1. **Problem Statement:** Determine the Z-transform of the finite duration signals given by: (e) $x_5(n) = \delta(n)$ (f) $x_6(n) = \delta(n - k), k > 0$ (g) $x_7(n) = \delta(n + k), k > 0$ 2. **Recall the Z-transform definition:** The Z-transform of a discrete-time signal $x(n)$ is defined as: $$X(z) = \sum_{n=-\infty}^{\infty} x(n) z^{-n}$$ 3. **Important property of the delta function:** The delta function $\delta(n - n_0)$ is zero everywhere except at $n = n_0$, where it equals 1. 4. **Calculate each Z-transform:** (e) For $x_5(n) = \delta(n)$: $$X_5(z) = \sum_{n=-\infty}^{\infty} \delta(n) z^{-n} = z^{0} = 1$$ (f) For $x_6(n) = \delta(n - k), k > 0$: $$X_6(z) = \sum_{n=-\infty}^{\infty} \delta(n - k) z^{-n} = z^{-k}$$ (g) For $x_7(n) = \delta(n + k), k > 0$: $$X_7(z) = \sum_{n=-\infty}^{\infty} \delta(n + k) z^{-n}$$ Since $\delta(n + k)$ is 1 at $n = -k$, $$X_7(z) = z^{k}$$ 5. **Summary:** - $X_5(z) = 1$ - $X_6(z) = z^{-k}$ - $X_7(z) = z^{k}$ These results show how the delta function shifts affect the powers of $z$ in the Z-transform.