15.5 The Convolution Integral

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15.5 The Convolution Integral

The term convolution means “folding.” Convolution is an invaluable tool to 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 e xcitation x(t), knowing the system impulse response h(t). This is achieved through the convolution integral, defined as

$$y(t) = \int_{-\infty}^{\infty} x(\lambda)h(t - \lambda) d\lambda$$

(15.66)

or simply

$$y(t) = x(t) * h(t)$$

(15.67)

where λ is a dummy v ariable and the asterisk denotes con volution. Equation (15.66) or (15.67) states that the output is equal to the input convolved with the unit impulse response. The convolution process is commutative:

$$y(t) = x(t) * h(t) = h(t) * x(t)$$

\n(15.68a)

or

$$y(t) = \int_{-\infty}^{\infty} x(\lambda)h(t-\lambda) d\lambda = \int_{-\infty}^{\infty} h(\lambda)x(t-\lambda) d\lambda$$

(15.68b)

This implies that the order in which the tw o 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.

Find g(t) given that Practice Problem 15.11

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 Eq. (15.66) 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

$$y(t) = \int_{-\infty}^{\infty} x(\lambda)h(t-\lambda) \, d\lambda = \int_{0}^{\infty} x(\lambda)h(t-\lambda) \, d\lambda \tag{15.69}$$

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 Eq. (15.69) becomes

$$y(t) = h(t) * x(t) = \int_0^t x(\lambda)h(t - \lambda)d\lambda$$

(15.70)

Here are some properties of the convolution integral.

1. x(t) * h(t) = h(t) * x(t) (Commutative) 2. f (t) * [x(t) + y(t)] = f (t) * x(t) + f (t) * y(t) (Distributive) 3. f (t) * [x(t) * y(t)] = [ f(t) * x(t)] * y(t) (Associative) 4. f (t) * δ(t) = ∫ −∞ ∞ f (λ)δ(t − λ) dλ = f (t) 5. f (t) * δ(t − to) = f (t − to) 6. f (t) * δ′(t) = ∫ −∞ ∞ f (λ)δ′(t − λ) dλ = f ′(t) 7. f (t) * u(t) = ∫ −∞ ∞ f (λ)u(t − λ) dλ = ∫ −∞ t f (λ) dλ

Before learning ho w to e valuate the con volution inte gral in Eq. (15.70), 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

$$f(t) = f_1(t) * f_2(t) = \int_0^t f_1(\lambda) f_2(t - \lambda) d\lambda$$

(15.71)

Taking the Laplace transform gives

$$F(s) = \mathcal{L}[f_1(t) * f_2(t)] = F_1(s)F_2(s)$$

\n(15.72)

To prove that Eq. (15.72) is true, we begin with the fact that F1(s) is defined as

$$F_1(s) = \int_{0^-}^{\infty} f_1(\lambda) e^{-s\lambda} d\lambda \qquad (15.73)$$

Multiplying this with F2(s) gives

$$F_1(s)F_2(s) = \int_{0^-}^{\infty} f_1(\lambda)[F_2(s)e^{-s\lambda}] d\lambda$$

(15.74)

We recall from the time shift property in Eq. (15.17) that the term in brackets can be written as

$$F_2(s)e^{-s\lambda} = \mathcal{L}[f_2(t-\lambda)u(t-\lambda)]$$

=

$$\int_0^\infty f_2(t-\lambda)u(t-\lambda)e^{-st} dt$$

(15.75)

Substituting Eq. (15.75) into Eq. (15.74) gives

$$F_1(s)F_2(s) = \int_0^\infty f_1(\lambda) \left[ \int_0^\infty f_2(t-\lambda)u(t-\lambda)e^{-st} dt \right] d\lambda \qquad (15.76)$$

Interchanging the order of integration results in

$$F_1(s)F_2(s) = \int_0^\infty \left[ \int_0^t f_1(\lambda)f_2(t-\lambda) d\lambda \right] e^{-st} dt \qquad (15.77)$$

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 Eq. (15.71). Hence,

$$F_1(s)F_2(s) = \mathcal{L}[f_1(t) * f_2(t)]$$

(15.78)

as desired. This indicates that convolution in the time domain is equivalent to multiplication in the s-domain. For example, if x(t) = 4e^−t and h(t) = 5e⁻2*^t* , applying the property in Eq. (15.78), we get

$$h(t) * x(t) = \mathcal{L}^{-1}[H(s)X(s)] = \mathcal{L}^{-1}\left[\left(\frac{5}{s+2}\right)\left(\frac{4}{s+1}\right)\right]$$ $$= \mathcal{L}^{-1}\left[\frac{20}{s+1} + \frac{-20}{s+2}\right]$$

(15.79)

$$= 20(e^{-t} - e^{-2t}), \qquad t \ge 0$$

Although we can find the convolution of tw o signals using Eq. (15.78), 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 a vailable in the form of e xperimental 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 Eq. (15.70) usually in volves 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: F or a gi ven 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 sk etch x(λ) and h(t − λ). To get x(λ) from the original function x(t) involves merely replacing every t with λ. Sketching h(t − λ) is the k ey to the con volution 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. Given that con volution is commutati ve, it may be more con venient to apply steps 1 and 2 to x(t) instead of h(t). The best way to illustrate the procedure is with some examples.

Example 15.12 Find the convolution of the two signals in Fig. 15.10.

Solution:

We follow the four steps to get y(t) = x1(t) * x2(t). First, we fold x1(t) as shown in Fig. 15.11(a) and shift it by t as sho wn in Fig. 15.11(b). F or different values of t, we now multiply the two functions and integrate to determine the area of the overlapping region.

For 0 < t < 1, there is no overlap of the two functions, as shown in Fig. 15.12(a). Hence,

$$y(t) = x_1(t) * x_2(t) = 0, \qquad 0 < t < 1 \tag{15.12.1}$$

For 1 < t < 2, the tw o signals o verlap between 1 and t, as sho wn in Fig. 15.12(b).

$$y(t) = \int_1^t (2)(1) \, d\lambda = 2\lambda \bigg|_1^t = 2(t - 1), \qquad 1 < t < 2 \tag{15.12.2}$$

For 2 < t < 3, the two signals completely overlap between (t − 1) and t, as shown in Fig. 15.12(c). It is easy to see that the area under the curve is 2. Or

$$y(t) = \int_{t-1}^{t} (2)(1) \, d\lambda = 2\lambda \Big|_{t-1}^{t} = 2, \qquad 2 < t < 3 \tag{15.12.3}$$

For 3 < t < 4, the two signals overlap between (t − 1) and 3, as sho wn in Fig. 15.12(d).

$$y(t) = \int_{t-1}^{3} (2)(1) \, d\lambda = 2\lambda \Big|_{t-1}^{3}$$

\n

$$= 2(3 - t + 1) = 8 - 2t, \qquad 3 < t < 4 \tag{15.12.4}$$

For t > 4, the two signals do not overlap [Fig. 15.12(e)], and

$$y(t) = 0, \qquad t > 4 \tag{15.12.5}$$

Combining Eqs. (15.12.1) to (15.12.5), we obtain

$$y(t) = \begin{cases} 0, & 0 \le t \le 1 \\ 2t - 2, & 1 \le t \le 2 \\ 2, & 2 \le t \le 3 \\ 8 - 2t, & 3 \le t \le 4 \\ 0, & t \ge 4 \end{cases}$$

(15.12.6)

0 1 t t

0 1 2 3

1

Figure 15.11

x1(t) x2(t)

Figure 15.10 For Example 15.12.

2

which is sketched in Fig. 15.13. 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. The result in Eq. (15.12.6) can be obtained without using the graphical procedure—by directly using Eq. (15.70) and the properties of step functions. This will be illustrated in Example 15.14.

Convolution of signals x1(t) and x2(t) in Fig. 15.10.

Graphically con volve the two functions in Fig. 15.14. To sho w ho w Practice Problem 15.12 powerful working in the s-domain is, verify your answer by performing the equivalent operation in the s-domain.

Answer: The result of the convolution y(t) is shown in Fig. 15.15, where

For Example 15.13.

Example 15.13 Graphically convolve g(t) and u(t) shown in Fig. 15.16.

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 Fig. 15.17(a), and shift it by t, as in Fig. 15.17(b). Since g(t) = t, 0 < t < 1 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 t < 0 so that y(0) = 0 for this case.

Figure 15.17 Convolution of g(t) and u(t) in Fig. 15.16 with g(t) folded.

For 0 < t < 1, g(t − λ) and u(λ) o verlap from 0 to t, as e vident in Fig. 15.17(b). Therefore,

$$y(t) = \int_0^t (1)(t - \lambda) d\lambda = \left(t\lambda - \frac{1}{2}\lambda^2\right)\Big|_0^t$$

= $t^2 - \frac{t^2}{2} = \frac{t^2}{2}$ , $0 \le t \le 1$ (15.13.1)

For t > 1, the two functions overlap completely between (t − 1) and t [see Fig. 15.17(c)]. Hence,

$$y(t) = \int_{t-1}^{t} (1)(t - \lambda) d\lambda$$

= $\left(t\lambda - \frac{1}{2}\lambda^2\right)\Big|_{t-1}^{t} = \frac{1}{2}, \qquad t \ge 1$ (15.13.2)

Thus, from Eqs. (15.13.1) and (15.13.2),

$$y(t) = \begin{cases} \frac{1}{2}t^2, & 0 \le t \le 1\\ \frac{1}{2}, & t \ge 1 \end{cases}$$

■ METHOD 2 Instead of folding g, suppose we fold the unit step function u(t), as in Fig. 15.18(a), and then shift it by t, as in Fig. 15.18(b). Because 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

$$y(t) = \int_0^t (1)\lambda \, d\lambda = \frac{1}{2}\lambda^2 \bigg|_0^t = \frac{t^2}{2}, \qquad 0 \le t \le 1 \quad (15.13.3)$$

Figure 15.18

Convolution of g(t) and u(t) in Fig. 15.16 with u(t) folded.

For t > 1, the tw o functions o verlap between 0 and 1, as sho wn in Fig. 15.18(c). Hence,

$$y(t) = \int_0^1 (1)\lambda \, d\lambda = \frac{1}{2}\lambda^2 \bigg|_0^1 = \frac{1}{2}, \qquad t \ge 1 \tag{15.13.4}$$

And, from Eqs. (15.13.3) and (15.13.4),

$$y(t) = \begin{cases} \frac{1}{2}t^2, & 0 \le t \le 1\\ \frac{1}{2}, & t \ge 1 \end{cases}$$

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 15.19 shows y(t).

Result of Example 15.13.

Given g(t) and f(t) in Fig. 15.20, graphically find y(t) = g(t) * ^f(t). Practice Problem 15.13

Answer:

$$y(t) = \begin{cases} 3(1 - e^{-t}), & 0 \le t \le 1 \\ 3(e - 1)e^{-t}, & t \ge 1 \\ 0, & \text{elsewhere.} \end{cases}$$

For the RL circuit in Fig. 15.21(a), use the convolution integral to find the Example 15.14 response io(t) due to the excitation shown in Fig. 15.21(b).

Solution:

1. Define. The problem is clearly stated and the method of solution is also specified.

Figure 15.21 For Example 15.14.

For the circuit in Fig. 15.21(a): (a) its s-domain equivalent, (b) its impulse response.

• 2. Present. We are to use the con volution inte gral to solv e for the response io(t) due to is(t) shown in Fig. 15.21(b).
• 3. Alternative. We have learned to do con volution by using the con volution integral and how to do it graphically. In addition, we could always work in the s-domain to solve for the current. We will solve for the current using the convolution integral and then check it using the graphical approach.
• 4. Attempt. As we stated, this problem can be solv ed in tw o ways: directly using the con volution inte gral 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 Fig. 15.22(a) gives

$$I_o = \frac{1}{s+1} I_s$$

Hence,

$$H(s) = \frac{I_o}{I_s} = \frac{1}{s+1}$$

(15.14.1)

and the inverse Laplace transform of this gives

$$h(t) = e^{-t} u(t)$$

(15.14.2)

Figure 15.22(b) shows the impulse response h(t) of the circuit.

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

$$I_o(s) = H(s)I_s(s)$$

With the given is(t) in Fig. 15.21(b),

$$i_s(t) = u(t) - u(t-2)$$

so that

$$i_o(t) = h(t) * i_s(t) = \int_0^t i_s(\lambda)h(t - \lambda) d\lambda$$

=

$$\int_0^t [u(\lambda) - u(\lambda - 2)]e^{-(t - \lambda)} d\lambda$$

(15.14.3)

Since u(λ − 2) = 0 for 0 < λ < 2, the inte grand involving u(λ) is nonzero for all λ > 0, whereas the inte grand 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,

$$i'_{o}(t) = \int_{0}^{t} (1)e^{-(t-\lambda)} d\lambda = e^{-t} \int_{0}^{t} (1)e^{\lambda} d\lambda$$

= $e^{-t}(e^{t} - 1) = 1 - e^{-t}, \qquad 0 < t < 2$ (15.14.4)

For t > 2,

$$i''_o(t) = \int_2^t (1)e^{-(t-\lambda)} d\lambda = e^{-t} \int_2^t e^{\lambda} d\lambda$$

= $e^{-t}(e^t - e^2) = 1 - e^2 e^{-t}, \qquad t > 2$ (15.14.5)

Substituting Eqs. (15.14.4) and (15.14.5) into Eq. (15.14.3) gives

$$i_o(t) = i'_o(t) - i''_o(t)$$

= $(1 - e^{-t})[u(t - 2) - u(t)] - (1 - e^2 e^{-t})u(t - 2)$
= $\begin{cases} 1 - e^{-t}A, & 0 < t < 2 \\ (e^2 - 1)e^{-t}A, & t > 2 \end{cases}$ (15.14.6)

5. Evaluate. To use the graphical technique, we may fold is(t) in Fig. 15.21(b) and shift by t, as shown in Fig. 15.23(a). For 0 < t < 2, the overlap between is(t − λ) and h(λ) is from 0 to t, so that

$$i_o(t) = \int_0^t (1)e^{-\lambda} d\lambda = -e^{-\lambda} \Big|_0^t = (1 - e^{-t}) \mathbf{A}, \qquad 0 \le t \le 2 \quad (15.14.7)$$

For t > 2, the tw o functions o verlap between ( t − 2) and t, as in Fig. 15.23(b). Hence,

$$i_o(t) = \int_{t-2}^{t} (1)e^{-\lambda} d\lambda = -e^{-\lambda} \Big|_{t-2}^{t} = -e^{-t} + e^{-(t-2)}$$

= $(e^2 - 1)e^{-t} A$ , $t \ge 0$ (15.14.8)

From Eqs. (15.14.7) and (15.14.8), the response is

15.14.7) and (15.14.8), the response is
\n

$$i_o(t) = \begin{cases} 1 - e^{-t} A, & 0 \le t \le 2 \\ (e^2 - 1)e^{-t} A, & t \ge 2 \end{cases}$$

\n(15.14.9)

which is the same as in Eq. (15.14.6). Thus, the response io(t) along the excitation is(t) is as shown in Fig. 15.24.

6. Satisfactory? We have satisf actorily solv ed the problem and can present the results as a solution to the problem.

Use convolution to find vo(t) in the circuit of Fig. 15.25(a) when the exci- Practice Problem 15.14 tation is the signal shown in Fig. 15.25(b). To show how powerful working in the s-domain is, verify your answer by performing the equivalent operation in the s-domain.

Answer:

$$20(e^{-t} - e^{-2t})u(t)$$

V.