Z-transform of ramp-and-hold continuous signal

Reading the figure: the tick labels are offset. "2" sits at the origin, so the ramp starts at $t = 0$ and reaches 1.0 four divisions later (the tick labelled 6), then stays at 1.0. Each division is taken as one sampling period, $T = 1$ s. So $x(t)$ rises linearly from 0 to 1 in 4 s and then holds at 1. This is the standard textbook signal $x(t) = \tfrac14 t - \tfrac14(t-4)1(t-4)$.

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

The ramp has slope $\dfrac{1}{4}$ and reaches 1 at $t = 4$ s, after which it holds at 1:

$$x(t) = \frac{1}{4}t\cdot1(t) - \frac{1}{4}(t-4)\cdot1(t-4)$$

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

Step 3: Z-transform

With $\mathcal{Z}[t] = \dfrac{Tz^{-1}}{(1-z^{-1})^2}$ ($T = 1$) and delay of 4 samples $\to z^{-4}$:

$$\begin{aligned} X(z) &= \frac{1}{4}\cdot\frac{z^{-1}}{(1-z^{-1})^2} - \frac{1}{4}z^{-4}\cdot\frac{z^{-1}}{(1-z^{-1})^2} \\ &= \frac{1}{4}\cdot\frac{z^{-1}(1 - z^{-4})}{(1-z^{-1})^2} \end{aligned}$$

Using $1 - z^{-4} = (1-z^{-1})(1+z^{-1}+z^{-2}+z^{-3})$:

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

Step 4: Check

Series: $X(z) = 0.25z^{-1} + 0.5z^{-2} + 0.75z^{-3} + z^{-4} + z^{-5} + \dots$, which matches the table.

Final value: $\lim_{z\to1}(1-z^{-1})X(z) = \dfrac{1\cdot 4}{4} = 1$, which is correct.

If the signal were instead read as reaching 1 after $n$ sampling periods, the same method gives $X(z) = \dfrac{z^{-1}(1-z^{-n})}{n(1-z^{-1})^2}$.

Answer: $X(z) = \dfrac{z^{-1}(1-z^{-4})}{4(1-z^{-1})^2} = \dfrac{z^3+z^2+z+1}{4z^3(z-1)}$ (with $T = 1$ s).