Question

Determine the Laplace transform of \(x(t) = (t^2 - 2t)u(t - 1)\)

• A. \(\dfrac{2e^{-s}(1 - s)}{s^3}\)
• B. \(\dfrac{e^{-2s}(1 - s)}{s^2}\)
• C. \(\dfrac{2e^{-2s}(1 + s)}{s^3}\)
• D. \(\dfrac{e^{-s}(1 - s)}{s^2}\)

Hint:

Use time-shifting property for \(u(t - a)\): replace \(t\) by \(t - a\), and multiply Laplace by \(e^{-as}\).

Answer 1

The Correct Option is A

Given: \[ x(t) = (t^2 - 2t)u(t - 1) \] We use the time-shifting property of the Laplace Transform: \[ \mathcal{L}\{f(t - a)u(t - a)\} = e^{-as}F(s) \] Let: \[ f(t) = t^2 - 2t \Rightarrow f(t - 1) = (t - 1)^2 - 2(t - 1) = t^2 - 2t + 1 - 2t + 2 = t^2 - 4t + 3 \] We will take Laplace of \(x(t) = f(t - 1)u(t - 1)\) \[ \Rightarrow \mathcal{L}\{x(t)\} = e^{-s} \cdot \mathcal{L}\{t^2 - 4t + 3\} \] Now compute Laplace of each term: \[ \mathcal{L}\{t^2\} = \dfrac{2}{s^3}, \mathcal{L}\{t\} = \dfrac{1}{s^2}, \mathcal{L}\{1\} = \dfrac{1}{s} \] \[ \Rightarrow \mathcal{L}\{x(t)\} = e^{-s} \left(\dfrac{2}{s^3} - \dfrac{4}{s^2} + \dfrac{3}{s} \right) = \dfrac{e^{-s}(2 - 4s + 3s^2)}{s^3} = \dfrac{2e^{-s}(1 - s)}{s^3} \]