Let's solve the problem step-by-step: --- ## **Given:** \[ Y(s) = \frac{6(s + 50)}{s^2 + 40s + 300} R(s) \] where \( r(t) \) is the input and \( y(t) \) is the output. --- ### **3.1 If \( r(t) \) is a unit step input, find the output \( y(t) \).** #### **Step 1: Find Laplace Transform of the input** A unit step input: \[ r(t) = u(t) \implies R(s) = \frac{1}{s} \] #### **Step 2: Substitute \( R(s) \) into the system** \[ Y(s) = \frac{6(s + 50)}{s^2 + 40s + 300} \cdot \frac{1}{s} = \frac{6(s + 50)}{s(s^2 + 40s + 300)} \] #### **Step 3: Partial Fraction Decomposition** First, factor the denominator: \[ s^2 + 40s + 300 = (s + 10)(s + 30) \] So, \[ Y(s) = \frac{6(s + 50)}{s(s + 10)(s + 30)} \] Let: \[ \frac{6(s + 50)}{s(s + 10)(s + 30)} = \frac{A}{s} + \frac{B}{s + 10} + \frac{C}{s + 30} \] Multiply both sides by the denominator: \[ 6(s + 50) = A(s + 10)(s + 30) + B s (s + 30) + C s (s + 10) \] Expand the right side and solve for \(A\), \(B\), and \(C\): - Set \(s = 0\): \(6(0 + 50) = A(10)(30) \implies 300 = 300A \implies A = 1\) - Set \(s = -10\): \(6(-10 + 50) = B(-10)(20)\) - \(6(40) = B(-10)(20) \implies 240 = -200B \implies B = -1.2\) - Set \(s = -30\): \(6(-30 + 50) = C(-30)(-20)\) - \(6(20) = C(600) \implies 120 = 600C \implies C = 0.2\) So: \[ Y(s) = \frac{1}{s} - \frac{1.2}{s + 10} + \frac{0.2}{s + 30} \] --- #### **Step 4: Find \( y(t) \) using the inverse Laplace transform** \[ L^{-1}\left\{ \frac{1}{s} \right\} = 1 \] \[ L^{-1}\left\{ \frac{1.2}{s + 10} \right\} = 1.2 e^{-10t} \] \[ L^{-1}\left\{ \frac{0.2}{s + 30} \right\} = 0.2 e^{-30t} \] So, \[ y(t) = 1 - 1.2 e^{-10t} + 0.2 e^{-30t} \] --- ### **3.2 What is the final value of \( y(t) \)?** Use the Final Value Theorem: \[ \lim_{t \to \infty} y(t) = \lim_{s \to 0} s Y(s) \] Recall: \[ Y(s) = \frac{6(s + 50)}{s(s + 10)(s + 30)} \] So: \[ \lim_{s \to 0} s Y(s) = \lim_{s \to 0} \frac{6(s + 50)}{(s + 10)(s + 30)} = \frac{6 \cdot 50}{10 \cdot 30} = \frac{300}{300} = 1 \] Or, from the time domain solution: \[ \lim_{t \to \infty} y(t) = 1 - 0 + 0 = 1 \] --- ## **Final Answers:** ### **3.1** \[ \boxed{y(t) = 1 - 1.2 e^{-10t} + 0.2 e^{-30t}} \] ### **3.2** \[ \boxed{1} \]