Example 3: Piecewise Convolution

This section illustrates the process of piecewise convolution using functions defined over specific intervals, detailing each step involved in the calculations.

In this section, we explore the example of piecewise convolution between two functions: f(t), defined as 1 for 0≤t≤1, and g(t), defined as t for the same interval. The convolution is computed for two cases, where the ranges of t differ, illustrating the integral calculations distinctly for the two scenarios.

Example 3: Piecewise Convolution

In this section, we analyze the convolution of two piecewise functions, f(t) and g(t), specifically designed to demonstrate how piecewise definitions impact the convolution process:

• f(t) is defined as:

  $$f(t) = \begin{cases} 1, & 0 \leq t \leq 1 \ 0, & t > 1 \end{cases}$$

• g(t) is defined as:

  $$g(t) = t$$

The goal is to compute the convolution of f and g, symbolized as (f ∗ g)(t).

Finding (f ∗ g)(t)

The formula for convolution is given by:

In solving this, we need to consider two scenarios based on the piecewise nature of f(t):

1. Case 1: 0 ≤ t ≤ 1
   • For this scenario, f(\tau) equals 1 for all \tau within the limits of integration:

• Evaluating this integral yields:

2. Case 2: t > 1
   • In this case, f(\tau) remains equal to 1 only from τ=0 to τ=1. Therefore, the limits of the integral adjust accordingly:

• This results in:

Piecewise Result

Combining both cases, we can express the final result of the convolution as a piecewise function:

This example illustrates the importance of considering the nature of the functions when performing convolution, particularly when applying it to real-world engineering scenarios.