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Section 5 Mathematical Formulas5-3-3 Laplace Transform of Combolution When the functions , ''$f (x)$'' & ''$g (x)$'' are given, the convolution, ''$f * g$'' is defined as the following equation. \begin{eqnarray} f*g(t)=\int_0^tf(t-\tau)g(\tau)d\tau \nonumber \end{eqnarray}Assuming that the Laplace transforms of the respective functions are \begin{eqnarray} \mathcal{L}[f(t)]=F(s)\;\;\;\;\;\mathcal{L}[g(t)]=G(s) \nonumber \end{eqnarray}the following equation is indicated. \begin{eqnarray} \mathcal{L}[f*g(t)]=F(s)G(s) \nonumber \end{eqnarray}【Example-1: Inverse Laplace transform of convolution 】 Using the sum of the products formula of the trigonometric function, the convolution of the ''$sin t$ & $cos t$'' can be shown in the following equation. \begin{eqnarray} (sint)*(cost)&=&\int_0^tsin(t-\tau)cos\tau d\tau \nonumber \\ &=&\frac{1}{2}\int_0^t(sint+sin(t-2\tau))d\tau \nonumber \\ &=&[\frac{\tau sint}{2}]_0^t+[\frac{cos(t-2\tau)}{4}]_0^t \nonumber \\ &=&\frac{tsint}{2} \nonumber \end{eqnarray}The Laplace transforms of the ''$sin t$ & $cos t$'' are indicated as follows respectively. \begin{eqnarray} \mathcal{L}[sint]=\frac{1}{s^2+1}\;\;\;\;\;\mathcal{L}[cost]=\frac{s}{s^2+1} \nonumber \end{eqnarray}Accordingly, the following result of the Laplace transform can be obtained. \begin{eqnarray} \mathcal{L}[sint*cost]=\mathcal{L}[\frac{tsint}{2}]=\frac{s}{(s^2+1)^2} \nonumber \end{eqnarray}【Example-2: Inverse Laplace transform of convolution 】 We figure out the following Laplace inverse transform. \begin{eqnarray} \mathcal{L}^{-1}\{\frac{1}{s(s^2+1)}\} \nonumber \end{eqnarray} \begin{eqnarray} \frac{1}{s(s^2+1)}=\frac{1}{s}\frac{1}{s^2+1}=F(s)G(s) \nonumber \end{eqnarray} \begin{eqnarray} \mathcal{L}^{-1}[F(s)]=\mathcal{L}^{-1}\{\frac{1}{s}\}=1\nonumber \end{eqnarray} \begin{eqnarray} \mathcal{L}^{-1}[G(s)]=\mathcal{L}^{-1}\{\frac{1}{s^2+1}\}=sint \nonumber \end{eqnarray} \begin{eqnarray} \mathcal{L}^{-1}\{\frac{1}{s(s^2+1)}\}&=&1*sint \nonumber \\ &=&\int_0^tsin\tau d\tau \nonumber \\ &=&1-cost \nonumber \end{eqnarray}【Example-3: Inverse Laplace transform of convolution 】 We figure out the following Laplace inverse transform. \begin{eqnarray} \mathcal{L}^{-1}\{\frac{F(s)\omega_d}{(s+\zeta \omega_n)^2+\omega_d^2}\} \nonumber \end{eqnarray} \begin{eqnarray} \frac{F(s)\omega_d}{(s+\zeta \omega_n)^2+\omega_d^2}&=&F(s)\frac{\omega_d}{(s+\zeta\omega_n)^2+\omega_d^2} \nonumber \\ &=&F(s)G(s) \nonumber \end{eqnarray} \begin{eqnarray} \mathcal{L}^{-1}[F(s)]&=&f(t) \nonumber \\ \mathcal{L}^{-1}[G(s)]&=&\mathcal{L}^{-1}\{\frac{\omega_d}{(s+\zeta \omega_n)^2+\omega_d^2}\} \nonumber \\ &=&e^{-\zeta \omega_n t}sin \omega_dt \nonumber \end{eqnarray} \begin{eqnarray} \mathcal{L}^{-1}\{F(s)G(s)\}&=&f(t)*g(t) \nonumber \\ &=&\int_0^tf(\tau)g(t-\tau)d\tau \nonumber \\ &=&\int_0^tf(\tau)e^{-\zeta \omega(t-\tau)}sin\omega_d(t-\tau) d\tau \nonumber \\ &=&e^{-\zeta \omega_n t}\{\int_0^tf(\tau)e^{\zeta \omega_n \tau}sin\omega_d t \;cos\omega_d \tau d\tau \nonumber \\ &\;&-\int_0^tf(\tau)e^{\zeta \omega_n \tau}cos\omega_dt \;sin\omega_d \tau d\tau\} \nonumber \\ &=&e^{-\zeta \omega_n t}\{sin\omega_dt\int_0^tf(\tau)e^{\zeta \omega_n \tau}cos\omega_d\tau d\tau \nonumber \\ &\;&-cos\omega_dt\int_0^tf(\tau)e^{\zeta \omega_n \tau}sin\omega_d \tau d\tau\} \nonumber \end{eqnarray}Substituting the integral variable, ''$\tau$''with ''$t$'' here, the solution can be shown in the following equation. \begin{eqnarray} \mathcal{L}^{-1}\{F(s)G(s)\}&=&f(t)*g(t) \nonumber \\ &=&e^{-\zeta \omega_n t}\{sin\omega_dt\int_0^tf(t)e^{\zeta \omega_n t}cos\omega_dt dt \nonumber \\ &\;&-cos\omega_dt\int_0^tf(t)e^{\zeta \omega_n t}sin\omega_d t dt\} \nonumber \end{eqnarray} |