18.5: A.5- Derivation of the Laplace Transform of the Convolution Integral
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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 \(\ref{eqn:6.2}\).