X(z) from convolution integral: 1/(s²(s+1))

Method: convolution integral in the left half-plane

When the denominator of $X(s)$ has at least two more poles than the numerator has zeros, the z-transform of the sampled $x(t)$ is the sum of residues at the poles of $X(s)$:

$$X(z) = \sum_{\text{poles of } X(s)} \text{Res}\left[X(s)\,\frac{z}{z - e^{sT}}\right]$$

For a pole of order $m$ at $s = s_i$:

$$\text{Res} = \frac{1}{(m-1)!}\lim_{s\to s_i}\frac{d^{m-1}}{ds^{m-1}}\left[(s-s_i)^m X(s)\frac{z}{z-e^{sT}}\right]$$

Apply to $X(s) = \dfrac{1}{s^2(s+1)}$

Poles: a double pole at $s = 0$ and a simple pole at $s = -1$.

Simple pole $s = -1$:

$$R_1 = \left[\frac{1}{s^2}\cdot\frac{z}{z - e^{sT}}\right]_{s=-1} = \frac{z}{z - e^{-T}}$$

Double pole $s = 0$:

$$\begin{aligned} R_2 &= \frac{d}{ds}\left[\frac{1}{s+1}\cdot\frac{z}{z - e^{sT}}\right]_{s=0} \\ &= \left[-\frac{1}{(s+1)^2}\cdot\frac{z}{z-e^{sT}} + \frac{1}{s+1}\cdot\frac{zTe^{sT}}{(z-e^{sT})^2}\right]_{s=0} \\ &= -\frac{z}{z-1} + \frac{Tz}{(z-1)^2} \end{aligned}$$

Sum of residues

$$X(z) = \frac{Tz}{(z-1)^2} - \frac{z}{z-1} + \frac{z}{z - e^{-T}}$$

Over the common denominator:

$$X(z) = \frac{z\left[(T - 1 + e^{-T})z + (1 - e^{-T} - Te^{-T})\right]}{(z-1)^2(z - e^{-T})}$$

Check

$x(t) = t - 1 + e^{-t}$ (from partial fractions $\frac{1}{s^2} - \frac{1}{s} + \frac{1}{s+1}$). Its z-transform is $\frac{Tz}{(z-1)^2} - \frac{z}{z-1} + \frac{z}{z-e^{-T}}$, which is the same.

For $T = 1$ s: $X(z) = \dfrac{z(0.3679z + 0.2642)}{(z-1)^2(z-0.3679)}$.

Answer: $X(z) = \dfrac{Tz}{(z-1)^2} - \dfrac{z}{z-1} + \dfrac{z}{z-e^{-T}} = \dfrac{z[(T-1+e^{-T})z + (1-e^{-T}-Te^{-T})]}{(z-1)^2(z-e^{-T})}$.