Inverse Z-transform by partial fractions

The function is solved by partial fractions of $X(z)/z$, because every entry in the z-transform table has a factor $z$ in its numerator. The question is solved as printed, with $e^{aT}$. (The usual textbook form has $e^{-aT}$. That version is given at the end.)

Let $c = e^{aT}$:

$$X(z) = \frac{(1-c)\,z}{(z-1)(z-c)}$$

Step 1: Divide by $z$ and expand

$$\frac{X(z)}{z} = \frac{1-c}{(z-1)(z-c)} = \frac{A}{z-1} + \frac{B}{z-c}$$ $$\begin{aligned} A &= \left[\frac{1-c}{z-c}\right]_{z=1} = \frac{1-c}{1-c} = 1 \\ B &= \left[\frac{1-c}{z-1}\right]_{z=c} = \frac{1-c}{c-1} = -1 \end{aligned}$$

Step 2: Multiply back by $z$

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

Step 3: Use standard pairs

$$x(k) = 1 - e^{akT}, \qquad k = 0, 1, 2, \dots$$

Check by long division

$X(z) = \dfrac{(1-c)z}{z^2 - (1+c)z + c} = (1-c)z^{-1} + (1-c)(1+c)z^{-2} + \dots$

• $x(0) = 1 - 1 = 0$ (matches)
• $x(1) = 1 - c$ (matches)
• $x(2) = 1 - c^2 = (1-c)(1+c)$ (matches)

Note on the usual form

For $a > 0$ the printed form grows without bound: the pole $z = e^{aT}$ lies outside the unit circle. The standard textbook function is $X(z) = \dfrac{(1-e^{-aT})z}{(z-1)(z-e^{-aT})}$, which is $\mathcal{Z}\left[\dfrac{a}{s(s+a)}\right]$. The same steps give $A = 1$, $B = -1$ and

$$x(k) = 1 - e^{-akT}$$

This is a sampled exponential rise toward 1.

Answer (as printed): $x(k) = 1 - e^{akT}$ for $k \ge 0$. With $e^{-aT}$: $x(k) = 1 - e^{-akT}$.