Let's solve the convolution step-by-step as requested: --- ### **Step 1: Understand the signals \( f(t) \) and \( g(t) \)** - \( f(t) \) is a rectangular pulse from \(-0.5\) to \(0.5\) with height 1. - \( g(t) \) is a rectangular pulse from \(-1\) to \(1\) with height 0.5. **Explanation:** Both functions are simple rectangular pulses, but \( g(t) \) is wider and shorter than \( f(t) \). --- ### **Step 2: Set up the convolution integral** The convolution is given by: \[ h(t) = \int_{-\infty}^{\infty} f(\tau)g(t-\tau)d\tau \] Since both functions are zero outside their respective intervals, the limits of integration are determined by the overlap of the two pulses as one slides past the other: - \( f(\tau) \) is nonzero for \( -0.5 \leq \tau \leq 0.5 \) - \( g(t-\tau) \) is nonzero for \( -1 \leq t-\tau \leq 1 \implies t-1 \leq \tau \leq t+1 \) So, the overlap occurs for: \[ \max(-0.5, t-1) \leq \tau \leq \min(0.5, t+1) \] Thus, \[ h(t) = \int_{\max(-0.5, t-1)}^{\min(0.5, t+1)} f(\tau)g(t-\tau)d\tau \] Since \(f(\tau) = 1\) and \(g(t-\tau) = 0.5\) within their intervals, \[ h(t) = \int_{\text{overlap}} 0.5\, d\tau = 0.5 \times (\text{length of overlap}) \] **Explanation:** The value of \( h(t) \) at each \( t \) is half the length of the overlap between the two rectangular pulses. --- ### **Step 3: Calculate and sketch \( h(t) \), and analyze widths** #### **Find the overlap as a function of \( t \):** - For \( t < -1.5 \): No overlap, \( h(t) = 0 \) - For \( -1.5 \leq t < -0.5 \): Overlap from \( -0.5 \) to \( t+1 \) Length: \( (t+1) - (-0.5) = t + 1.5 \) So, \( h(t) = 0.5 \times (t+1.5) \) - For \( -0.5 \leq t < 0.5 \): Full overlap, length = \(0.5 - (-0.5) = 1\) So, \( h(t) = 0.5 \times 1 = 0.5 \) - For \( 0.5 \leq t < 1.5 \): Overlap from \( t-1 \) to \( 0.5 \) Length: \( 0.5 - (t-1) = 1.5 - t \) So, \( h(t) = 0.5 \times (1.5 - t) \) - For \( t \geq 1.5 \): No overlap, \( h(t) = 0 \) #### **Sketch:** - \( h(t) \) is a triangle: rises from 0 at \( t = -1.5 \) to 0.5 at \( t = -0.5 \), stays at 0.5 until \( t = 0.5 \), then falls to 0 at \( t = 1.5 \). #### **Relative widths:** - \( f(t) \) width: 1 - \( g(t) \) width: 2 - \( h(t) \) width: 3 (from \( -1.5 \) to \( 1.5 \)), i.e., sum of the widths of \( f(t) \) and \( g(t) \). **Explanation:** The convolution of two rectangles results in a "tent" (trapezoid/triangle) whose width is the sum of the widths of the two original pulses. --- ### **Final Answer:** \[ h(t) = \begin{cases} 0 & |t| > 1.5 \\ 0.5 \times (t + 1.5) & -1.5 < t < -0.5 \\ 0.5 & -0.5 \leq t \leq 0.5 \\ 0.5 \times (1.5 - t) & 0.5 < t < 1.5 \\ 0 & \text{otherwise} \end{cases} \] - The width of \( h(t) \) is the sum of the widths of \( f(t) \) and \( g(t) \). - The maximum value of \( h(t) \) is 0.5. - Sketch \( h(t) \) as a triangle/trapezoid from \( t = -1.5 \) to \( t = 1.5 \), peaking at 0.5 between \( t = -0.5 \) and \( t = 0.5 \). --- **Observation:** The convolution of two rectangular pulses produces a tent-shaped function whose width is the sum of the original pulse widths.