How to Verify a Convolution Integral Problem Numerically

To arrive at the analytical solution, you need to break the problem down into cases, or intervals of
time t where you can evaluate the integral to form a piecewise contiguous solution. When you put these contiguous intervals together, you have a complete solution for all values of t:

 you can figure out how many cases of convolution are formed by analyzing signals.

• if both signals are finite, then there are 4 or 5 cases (e.g no overlap, entering overlap, full overlap, leaving overlap and no overlap)

• if both signals are infinite, then there are only two cases usually (e.g. no overlap and partial overlap)

• if one of signal is finite while other is infinite, then there are three cases (e.g. no overlap, partial overlap and full overlap)

• Note that the output/result of convolution of two signals support interval runs from the sum of the starting values for x(t) and h(t) and ends at the ending values.

Consider the convolution integral for two continuous-time finite signals x(t) and h(t) shown.

cases in this convolution are described in following GIF image

In this example, each case is treat as a step in the solution that follows:

• Case 1: You start with t + 2 < 0, or equivalently t < –2. The product of waveforms h(λ) and x(t – λ) don’t overlap in the convolution integral integrand,
  

  

  so for Case 1 the integral is just y(t) = 0 for t < –2.
• Case 2: Consider the next interval to the right of Case 1. This case is formed by the joint condition of the signal edge at t + 2 ≥ 0 and t + 2 < 3, which is equivalent to writing –2 ≤ t < 1.
  

  

  The integrand of the convolution integral is 2x3 with the integration limits running from 0 to t + 2.
  You find y(t) on this interval by evaluating the convolution integral:

• Case 3: The next interval in the series is t + 2≥ 3 and t – 3 < 0 or 1 ≤ t < 3.
  

  

  The integrand is again 2 x 3, but now the integration limits run from 0 to 3. Evaluating the convolution integral yields the following:

• Case 4: The final interval involving signal overlap occurs when t – 3 ≥ 0 and t – 3 < 3, which implies that 3 ≤ t ≤ 6. The integration limits run from t – 4 to 3, 
  

  

  so the convolution integral result is

• Case 5: You can see that when t – 3 > 3 or t > 6, 
  

  

  no overlap occurs between the signals of the integrand, so the output is y(t) = 0, t > 6.

Collecting the various pieces of the solution, you have the following: