
# 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 $$y={\cal L}^{-1}(Y)$$. 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 $${\cal L}^{-1}(FG)$$, 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

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

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

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

respectively. Therefore

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

in this case.

This motivates the next definition.

Theorem $$\PageIndex{1}$$: 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 $${\cal L}^{-1}(FG)=f*g$$ in the special case where $$g(t)=e^{at}$$. This next theorem states that this is true in general.

Theorem $$\PageIndex{2}$$: The Convolution Theorem

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

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

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,

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

This iterated integral equals a double integral over the region shown in Figure $$\PageIndex{1}$$. Reversing the order of integration yields

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

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*} {\cal 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

Example $$\PageIndex{1}$$

Let

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

Verify that $${\cal L}(f\ast g)={\cal L}(f){\cal L}(g)$$, 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} {\cal 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] &={\cal L}(e^{at}){\cal L}(e^{bt})={\cal L}(f){\cal 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 $$\PageIndex{2}$$

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,

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

Since

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

the convolution theorem implies that

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

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 $$\PageIndex{3}$$

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,

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

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

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

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 $$\PageIndex{4}$$

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,

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

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

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

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 $$\int_0^t u(\tau)v(t-\tau)\,d\tau$$ is a convolution integral. The convolution theorem provides a convenient way to evaluate convolution integrals.

Example $$\PageIndex{5}$$

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 $$\PageIndex{6}$$

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 $$y={\cal L}^{-1}(Y)$$.

Example $$\PageIndex{7}$$

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 $${\cal L}(f)$$ exists. In Exercises 8.6.11-8.6.14 it is shown that the formula is valid under much weaker conditions on $$f$$.

Theorem $$\PageIndex{3}$$

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 $$\PageIndex{4}$$, 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 $$\PageIndex{4}$$ 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 $$\PageIndex{2}$$ and $$\PageIndex{3}$$, 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 $$W={\cal L}(w)$$ 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.