18.5: A.5- Derivation of the Laplace Transform of the Convolution Integral

We consider two physically realistic functions of time, \(f_{1}(t)\) and \(f_{2}(t)\), that are zero for all time \(t<0\) and non-zero only for \(t \geq 0\). The convolution integral is defined to be a definite integral involving \(f_{1}(t)\) and \(f_{2}(t)\) in either of the following forms:

\[C I(t)=\int_{\tau=0}^{\tau=t} f_{1}(\tau) f_{2}(t-\tau) d \tau=\int_{\tau=0}^{\tau=t} f_{1}(t-\tau) f_{2}(\tau) d \tau\label{eqn:6.1} \]

In these definite integrals, \(\tau\) is the dummy variable of integration, and time \(t\) appears both in the upper limit of the integral and in argument (\(t-\tau\)) of the integrand.

The Laplace transform \(L[C I(t)]\) is called the convolution transform. Let us assume that the Laplace transforms of functions \(f_{1}(t)\) and \(f_{2}(t)\) exist: \(F_{1}(s)=L\left[f_{1}(t)\right]\) and \(F_{2}(s)= L\left[f_{2}(t)\right]\). The derivation to follow will show that the product of these two transforms equals the convolution transform:

\[F_{1}(s) \times F_{2}(s)=L[C I(t)]=L\left[\int_{\tau=0}^{\tau=t} f_{1}(\tau) f_{2}(t-\tau) d \tau\right]\label{eqn:6.2} \]

Equation \(\ref{eqn:6.2}\) is certainly not an intuitively obvious result. The short, formal derivation (Meirovitch, 1967, p. 534 and Ogata, 1998, pp. 43-44) involves interchanging of orders of integration. The first step is to revise the upper limit in the convolution integral since, by definition, \(f_{2}(t-\tau)=0\) for \(t-\tau<0\), that is, for \(\tau>t\):

\[C I(t)=\int_{\tau=0}^{\tau=t} f_{1}(\tau) f_{2}(t-\tau) d \tau=\int_{\tau=0}^{\tau=\infty} f_{1}(\tau) f_{2}(t-\tau) d \tau \nonumber \]

Next, use the standard definition of a Laplace transform:

\[L[C I(t)]=L\left[\int_{\tau=0}^{\tau=\infty} f_{1}(\tau) f_{2}(t-\tau) d \tau\right]=\int_{t=0}^{t=\infty} e^{-s t}\left[\int_{\tau=0}^{\tau=\infty} f_{1}(\tau) f_{2}(t-\tau) d \tau\right] d t \nonumber \]

Now interchange the orders of integration between \(t\) and \(\tau\), an operation permitted by the assumed convergence of the integrals, and re-arrange the terms within the integrands:

\[L[C I(t)]=\int_{\tau=0}^{\tau=\infty} f_{1}(\tau)\left[\int_{t=0}^{t=\infty} e^{-s t} f_{2}(t-\tau) d t\right] d \tau \nonumber \]

In the inner integral, change the integration variable to \(\lambda=t-\tau\), so that \(t=\tau+\lambda\) and \(dt = d\lambda\), since \(\tau\) is regarded as a constant within this integration:

\[L[C I(t)]=\int_{\tau=0}^{\tau=\infty} f_{1}(\tau) d \tau\left[\int_{\lambda=-\tau}^{\lambda=\infty} e^{-s(\tau+\lambda)} f_{2}(\lambda) d \lambda\right] \nonumber \]

Now, since \(f_{2}(t)=0\) for \(\lambda<0\), in the second integral we set the lower limit to zero. Also, again re-arrange terms within the integrands, to find

\[L[C I(t)]=\int_{\tau=0}^{\tau=\infty} e^{-s \tau} f_{1}(\tau) d \tau\left[\int_{\lambda=0}^{\lambda=\infty} e^{-s \lambda} f_{2}(\lambda) d \lambda\right] \equiv F_{1}(s) \times F_{2}(s) \nonumber \]

This completes the derivation of the convolution transform, Equation \(\ref{eqn:6.2}\).