Z-transform of k² e^(−ak) from unit step

Start from the unit step and build the result in three steps: multiply by $e^{-ak}$ (scaling in the z-domain), then multiply by $k$ twice (differentiation in the z-domain).

Properties used

1. Unit step: $\mathcal{Z}[1(k)] = \displaystyle\sum_{k=0}^{\infty}z^{-k} = \frac{z}{z-1}$, $|z| > 1$
2. Scaling (multiplication by $b^k$): $\mathcal{Z}[b^k x(k)] = X\left(\dfrac{z}{b}\right)$
3. Multiplication by $k$: $\mathcal{Z}[k\,x(k)] = -z\dfrac{d}{dz}X(z)$

Proof of property 3: $-z\dfrac{d}{dz}\sum x(k)z^{-k} = -z\sum(-k)x(k)z^{-k-1} = \sum k\,x(k)z^{-k}$.

Step 1: $e^{-ak}$

Use property 2 with $b = e^{-a}$, writing $E = e^{-a}$ for short:

$$\mathcal{Z}[e^{-ak}] = \frac{z/E}{z/E - 1} = \frac{z}{z - E} = \frac{z}{z - e^{-a}}$$

Step 2: $k\,e^{-ak}$

$$\mathcal{Z}[k\,e^{-ak}] = -z\frac{d}{dz}\left(\frac{z}{z-E}\right) = -z\cdot\frac{(z-E) - z}{(z-E)^2} = \frac{Ez}{(z-E)^2}$$

Step 3: $k^2e^{-ak}$

$$\begin{aligned} \mathcal{Z}[k^2e^{-ak}] &= -z\frac{d}{dz}\left[\frac{Ez}{(z-E)^2}\right] \\ &= -z\cdot\frac{E(z-E)^2 - Ez\cdot2(z-E)}{(z-E)^4} \\ &= -z\cdot\frac{E(z-E) - 2Ez}{(z-E)^3} = -z\cdot\frac{-E(z+E)}{(z-E)^3} \\ &= \frac{Ez(z+E)}{(z-E)^3} \end{aligned}$$

Result

$$\mathcal{Z}[k^2e^{-ak}] = \frac{e^{-a}z\,(z + e^{-a})}{(z - e^{-a})^3} = \frac{e^{-a}z^{-1}(1 + e^{-a}z^{-1})}{(1 - e^{-a}z^{-1})^3}, \quad |z| > e^{-a}$$

Check

Take $e^{-a} = 0.5$. Then $X(z) = \dfrac{0.5z(z+0.5)}{(z-0.5)^3}$, and its series is $0.5z^{-1} + z^{-2} + 1.125z^{-3} + z^{-4} + \dots$. Direct values of $k^2(0.5)^k$ for $k = 0$ to $4$ are $0, 0.5, 1, 1.125, 1$, which match.

As a special case, $a = 0$ gives $\mathcal{Z}[k^2] = \dfrac{z(z+1)}{(z-1)^3}$, the known result.

Answer: $\mathcal{Z}[k^2e^{-ak}] = \dfrac{e^{-a}z(z + e^{-a})}{(z - e^{-a})^3}$.