8.6: Convolution

In this section we consider the problem of finding the inverse Laplace transform of a product $H(s)=F(s)G(s)$, where $F$ and $G$ are the Laplace transforms of known functions $f$ and $g$. To motivate our interest in this problem, consider the initial value problem

$$ay''+by'+cy=f(t),\quad y(0)=0,\quad y'(0)=0.\nonumber$$

Taking Laplace transforms yields

$$(as^2+bs+c)Y(s)=F(s),\nonumber$$

so

$$\label{eq:8.6.1} Y(s)=F(s)G(s),$$

where

$$G(s)={1\over as^2+bs+c}.\nonumber$$

Until now wen’t been interested in the factorization indicated in Equation $\ref{eq:8.6.1}$, since we dealt only with differential equations with specific forcing functions. Hence, we could simply do the indicated multiplication in Equation $\ref{eq:8.6.1}$ and use the table of Laplace transforms to find $y={\mathscr L}^{-1}(Y)$. However, this isn’t possible if we want a formula for $y$ in terms of $f$, which may be unspecified.

To motivate the formula for ${\mathscr L}^{-1}(FG)$, consider the initial value problem

$$\label{eq:8.6.2} y'-ay=f(t),\quad y(0)=0,$$

which we first solve without using the Laplace transform. The solution of the differential equation in Equation $\ref{eq:8.6.2}$ is of the form $y=ue^{at}$ where

$$u'=e^{-at}f(t).\nonumber$$

Integrating this from $0$ to $t$ and imposing the initial condition $u(0)=y(0)=0$ yields

$$u=\int_0^t e^{-a\tau}f(\tau)\,d\tau.\nonumber$$

Therefore

$$\label{eq:8.6.3} y(t)=e^{at}\int_0^t e^{-a\tau}f(\tau)\,d\tau=\int_0^t e^{a(t-\tau)}f(\tau)\,d\tau.$$

Now we’ll use the Laplace transform to solve Equation $\ref{eq:8.6.2}$ and compare the result to Equation $\ref{eq:8.6.3}$. Taking Laplace transforms in Equation $\ref{eq:8.6.2}$ yields

$$(s-a)Y(s)=F(s),\nonumber$$

so

$$Y(s)=F(s) {1\over s-a},\nonumber$$

which implies that

$$\label{eq:8.6.4} y(t)= {\mathscr L}^{-1}\left(F(s){1\over s-a}\right).$$

If we now let $g(t)=e^{at}$, so that

$$G(s)={1\over s-a},\nonumber$$

then Equation $\ref{eq:8.6.3}$ and Equation $\ref{eq:8.6.4}$ can be written as

$$y(t)=\int_0^tf(\tau)g(t-\tau)\,d\tau\nonumber$$

and

$$y={\mathscr L}^{-1}(FG),\nonumber$$

respectively. Therefore

$$\label{eq:8.6.5} {\mathscr L}^{-1}(FG)=\int_0^t f(\tau)g(t-\tau)\,d\tau$$

in this case.

This motivates the next definition.

Definition 8.6.1 : Convolution

The convolution $f*g$ of two functions $f$ and $g$ is defined by

$$(f*g)(t)=\int_0^t f(\tau)g(t-\tau)\,d\tau.\nonumber$$

It can be shown (Exercise 8.6.6) that $f\ast g=g\ast f$; that is,

$$\int_0^tf(t-\tau)g(\tau)\,d\tau=\int_0^tf(\tau)g(t-\tau)\,d\tau. \nonumber$$

Equation $\ref{eq:8.6.5}$ shows that ${\mathscr L}^{-1}(FG)=f*g$ in the special case where $g(t)=e^{at}$. This next theorem states that this is true in general.

Theorem 8.6.2 : The Convolution Theorem

If ${\mathscr L}(f)=F$ and ${\mathscr L}(g)=G,$ then

$${\mathscr L}(f*g)=FG.\nonumber$$

A complete proof of the convolution theorem is beyond the scope of this book. However, we’ll assume that $f\ast g$ has a Laplace transform and verify the conclusion of the theorem in a purely computational way. By the definition of the Laplace transform,

$${\mathscr L}(f\ast g)=\int_0^\infty e^{-st}(f\ast g)(t)\,dt=\int_0^\infty e^{-st} \int_0^t f(\tau)g(t-\tau)\,d\tau\,dt.\nonumber$$

This iterated integral equals a double integral over the region shown in Figure 8.6.1 . Reversing the order of integration yields

$$\label{eq:8.6.6} {\mathscr L}(f*g)=\int_0^\infty f(\tau)\int^\infty_\tau e^{-st}g(t-\tau)\, dt \,d\tau.$$

However, the substitution $x=t-\tau$ shows that

$$\begin{align*} \int^\infty_\tau e^{-st}g(t-\tau)\,dt&=\int_0^\infty e^{-s(x+\tau)}g(x)\,dx\\[4pt] &= e^{-s\tau}\int_0^\infty e^{-sx}g(x)\,dx=e^{-s\tau}G(s).\end{align*}\nonumber$$

Substituting this into Equation $\ref{eq:8.6.6}$ and noting that $G(s)$ is independent of $\tau$ yields

$$\begin{align*} {\mathscr L}(f\ast g)&= \int_0^\infty e^{-s\tau} f(\tau)G(s)\,d\tau\\[4pt] & =G(s)\int_0^\infty e^{-st}f(\tau)\,d\tau=F(s)G(s).\end{align*}\nonumber$$

Figure 8.6.1

Example 8.6.1

Let

$$f(t)=e^{at}\quad \text{and} \quad g(t)=e^{bt}\qquad (a\ne b).\nonumber$$

Verify that ${\mathscr L}(f\ast g)={\mathscr L}(f){\mathscr L}(g)$, as implied by the convolution theorem.

Solution

We first compute

$$\begin{aligned} (f\ast g) &= \int_{0}^{t}e^{a\tau }e^{b(t-\tau )}d\tau &=e^{bt}\int_{0}^{t}e^{(a-b)\tau }d\tau \\ &= \left. e^{bt}\frac{e^{a-b}\tau }{a-b} \right|_{0}^{t} &=\frac{e^{bt}[e^{(a-b)t}-1] }{a-b} \\ &=\frac{e^{at}-e^{bt}}{a-b} \end{aligned}\nonumber$$

Since

$$e^{at}\leftrightarrow {1\over s-a}\quad\mbox{ and }\quad e^{bt}\leftrightarrow {1\over s-b},\nonumber$$

it follows that

$$\begin{aligned} {\mathscr L}(f\ast g)&={1\over a-b}\left[{1\over s-a}-{1\over s-b}\right]\\[5pt] &={1\over(s-a)(s-b)}\\[5pt] &={\mathscr L}(e^{at}){\mathscr L}(e^{bt})={\mathscr L}(f){\mathscr L}(g).\end{aligned}\nonumber$$

A Formula for the Solution of an Initial Value Problem

The convolution theorem provides a formula for the solution of an initial value problem for a linear constant coefficient second order equation with an unspecified. The next three examples illustrate this.

Example 8.6.2

Find a formula for the solution of the initial value problem

$$\label{eq:8.6.7} y''-2y'+y=f(t),\quad y(0)=k_0,\quad y'(0)=k_1.$$

Solution

Taking Laplace transforms in Equation $\ref{eq:8.6.7}$ yields

$$(s^2-2s+1)Y(s)=F(s)+(k_1+k_0s)-2k_0.\nonumber$$

Therefore

$$\begin{align*} Y(s)&= {1\over(s-1)^2}F(s)+{k_1+k_0s-2k_0\over(s-1)^2}\\[5pt] &= {1\over(s-1)^2}F(s)+{k_0\over s-1}+{k_1-k_0\over(s-1)^2}.\end{align*}\nonumber$$

From the table of Laplace transforms,

$${\mathscr L}^{-1}\left({k_0\over s-1}+{k_1-k_0\over(s-1)^2}\right) =e^t\left(k_0+(k_1-k_0)t\right). \nonumber$$

Since

$${1\over(s-1)^2}\leftrightarrow te^t\quad \text{and} \quad F(s) \leftrightarrow f(t), \nonumber$$

the convolution theorem implies that

$${\mathscr L}^{-1} \left({1\over(s-1)^2}F(s)\right)= \int_0^t\tau e^\tau f(t-\tau)\,d\tau. \nonumber$$

Therefore the solution of Equation $\ref{eq:8.6.7}$ is

$$y(t)=e^t\left(k_0+(k_1-k_0)t\right)+\int_0^t\tau e^\tau f(t-\tau)\, d\tau. \nonumber$$

Example 8.6.3

Find a formula for the solution of the initial value problem

$$\label{eq:8.6.8} y''+4y=f(t),\quad y(0)=k_0,\quad y'(0)=k_1.$$

Solution

Taking Laplace transforms in Equation $\ref{eq:8.6.8}$ yields

$$(s^2+4)Y(s) =F(s)+k_1+k_0s. \nonumber$$

Therefore

$$Y(s) ={1\over(s^2+4)}F(s)+{k_1+k_0s\over s^2+4}. \nonumber$$

From the table of Laplace transforms,

$${\mathscr L}^{-1}\left(k_1+k_0s\over s^2+4\right)=k_0\cos 2t+{k_1\over 2}\sin 2t. \nonumber$$

Since

$${1\over(s^2+4)}\leftrightarrow {1\over 2}\sin 2t\quad \text{and} \quad F(s)\leftrightarrow f(t), \nonumber$$

the convolution theorem implies that

$${\mathscr L}^{-1}\left({1\over(s^2+4)}F(s)\right)= {1\over 2}\int_0^t f(t-\tau)\sin 2\tau\, d\tau. \nonumber$$

Therefore the solution of Equation $\ref{eq:8.6.8}$ is

$$y(t)=k_0\cos 2t+{k_1\over 2}\sin 2t+{1\over 2}\int_0^tf(t-\tau)\sin 2\tau\,d\tau. \nonumber$$

Example 8.6.4

Find a formula for the solution of the initial value problem

$$\label{eq:8.6.9} y''+2y'+2y=f(t),\quad y(0)=k_0,\quad y'(0)=k_1.$$

Solution

Taking Laplace transforms in Equation $\ref{eq:8.6.9}$ yields

$$(s^2+2s+2)Y(s)=F(s)+k_1+k_0s+2k_0.\nonumber$$

Therefore

$$\begin{aligned} Y(s)&=\frac{1}{(s+1)^{2}+1}F(s)+\frac{k_{1}+k_{0}s+2k_{0}}{(s+1)^{2}+1} \\ &=\frac{1}{(s+1)^{2}+1}F(s)+\frac{(k_{1}+k_{0})+k_{0}(s+1)}{(s+1)^{2}+1} \end{aligned} \nonumber$$

From the table of Laplace transforms,

$${\mathscr L}^{-1}\left((k_1+k_0)+k_0(s+1)\over(s+1)^2+1\right)= e^{-t}\left((k_1+k_0)\sin t+k_0\cos t\right). \nonumber$$

Since

$${1\over(s+1)^2+1}\leftrightarrow e^{-t}\sin t \quad \text{and} \quad F(s)\leftrightarrow f(t), \nonumber$$

the convolution theorem implies that

$${\mathscr L}^{-1}\left({1\over(s+1)^2+1}F(s)\right)= \int_0^t f(t-\tau)e^{-\tau}\sin\tau \,d\tau. \nonumber$$

Therefore the solution of Equation $\ref{eq:8.6.9}$ is

$$\label{eq:8.6.10} y(t)=e^{-t}\left((k_1+k_0)\sin t+k_0\cos t\right)+\int_0^tf(t-\tau)e^{-\tau}\sin\tau\,d\tau.$$

Evaluating Convolution Integrals

We’ll say that an integral of the form $\displaystyle \int_0^t u(\tau)v(t-\tau)\,d\tau$ is a convolution integral. The convolution theorem provides a convenient way to evaluate convolution integrals.

Example 8.6.5

Evaluate the convolution integral

$$h(t)=\int_0^t(t-\tau)^5\tau^7 d\tau. \nonumber$$

Solution

We could evaluate this integral by expanding $(t-\tau)^5$ in powers of $\tau$ and then integrating. However, the convolution theorem provides an easier way. The integral is the convolution of $f(t)=t^5$ and $g(t)=t^7$. Since

$$t^5\leftrightarrow {5!\over s^6}\quad\mbox{ and }\quad t^7 \leftrightarrow {7!\over s^8},\nonumber$$

the convolution theorem implies that

$$h(t)\leftrightarrow {5!7!\over s^{14}}={5!7!\over 13!}\, {13! \over s^{14}},\nonumber$$

where we have written the second equality because

$${13!\over s^{14}}\leftrightarrow t^{13}.\nonumber$$

Hence,

$$h(t)={5!7!\over 13!}\, t^{13}.\nonumber$$

Example 8.6.6

Use the convolution theorem and a partial fraction expansion to evaluate the convolution integral

$$h(t)=\int_0^t\sin a(t-\tau)\cos b\tau\,d\tau\quad (|a|\ne |b|).\nonumber$$

Solution

Since

$$\sin at\leftrightarrow {a\over s^2+a^2}\quad\mbox{and}\quad \cos bt\leftrightarrow {s\over s^2+b^2},\nonumber$$

the convolution theorem implies that

$$H(s)={a\over s^2+a^2}{s\over s^2+b^2}.\nonumber$$

Expanding this in a partial fraction expansion yields

$$H(s)={a\over b^2-a^2}\left[{s\over s^2+a^2}-{s\over s^2+b^2}\right].\nonumber$$

Therefore

$$h(t)={a\over b^2-a^2}\left(\cos at-\cos bt\right).\nonumber$$

Volterra Integral Equations

An equation of the form

$$\label{eq:8.6.11} y(t)=f(t)+\int_0^t k(t-\tau) y(\tau)\,d\tau$$

is a Volterra integral equation. Here $f$ and $k$ are given functions and $y$ is unknown. Since the integral on the right is a convolution integral, the convolution theorem provides a convenient formula for solving Equation $\ref{eq:8.6.11}$. Taking Laplace transforms in Equation $\ref{eq:8.6.11}$ yields

$$Y(s)=F(s)+K(s) Y(s),\nonumber$$

and solving this for $Y(s)$ yields

$$Y(s)={F(s)\over 1-K(s)}.\nonumber$$

We then obtain the solution of Equation $\ref{eq:8.6.11}$ as $y={\mathscr L}^{-1}(Y)$.

Example 8.6.7

Solve the integral equation

$$\label{eq:8.6.12} y(t)=1+2\int_0^t e^{-2(t-\tau)} y(\tau)\,d\tau.$$

Solution

Taking Laplace transforms in Equation $\ref{eq:8.6.12}$ yields

$$Y(s)={1\over s}+{2\over s+2} Y(s),\nonumber$$

and solving this for $Y(s)$ yields

$$Y(s)={1\over s}+{2\over s^2}.\nonumber$$

Hence,

$$y(t)=1+2t.\nonumber$$

Transfer Functions

The next theorem presents a formula for the solution of the general initial value problem

$$ay''+by'+cy=f(t),\quad y(0)=k_0,\quad y'(0)=k_1,\nonumber$$

where we assume for simplicity that $f$ is continuous on $[0,\infty)$ and that ${\mathscr L}(f)$ exists. In Exercises 8.6.11-8.6.14 it is shown that the formula is valid under much weaker conditions on $f$.

Theorem 8.6.3

Suppose $f$ is continuous on $[0,\infty)$ and has a Laplace transform. Then the solution of the initial value problem

$$\label{eq:8.6.13} ay''+by'+cy=f(t),\quad y(0)=k_0,\quad y'(0)=k_1,$$

is

$$\label{eq:8.6.14} y(t)=k_0y_1(t)+k_1y_2(t)+\int_0^tw(\tau)f(t-\tau)\,d\tau,$$

where $y_1$ and $y_2$ satisfy

$$\label{eq:8.6.15} ay_1''+by_1'+cy_1=0,\quad y_1(0)=1,\quad y_1'(0)=0,$$

and

$$\label{eq:8.6.16} ay_2''+by_2'+cy_2=0,\quad y_2(0)=0,\quad y_2'(0)=1,$$

and

$$\label{eq:8.6.17} w(t)={1\over a}y_2(t).$$

Proof

Taking Laplace transforms in Equation $\ref{eq:8.6.13}$ yields

$$p(s)Y(s)=F(s)+a(k_1+k_0s)+bk_0,\nonumber$$

where

$$p(s)=as^2+bs+c.\nonumber$$

Hence,

$$\label{eq:8.6.18} Y(s)=W(s)F(s)+V(s)$$

with

$$\label{eq:8.6.19} W(s)={1\over p(s)}$$

and

$$\label{eq:8.6.20} V(s)={a(k_1+k_0s)+bk_0\over p(s)}.$$

Taking Laplace transforms in Equation $\ref{eq:8.6.15}$ and Equation $\ref{eq:8.6.16}$ shows that

$$p(s)Y_1(s)=as+b\quad\mbox{and}\quad p(s)Y_2(s)=a.\nonumber$$

Therefore

$$Y_1(s)={as+b\over p(s)}\nonumber$$

and

$$\label{eq:8.6.21} Y_2(s)={a\over p(s)}.$$

Hence, Equation $\ref{eq:8.6.20}$ can be rewritten as

$$V(s)=k_0Y_1(s)+k_1Y_2(s).\nonumber$$

Substituting this into Equation $\ref{eq:8.6.18}$ yields

$$Y(s)=k_0Y_1(s)+k_1Y_2(s)+{1\over a}Y_2(s)F(s).\nonumber$$

Taking inverse transforms and invoking the convolution theorem yields Equation $\ref{eq:8.6.14}$. Finally, Equation $\ref{eq:8.6.19}$ and Equation $\ref{eq:8.6.21}$ imply Equation $\ref{eq:8.6.17}$.

It is useful to note from Equation $\ref{eq:8.6.14}$ that $y$ is of the form

$$y=v+h,\nonumber$$

where

$$v(t)=k_0y_1(t)+k_1y_2(t)\nonumber$$

depends on the initial conditions and is independent of the forcing function, while

$$h(t)=\int_0^tw(\tau)f(t-\tau)\, d\tau\nonumber$$

depends on the forcing function and is independent of the initial conditions. If the zeros of the characteristic polynomial

$$p(s)=as^2+bs+c\nonumber$$

of the complementary equation have negative real parts, then $y_1$ and $y_2$ both approach zero as $t\to\infty$, so $\lim_{t\to\infty}v(t)=0$ for any choice of initial conditions. Moreover, the value of $h(t)$ is essentially independent of the values of $f(t-\tau)$ for large $\tau$, since $\lim_{\tau\to\infty}w(\tau)=0$. In this case we say that $v$ and $h$ are transient and steady state components, respectively, of the solution $y$ of Equation $\ref{eq:8.6.13}$. These definitions apply to the initial value problem of Example 8.6.4 , where the zeros of

$$p(s)=s^2+2s+2=(s+1)^2+1\nonumber$$

are $-1\pm i$. From Equation $\ref{eq:8.6.10}$, we see that the solution of the general initial value problem of Example 8.6.4 is $y=v+h$, where

$$v(t)=e^{-t}\left((k_1+k_0)\sin t+k_0\cos t\right)\nonumber$$

is the transient component of the solution and

$$h(t)=\int_0^t f(t-\tau)e^{-\tau}\sin\tau\,d\tau\nonumber$$

is the steady state component. The definitions don’t apply to the initial value problems considered in Examples 8.6.2 and 8.6.3 , since the zeros of the characteristic polynomials in these two examples don’t have negative real parts.

In physical applications where the input $f$ and the output $y$ of a device are related by Equation $\ref{eq:8.6.13}$, the zeros of the characteristic polynomial usually do have negative real parts. Then $W={\mathscr L}(w)$ is called the transfer function of the device. Since

$$H(s)=W(s)F(s),\nonumber$$

we see that

$$W(s)={H(s)\over F(s)}\nonumber$$

is the ratio of the transform of the steady state output to the transform of the input.

Because of the form of

$$h(t)=\int_0^tw(\tau)f(t-\tau)\,d\tau,\nonumber$$

$w$ is sometimes called the weighting function of the device, since it assigns weights to past values of the input $f$. It is also called the impulse response of the device, for reasons discussed in the next section.

Formula Equation $\ref{eq:8.6.14}$ is given in more detail in Exercises 8.6.8-8.6.10 for the three possible cases where the zeros of $p(s)$ are real and distinct, real and repeated, or complex conjugates, respectively.