1. **Problem statement:** Find the z-transform of the given Laplace domain 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 $$X(z) = \sum_{n=0}^\infty x[n] z^{-n}$$ 3. **Relation between Laplace and z-transform:** The z-transform is typically used for discrete signals, while Laplace transform is for continuous signals. To find the z-transform from Laplace domain expressions, we often consider the inverse Laplace transform to get the time-domain function and then sample it to get the discrete sequence for z-transform. 4. **Part (i):** Given $$F(s) = \frac{1}{(s+a)^2}$$ The inverse Laplace transform is known: $$f(t) = t e^{-a t} u(t)$$ where $u(t)$ is the unit step function. 5. **Sampling:** Sample at $t = nT$ (assuming sampling period $T=1$ for simplicity): $$f[n] = n e^{-a n}$$ for $n \geq 0$ 6. **z-transform of $f[n]$:** $$X(z) = \sum_{n=0}^\infty n e^{-a n} z^{-n}$$ 7. **Use the formula for sum:** Recall that $$\sum_{n=0}^\infty n r^n = \frac{r}{(1-r)^2}$$ for $|r|<1$. Here, $$r = e^{-a} z^{-1}$$ So, $$X(z) = \sum_{n=0}^\infty n (e^{-a} z^{-1})^n = \frac{e^{-a} z^{-1}}{(1 - e^{-a} z^{-1})^2}$$ 8. **Simplify:** $$X(z) = \frac{e^{-a} z^{-1}}{(1 - e^{-a} z^{-1})^2}$$ This is the z-transform of the sampled function. --- 9. **Part (ii):** Given $$F(s) = \frac{10(1 - e^{-s})}{s(s+2)}$$ 10. **Rewrite:** $$F(s) = 10 \cdot \frac{1 - e^{-s}}{s(s+2)}$$ 11. **Inverse Laplace transform:** Recall that $$\mathcal{L}^{-1}\left\{ \frac{1}{s(s+2)} \right\} = \frac{1}{2} (1 - e^{-2t}) u(t)$$ Also, multiplication by $(1 - e^{-s})$ in Laplace corresponds to the difference of the function at $t$ and $t-1$ (unit delay). 12. **Therefore,** $$f(t) = 10 \left[ \frac{1}{2} (1 - e^{-2t}) u(t) - \frac{1}{2} (1 - e^{-2(t-1)}) u(t-1) \right] = 5 \left[ (1 - e^{-2t}) u(t) - (1 - e^{-2(t-1)}) u(t-1) \right]$$ 13. **Sampling at $t=n$:** $$f[n] = 5 \left[ (1 - e^{-2n}) - (1 - e^{-2(n-1)}) u(n-1) \right]$$ For $n=0$, $u(-1)=0$, so $$f[0] = 5 (1 - 1) = 0$$ For $n \geq 1$, $$f[n] = 5 \left[ (1 - e^{-2n}) - (1 - e^{-2(n-1)}) \right] = 5 (e^{-2(n-1)} - e^{-2n})$$ 14. **Simplify:** $$f[n] = 5 e^{-2(n-1)} (1 - e^{-2})$$ for $n \geq 1$, and $f[0] = 0$ 15. **z-transform:** $$X(z) = \sum_{n=0}^\infty f[n] z^{-n} = 0 + \sum_{n=1}^\infty 5 e^{-2(n-1)} (1 - e^{-2}) z^{-n}$$ 16. **Rewrite sum:** Let $m = n-1$, then $$X(z) = 5 (1 - e^{-2}) z^{-1} \sum_{m=0}^\infty (e^{-2} z^{-1})^m = 5 (1 - e^{-2}) z^{-1} \cdot \frac{1}{1 - e^{-2} z^{-1}}$$ 17. **Final expression:** $$X(z) = \frac{5 (1 - e^{-2}) z^{-1}}{1 - e^{-2} z^{-1}}$$ --- **Final answers:** (i) $$X(z) = \frac{e^{-a} z^{-1}}{(1 - e^{-a} z^{-1})^2}$$ (ii) $$X(z) = \frac{5 (1 - e^{-2}) z^{-1}}{1 - e^{-2} z^{-1}}$$