Convolution

Section 11.10 Convolution

Let \(F(s)\) and \(G(s)\) denote the transforms of \(f(t)\) and \(g(t)\text{,}\) respectively. Then the inverse of the product \(F(s)G(s)\) is given by the function \(h(t) =(f\ast g)(t)\) and is called the convolution of \(f(t)\) and \(g(t)\) and can be regarded as a generalized product of \(f(t)\) and \(g(t)\text{.}\) Convolutions are helpful in solving integral equations.

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Theorem 11.10.1. Convolution Theorem.

Let \(F(s)\) and \(G(s)\) denote the Laplace transforms of \(f(t)\) and \(g(t)\text{,}\) respectively. Then the product given by \(H(s)=F(s)G(s)\) is the Laplace transformation of the convolution of \(f\) and \(g\text{.}\) It is denoted by \(h(t)=(f\ast g)(t)\text{,}\) and has the integral representation

\begin{align} h(t)\amp=(f\ast g)(t) = \int_0^tf(\tau)g(t-\tau)\,d\tau, \quad \text{or}\tag{11.10.1}\\ h(t)\amp=(g\ast f) (t) =\int_0^tg(\tau)f(t-\tau )\,d\tau\text{.}\tag{11.10.2} \end{align}

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Proof.

The following proof is given for the special case when \(s\) is a real number. The general case is covered in advanced texts. Using the dummy variables \(\sigma\) and \(\tau\) and the integrals defining the transforms, we can express their product as

\begin{equation*} F(s)G(s) = \left[\int_0^{\infty}f(\sigma) e^{-s\sigma}\,d\sigma\right]\left[\int_0^{\infty}g(\tau)e^{-s\tau}\,d\tau\right]\text{.} \end{equation*}

The product of integrals in this equation can be written as an iterated integral:

\begin{equation*} F(s)G(s) = \int_0^{\infty}\left[\int_0^{\infty}f(\sigma)e^{-s(\sigma+\tau)}\,d\sigma\right]g(\tau)\,d\tau\text{.} \end{equation*}

Hold \(\tau\) fixed, and use the change of variables \(t=\sigma +\tau\) and \(dt=d\sigma\text{.}\) Then the inner integral in the equation can be rewritten to obtain

\begin{align*} F(s)G(s) \amp = \int_0^{\infty}\left[\int_{\tau }^{\infty}f(t-\tau)e^{-st}\,dt\right]g(\tau)\,d\tau\\ \amp = \int_0^{\infty}\left[\int_{\tau }^{\infty}f(t-\tau)g(\tau)e^{-st}dt\right]d\tau\text{.} \end{align*}

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The region of integration for this last iterated integral is the wedge-shaped region in the \((t,\tau)\) plane shown in Figure 11.10.2.

Figure 11.10.2. The region of integration in the convolution theorem
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The order of integration in the integral can be reversed to get:

\begin{equation*} F(s)G(s) = \int_0^{\infty}\left[\int_0^tf(t-\tau)g(\tau)e^{-st}\,d\tau\right]dt\text{.} \end{equation*}

This equation can be written as

\begin{align*} F(s)G(s) \amp = \int_0^{\infty}\left[\int_0^tf(t-\tau)g(\tau)\,d\tau\right]e^{-st}\,dt\\ \amp = \mathcal{L}^{-1}\left(\int_0^tf(t-\tau)g(\tau)\,d\tau\right)\text{,} \end{align*}

which establishes Equation (11.10.2). Since we can interchange the role of the functions \(f(t)\) and \(g(t)\text{,}\) Equation (11.10.1) follows immediately.

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Table 11.10.3 summarizes some important convolution properties.

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Table 11.10.3. Convolution Properties

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Commutativity	\(f\ast g=g\ast f\)
Distributivity	\(f\ast (g+h) =f\ast g+f\ast h\)
Associativity	\((f\ast g) \ast h=f\ast (g\ast h)\)
Zero	\(f\ast 0=0\)

Example 11.10.4.

Show that \(\mathcal{L}^{-1}\left(\frac{2s}{(s^2+1)^2}\right) = t\sin t\text{.}\)

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Solution.

Let \(F(s) = \frac{1}{s^2+1}\text{,}\) \(G(s) =\frac{2s}{s^2+1}, \, f(t) = \sin t, \, g(t) =2\cos t\text{,}\) respectively. Applying the convolution theorem we get

\begin{align*} \mathcal{L}^{-1}\left(\frac{1}{s^2+1}\frac{2s}{s^2+1}\right) \amp = \mathcal{L}^{-1}\big(F(s)G(s)\big)\\ \amp = \int_0^t2\sin(t-\tau) \cos \tau\,d\tau\\ \amp = \int_0^t[2\sin t\cos^2\tau -2\cos t\sin \tau \cos \tau]\,d\tau\\ \amp = \left[\sin t(\tau +\sin \tau \cos \tau ) - \cos t\sin^2\tau\right]\Big|_{\tau =0}^{\tau =t}\\ \amp = t\sin t+\sin^2t\cos t-\cos t\sin^2t\\ \amp = t\sin t\text{.} \end{align*}

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Example 11.10.5.

Use the convolution theorem to solve the integral equation

\begin{equation*} f(t) = 2\cos t-\int_0^t(t-\tau)f(\tau)\,d\tau\text{.} \end{equation*}

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Solution.

Letting \(F(s) =\mathcal{L}\big(f(t)\big)\) and using \(\mathcal{L}(t) = \frac{1}{s^2}\) in the convolution theorem we obtain

\begin{equation*} F(s) = \frac{2s}{s^2+1} - \frac{1}{s^2}F(s)\text{.} \end{equation*}

Solving for \(F(s)\) we get

\begin{equation*} F(s) = \frac{2s^3}{(s^2+1)^2} = \frac{2s}{s^2+1} - \frac{2s}{(s^2+1)^2}\text{,} \end{equation*}

and the solution is

\begin{equation*} f(t) = 2\cos t - t\sin t\text{.} \end{equation*}

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Engineers and physicists sometimes consider forces that produce large effects that are applied over a very short time interval. The force acting at the time an earthquake starts is an example. This leads to the idea of a unit impulse function \(\delta(t)\text{.}\) Consider a small positive constant \(a\text{.}\) The function \(\delta_a(t)\) is defined by

\begin{equation*} \delta_a(t) = \begin{cases}\frac{1}{a} \amp \text{for} \quad 0\lt t\lt a,\\ \, 0 \amp \text{otherwise}. \end{cases} \end{equation*}

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The unit impulse function is obtained by letting the interval width go to zero, i.e.,

\begin{equation*} \delta(t) = \lim_{a \to 0}\delta_a(t). \end{equation*}

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Figure 11.10.6 shows the graph of \(\delta_a(t)\) for \(a=10, \, 40\text{,}\) and \(100\text{.}\) Although \(\delta(t)\) is called the Dirac delta function, it is not an ordinary function. To be precise it is a distribution, and the theory of distributions permits manipulations of \(\delta(t)\) as though it were a function. For our work, we will treat \(\delta(t)\) as a function and investigate its properties.

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Figure 11.10.6. Graphs of \(y=\delta_a(t)\) for \(a=10\text{,}\) \(40\text{,}\) and \(100\)
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Example 11.10.7.

Show that \(\mathcal{L}\big(\delta(t)\big) = 1\text{.}\)

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Solution.

By definition, the Laplace transform of \(\delta_a(t)\) is

\begin{equation*} \mathcal{L}\big(\delta_a(t)\big) = \int_0^{\infty}\delta_a(t)e^{-st}\,dt = \int_0^{a}\frac{1}{a}e^{-st}\,dt = \frac{1-e^{-sa}}{sa}\text{.} \end{equation*}

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Letting \(a \to 0\) in equation, and using L’Hôpital’s rule, we obtain

\begin{equation*} \mathcal{L}\big(\delta(t)\big) = \lim_{a \to 0}\mathcal{L}\big(\delta_a(t)\big) = \lim\limits_{a \to 0}\frac{1-e^{-sa}}{sa}=\lim\limits_{a \to 0}\frac{0+se^{-sa}}{s} = 1\text{.} \end{equation*}

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We now turn our attention to the unit impulse function. First, consider the function \(f_a(t)\) obtained by integrating \(\delta_a(t)\text{:}\)

\begin{equation*} f_a(t) = \int_0^t\delta_a(\tau )\,d\tau = \begin{cases} \, 0 \amp \text{for} \quad t \lt 0,\\ \frac{t}{a} \amp \text{for} \quad 0 \le t \le a,\\ \, 1 \amp \text{for} \quad a \lt t. \end{cases} \end{equation*}

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Then it is easy to see that \(U_0(t) =\lim\limits_{a \to 0}f_a(t)\) (see Figure 11.10.8).

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Figure 11.10.8. The integral of \(\delta_a(t)\) is \(f_a(t)\text{,}\) which becomes \(U_0(T)\text{,}\) which becomes \(U_0(t)\) when \(a \to 0\)
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The response of a system to the unit impulse function is illustrated in the next example.

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Example 11.10.9.

Solve the initial value problem

\begin{equation*} y\,''(t)+4y\,'(t)+13y(t)=3\delta(t), \quad \text{with} \quad y(0) =0 \quad \text{and} \quad y\,'(0^-) = 0\text{.} \end{equation*}

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Solution.

Taking transforms results in \((s^2+4s+13)Y(s) = 3\mathcal{L}(\delta (t)) = 3\text{,}\) so that

\begin{equation*} Y(s) = \frac{3}{s^2+4s+13}=\frac{3}{(s+2)^2+3^2}\text{,} \end{equation*}

so the solution iso

\begin{equation*} y(t)=e^{-2t}\sin 3t\text{.} \end{equation*}

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Note: The condition \(y\,'(0^-) =0\) is not satisfied by the “solution” \(y(t)\text{.}\) Recall that all solutions using the Laplace transform are to be considered zero for values of \(t\lt 0\text{.}\) Hence the graph of \(y(t)\) is given in Figure 11.10.10.

Figure 11.10.10. The solution \(y=y(t)\) to Example 11.10.9
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We see that \(y,'(t)\) has a jump discontinuity of magnitude \(+3\) at the origin. This happens because either \(y(t)\) or \(y\,'(t)\) must have a jump discontinuity at the origin whenever the Dirac delta function occurs as part of the input or driving function.

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We now illustrate how the convolution method can be used to solve initial value problems.

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Example 11.10.11. IVP Convolution Method.

Show initial value problem

\begin{equation*} ay\,''(t) + by\,'(t) + cy(t) = g(t), \quad \text{with} \quad y(0) = y_0 \quad \text{and} \quad y\,'(0) = y_1 \end{equation*}

has the unique solution

\begin{equation*} y(t) = u(t) + (h\ast g)(t)\text{,} \end{equation*}

where \(u(t)\) is the solution to the homogeneous equation

\begin{equation*} au\,''(t) + bu\,'(t) + cu(t) = 0, \quad \text{with} \quad u(0) = y_0 \quad \text{and} \quad u'(0) = y_1\text{,} \end{equation*}

and \(h(t)\) is the function whose Laplace transform given by \(H(s) = \frac{1}{as^2+bs+c}\text{.}\)

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Solution.

The particular solution is found by solving the equation

\begin{equation*} av\,''(t) + bv\,'(t) + cv(t) = g(t), \quad \text{with} \quad v(0) = 0 \quad \text{and} \quad v\,'(0) = 0\text{.} \end{equation*}

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Taking the Laplace transform of both sides of this equation produces

\begin{equation*} as^2V(s) + bsV(s) + cV(s) = G(s)\text{.} \end{equation*}

Solving for \(V(s)\) in the latter equation yields \(V(s)=\frac{1}{as^2+bs+c}G(s)\text{.}\) If we set \(H(s) =\frac{1}{as^2+bs+c}\text{,}\) then \(V(s) =H(s)G(s)\text{,}\) and the particular solution is given by the convolution

\begin{equation*} v(t) = (h\ast g)(t)\text{.} \end{equation*}

The general solution is \(y(t) = u(t) + v(t) = u(t) + (h\ast g)(t)\text{.}\) To verify that the initial conditions are met we compute

\begin{align*} y(0) \amp = u(0) + v(0) = y_0 + i0 = y_0, \quad \text{and}\\ y\,'(0) \amp = u\,'(0) +v\,'(0) = y_1 + 0 = y_1\text{.} \end{align*}

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Example 11.10.12.

Use the convolution method to solve the IVP

\begin{equation*} y\,''(t) + y(t) = \tan t, \quad \text{with} \quad y(0) = 1 \quad \text{and} \quad y\,'(0) = 2\text{.} \end{equation*}

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Solution.

First solve \(u''(t) + u(t) = 0\) with \(u(0) = 1\) and \(u'(0) = 2\text{.}\) Taking the Laplace transform yields \(s^2U(s) - s- 2 + U(s) = 0\text{.}\) Solving for \(U(s)\) gives \(U(s) = \frac{s+2}{s^2+1}\text{,}\) and it follows that

\begin{equation*} u(t) = \cos t + 2\sin t\text{.} \end{equation*}

Second, observe that \(H(s) = \frac{1}{s^2+1}\) and \(h(t) = \sin t\text{,}\) so that

\begin{align*} v(t) \amp = (h\ast g) (t) = \int_0^t\sin(t-s) \tan(s)\,ds\\ \amp = \left.\left[\cos (t) \ln \frac{\cos s}{1+\sin s}-\sin (t-s)\right]\right|_{s=0}^{s=t}\\ \amp = \cos (t) \ln \frac{\cos t}{1+\sin t}+\sin (t)\text{.} \end{align*}

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Therefore, the solution is

\begin{equation*} y(t) = u(t) + v(t) = \cos t + 3\sin t + \cos(t) \ln \frac{\cos t}{1+\sin t}\text{.} \end{equation*}

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Exercises Exercises

1.

Find the indicated convolution for each of the following:

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(a)

\(t \ast t\text{;}\)

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Solution.

\(f(t)=t, \; g(t)=t, \quad\) and \((f \ast g)(t) = \int_0^tf(\tau)g(t-\tau)\,d\tau = \int_0^t\tau(t-\tau)\,d\tau = \frac{t^3}{6}\text{.}\)

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(b)

\(t \ast \sin t\text{;}\)

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(c)

\(e^t\ast e^{2t}\text{;}\)

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Solution.

\(f(t)=e^t, \, g(t)=e^{2t}, \quad\) and \(\quad (g \ast f)(t) = \int_0^te^{2\tau }e^{t-\tau}\,d\tau = -e^t+e^{2t} \text{.}\)

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(d)

\(\sin t\ast \sin 2t\text{.}\)

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2.

Use convolution to find \(\mathcal{L}^{-1}\big(F(s)\big)\) for each of the following:

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(a)

\(F(s) =\frac{2}{(s-1)(s-2)}\text{;}\)

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Solution.

\(f(t)=\mathcal{L}^{-1}\Big(\frac{2}{(s-1)(s-2)}\Big)=-2e^t+2e^{2t}\text{.}\)

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(b)

\(F(s) = \frac{6}{s^3}\text{;}\)

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(c)

\(F(s) = \frac{1}{s(s^2+1)}\text{;}\)

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Solution.

\(f(t)=\mathcal{L}^{-1}\Big(\frac{1}{s(s^2+1)}\Big) = 1 - \cos t\text{.}\)

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(d)

\(F(s) = \frac{s}{(s^2+1)(s^2+4)}\text{.}\)

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3.

Prove the distributive law for convolution: \(f\ast (g+h) = f\ast g+f\ast g\text{.}\)

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Solution.

Proof:

\begin{align*} f \ast(g+h) \amp = \int_0^tf(\tau)(g+h)(t-\tau)\,d\tau\\ \amp = \int_0^tf(\tau)g(t-\tau)\,d\tau + \int_{0}^tf(\tau)h(t-\tau )d\tau\\ \amp = f\ast g+f\ast g. \end{align*}

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4.

Use the convolution theorem and mathematical induction to show that

\begin{equation*} \mathcal{L}^{-1}\left(\frac{1}{(s-a)^n}\right) = \frac{1}{(n-1)!}t^{n-1}e^{at}. \end{equation*}

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5.

Find \(\mathcal{L}^{-1}(\frac{s}{s-1})\text{.}\)

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Solution.

\(f(t) = \mathcal{L}^{-1}\big(\frac{s}{s-1}) = e^t+\delta (t)\text{.}\)

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6.

Find \(\mathcal{L}^{-1}(\frac{s^2}{s^2+1})\text{.}\)

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7.

Use the convolution theorem to solve the initial value problem

\begin{equation*} y\,''(t) +y(t) = 2\sin t, \quad \text{with} \quad y(0) = 0 \quad \text{and} \quad y\,'(0) = 0 \text{.} \end{equation*}

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Solution.

\(y(t) = -t\cos t+\sin t\text{.}\)

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8.

Use the convolution theorem to show that the solution to the initial value problem

\begin{equation*} y\,''(t) + \omega^2y(t) = f(t),\quad \text{with} \quad y(0) = 0 \quad \text{and} \quad y\,'(0)=0 \end{equation*}

\begin{equation*} y(t) = \frac{1}{\omega}\int_0^tf(\tau)\sin[\omega(t-\tau )]\,d\tau. \end{equation*}

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9.

Find \(\mathcal{L}\left(\int_0^te^{-\tau }\cos (t-\tau)\,d\tau\right)\text{.}\)

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Solution.

\(\mathcal{L}\left(\int_{0}^te^{-\tau}\cos(t-\tau)\,d\tau \right) = \mathcal{L}(e^t) \mathcal{L}\big(\cos(t)\big) = \frac{s}{(s+1)(s^2+1)}.\)

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10.

Find \(\mathcal{L}\left(\int_0^t(t-\tau )^2e^{\tau}\,d\tau\right)\text{.}\)

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11.

Let \(F(s) =\mathcal{L}\big(f(t)\big)\text{.}\) Use convolution to show that

\begin{equation*} \mathcal{L}^{-1}\left(\frac{F(s)}{s}\right) = \int_0^tf(\tau)\,d\tau. \end{equation*}

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Solution.

Given \(F(s) = \mathcal{L}\big(f(t)\big)\) and \(G(s)=\mathcal{L}(1) = \frac{1}{s}\) and \(g(t)=1\text{,}\) we have

\begin{align*} \mathcal{L}^{-1}\left(\frac{F(s)}{s}\right) \amp = F(s)G(s)\\ \amp = (f \ast g)(t)\\ \amp = \int_0^tf(\tau)g(t-\tau)\,d\tau\\ \amp = \int_{0}^tf(\tau)\,d\tau. \end{align*}

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