Z-transforms of t²e^(−at) and t·x(t−nT)

(i) $x(t) = t^2e^{-at}$

Sampled: $x(kT) = T^2k^2(e^{-aT})^k$. Let $r = e^{-aT}$.

From $\mathcal{Z}[1] = \frac{z}{z-1}$ and $\mathcal{Z}[k\,x(k)] = -z\frac{dX}{dz}$:

$$\mathcal{Z}[k] = \frac{z}{(z-1)^2}, \qquad \mathcal{Z}[k^2] = \frac{z(z+1)}{(z-1)^3}$$

Multiplying by $r^k$ replaces $z$ by $z/r$:

$$\mathcal{Z}[k^2r^k] = \frac{(z/r)(z/r+1)}{(z/r-1)^3} = \frac{rz(z+r)}{(z-r)^3}$$ $$X(z) = \frac{T^2e^{-aT}z(z+e^{-aT})}{(z-e^{-aT})^3}$$

(ii) $y(t) = t\cdot x(t-nT)$

Assume $x(t) = 0$ for $t < 0$ and $\mathcal{Z}[x(t)] = X(z)$.

Step 1 (real translation): $\mathcal{Z}[x(t-nT)] = z^{-n}X(z)$.

Step 2 (multiplication by $t$): for sampled signals $t = kT$, and $\mathcal{Z}[kT\,f(kT)] = -Tz\frac{d}{dz}F(z)$.

$$\begin{aligned} Y(z) &= -Tz\frac{d}{dz}\left[z^{-n}X(z)\right] \\ &= -Tz\left[-nz^{-n-1}X(z) + z^{-n}\frac{dX(z)}{dz}\right] \\ &= z^{-n}\left[nT\,X(z) - Tz\frac{dX(z)}{dz}\right] \end{aligned}$$

Example: if $x(t) = 1(t)$, $X(z) = \frac{z}{z-1}$ and $\frac{dX}{dz} = \frac{-1}{(z-1)^2}$, so $Y(z) = z^{-n}\left[\frac{nTz}{z-1} + \frac{Tz}{(z-1)^2}\right]$, the transform of $kT$ for $k \ge n$.

Answers: (i) $\dfrac{T^2e^{-aT}z(z+e^{-aT})}{(z-e^{-aT})^3}$; (ii) $z^{-n}\left[nTX(z) - Tz\dfrac{dX(z)}{dz}\right]$.