Question

Use the Laplace transform table to find $\mathrm{f}\left(\mathrm{t}\right)={\int }_0^{\mathrm{t}}{\mathrm{e}}^{-\mathrm{\tau }}\sin \left(\mathrm{t}-\mathrm{\tau }\right)\mathrm{d\tau }$. Hint: In L34, let$\mathit{g}\left(t\right)\mathbf{=}{\mathit{e}}^{\mathbf{-}\mathbf{t}}$and$\mathit{h}\left(t\right)\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathit{t}$, and find$\mathit{G}\left(p\right)\mathit{H}\left(p\right)$which is the Laplace transform of the integral you want. Break the result into partial fractions and look up the inverse transforms.

Short answer

Evaluation for the given convolution integral is$=\frac{1}{2}{e}^{-t}-\frac{1}{2}\cos \left(t\right)+\frac{1}{2}\sin \left(t\right)$

Step-by-step solution

Given information

Given the convolution integral,

$f\left(t\right)={\int }_0^t{e}^{-\tau }\sin \left(t-\tau \right)d\tau$

Where, the function $g\left(t\right)$and h(t) are given by

$g\left(t\right)=e^{-t}h\left(t\right)=\sin \left(t\right)$

And, from the Laplace transformation, look up for the identity L.2 and L.3, know that the Laplace transform for the function $g\left(t\right)$ and $h\left(t\right)$is given by

$L\left(g\left(t\right)\right)=\frac{1}{\left(p+1\right)}L\left(h\left(t\right)\right)=\frac{1}{p^2+1}$

Thus,

$G\left(p\right)=\frac{1}{\left(p+1\right)}H\left(p\right)=\frac{1}{p^2+1}$

Convolution of two function g and h

The convolution of two function $\mathit{g}$and$\mathit{h}$, are defined as follows

$$\begin{gathered} \mathit{g}\mathbf{*}\mathit{h}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{t}}\mathit{g}\left(\tau \right)\mathit{h}\left(t-\tau \right)\mathit{d}\mathit{\tau } \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{t}}\mathit{g}\left(t-\tau \right)\mathit{h}\left(\tau \right)\mathit{d}\mathit{\tau } \end{gathered}$$

Laplace of the convolution

The Laplace of the convolution of the two function $h$and $g$, are the product of the Laplace of the function $h$and $g$, such that

$L\left(g*h\right)=G\left(p\right)H\left(p\right)$

Thus, if a convolution integral, find the functions and their Laplace transform, solve the integration where the product

$G\left(p\right)H\left(p\right)$

Would be simplified partial fractions, and then take the Laplace inverse transform of the fractions evaluate the integral, here

$L^{-1}\left(G\left(p\right)H\left(p\right)\right)=g*h$

Thus,

$$\begin{gathered} G\left(p\right)=\frac{1}{\left(p+1\right)} \\ H\left(p\right)=\frac{1}{p^2+1} \end{gathered}$$

And hence, the Laplace of the convolution is

$L\left(g*h\right)=\frac{1}{\left(p+1\right)\left(p^2+1\right)}$

Find the constants A,B and C

Now, find the convolution integral by the use of Laplace

$\frac{1}{\left(p+1\right)\left(p^2+1\right)}=\frac{A}{\left(p+1\right)}+\frac{Bp+C}{\left(p^2+1\right)}$

Multiply the whole equation by $\left(p+1\right)\left(p^2+1\right)$, thus

$1=A\left(p^2+1\right)+\left(Bp+C\right)\left(p+1\right)$

Now, find the constants A,B and C, start to set $p=-1$, this would work on vanish the second term, and hence reduce the number of unknowns in the equation into one unknown, and thus

$$\begin{gathered} 1=A\left({\left(-1\right)}^2+1\right)+\left(Bp+C\right)\left(-1+1\right) \\ 1=A\left(1+1\right)+\left(Bp+C\right)\left(0\right) \\ 2A=1 \\ A=\frac{1}{2} \end{gathered}$$

Find constants

The value of constant C is given by

Set the value of p to be zero, notice that the Constant Bwould vanish, and hence

$$\begin{gathered} 1=A\left({\left(0\right)}^2+1\right)+\left(B\left(0\right)+C\right)\left(\left(0\right)+1\right) \\ 1=A+\frac{0+C}{1} \\ =A+C \end{gathered}$$

And substitute the value of the constant A,

$$\begin{gathered} 1=\left(\frac{1}{2}\right)+C \\ C=\frac{1}{2} \end{gathered}$$

Know, the value of the constant A and C , chose any suitable value for p, in order to solve for B, may for example set the value of p to be 1 , hence

$$\begin{gathered} 1=A\left({\left(1\right)}^2+1\right)+\frac{B+C}{2} \\ =2A+\frac{B}{2}+\frac{C}{2} \end{gathered}$$

And substitute by the value of the constants A and C,

$$\begin{gathered} 1=2\left(\frac{1}{2}\right)+\frac{B}{2}+\frac{1/2}{2} \\ 1=1+\frac{1}{4}+\frac{B}{2} \\ 1=\frac{5}{4}+\frac{B}{2} \end{gathered}$$

Solve further the equation

$$\begin{gathered} \frac{B}{2}=1-\frac{5}{4} \\ =-\frac{1}{4} \\ B=-\frac{1}{2} \end{gathered}$$

Evaluation for the given convolution integral

The partial fraction, for fraction is

$$\begin{gathered} \frac{1}{\left(p+1\right)\left(p^2+1\right)}=\frac{1/2}{\left(p+1\right)}+\frac{\left(-1/2\right)p+\left(1/2\right)}{\left(p^2+1\right)} \\ =\frac{1}{2}\frac{1}{p+1}-\frac{1}{2}\frac{p}{p^2+1}+\frac{1}{2}\frac{1}{p^2+1} \end{gathered}$$

Now, know a simpler form for the product of the function $G\left(p\right)$and $H\left(p\right)$, easily find their Laplace inverse transform which would be the value for the integral and thus

$L^{-1}\left(G\left(p\right)H\left(p\right)\right)=\frac{1}{2}{L}^{-1}\left(\frac{1}{p+1}\right)-\frac{1}{2}{L}^{-1}\left(\frac{p}{p^2+1}\right)+\frac{1}{2}{L}^{-1}\left(\frac{1}{p^2+1}\right)$

And, use the identity L.2, L.4 and L.3 respectively, from the Laplace transformation table,

$=\frac{1}{2}\left(e^{-t}\right)-\frac{1}{2}\left(\cos \left(t\right)\right)+\frac{1}{2}\left(\sin \left(t\right)\right)$

Which, is evaluation for the given convolution integral, i.e.

$$\begin{gathered} f\left(t\right)={\int }_0^t{e}^{-\tau }\sin \left(t-\tau \right)d\tau \\ =L^{-1}\left(G\left(p\right)H\left(p\right)\right) \\ =\frac{1}{2}{e}^{-t}-\frac{1}{2}\cos \left(t\right)+\frac{1}{2}\sin \left(t\right) \end{gathered}$$