Question

Use L34 and L2 to find the inverse transform of $\mathbf{G}\left(p\right)\mathbf{H}\left(p\right)$whenand $\mathbf{G}\left(p\right)\mathbf{=}\mathbf{1}\mathbf{/}\left(p+a\right)\mathbf{}\mathbf{and}\mathbf{}\mathbf{H}\left(p\right)\mathbf{=}\mathbf{1}\mathbf{/}\left(p+b\right)$; your result should be L7 .

Short answer

The inverse transforms of is $G\left(p\right)H\left(p\right)is\frac{1}{b-a}\left(e^{-at}-e^{-b\tau }\right).$

Step-by-step solution

Given information.

The given expressions are and$G\left(p\right)H\left(p\right)is\frac{1}{\left(p+a\right)}andH\left(p\right)is\frac{1}{\left(p+b\right)}$ .

Inverse transform.

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

$$\begin{gathered} \mathbf{L}\left\{f\right\}\left(s\right)\mathbf{=}\mathbf{L}\left\{\left\{f\right\}\right\}\left(s\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{F}\left(s\right) \end{gathered}$$

where L is the Laplace transform.

Find the inverse transform of G(p)H(p). .

Calculate Laplace inverse.

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

Calculate Laplace inverse of $\mathrm{G}\left(\mathrm{p}\right).$

role="math" localid="1659264162995" $$\begin{gathered} L^{-1}G\left(p\right)=L^{-1}\left(\frac{1}{p+a}\right) \\ g\left(t\right)=e^{-at} \\ g\left(t-\tau \right)=e^{-a\left(t-\tau \right)} \end{gathered}$$

Calculate Laplace inverse of H(p).

$$\begin{gathered} L^{-1}H\left(p\right)=L^{-1}\left(\frac{1}{p+b}\right) \\ h\left(t\right)=e^{-bt} \\ h\left(t-\tau \right)=e^{-b\tau } \end{gathered}$$

Calculate Laplace inverse of G(p)H(p) .

$L^{-1}\left[\left(\frac{1}{p+a}\right)\left(\frac{1}{p+b}\right)\right]=\frac{1}{b-a}\left(e^{-at}-e^{-b\tau }\right)$

Thus, the Laplace inverse of is $G\left(p\right)H\left(p\right)is\frac{1}{b-a}\left(e^{-at}-e^{-b\tau }\right).$.