Actions speak louder than words

Finding \( h(t) \) from Transfer Function \( H(s) \) Find \( h(t) \) from \( H(s) = \frac{s^2}{s^3 + 4s^2 + 4s} \) Answer: \[ h(t) = e^{-2t} - 2t e^{-2t} \] Discussion \( Graph: h(t) = e^{-2t} - 2t e^{-2t} \)   [05220240605]

Finding \( h(t) \) from Transfer Function \( H(s) \) Find \( h(t) \) from \( H(s) = \frac{s^2}{s^3 + 4s^2 + 4s} \) Discussion: We need to perform the inverse Laplace transform. Here are the steps to follow to obtain \( h(t) \) from the transfer function \( H(s) \): Step 1: Factorize the denominator of \( H(s) \) \[ H(s) = \frac{s^2}{s^3 + 4s^2 + 4s} = \frac{s^2}{s(s^2 + 4s + 4)} = \frac{s^2}{s(s + 2)^2} \] Step 2: Convert the fraction into a simpler partial fraction form for easier inversion \[ H(s) = \frac{s^2}{s(s + 2)^2} = \frac{A}{s} + \frac{B}{s + 2} + \frac{C}{(s + 2)^2} \] \[ s^2 = A(s + 2)^2 + Bs(s + 2) + Cs \] \[ s^2 = A s^2 + 4A s + 4A + B s^2 + 2B s + C s \] \[ s^2 = (A + B) s^2 + (4A + 2B + C) s + 4A \] Step 3: Determine the Coefficients \[ s^2 = (A + B) s^2 + (4A + 2B + C) s + 4A \] By comparing coefficients, we get: