Chapter 6.6, Problem 1P

(a)

To determine

To establish: The commutative law of the convolution integral.

Explanation of Solution

Theorem used:

If $F\left(s\right)=L\left\{f\left(t\right)\right\}$ and $G\left(s\right)=L\left\{g\left(t\right)\right\}$ both exist for $s>a\ge 0$, then

$H\left(s\right)=F\left(s\right)G\left(s\right)=L\left\{h\left(t\right)\right\},s>a\left(1\right)$

Where $h\left(t\right)={\displaystyle {\int }_0^{t}f\left(t-\tau \right)}g\left(\tau \right)d\tau ={\displaystyle {\int }_0^{t}f\left(\tau \right)}g\left(t-\tau \right)d\tau \left(2\right)$

The function h is known as the convolution of f and g; the integrals in equation (2) are called convolution integrals.

Calculation:

Consider left hand side $\left(f\ast g\right)\left(t\right)={\displaystyle {\int }_0^{t}f\left(t-\tau \right)}g\left(\tau \right)d\tau$.

Let $u=t-\tau$ then $du=-d\tau$.

$\begin{array}{c}\left(f\ast g\right)\left(t\right)={\displaystyle {\int }_t^{0}f\left(u\right)}g\left(t-u\right)\left(-du\right)\\ ={\displaystyle {\int }_0^{t}f\left(u\right)}g\left(t-u\right)du\\ =\left(g\ast f\right)\left(t\right)\end{array}$

Hence, the commutative law is proved.

(b)

To determine

To establish: The distributive law of the convolution integral.

Explanation of Solution

Consider left hand side $\left(f\ast \left(g_1+g_2\right)\right)\left(t\right)={\displaystyle {\int }_0^{t}f\left(t-\tau \right)}\left(g_1+g_2\right)\left(\tau \right)d\tau$.

$\begin{array}{c}\left(f\ast \left(g_1+g_2\right)\right)\left(t\right)={\displaystyle {\int }_0^{t}f\left(t-\tau \right)}\left(g_1\left(\tau \right)+g_2\left(\tau \right)\right)d\tau \\ ={\displaystyle {\int }_0^t\left[f\left(t-\tau \right)g_1\left(\tau \right)+f\left(t-\tau \right)g_2\left(\tau \right)\right]}d\tau \\ ={\displaystyle {\int }_0^{t}f\left(t-\tau \right)g_1\left(\tau \right)d\tau }+{\displaystyle {\int }_0^{t}f\left(t-\tau \right)g_2\left(\tau \right)d\tau }\\ =\left(f\ast g_1\right)\left(t\right)+\left(f\ast g_2\right)\left(t\right)\end{array}$

Hence, the distributive law is proved.

(c)

To determine

To establish: The associative law of the convolution integral.

Explanation of Solution

Consider left hand side $\left(f\ast \left(g\ast h\right)\right)\left(t\right)={\displaystyle {\int }_0^{t}f\left(t-\tau \right)}\left(g\ast h\right)\left(\tau \right)d\tau$.

$\begin{array}{c}\left(f\ast \left(g\ast h\right)\right)\left(t\right)={\displaystyle {\int }_0^t{\displaystyle {\int }_0^{\tau }\left[f\left(t-\tau \right)g\left(\tau -\eta \right)h\left(\eta \right)\right]}}d\tau d\eta \\ ={\displaystyle {\int }_0^t{\displaystyle {\int }_{\eta }^t\left[f\left(t-\tau \right)g\left(\tau -\eta \right)h\left(\eta \right)\right]}}d\tau d\eta \end{array}$

Let $u=\tau -\eta$ then $d\tau =du$.

$\begin{array}{c}\left(f\ast \left(g\ast h\right)\right)\left(t\right)={\displaystyle {\int }_0^t{\displaystyle {\int }_0^{t-\eta }\left[f\left(t-\eta -u\right)g\left(u\right)du\right]}}d\eta h\eta \\ ={\displaystyle {\int }_0^t\left(f\ast g\right)}\left(t-\eta \right)h\left(\eta \right)d\left(\eta \right)\\ =\left(\left(f\ast g\right)\ast h\right)\left(t\right)\end{array}$

Hence, the associative law is proved.