# 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\left(t\right)$$f(t)$ and $g\left(t\right)$$g(t)$ be piecewise continuous on $\left[0,\mathrm{\infty }\right)$$[0,\infty)$. The convolution of $f\left(t\right)$$f(t)$ and $g\left(t\right)$$g(t)$, $\left(f\ast g\right)\left(t\right)$$(f\ast g)(t)$, is defined by $\left(f\ast g\right)\left(t\right)={\int }_{0}^{t}f\left(t-v\right)g\left(v\right)\phantom{\rule{0.167em}{0ex}}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 ${e}^{2t}\ast t$$\displaystyle e^{2t}\ast t$.

Solution:

Use convolution definition:

${e}^{2t}\ast t={\int }_{0}^{t}{e}^{2\left(t-v\right)}v\phantom{\rule{0.167em}{0ex}}dv$$e^{2t}\ast t=\int_0^t{e^{2(t-v)}v\,dv}$

Rewrite using algebra:

$⇒{e}^{2t}\ast t={\int }_{0}^{t}{e}^{2t-2v}v\phantom{\rule{0.167em}{0ex}}dv$$\Rightarrow e^{2t}\ast t=\int_0^t{e^{2t-2v}v\,dv}$

Find antiderivative using integration by parts:

$⇒{e}^{2t}\ast t={\left(-\frac{1}{2}v{e}^{2t-2v}-\frac{1}{4}{e}^{2t-2v}\right)|}_{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:

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

$⇒{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 \mathrm{sin}\left(t\right)$$t\ast \sin{(t)}$

#### Properties of Convolution

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

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

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

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

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

#### Investigation 02

Prove the above properties of the convolution.

### Theorem: Convolution Theorem

Let $f\left(t\right)$$f(t)$ and $g\left(t\right)$$g(t)$ be piecewise continuous on $\left[0,\mathrm{\infty }\right)$$[0,\infty)$ and of exponential order $\alpha$$\alpha$. Let $F\left(s\right)=\mathcal{L}\left\{f\left(t\right)\right\}\left(s\right)$$F(s)=\mathcal{L}\left\{f(t)\right\}(s)$ and $G\left(s\right)=\mathcal{L}\left\{g\left(t\right)\right\}\left(s\right)$$G(s)=\mathcal{L}\left\{g(t)\right\}(s)$, then $\mathcal{L}\left\{f\left(t\right)\ast g\left(t\right)\right\}\left(s\right)=F\left(s\right)G\left(s\right)$$\mathcal{L}\left\{f(t)\ast g(t)\right\}(s)=F(s)G(s)$ or, equivalently, ${\mathcal{L}}^{-1}\left\{F\left(s\right)G\left(s\right)\right\}\left(t\right)=f\left(t\right)\ast g\left(t\right)$$\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 ${\mathcal{L}}^{-1}\left\{\frac{1}{{s}^{2}-3s-10}\right\}\left(t\right)$$\displaystyle \mathcal{L}^{-1}\left\{{\color{gsured}\frac{1}{s^2-3s-10}}\right\}(t)$.

Solution:

Rewrite by factoring:

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

$⇒{\mathcal{L}}^{-1}\left\{\frac{1}{{s}^{2}-3s-10}\right\}\left(t\right)={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:

$⇒{\mathcal{L}}^{-1}\left\{\frac{1}{{s}^{2}-3s-10}\right\}\left(t\right)={\int }_{0}^{t}{e}^{5\left(t-v\right)}{e}^{2v}\phantom{\rule{0.167em}{0ex}}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:

$⇒{\mathcal{L}}^{-1}\left\{\frac{1}{{s}^{2}-3s-10}\right\}\left(t\right)={\int }_{0}^{t}{e}^{5t-7v}\phantom{\rule{0.167em}{0ex}}dv$$\Rightarrow\mathcal{L}^{-1}\left\{\dfrac{1}{s^2-3s-10}\right\}(t)=\int_0^t{e^{5t-7v}\,dv}$

Find antiderivative:

$⇒{\mathcal{L}}^{-1}\left\{\frac{1}{{s}^{2}-3s-10}\right\}\left(t\right)={\left(-\frac{1}{7}{e}^{5t-7v}\right)|}_{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:

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

$⇒{\mathcal{L}}^{-1}\left\{\frac{1}{{s}^{2}-3s-10}\right\}\left(t\right)=-\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 ${\mathcal{L}}^{-1}\left\{\frac{1}{{s}^{2}-10s+21}\right\}\left(t\right)$$\displaystyle \mathcal{L}^{-1}\left\{\frac{1}{s^2-10s+21}\right\}(t)$

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

3. Find ${\mathcal{L}}^{-1}\left\{\frac{s+1}{\left({s}^{2}+1{\right)}^{2}}\right\}\left(t\right)$$\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. ${y}^{\phantom{\rule{0.167em}{0ex}}\prime \prime }\left(t\right)-5{y}^{\phantom{\rule{0.167em}{0ex}}\prime }\left(t\right)+6y\left(t\right)=\mathrm{cos}\left(2t\right)$$\displaystyle y^{\,\prime\prime}(t)-5y^{\,\prime}(t)+6y(t)=\cos{(2t)}$, $y\left(0\right)=0$$y(0)=0$, ${y}^{\phantom{\rule{0.167em}{0ex}}\prime }\left(0\right)=0$$y^{\,\prime}(0)=0$.

2. ${y}^{\phantom{\rule{0.167em}{0ex}}\prime \prime }\left(t\right)+2{y}^{\phantom{\rule{0.167em}{0ex}}\prime }\left(t\right)+2y\left(t\right)=\mathrm{sin}\left(2t\right)$$\displaystyle y^{\,\prime\prime}(t)+2y^{\,\prime}(t)+2y(t)=\sin{(2t)}$, $y\left(0\right)=1$$y(0)=1$, ${y}^{\phantom{\rule{0.167em}{0ex}}\prime }\left(0\right)=-1$$y^{\,\prime}(0)=-1$.