Inverse Z-transform of (1+z⁻¹−z⁻²)/(1−z⁻¹)

Split $X(z)$ into a polynomial part plus a proper part, then read each term from the z-transform table.

Step 1: Divide the numerator by the denominator (in powers of $z^{-1}$)

Write $1 + z^{-1} - z^{-2} = (1 - z^{-1})(\alpha + \beta z^{-1}) + \gamma$:

$$(1 - z^{-1})(\alpha + \beta z^{-1}) + \gamma = (\alpha + \gamma) + (\beta - \alpha)z^{-1} - \beta z^{-2}$$

Comparing coefficients:

• $z^{-2}$: $-\beta = -1 \Rightarrow \beta = 1$
• $z^{-1}$: $\beta - \alpha = 1 \Rightarrow \alpha = 0$
• constant: $\alpha + \gamma = 1 \Rightarrow \gamma = 1$

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

Step 2: Inverse transform

• $z^{-1} \;\leftrightarrow\; \delta(k-1)$ (a unit pulse at $k = 1$)
• $\dfrac{1}{1-z^{-1}} \;\leftrightarrow\; 1(k)$ (a unit step)

$$x(k) = 1(k) + \delta(k-1)$$

So:

Check by direct division

$(1 + z^{-1} - z^{-2}) \div (1 - z^{-1}) = 1 + 2z^{-1} + z^{-2} + z^{-3} + \dots$, which gives the same sequence.

Answer: $x(0) = 1$, $x(1) = 2$, $x(k) = 1$ for $k \ge 2$, i.e. $x(k) = 1(k) + \delta(k-1)$.