Inverse Z-transform of z⁻³/((1−z⁻¹)(1−0.2z⁻¹))

Remove the pure delay $z^{-3}$ first. Invert the remaining part by partial fractions, then shift the result by 3 samples.

Step 1: Write $X(z) = z^{-3}F(z)$

$$F(z) = \frac{1}{(1-z^{-1})(1-0.2z^{-1})} = \frac{z^2}{(z-1)(z-0.2)}$$

Step 2: Partial fractions of $F(z)/z$

$$\frac{F(z)}{z} = \frac{z}{(z-1)(z-0.2)} = \frac{A}{z-1} + \frac{B}{z-0.2}$$ $$A = \frac{1}{1-0.2} = 1.25, \qquad B = \frac{0.2}{0.2-1} = -0.25$$ $$F(z) = \frac{1.25z}{z-1} - \frac{0.25z}{z-0.2} \;\Rightarrow\; f(k) = 1.25 - 0.25(0.2)^k = 1.25\left[1 - (0.2)^{k+1}\right]$$

Step 3: Apply the shift theorem

$z^{-3}F(z) \leftrightarrow f(k-3)\,1(k-3)$:

$$x(k) = \begin{cases} 0, & k = 0, 1, 2 \\[4pt] 1.25\left[1 - (0.2)^{k-2}\right], & k \ge 3 \end{cases}$$

Equivalently $x(k) = \left[1.25 - 0.25(0.2)^{k-3}\right]1(k-3)$.

Values

Check

• Long division: $X(z) = z^{-3} + 1.2z^{-4} + 1.24z^{-5} + 1.248z^{-6} + \dots$, which matches.
• Final value theorem: $\lim_{z\to1}(1-z^{-1})X(z) = \dfrac{1}{1-0.2} = 1.25$, which matches.

Answer: $x(k) = 1.25\left[1 - (0.2)^{k-2}\right]$ for $k \ge 3$, and $x(k) = 0$ for $k < 3$.