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
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
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
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 8.6.2
: 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,
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
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.
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
\[t^5\leftrightarrow {5!\over s^6}\quad\mbox{ and }\quad t^7 \leftrightarrow {7!\over s^8},\nonumber \]
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 \(y={\cal L}^{-1}(Y)\).
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 8.6.3
Suppose \(f\) is continuous on \([0,\infty)\) and has a Laplace transform. Then the solution of the initial value problem
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
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 \(W={\cal L}(w)\) 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.
\(w\) 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.