Let's solve each part step by step: --- ### (a) \( F(s) = \frac{e^{-s}}{s^2 - 3} + \frac{6}{s^3} \) #### **Step 1: Inverse Laplace of each term** **First Term:** \( \frac{e^{-s}}{s^2 - 3} \) - \( \mathcal{L}^{-1}\left\{\frac{1}{s^2 - a^2}\right\} = \frac{1}{a} \sinh(at) \) - \( e^{-as}F(s) \xrightarrow{\mathcal{L}^{-1}} f(t-a)u(t-a) \) (shifting property) Here, \( a = \sqrt{3} \): So, \[ \mathcal{L}^{-1}\left\{\frac{1}{s^2 - 3}\right\} = \frac{1}{\sqrt{3}} \sinh(\sqrt{3}t) \] Apply shifting \( e^{-s} \): \[ \mathcal{L}^{-1}\left\{\frac{e^{-s}}{s^2 - 3}\right\} = \frac{1}{\sqrt{3}} \sinh(\sqrt{3}(t-1))u(t-1) \] **Second Term:** \( \frac{6}{s^3} \) - \( \mathcal{L}^{-1}\left\{\frac{n!}{s^{n+1}}\right\} = t^n \) - So, \( \frac{6}{s^3} = \frac{3!}{s^3} \Rightarrow t^2 \) \[ \mathcal{L}^{-1}\left\{\frac{6}{s^3}\right\} = t^2 \] #### **Final Answer for (a):** \[ f(t) = \frac{1}{\sqrt{3}} \sinh(\sqrt{3}(t-1)) \, u(t-1) + t^2 \] --- ### (b) \( F(s) = \frac{s-3}{s^2 + 2s + 5} \) #### **Step 1: Complete the square in denominator** \[ s^2 + 2s + 5 = (s+1)^2 + 2^2 \] #### **Step 2: Express numerator in terms of denominator** \[ s - 3 = (s+1) - 4 \] So, \[ F(s) = \frac{(s+1) - 4}{(s+1)^2 + 4} = \frac{s+1}{(s+1)^2 + 4} - \frac{4}{(s+1)^2 + 4} \] #### **Step 3: Inverse Laplace** - \( \mathcal{L}^{-1}\left\{\frac{s-a}{(s-a)^2 + \omega^2}\right\} = e^{at}\cos(\omega t) \) - \( \mathcal{L}^{-1}\left\{\frac{\omega}{(s-a)^2 + \omega^2}\right\} = e^{at}\sin(\omega t) \) - Here, \( a = -1 \), \( \omega = 2 \): So, \[ \mathcal{L}^{-1}\left\{\frac{s+1}{(s+1)^2 + 4}\right\} = e^{-t}\cos(2t) \] \[ \mathcal{L}^{-1}\left\{\frac{4}{(s+1)^2 + 4}\right\} = 2e^{-t}\sin(2t) \] #### **Final Answer for (b):** \[ f(t) = e^{-t}\cos(2t) - 2e^{-t}\sin(2t) \] --- ### (c) \( H(s) = \frac{s}{(s-1)(4s-6)} \) #### **Step 1: Factor and Partial Fractions** \( 4s - 6 = 4(s-1.5) \) Write: \[ \frac{s}{(s-1)(4s-6)} = \frac{s}{4(s-1)(s-1.5)} \] Set up partial fractions: \[ \frac{s}{4(s-1)(s-1.5)} = \frac{A}{s-1} + \frac{B}{s-1.5} \] \[ \frac{s}{4(s-1)(s-1.5)} = \frac{A(s-1.5) + B(s-1)}{4(s-1)(s-1.5)} \] Equate numerators: \[ s = A(s-1.5) + B(s-1) \] Let \( s=1 \): \( 1 = A(1-1.5) + B(0) \rightarrow 1 = -0.5A \Rightarrow A = -2 \) Let \( s=1.5 \): \( 1.5 = A(0) + B(1.5-1) \rightarrow 1.5 = 0.5B \Rightarrow B = 3 \) So, \[ \frac{s}{4(s-1)(s-1.5)} = \frac{-2}{4(s-1)} + \frac{3}{4(s-1.5)} = -\frac{1}{2(s-1)} + \frac{3}{4(s-1.5)} \] #### **Step 2: Inverse Laplace** - \( \mathcal{L}^{-1}\left\{ \frac{1}{s-a} \right\} = e^{at} \) So, \[ h(t) = -\frac{1}{2}e^{t} + \frac{3}{4}e^{1.5t} \] --- ## **Final Answers** **(a)** \[ f(t) = \frac{1}{\sqrt{3}} \sinh(\sqrt{3}(t-1))\, u(t-1) + t^2 \] **(b)** \[ f(t) = e^{-t}\cos(2t) - 2e^{-t}\sin(2t) \] **(c)** \[ h(t) = -\frac{1}{2}e^{t} + \frac{3}{4}e^{1.5t} \]