## Step 1: Express \( f(t) \) Using Unit Step Functions The unit step function \( u_a(t) \) is defined as: \[ u_a(t) = \begin{cases} & \text{if } t < a \\ 1 & \text{if } t \geq a \end{cases} \] Given the piecewise function: \[ f(t) = \begin{cases} 1 & < t < \frac{\pi}{2} \\ 2\sin t & \frac{\pi}{2} < t < \frac{3\pi}{2} \\ t & t > \frac{3\pi}{2} \end{cases} \] We can write \( f(t) \) as: \[ \begin{aligned} f(t) &= 1 \cdot [u_(t) - u_{\frac{\pi}{2}}(t)] + 2\sin t \cdot [u_{\frac{\pi}{2}}(t) - u_{\frac{3\pi}{2}}(t)] + t \cdot u_{\frac{3\pi}{2}}(t) \\ &= 1 \cdot u_(t) + [2\sin t - 1]u_{\frac{\pi}{2}}(t) + [t - 2\sin t]u_{\frac{3\pi}{2}}(t) \end{aligned} \] **Explanation:** Each interval is represented using differences of unit step functions to "turn on" and "off" each segment at the appropriate bounds. --- ## Step 2: Find the Laplace Transform Recall: - \( \mathcal{L}\{f(t)\} = F(s) \) - \( \mathcal{L}\{f(t)u_a(t)\} = e^{-as} \mathcal{L}\{f(t+a)\} \) Let's compute the Laplace transform term by term: ### 1. \( 1 \cdot u_(t) \) \[ \mathcal{L}\{1\} = \frac{1}{s} \] ### 2. \( [2\sin t - 1]u_{\frac{\pi}{2}}(t) \) \[ \mathcal{L}\{f(t)u_a(t)\} = e^{-as} \mathcal{L}\{f(t+a)\} \] Here, \( f(t) = 2\sin t - 1 \), \( a = \frac{\pi}{2} \): \[ \mathcal{L}\{2\sin t - 1\} = 2 \cdot \frac{1}{s^2 + 1} - \frac{1}{s} \] So, \[ e^{-\frac{\pi}{2}s} \left[2 \cdot \frac{1}{s^2+1} - \frac{1}{s}\right] \] ### 3. \( [t - 2\sin t]u_{\frac{3\pi}{2}}(t) \) Shift by \( a = \frac{3\pi}{2} \): \[ \mathcal{L}\{t\} = \frac{1}{s^2} \] \[ \mathcal{L}\{2\sin t\} = 2 \cdot \frac{1}{s^2 + 1} \] So, \[ e^{-\frac{3\pi}{2}s} \left[ \frac{1}{s^2} - 2 \cdot \frac{1}{s^2+1} \right] \] --- ## Final Answer \[ \boxed{ \begin{aligned} \mathcal{L}\{f(t)\} =\ & \frac{1}{s} + e^{-\frac{\pi}{2}s}\left( \frac{2}{s^2+1} - \frac{1}{s} \right)\\ &+ e^{-\frac{3\pi}{2}s}\left( \frac{1}{s^2} - \frac{2}{s^2+1} \right) \end{aligned} } \] **Explanation:** Each segment is written with unit step functions, and Laplace transforms are found using the shifting theorem and basic Laplace properties. The final answer is a sum of the Laplace transforms of each segment, properly shifted.