Question

Verify L28 in the table by using L27 and the convolution integral.

Short answer

$L_{28}:L^{-1}[e^{-pu}.G(p\left)\right]=g(t-a)$

Step-by-step solution

Given information

Prove that L_2yand L_2sby convolution integral

Definition of Laplace Transformation

A transformation of a function into the function that is useful especially in reducing the solution of an ordinary linear differential equation with constant coefficients to the solution of a polynomial equation.

The inverse Laplace transform of a function F(s)is the piecewise-continuous and exponentially-restricted real function f(t)

Verify the given function

Laplace transform of L₂₇and L₂₈

$$\begin{gathered} L_{28}:L^{-1}[e^{pu}.G(p\left)\right]=g(t-a,)t>a>0=0,t<a \\ {L}_{27}:L^{-1}\left(e^{pu}\right)=\delta (t-a);a\ge 0 \\ {L}^{-1}[e^{-pa}-G(p\left)\right]=L^{-1}\left(G\right(p).e^{-pu}) \\ =L^{-1}\left[G\right(p).H(P\left)\right] \end{gathered}$$

Solve further

$$\begin{gathered} =g*h \\ ={\int }_0^{1}g\left(\tau \right).h(t-\tau )d\tau \\ ={\int }_0^{1}g(t-\tau ).h\left(t\right)d\tau \end{gathered}$$

$$\begin{gathered} H\left(p\right)=e^{-pa} \\ L\left(h\right(r\left)\right)=e^{-pa}h\left(t\right) \\ =L^{-1}\left(e^{-pa}\right) \\ h\left(t\right)=\delta (t-a) \end{gathered}$$

$h\left(t\right)=\delta (\tau -a),a\ge 0$

$$\begin{gathered} L^{-1}(e^{-pa}-G(p\left)\right)={\int }_0^{1}g(t-\tau ).\delta (r-a) \\ {L}^{-1}({\epsilon }^{-pa}-G(p\left)\right)=g(t-a),0<a<t \\ =0,t<\mathrm{an}L<1(e^{-pa}-G(p\left)\right) \\ =g(t-a),t>a>0=0,t<a \end{gathered}$$