Let's solve the problem step by step: --- ### **Step 1: Simplify the function using Heaviside functions** Given: \[ f(t) = (2-t)[h(t-4) - h(t-6)] \] Recall: - \( h(t-a) \) is the Heaviside (unit step) function. - \( h(t-4) - h(t-6) \) is 1 when \( 4 \leq t < 6 \), 0 otherwise. So, over \( 4 \leq t < 6 \): \[ f(t) = 2-t \] We can write: \[ f(t) = (2-t)[h(t-4) - h(t-6)] = (2-t)h(t-4) - (2-t)h(t-6) \] But we want each term as a function of \( t \) shifted by \( a \): Let’s use the property: \[ f(t)h(t-a) = f(t+a)h(t-a) \] So, - \( (2-t)h(t-4) = (2-(t))h(t-4) = [2-(t-4+4)]h(t-4) = [2-(t-4)-4]h(t-4) \) - Actually, the correct property is: for Laplace, \( f(t)h(t-a) \to e^{-as}\mathcal{L}\{f(t+a)\} \) Instead, let's keep: \[ f(t) = (2-t)h(t-4) - (2-t)h(t-6) \] --- **Explanation for Step 1:** We rewrote \( f(t) \) in terms of Heaviside functions to identify the intervals for Laplace transform and set up for shifting properties. --- ### **Step 2: Apply Laplace Transform Using Shifting Property** Recall Laplace property: \[ \mathcal{L}\{f(t-a)h(t-a)\} = e^{-as}F(s) \] Let's write \( f(t) \) in shifted form: - For \( (2-t)h(t-4) \), let \( \tau = t-4 \Rightarrow t = \tau+4 \): \[ (2-t)h(t-4) = [2-(\tau+4)]h(\tau) = (-\tau-2)h(\tau) \] So, \( f_1(\tau) = -(\tau+2) \). Now Laplace of \( f_1(\tau) \): \[ \mathcal{L}\{-(t+2)\} = -\mathcal{L}\{t\} - 2\mathcal{L}\{1\} = -\frac{1}{s^2} - \frac{2}{s} \] So: \[ \mathcal{L}\{(2-t)h(t-4)\} = e^{-4s}\left(-\frac{1}{s^2} - \frac{2}{s}\right) \] Similarly, for \( (2-t)h(t-6) \), shift by 6: - Let \( \tau = t-6 \Rightarrow (2-t)h(t-6) = [2-(\tau+6)]h(\tau) = (-\tau-4)h(\tau) \) So Laplace is: \[ \mathcal{L}\{(2-t)h(t-6)\} = e^{-6s}\left(-\frac{1}{s^2} - \frac{4}{s}\right) \] **Now combine:** \[ F(s) = \mathcal{L}\{f(t)\} = e^{-4s}\left(-\frac{1}{s^2} - \frac{2}{s}\right) - e^{-6s}\left(-\frac{1}{s^2} - \frac{4}{s}\right) \] --- **Explanation for Step 2:** We used the Laplace shifting theorem to transform each term, shifted by 4 and 6, and subtracted to get the final Laplace transform. --- ## **Final Answer:** \[ \boxed{ F(s) = -\left(\frac{1}{s^2} + \frac{2}{s}\right)e^{-4s} + \left(\frac{1}{s^2} + \frac{4}{s}\right)e^{-6s} } \] Or, equivalently: \[ \boxed{ F(s) = \left[-\frac{1}{s^2}-\frac{2}{s}\right]e^{-4s} + \left[\frac{1}{s^2}+\frac{4}{s}\right]e^{-6s} } \]