Inverse Z-transform by partial fractions

Since $X(z)$ has a zero at $z=0$, expand $X(z)/z$ in partial fractions.

$$\frac{X(z)}{z} = \frac{2z^2+1}{(z-2)^2(z-1)} = \frac{A}{z-1} + \frac{B}{(z-2)^2} + \frac{C}{z-2}$$

Coefficients

$$\begin{aligned} A &= \left.\frac{2z^2+1}{(z-2)^2}\right|_{z=1} = \frac{2+1}{1} = 3 \\ B &= \left.\frac{2z^2+1}{z-1}\right|_{z=2} = \frac{8+1}{1} = 9 \\ C &= \left.\frac{d}{dz}\left[\frac{2z^2+1}{z-1}\right]\right|_{z=2} = \left.\frac{4z(z-1) - (2z^2+1)}{(z-1)^2}\right|_{z=2} \\ &= \frac{8 - 9}{1} = -1 \end{aligned}$$

So

$$X(z) = \frac{3z}{z-1} + \frac{9z}{(z-2)^2} - \frac{z}{z-2}$$

Inverse transforms used

• $\mathcal{Z}^{-1}\left[\frac{z}{z-1}\right] = 1$
• $\mathcal{Z}^{-1}\left[\frac{z}{z-a}\right] = a^k$
• $\mathcal{Z}^{-1}\left[\frac{z}{(z-a)^2}\right] = k\,a^{k-1}$

Result

$$x(k) = 3 + 9k\,2^{k-1} - 2^k, \quad k = 0, 1, 2, \dots$$

Check (first few values)

$x(0) = \lim_{z\to\infty}X(z) = 2$ (initial value theorem), which agrees. Long division of $X(z) = \frac{2z^3+z}{z^3-5z^2+8z-4}$ also gives $2 + 10z^{-1} + 35z^{-2} + 103z^{-3} + \cdots$

Answer: $x(k) = 3 + 9k(2^{k-1}) - 2^k$, which can also be written $x(k) = 3 + (4.5k - 1)\,2^k$.