Let's solve each convolution step by step. ### **Preliminaries** Let \( p(t) \) be the **unit step function**: \[ p(t) = \begin{cases} 1, & t \geq \\ , & t < \end{cases} \] The **convolution** of \( f(t) \) and \( g(t) \) is defined as: \[ (f \ast g)(t) = \int_{-\infty}^{\infty} f(\tau)g(t-\tau) d\tau \] For causal functions (zero for \( t< \)), this simplifies to: \[ (f \ast g)(t) = \int_{}^{t} f(\tau)g(t-\tau) d\tau \] --- ## **A. \( p(t) \ast p(t-1) \)** ### **Step 1: Write the convolution integral** \[ (p(t) \ast p(t-1))(t) = \int_{}^{t} p(\tau) p(t-\tau-1) d\tau \] ### **Step 2: Analyzing the functions** - \( p(\tau) = 1 \) for \( \tau \geq \) - \( p(t-\tau-1) = 1 \) for \( t-\tau-1 \geq \implies \tau \leq t-1 \) ### **Step 3: Limits of integration** - Both are 1 only if \( \leq \tau \leq t-1 \) - But \( \tau \leq t \), so upper limit is \( \min(t, t-1) \) - For \( t < 1 \), \( t-1 < \), so integration is over an empty interval: result is . ### **Step 4: Evaluate the integral** For \( t \geq 1 \): \[ \int_{}^{t-1} 1 \, d\tau = [\tau]_{}^{t-1} = t-1 \] For \( t < 1 \): \[ \int_{}^{} 1 \, d\tau = \] ### **Step 5: Combine** \[ (p(t) \ast p(t-1))(t) = \begin{cases} , & t < 1 \\ t-1, & t \geq 1 \end{cases} \] Or, \[ \boxed{(p(t) \ast p(t-1))(t) = (t-1) p(t-1)} \] --- ## **B. \( e^{-t} p(t) \ast p(t) \)** ### **Step 1: Write the convolution** \[ (e^{-t} p(t) \ast p(t))(t) = \int_{}^{t} e^{-\tau} p(\tau) p(t-\tau) d\tau \] But both step functions are 1 for \( \leq \tau \leq t \): \[ = \int_{}^{t} e^{-\tau} d\tau \] ### **Step 2: Evaluate the integral** \[ = \left[ -e^{-\tau} \right]_{}^{t} = -(e^{-t} - 1) = 1 - e^{-t} \] ### **Step 3: Include causality** \[ \boxed{(e^{-t} p(t) \ast p(t))(t) = (1 - e^{-t}) p(t)} \] --- ## **C. \( t p(t) \ast p(t) \)** ### **Step 1: Write the convolution** \[ (t p(t) \ast p(t))(t) = \int_{}^{t} \tau \cdot 1 \, d\tau = \int_{}^{t} \tau \, d\tau \] ### **Step 2: Evaluate** \[ = \left[ \frac{1}{2} \tau^2 \right]_{}^{t} = \frac{1}{2} t^2 \] ### **Step 3: Include causality** \[ \boxed{(t p(t) \ast p(t))(t) = \frac{1}{2} t^2 p(t)} \] --- ## **Final Answers** \[ \boxed{ \begin{aligned} \text{A.}\quad & (p(t) \ast p(t-1))(t) = (t-1) p(t-1) \\ \text{B.}\quad & (e^{-t} p(t) \ast p(t))(t) = (1 - e^{-t}) p(t) \\ \text{C.}\quad & (t p(t) \ast p(t))(t) = \frac{1}{2} t^2 p(t) \end{aligned} } \]