6.6: The convolution integral
- Page ID
- 155605
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)The convolution of \(f\) and \(g\) is the function \(f * g\) defined by
\[ (f*g)(t) = \int_0^t f(t-s)g(s) ds = \int_0^t f(x)g(t-x) dx \notag\]
The convolution integral has the following properties:
- Commutative: \(f * g = g*f\)
- Associative: \((f * g)*h = f*(g*h)\)
- Distributive with respect to addition: \(f *(g_1 +g_2) = f*g_1 + f*g_2\)
- \(f*0 = 0*f = 0\)
The convolution integral can be used instead of partial fractions when computing the inverse LaPlace transform.
Compute \(\cos(t) *1\)
Solution
From \((6.6.1)\)we have \(\cos(t) *1 = \int_0^t \cos(t-s) \cdot 1 ds = \int_0^t \cos(x) \cdot 1 dx = \sin(x) \Big|_0^t = \sin(t) -\sin(0) = \sin(t) \)
\(\mathcal{L}((f * g)(t))=\mathcal{L}(f(t)) \cdot \mathcal{L}(g(t))=F(s) G(s)\)
\[\notag\begin{array}{l}
\mathcal{L}(f(t)) \mathcal{L}(g(t))=\int_0^{\infty} e^{-s y} f(y) d y \int_0^{\infty} e^{-s x} g(x) d x \\
=\int_0^{\infty}\left[\int_0^{\infty} e^{-s y} f(y) d y\right] e^{-s x} g(x) d x \\
=\int_0^{\infty}\left[\int_0^{\infty} e^{-s y} f(y) e^{-s x} g(x) d y\right] d x \\
=\int_0^{\infty}\left[\int_0^{\infty} e^{-s(y+x)} f(y) g(x) d y\right] d x \\
=\int_0^{\infty}\left[\int_0^{\infty} e^{-s(y+x)} f(y) g(x) d x\right] d y
\end{array}\]
The we let \( t=x+y, d t=d x\)
\[\notag\begin{array}{l}
=\int_0^{\infty}\left[\int_y^{\infty} e^{-s t} f(y) g(t-y) d t\right] d y \\
=\int_0^{\infty}\left[\int_0^t e^{-s t} f(y) g(t-y) d y\right] d t \\
=\int_0^{\infty} e^{-s t}\left[\int_0^t f(y) g(t-y) d y\right] d t \\
=\int_0^{\infty} e^{-s t}(f * g)(t) d t
=\mathcal{L}(f*g)
\end{array}\]
\( \mathcal{L}^{-1}(F(s) \cdot G(s)) = (f*g)(t) = \int_0^t f(t-x)g(x)dx\)
Compute \(\mathcal{L}^{-1}\left(\frac{1}{s(s-a)} \right)\)
Solution
\(\mathcal{L}^{-1}\left(\frac{1}{s(s-a)}\right) = \mathcal{L}^{-1}\left(\frac{1}{s}\frac{1}{(s-a)}\right) = 1 * e^{at} = \int_0^t 1 \cdot e^{ax} dx = \frac{1}{a} e^{ax} \Big|_0^t = \frac{e^{at}}{a} - \frac{1}{a}.\)
It is highly beneficial for engineering majors, particularly those in electrical and computer engineering, to review the convolution integral. This foundational concept will be extensively expanded upon in future courses. In advanced classes such as Linear Systems I and II, the convolution integral plays a critical role in understanding system responses, signal processing, and the behavior of linear time-invariant (LTI) systems.
To see more examples of the convolution integral related to engineering visit: https://lpsa.swarthmore.edu/Convolution/ConvolvePrint.html