1. **Problem Statement:** Find the z-transform of the given functions: (i) $$F(s) = \frac{1}{(s+a)^2}$$ (ii) $$F(s) = \frac{10(1-e^{-s})}{s(s+2)}$$ 2. **Recall the z-transform definition:** The z-transform of a discrete-time signal $x[n]$ is defined as: $$X(z) = \sum_{n=0}^\infty x[n] z^{-n}$$ However, here we are given Laplace domain functions $F(s)$, so we need to relate the Laplace transform to the z-transform. 3. **Relation between Laplace and z-transform:** If $s = \frac{1}{T} \ln z$, where $T$ is the sampling period, then the z-transform can be related to the Laplace transform by substituting $s$ accordingly. 4. **(i) For** $$F(s) = \frac{1}{(s+a)^2}$$ This is the Laplace transform of the function $f(t) = t e^{-a t} u(t)$, where $u(t)$ is the unit step. The discrete-time sequence sampled at $t = nT$ is: $$x[n] = n T e^{-a n T}$$ The z-transform is: $$X(z) = \sum_{n=0}^\infty n T e^{-a n T} z^{-n}$$ Rewrite $e^{-a T} = r$, then: $$X(z) = T \sum_{n=0}^\infty n (r z^{-1})^n$$ Using the formula for the sum: $$\sum_{n=0}^\infty n x^n = \frac{x}{(1-x)^2}$$ for $|x|<1$, we get: $$X(z) = T \cdot \frac{r z^{-1}}{(1 - r z^{-1})^2} = \frac{T r z^{-1}}{(1 - r z^{-1})^2}$$ 5. **(ii) For** $$F(s) = \frac{10(1-e^{-s})}{s(s+2)}$$ Rewrite numerator: $$1 - e^{-s}$$ corresponds to a difference in Laplace domain, which relates to a discrete difference in time domain. We can write: $$F(s) = 10 \left( \frac{1}{s(s+2)} - \frac{e^{-s}}{s(s+2)} \right)$$ The inverse Laplace of $$\frac{1}{s(s+2)}$$ is: $$f_1(t) = \frac{1}{2} t - \frac{1}{2} + \frac{1}{2} e^{-2 t}$$ The term $$e^{-s}$$ corresponds to a delay of 1 unit in time, so: $$f_2(t) = f_1(t-1) u(t-1)$$ Therefore, the time domain function is: $$f(t) = 10 (f_1(t) - f_1(t-1) u(t-1))$$ Sampling at $t = nT$ and taking the z-transform: $$X(z) = 10 \left( X_1(z) - z^{-1} X_1(z) \right) = 10 X_1(z) (1 - z^{-1})$$ Where $X_1(z)$ is the z-transform of $f_1(nT)$. The z-transform of $f_1(nT)$ can be computed by known transforms or numerical methods depending on $T$. **Final answers:** (i) $$X(z) = \frac{T e^{-a T} z^{-1}}{(1 - e^{-a T} z^{-1})^2}$$ (ii) $$X(z) = 10 X_1(z) (1 - z^{-1})$$ where $X_1(z)$ is the z-transform of $f_1(nT) = \frac{1}{2} n T - \frac{1}{2} + \frac{1}{2} e^{-2 n T}$.