The Evaluation of the Convolution Integral

Contents

Introduction

This page goes through an example that describes how to evaluate the convolution integral for a piecewise function (the recatngular pulse).  The key idea is to split the integral up into distinct regions where the integral can be evaluated.

Consider the system described by the differential equation:

which has an impulse response given  by

       

We will use convolution to find the zero input response of this system to the input given by a rectangular pulse, which we define piecewise by three distinction sections.

       

The next section reiterates the development of the page deriving the convolution integral.  If you feel you know that material, you can skip ahead to the mechanics of using the convolution integral.

Review: Convolution as sum of impulse responses

For continuity with the page deriving the convolution integral
we can approximate the input by a series of impulses...
   
plot  the response of the system to each of these impulses...
and plot the response as the sum of the individual responses

 (the red line on the plot shows is the exact solution).

 

Animation: The Convolution Integral

Likewise the convolution can be considered from the point of view of the convolution integral.

Click on picture below to see animation.

The Mechanics of the Convolution Integral

Now lets discuss how we can find an exact solution to this problem.   To find the output of the system with impulse response

to the input

we will use the convolution integral

Because the input function has three distinct regions t<0, 0<t<1 and 1<t, we will need to split up the integral into three parts.

Section 1: t<0

For t<0 the argument of the impulse function (t-λ) is always negative.  Since h(t-λ)=0 for (t-λ)<0, the result of the integral is zero for t<0. 

This situation is depicted graphically below (t=-0.2):

 So the result for the first part of our solution is

Section 2:  0<t<1

For 0<t<1 we need to evaluate the integral only from λ=0 to λ=t, since f(λ)=0 when λ<0, and h(t-λ)=0 when (t-λ)<0 (or, equivalently t<λ).  So the integral becomes, in effect: 

This situation is depicted graphically below (t=0.5):

We can now evaluate the integral of the solid black line:

Thus, the result for the second part of the solution  is

Section 3:  1<t

For 1<t we need to evaluate the integral only from λ=0 to λ=1, since f(λ)=0 when λ<0 and when λ>1.  So the integral becomes, in effect: 

This situation is depicted graphically below (t=1.2):

We can now evaluate the integral:

Thus, the result for the third part of the solution is:

The Complete Answer

We can get the results for all time by combining the solutions from the three parts.



This result is shown below.  Click on image to go to an animation of the procedure.

This problem is solved elsewhere using the Laplace Transform (which is a much simpler technique, computationally).


The convolution integral is a completely general method for finding the output of a linear system for any input.  The integral is often difficult to evaluate, but this page gives one example of how this can be accomplished for a relatively simple system. 


References

© Copyright 2005 to 2015 Erik Cheever    This page may be freely used for educational purposes.

Erik Cheever       Department of Engineering         Swarthmore College