Z-transform of a step delayed by 4

$x(t)$ is a unit step that switches on just after $t = 4$. No sampling period is given, so take $T = 1$ s (the usual choice for such problems).

Step 1: Sampled sequence

$x(kT) = 1$ only when $kT > 4$. At $t = 4$ exactly the value is 0 (since $x = 0$ for $t \le 4$).

So $x(k) = 1(k-5)$, a unit step delayed by 5 samples.

Step 2: Z-transform from the definition

$$\begin{aligned} X(z) &= \sum_{k=0}^{\infty}x(k)z^{-k} = z^{-5} + z^{-6} + z^{-7} + \dots \\ &= z^{-5}\left(1 + z^{-1} + z^{-2} + \dots\right) = \frac{z^{-5}}{1 - z^{-1}}, \quad |z| > 1 \end{aligned}$$

The same result follows from the shift theorem: $\mathcal{Z}[1(k-5)] = z^{-5}\mathcal{Z}[1(k)]$.

$$X(z) = \frac{z^{-5}}{1-z^{-1}} = \frac{1}{z^4(z-1)}$$

Remarks

• If the step were defined as 1 for $t \ge 4$, the sample at $k = 4$ would be 1, and $X(z) = \dfrac{z^{-4}}{1-z^{-1}}$. The strict inequality $t > 4$ is what moves the first nonzero sample to $k = 5$.
• For a general $T$ with $4 = nT$ ($n$ an integer), the first nonzero sample is $k = n+1$, so $X(z) = \dfrac{z^{-(n+1)}}{1-z^{-1}}$.

Answer ($T = 1$ s): $X(z) = \dfrac{z^{-5}}{1 - z^{-1}} = \dfrac{1}{z^4(z-1)}$.