Convolution

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 7.7 – Convolution

Expected Educational Results

• Objective 24–1: I understand the properties of the convolution of two functions.

• Objective 24–2: I can compute the convolution of two functions.

• Objective 24–3: I can use the convolution of two functions to find inverse Laplace transforms.

Convolution

Definition: Convolution

Let $f(t)$$f(t)$ and $g(t)$$g(t)$ be piecewise continuous on $[0,\mathrm{\infty })$$[0,\infty)$. The convolution of $f(t)$$f(t)$ and $g(t)$$g(t)$, $(f\ast g)(t)$$(f\ast g)(t)$, is defined by ${\displaystyle (f\ast g)(t)={\int }_0^{t}f(t-v)g(v)dv}$$\displaystyle (f\ast g)(t)=\int_0^t{f(t-v)g(v)\,dv}$

NOTE: In the definition of convolution, the “first” function is translated by $v$$v$ where $v$$v$ is the “dummy” variable for the integral.

Example 01: Evaluate ${\displaystyle e^{2t}\ast t}$$\displaystyle e^{2t}\ast t$.

Solution:

Use convolution definition:

$e^{2t}\ast t={\int }_0^t{e}^{2(t-v)}vdv$$e^{2t}\ast t=\int_0^t{e^{2(t-v)}v\,dv}$

Rewrite using algebra:

$\Rightarrow e^{2t}\ast t={\int }_0^t{e}^{2t-2v}vdv$$\Rightarrow e^{2t}\ast t=\int_0^t{e^{2t-2v}v\,dv}$

Find antiderivative using integration by parts:

$\Rightarrow e^{2t}\ast t={(-\frac{1}{2}v{e}^{2t-2v}-\frac{1}{4}{e}^{2t-2v})|}_{v=0}^{v=t}$$\Rightarrow e^{2t}\ast t=\left.\left(-\frac{1}{2}ve^{2t-2v}-\frac{1}{4}e^{2t-2v}\right)\right|_{v=0}^{v=t}$

Use FTC-II:

$\Rightarrow e^{2t}\ast t=-\frac{1}{2}t{e}^{2t-2t}-\frac{1}{4}{e}^{2t-2t}-(-\frac{1}{2}(0)e^{2t-2(0)}-\frac{1}{4}{e}^{2t-2(0)})$$\Rightarrow e^{2t}\ast t=-\frac{1}{2}te^{2t-2t}-\frac{1}{4}e^{2t-2t}-\left(-\frac{1}{2}(0)e^{2t-2(0)}-\frac{1}{4}e^{2t-2(0)}\right)$

Simplify:

$\Rightarrow e^{2t}\ast t=-\frac{1}{2}t-\frac{1}{4}+0+\frac{1}{4}{e}^{2t}$$\Rightarrow e^{2t}\ast t=-\frac{1}{2}t-\frac{1}{4}+0+\frac{1}{4}e^{2t}$

Investigation 01

Evaluate the following convolutions.

1. $t\ast t^2$$t\ast t^2$

2. $t\ast \sin (t)$$t\ast \sin{(t)}$

Properties of Convolution

Let $f(t)$$f(t)$, $g(t)$$g(t)$, and $h(t)$$h(t)$ be piecewise continuous on $[0,\mathrm{\infty })$$[0,\infty)$

1. $(f\ast g)=(g\ast f)$$(f\ast g)=(g\ast f)$

2. $(f\ast (g+h))=(f\ast g)+(f\ast h)$$(f\ast (g+h))=(f\ast g)+(f\ast h)$

3. $(f\ast g)\ast h=f\ast (g\ast h)$$(f\ast g)\ast h=f\ast (g\ast h)$

4. $(f\ast 0)=0$$(f\ast 0)=0$

Investigation 02

Prove the above properties of the convolution.

Theorem: Convolution Theorem

Let $f(t)$$f(t)$ and $g(t)$$g(t)$ be piecewise continuous on $[0,\mathrm{\infty })$$[0,\infty)$ and of exponential order $\alpha$$\alpha$. Let $F(s)=\mathcal{L}\{f(t)\}(s)$$F(s)=\mathcal{L}\left\{f(t)\right\}(s)$ and $G(s)=\mathcal{L}\{g(t)\}(s)$$G(s)=\mathcal{L}\left\{g(t)\right\}(s)$, then $\mathcal{L}\{f(t)\ast g(t)\}(s)=F(s)G(s)$$\mathcal{L}\left\{f(t)\ast g(t)\right\}(s)=F(s)G(s)$ or, equivalently, ${\mathcal{L}}^{-1}\{F(s)G(s)\}(t)=f(t)\ast g(t)$$\mathcal{L}^{-1}\left\{F(s)G(s)\right\}(t)=f(t)\ast g(t)$.

NOTE: The convolution theorem helps find inverse Laplace transforms when the method of partial fractions fails.

Investigation 03

Prove the Convolution Theorem.

Hint: You will need to switch the order of integration and use u-substitution.

Example 02: Evaluate ${\displaystyle {\mathcal{L}}^{-1}\left\{\frac{1}{s^2-3s-10}\right\}(t)}$$\displaystyle \mathcal{L}^{-1}\left\{{\color{gsured}\frac{1}{s^2-3s-10}}\right\}(t)$.

Solution:

Rewrite by factoring:

${\mathcal{L}}^{-1}\left\{{\displaystyle \frac{1}{s^2-3s-10}}\right\}(t)={\mathcal{L}}^{-1}\left\{{\displaystyle \frac{1}{(s-5)(s+2)}}\right\}(t)$$\mathcal{L}^{-1}\left\{\dfrac{1}{s^2-3s-10}\right\}(t)=\mathcal{L}^{-1}\left\{\dfrac{1}{(s-5)(s+2)}\right\}(t)$

Use Convolution Theorem:

$\Rightarrow {\mathcal{L}}^{-1}\left\{{\displaystyle \frac{1}{s^2-3s-10}}\right\}(t)=e^{5t}\ast e^{-2t}$$\Rightarrow\mathcal{L}^{-1}\left\{\dfrac{1}{s^2-3s-10}\right\}(t)=e^{5t}\ast e^{-2t}$

Use the definition of convolution:

$\Rightarrow {\mathcal{L}}^{-1}\left\{{\displaystyle \frac{1}{s^2-3s-10}}\right\}(t)={\int }_0^t{e}^{5(t-v)}{e}^{2v}dv$$\Rightarrow\mathcal{L}^{-1}\left\{\dfrac{1}{s^2-3s-10}\right\}(t)=\int_0^t{e^{5(t-v)}e^{2v}\,dv}$

Simplify using algebra:

$\Rightarrow {\mathcal{L}}^{-1}\left\{{\displaystyle \frac{1}{s^2-3s-10}}\right\}(t)={\int }_0^t{e}^{5t-7v}dv$$\Rightarrow\mathcal{L}^{-1}\left\{\dfrac{1}{s^2-3s-10}\right\}(t)=\int_0^t{e^{5t-7v}\,dv}$

Find antiderivative:

$\Rightarrow {\mathcal{L}}^{-1}\left\{{\displaystyle \frac{1}{s^2-3s-10}}\right\}(t)={(-\frac{1}{7}{e}^{5t-7v})|}_{v=0}^{v=t}$$\Rightarrow\mathcal{L}^{-1}\left\{\dfrac{1}{s^2-3s-10}\right\}(t)=\left.\left(-\frac{1}{7}e^{5t-7v}\right)\right|_{v=0}^{v=t}$

Use FTC-II:

$\Rightarrow {\mathcal{L}}^{-1}\left\{{\displaystyle \frac{1}{s^2-3s-10}}\right\}(t)=-\frac{1}{7}{e}^{5t-7t}-(-\frac{1}{5}{e}^{5t-7(0)})$$\Rightarrow\mathcal{L}^{-1}\left\{\dfrac{1}{s^2-3s-10}\right\}(t)=-\frac{1}{7}e^{5t-7t}-\left(-\frac{1}{5}e^{5t-7(0)}\right)$

Simplify using algebra:

$\Rightarrow {\mathcal{L}}^{-1}\left\{{\displaystyle \frac{1}{s^2-3s-10}}\right\}(t)=-\frac{1}{7}{e}^{-2t}+\frac{1}{7}{e}^{5t}$$\Rightarrow\mathcal{L}^{-1}\left\{\dfrac{1}{s^2-3s-10}\right\}(t)=-\frac{1}{7}e^{-2t}+\frac{1}{7}e^{5t}$

NOTE: Compare this result with the solution to Example 01 in CPT_22.

Investigation 04

Use the convolution to find the following inverse Laplace transform:

1. Find ${\displaystyle {\mathcal{L}}^{-1}\left\{\frac{1}{s^2-10s+21}\right\}(t)}$$\displaystyle \mathcal{L}^{-1}\left\{\frac{1}{s^2-10s+21}\right\}(t)$

2. Find ${\displaystyle {\mathcal{L}}^{-1}\left\{\frac{1}{s(s^2+1)}\right\}(t)}$$\displaystyle \mathcal{L}^{-1}\left\{\frac{1}{s(s^2+1)}\right\}(t)$

3. Find ${\displaystyle {\mathcal{L}}^{-1}\left\{\frac{s+1}{(s^2+1{)}^2}\right\}(t)}$$\displaystyle \mathcal{L}^{-1}\left\{\frac{s+1}{(s^2+1)^2}\right\}(t)$

Solving IVPs Using Convolution

Investigation 05

Solve the following IVPs:

1. ${\displaystyle y^{\prime \prime }(t)-5y^{\prime }(t)+6y(t)=\cos (2t)}$$\displaystyle y^{\,\prime\prime}(t)-5y^{\,\prime}(t)+6y(t)=\cos{(2t)}$, $y(0)=0$$y(0)=0$, $y^{\prime }(0)=0$$y^{\,\prime}(0)=0$.

2. ${\displaystyle y^{\prime \prime }(t)+2y^{\prime }(t)+2y(t)=\sin (2t)}$$\displaystyle y^{\,\prime\prime}(t)+2y^{\,\prime}(t)+2y(t)=\sin{(2t)}$, $y(0)=1$$y(0)=1$, $y^{\prime }(0)=-1$$y^{\,\prime}(0)=-1$.

CC BY-NC-SA 4.0

This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License [http://creativecommons.org/licenses/by-nc-sa/4.0/].

Last Modified: Sunday, 8 November 2020 22:22 EDT