Z-transform of a ramp-and-hold input (to 3)

From the figure, $x(t)$ is a unit-slope ramp that stops rising at $t = 3$ and then holds at 3. Take the sampling period $T = 1$ s, from the sample lines at $t = 1, 2, \dots$.

Step 1: Write $x(t)$ as a ramp minus a delayed ramp

$$x(t) = t\cdot 1(t) - (t-3)\cdot 1(t-3)$$

For $t \le 3$ only the first term acts, giving $x = t$. For $t > 3$ the result is $t - (t-3) = 3$.

Step 2: Sampled values ($T = 1$)

Step 3: Take the z-transform

Use $\mathcal{Z}[k] = \dfrac{z^{-1}}{(1-z^{-1})^2}$ (with $T = 1$) and the shift theorem $\mathcal{Z}[x(k-n)] = z^{-n}X(z)$:

$$\begin{aligned} X(z) &= \frac{z^{-1}}{(1-z^{-1})^2} - z^{-3}\frac{z^{-1}}{(1-z^{-1})^2} = \frac{z^{-1}(1 - z^{-3})}{(1-z^{-1})^2} \end{aligned}$$

Since $1 - z^{-3} = (1 - z^{-1})(1 + z^{-1} + z^{-2})$:

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

Step 4: Check by series

$$X(z) = z^{-1} + 2z^{-2} + 3z^{-3}\left(1 + z^{-1} + z^{-2} + \dots\right) = z^{-1} + 2z^{-2} + 3z^{-3} + 3z^{-4} + \dots$$

The coefficients $0, 1, 2, 3, 3, \dots$ match the samples.

Final value check: $\lim_{z\to1}(1-z^{-1})X(z) = 1\cdot(1+1+1) = 3$, which is correct.

Answer: $X(z) = \dfrac{z^{-1}(1 - z^{-3})}{(1 - z^{-1})^2} = \dfrac{z^2 + z + 1}{z^2(z-1)}$ (with $T = 1$ s).