Laplace Transform Convolution Integral Complete Examples

laplace transform convolution integral

The Laplace transform convolution integral is one of the applications of the Laplace transform.

The convolution of two signals consists of time-reversing one of the signals, shifting it, and multiplying it point by point with the second signal, and integrating the product.

Laplace Transform Convolution Integral

The term convolution means “folding.” A convolution is an invaluable tool for the engineer because it provides a means of viewing and characterizing physical systems.

For example, it is used in finding the response y(t) of a system to an excitation x(t), knowing the system impulse response h(t). This is achieved through the convolution integral, defined as

 laplace transform convolution intergral
(1)

or simply

 laplace transform convolution intergral
(2)

where λ is a dummy variable and the asterisk denotes convolution. Equations.(1) or (2) states that the output is equal to the input convolved with the unit impulse response. The convolution process is commutative:

 laplace transform convolution intergral
(3a)

or

 laplace transform convolution intergral
(3b)

This implies that the order in which the two functions are convolved is immaterial. We will see shortly how to take advantage of this commutative property when performing graphical computation of the convolution integral.

The convolution of two signals consists of time-reversing one of the signals, shifting it, and multiplying it point by point with the second signal, and integrating the product.

The convolution integral in Equation.(1) is the general one; it applies to any linear system. However, the convolution integral can be simplified if we assume that a system has two properties.

First, if
x(t) = 0 for
t < 0,
then

 laplace transform convolution intergral
(4)

Second, if the system’s impulse response is causal

i.e., h(t) = 0 for t < 0, then
h(t λ) = 0 for t λ < 0 or λ > t,
so that Equation.(4)
becomes

 laplace transform convolution intergral
(5)

Here are some properties of the convolution integral.

 laplace transform convolution intergral

Before learning how to evaluate the convolution integral in Equation.(5), let us establish the link between the Laplace transform and the convolution integral.

Given two functions f1(t) and f2(t) with Laplace transforms F1(s) and F2(s), respectively, their convolution is

 laplace transform convolution intergral
(6)

Taking the Laplace transform gives

 laplace transform convolution intergral
(7)

To prove that Equation.(7) is true, we begin with the fact that F1(s) is defined as

 laplace transform convolution intergral
(8)

Multiplying this with F2(s) gives

 laplace transform convolution intergral
(9)

We recall from the time shift property that the term in brackets can be written as

 laplace transform convolution intergral
(10)

Substituting Equation.(10) into Equation.(9) gives

 laplace transform convolution intergral
(11)

Interchanging the order of integration results in

 laplace transform convolution intergral
(12)

The integral in brackets extends only from 0 to t because the delayed unit step u(t λ) = 1 for λ < t and u(t λ) = 0 for λ > t. We notice that the integral is the convolution of f1(t) and f2(t) as in Equation.(6).

Hence,

 laplace transform convolution intergral
(13)

as desired. This indicates that convolution in the time domain is equivalent to multiplication in the s domain.

For example, if
x(t) = 4et and
h(t) = 5e−2t,
applying the property in Equation.(13), we get

 laplace transform convolution intergral
(14)

Although we can find the convolution of two signals using Equation.(13), as we have just done, if the product F1(s)F2(s) is very complicated, finding the inverse may be tough.

Also, there are situations in which f1(t) and f2(t) are available in the form of experimental data and there are no explicit Laplace transforms. In these cases, one must do the convolution in the time domain.

The process of convolving two signals in the time domain is better appreciated from a graphical point of view. The graphical procedure for evaluating the convolution integral in Equation.(5) usually involves four steps.

Steps to Evaluate the Convolution Integral:
1. Folding: Take the mirror image of h(λ) about the ordinate axis to
obtain h(λ).
2. Displacement: Shift or delay h(λ) by t to obtain h(t λ).
3. Multiplication: Find the product of h(t λ) and x(λ).
4. Integration: For a given time t, calculate the area under the
product h(t λ)x(λ) for 0 < λ < t to get y(t) at t.

The folding operation in step 1 is the reason for the term convolution. The function h(t λ) scans or slides over x(λ). In view of this superposition procedure, the convolution integral is also known as the superposition integral.

To apply the four steps, it is necessary to be able to sketch x(λ) and h(t λ). To get x(λ) from the original function x(t) involves merely replacing t with λ. Sketching h(t λ) is the key to the convolution process.

It involves reflecting h(λ) about the vertical axis and shifting it by t. Analytically, we obtain h(t λ) by replacing every t in h(t) by t λ.

Since convolution is commutative, it may be more convenient to apply steps 1 and 2 to x(t) instead of h(t). The best way to illustrate the procedure is with some examples.

Laplace Transform Convolution Integral Examples

Let’s review the Laplace transform convolution integral examples below:

Laplace Transform Convolution Integral Example 1

Find the convolution of the two signals in Figure.(1).

laplace transform convolution intergral
Figure 1

Solution:
We follow the four steps to get
y (t ) = x1(t) x2(t).
First, we fold x1(t)
as shown in Figure.(2a) and shift it by t as shown in Figure.(2b). For different values of t, we now multiply the two functions and integrate to determine the area of the overlapping region.

laplace transform convolution intergral.jpg
Figure 2. (a) Folding x1(λ), (b) shifting x1(−λ) by t.

For 0 < t < 1, there is no overlap of the two functions, as shown in Figure.(3a).

laplace transform convolution intergral
Figure 3. Overlapping of x1(t − λ) and x2(λ) for: (a) 0 < t < 1, (b) 1 < t < 2, (c) 2 < t < 3, (d) 3 < t < 4, (e) t > 4.

Hence,

 laplace transform convolution intergral
(1.1)

For 1 < t < 2, the two signals overlap between 1 and t, as shown in Figure.(3b).

 laplace transform convolution intergral
(1.2)

For 2 < t < 3, the two signals completely overlap between (t − 1) and t, as shown in Figure.(3c). It is easy to see that the area under the curve is 2. Or

 laplace transform convolution intergral
(1.3)

For 3 < t < 4, the two signals overlap between (t − 1) and 3, as shown in Figure.(3d).

 laplace transform convolution intergral
(1.4)

For t > 4, the two signals do not overlap [Figure.(3e)], and

 laplace transform convolution intergral
(1.5)

Combining Equations.(1.1) to (1.5), we obtain

 laplace transform convolution intergral
(1.6)

which is sketched in Figure.(4). Notice that y(t) in this equation is continuous. This fact can be used to check the results as we move from one range of t to another.

laplace transform convolution intergral
Figure 4. Convolution of signals x1(t) and x2(t) in Figure.(1).

The result in Equation.(1.6) can be obtained without using the graphical procedure—by directly using Equation.(5) and the properties of step functions.

Laplace Transform Convolution Integral Example 2

Graphically convolve g(t) and u(t) shown in Fig. 15.16.

laplace transform convolution intergral
Figure 5

Solution:
Let y(t) = g(t) u(t). We can find y(t) in two ways.

METHOD 1 Suppose we fold g(t), as in Figure.(6a), and shift it by t, as in Figure.(6b). Since originally, we expect that
g(t λ) = t λ, 0 < t λ < 1 or t − 1 < λ < t.
There is no overlap of the two functions when so that
for this case.

laplace transform convolution intergral
Figure 6. Convolution of g(t) and u(t) in Figure.(5) with g(t) folded.

For 0 < t < 1, g(t λ) and u(λ) overlap from 0 to t, as evident in Figure.(6b). Therefore,

 laplace transform convolution intergral
(2.1)

For t > 1, the two functions overlap completely between (t − 1) and t [see Figure.(6c)]. Hence,

 laplace transform convolution intergral
(2.2)

Thus, from Equations.(2.1) and (2.2),

 laplace transform convolution intergral

METHOD 2 Instead of folding g, suppose we fold the unit step function u(t), as in Figure.(7a), and then shift it by t, as in Figure.(7b).

laplace transform convolution intergral
Figure 7. Convolution of g(t) and u(t) in Figure.(5) with u(t) folded.

Since
u(t) = 1 for t > 0, u(t λ) = 1 for t λ > 0 or λ < t,
the
two functions overlap from 0 to t, so that

 laplace transform convolution intergral
(2.3)

For t > 1, the two functions overlap between 0 and 1, as shown in Figure.(7c). Hence,

 laplace transform convolution intergral
(2.4)

And, from Equations.(2.3) and (2.4),

 laplace transform convolution intergral

Although the two methods give the same result, as expected, notice
that it is more convenient to fold the unit step function u(t) than fold
g(t) in this example. Figure.(8) shows y(t).

laplace transform convolution intergral
Figure 8

Laplace Transform Convolution Integral Example 3

For the RL circuit in Figure.(9a), use the convolution integral to find the response io(t) due to the excitation shown in Figure.(9b).

laplace transform convolution intergral
Figure 9

This problem can be solved in two ways: directly using the convolution integral or using the graphical technique. To use either approach, we first need the unit impulse response h(t) of the circuit. In the s domain, applying the current division principle to the circuit in Figure.(10a) gives

 laplace transform convolution intergralHence,

 laplace transform convolution intergral
(3.1)

and the inverse Laplace transform of this gives

 laplace transform convolution intergral
(3.2)

Figure.(10b) shows the impulse response h(t) of the circuit.

 laplace transform convolution intergral
Figure 10. For the circuit in Figure.(1): (a) its s-domain equivalent, (b) its impulse response.

METHOD 1 : To use the convolution integral directly, recall that the response is given in the s domain as

 laplace transform convolution intergral

With the given is (t) in Figure.(9b),

 laplace transform convolution intergral

so that

 laplace transform convolution intergral
(3.3)

Since u(λ − 2) = 0 for 0 < λ < 2, the integrand involving u(λ) is nonzero for all λ > 0, whereas the integrand involving u(λ − 2) is nonzero only for λ > 2. The best way to handle the integral is to do the two parts separately.

For 0 < t < 2,

 laplace transform convolution intergral
(3.4)

For t > 2,

 laplace transform convolution intergral
(3.5)

Substituting Equations.(3.4) and (3.5) into (3.3) gives

 laplace transform convolution intergral
(3.6)

METHOD 2 : To use the graphical technique, we may fold is(t) in Figure.(9a) and shift by t, as shown in Figure.(11a).

 laplace transform convolution intergral
Figure 11

For 0 < t < 2, the overlap between
is(t λ) and h(λ) is from 0 to t, so that

 laplace transform convolution intergral
(3.7)

For t > 2, the two functions overlap between (t − 2) and t, as in Figure.(11b). Hence

 laplace transform convolution intergral
(3.8)

From Equations.(3.7) and (3.8), the response is

 laplace transform convolution intergral
(3.9)

which is the same as in Equation.(3.6). Thus, the response io(t) along the excitation is(t) is as shown in Figure.(12).

 laplace transform convolution intergral
Figure 12