Z-transform of ramp-and-hold curve (to 1 at t=2)

From the figure, $x(t)$ rises linearly with slope $0.5$ from 0 at $t = 0$ to 1 at $t = 2$, and then stays at 1. Take the sampling period $T = 1$ s, as marked on the axis. The general-$T$ form is also given.

Step 1: Express the curve with standard functions

$$x(t) = 0.5\,t\cdot1(t) - 0.5\,(t-2)\cdot1(t-2)$$

• For $0 \le t \le 2$: $x = 0.5t$
• For $t > 2$: $x = 0.5t - 0.5t + 1 = 1$

 x(t)
  1 |        ________________
    |       /
0.5 |     /
    |   /
  0 +--/-----+-----+-----+----> t
    0     1     2     3     4

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

Step 3: Z-transform

Use $\mathcal{Z}[t] = \dfrac{Tz^{-1}}{(1-z^{-1})^2}$ and the real translation theorem $\mathcal{Z}[x(t-nT)1(t-nT)] = z^{-n}X(z)$, with $n = 2/T$:

$$X(z) = 0.5\cdot\frac{Tz^{-1}}{(1-z^{-1})^2}\left(1 - z^{-2/T}\right)$$

With $T = 1$, $n = 2$:

$$\begin{aligned} X(z) &= \frac{0.5\,z^{-1}(1 - z^{-2})}{(1-z^{-1})^2} = \frac{0.5\,z^{-1}(1 + z^{-1})}{1 - z^{-1}} \\ &= \frac{0.5(z + 1)}{z(z-1)} \end{aligned}$$

Step 4: Verify by the definition

Directly from $X(z) = \sum x(k)z^{-k}$:

$$\begin{aligned} X(z) &= 0.5z^{-1} + z^{-2} + z^{-3} + z^{-4} + \dots \\ &= 0.5z^{-1} + \frac{z^{-2}}{1-z^{-1}} = \frac{0.5z^{-1}(1 - z^{-1}) + z^{-2}}{1 - z^{-1}} = \frac{0.5z^{-1} + 0.5z^{-2}}{1-z^{-1}} \end{aligned}$$

This is the same as Step 3.

Step 5: Checks with the limit theorems

• Initial value: $x(0) = \lim_{z\to\infty}X(z) = 0$, which is correct.
• Final value: $\lim_{z\to1}(1-z^{-1})X(z) = 0.5(1+1) = 1$, which is correct.

Note for other $T$

If $T = 0.5$ s, then $n = 4$ and $X(z) = \dfrac{0.25z^{-1}(1 - z^{-4})}{(1-z^{-1})^2}$, with samples $0, 0.25, 0.5, 0.75, 1, 1, \dots$

Answer ($T = 1$ s): $X(z) = \dfrac{0.5z^{-1}(1 - z^{-2})}{(1-z^{-1})^2} = \dfrac{0.5(z+1)}{z(z-1)}$.