Z-transform of ramp minus delayed ramp

$x(t)$ is a ramp of slope $\tfrac14$ minus the same ramp delayed by 4 s. It rises from 0 to 1 over 4 s and then stays at 1.

Step 1: Standard pairs

• $\mathcal{Z}[t\cdot1(t)] = \dfrac{Tz^{-1}}{(1-z^{-1})^2}$
• Real translation (shift) theorem: if $4 = nT$, then $\mathcal{Z}[x(t - nT)1(t-nT)] = z^{-n}X(z)$

Step 2: General sampling period $T$ (4 an integer multiple of $T$)

$$X(z) = \frac{1}{4}\cdot\frac{Tz^{-1}}{(1-z^{-1})^2}\left(1 - z^{-4/T}\right)$$

Step 3: With $T = 1$ s (usual case, $n = 4$)

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

Here $1 - z^{-4} = (1-z^{-1})(1+z^{-1}+z^{-2}+z^{-3})$ was used.

Step 4: Check with the samples

Long division gives $X(z) = 0.25z^{-1} + 0.5z^{-2} + 0.75z^{-3} + z^{-4} + z^{-5} + \dots$, which matches. The final value from the theorem, $\lim_{z\to1}(1-z^{-1})X(z) = 1$, is also correct.

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