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Textbook by: Carlos E. Davila. E-mail the author

# Evaluation of Convolution Integrals

Module by: Carlos E. Davila. E-mail the author

The key to evaluating a convolution integral such as

x ( t ) * h ( t ) = - x ( τ ) h ( t - τ ) d τ x ( t ) * h ( t ) = - x ( τ ) h ( t - τ ) d τ
(1)

is to realize that as far as the integral is concerned, the variable tt is a constant and the integral is over the variable ττ. Therefore, for each tt, we are finding the area of the product x(τ)h(t-τ)x(τ)h(t-τ). Let's look at an example that illustrates how this works.

Example 3.1 Find the convolution of x(t)=u(t)x(t)=u(t) and h(t)=e-tu(t)h(t)=e-tu(t). The convolution integral is given by

h ( t ) * x ( t ) = - e - τ u ( τ ) u ( t - τ ) d τ h ( t ) * x ( t ) = - e - τ u ( τ ) u ( t - τ ) d τ
(2)

Figure 1 shows the graph of e-τu(τ)e-τu(τ), e-tu(t)e-tu(t), and their product. From the graph of the product, it is easy to see the the convolution integral becomes

0 t e - τ d τ = 1 - e - t , t 0 0 , t < 0 0 t e - τ d τ = 1 - e - t , t 0 0 , t < 0
(3)

Signals which can be expressed in functional form should be convolved as in the above example. Other signals may not have an easy functional representation but rather may be piece-wise linear. In order to convolve such signals, one must evaluate the convolution integral over different intervals on the tt-axis so that each distinct interval corresponds to a different expression for x(t)*h(t)x(t)*h(t). The following example illustrates this:

Example 3.2 Suppose we attempt to convolve the unit step function x(t)=u(t)x(t)=u(t) with the trapezoidal function

h ( t ) = t , 0 t < 1 1 , 1 t < 2 0 , elsewhere h ( t ) = t , 0 t < 1 1 , 1 t < 2 0 , elsewhere
(4)

From Figure 2, it can be seen that on the interval 0t<10t<1, the product x(t-τ)h(τ)x(t-τ)h(τ) is an equilateral triangle with area t2/2t2/2. On the interval 1t<21t<2, the area of x(t-τ)h(τ)x(t-τ)h(τ) is t-1/2t-1/2. This latter area results by adding the area of an equilateral triangle having a base of 1, and the area of a rectangle having a base of t-1t-1 and a height of 1. For all values of tt greater than 2, the convolution is 1.5 since x(t-τ)h(τ)=h(τ)x(t-τ)h(τ)=h(τ) and h(τ)h(τ) is a trapezoid having an area of 1.5. Finally, for t<0t<0, the convolution is zero since x(t-τ)h(τ)=0x(t-τ)h(τ)=0.

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