 - 2p + p^2 = (1-p)^2$, so $x(1) = 1$. Long division gives $X(z) = z^{-1} + (2+p)z^{-2} + \cdots$ ✔
• $k=2$: $x(2) = \frac{2-3p+p^3}{(1-p)^2} = p + 2$ ✔

Answer: $x(k) = \dfrac{k - (k+1)e^{-aT} + e^{-a(k+1)T}}{(1-e^{-aT})^2}$, $k = 0, 1, 2, \dots$