Z-transform of a finite two-sided sequence

$x(k) = (1/2)^k = 2^{-k}$ is defined only for $-4 \le k \le 4$, so it is a finite-length, two-sided sequence. Use the two-sided definition $X(z) = \sum_k x(k)z^{-k}$.

Step 1: List the samples

Step 2: Write the sum

$$\begin{aligned} X(z) &= \sum_{k=-4}^{4}2^{-k}z^{-k} \\ &= 16z^4 + 8z^3 + 4z^2 + 2z + 1 + 0.5z^{-1} + 0.25z^{-2} + 0.125z^{-3} + 0.0625z^{-4} \end{aligned}$$

Step 3: Closed form (finite geometric series)

The ratio between successive terms is $r = (2z)^{-1}$, and the first term is $(2z)^4$:

$$X(z) = \sum_{k=-4}^{4}(2z)^{-k} = \frac{(2z)^4\left[1 - (2z)^{-9}\right]}{1 - (2z)^{-1}} = \frac{(2z)^4 - (2z)^{-5}}{1 - \frac{1}{2z}}$$

Multiplying by $\dfrac{2z}{2z}$:

$$X(z) = \frac{(2z)^9 - 1}{(2z)^4\,(2z - 1)} = \frac{512z^9 - 1}{16z^4(2z-1)}$$

The apparent pole at $z = \tfrac12$ is cancelled by a zero of the numerator, so $X(z)$ is really a finite polynomial in $z$ and $z^{-1}$.

Step 4: Region of convergence

A finite sum always converges except where a term becomes infinite:

• the positive powers $z^4, \dots, z$ blow up at $z = \infty$
• the negative powers $z^{-1}, \dots, z^{-4}$ blow up at $z = 0$

$$\text{ROC: } 0 < |z| < \infty$$

Check: at $z = 1$ the sum is $16+8+4+2+1+0.5+0.25+0.125+0.0625 = 31.9375$, and the closed form gives $\dfrac{511}{16} = 31.9375$.

Answer: $X(z) = 16z^4 + 8z^3 + 4z^2 + 2z + 1 + 0.5z^{-1} + 0.25z^{-2} + 0.125z^{-3} + 0.0625z^{-4} = \dfrac{512z^9 - 1}{16z^4(2z-1)}$, ROC $0 < |z| < \infty$.