91影视

Let$f\left(t\right)$and be piecewise continuous on$[0,\mathrm{鈭�})$and of exponential orderand

set,$F\left(s\right)=\mathcal{L}\left\{f\right\}\left(s\right)\text{and}G\left(s\right)=\mathcal{L}\left\{g\right\}\left(s\right)$ then,

$\mathcal{L}\{f鈭�g\}\left(s\right)=F\left(s\right)G\left(s\right),$or

${\mathcal{L}}^{鈭�1}\left\{F\right(s\left)G\right(s\left)\right\}\left(t\right)=(f鈭�g)\left(t\right)$

Consider the given function,

$\frac{s+1}{{(s^2+1)}^2}$

Let,$y\left(s\right)=\frac{s+1}{{(s^2+1)}^2}$

Take inverse Laplace transform on both sides,

role="math" localid="1664133565583" ${\mathcal{L}}^{鈭�1}\left[y\right(s\left)\right]={\mathcal{L}}^{鈭�1}\left[\frac{s+1}{{(s^2+1)}^2}\right]$$鈫�\left(1\right)$

Hence, the convolution formula is,${\mathcal{L}}^1\left[f\left(s\right)鈰�g\left(s\right)\right]=f鈰�g={鈭�}_0^{t}f(t鈭�v)g\left(v\right)dv$ , where and

$f\left(s\right)=\frac{1}{s^2+1}$And,$f\left(t\right)=\sin t$

$\begin{array}{c}g\left(s\right)=\frac{s+1}{s^2+1}\\ =\frac{s}{s^2+1}+\frac{1}{s^2+1}\end{array}$and$g\left(t\right)=\cos t+\sin t$

Thus, equation(1) becomes,

$\begin{array}{c}y\left(t\right)={鈭�}_0^t\sin (t鈭�v)鈰�[\cos v+\sin v]dv\\ ={鈭�}_0^t\sin (t鈭�v)\cos vdv+{鈭�}_0^t\sin (t鈭�v)\sin vdv\\ ={鈭�}_0^t\frac{1}{2}[\sin t+\sin (t鈭�2v\left)\right]dv+{鈭�}_0^t\frac{鈭�1}{2}[\cos t鈭�\cos (t鈭�2v\left)\right]dv\end{array}$

Use the formula,$\begin{array}{c}\sin (A+B)+\sin (A鈭�B)=2\sin A鈰�\cos B\\ \cos (A+B)鈭�\cos (A鈭�B)=鈭�2\sin A鈰�\sin B\end{array}$

$\begin{array}{c}y\left(t\right)={鈭�}_0^t\frac{1}{2}\sin tdv+{鈭�}_0^t\frac{1}{2}\sin (t鈭�2v)dv鈭�{鈭�}_0^t\frac{1}{2}\cos tdv+{鈭�}_0^t\frac{1}{2}\cos (t鈭�2v)dv\\ =\frac{1}{2}\sin t{\left[v\right]}_0^t+\frac{1}{4}\cos {[t鈭�2v]}_0^t鈭�\frac{1}{2}\cos {\left[v\right]}_0^t鈭�\frac{1}{4}\sin (t鈭�2v){|}_0^t\\ =\frac{1}{2}t\sin t鈭�0鈭�\frac{1}{2}t\cos t+\frac{1}{2}\sin t\\ =\frac{1}{2}[t\sin t鈭�t\cos t+\sin t]\end{array}$

Therefore, the inverse Laplace transform is.$y\left(t\right)=\frac{1}{2}[t\sin t鈭�t\cos t+\sin t]$