Question

Suppose that$\mathcal{L}\left\{f_1\left(t\right)\right\}=F_1\left(s\right)$ for $s>c_1$and that$\mathcal{L}\left\{f_2\left(t\right)\right\}=F_2\left(s\right)$ for $s>c_2$. When does

$\mathcal{L}\left\{f_1\left(t\right)+f_2\left(t\right)\right\}=F_1\left(s\right)+F_2\left(s\right)?$

Short answer

$\mathcal{L}\left\{f_1\left(t\right)+f_2\left(t\right)\right\}=F_1\left(s\right)+F_2\left(s\right)$$isvalidif$$s>\max \left\{c_1,c_2\right\}$

Step-by-step solution

Definition of Laplace Transform

Let f be a function defined for $t⩾0$. Then the integral

$\mathcal{L}\left\{f\right(t\left)\right\}={\int }_0^x{e}^{-xt}f\left(t\right)dt$

is said to be the Laplace transform of f, provided that the integral converges.

Use Laplace Transform

The formula for Laplace transform.

$\text{L}\left\{f\right(t\left)\right\}={\int }_0^{\infty }{e}^{st}f\left(t\right)\text{d}t$

$\mathcal{L}\left\{f_1\left(t\right)\right\}$converges for$s>c_1$and$\mathcal{L}\left\{f_2\left(t\right)\right\}$converges for$s>c_2$$s>c_2$.

$\mathcal{L}\left\{f_1\left(t\right)+f_2\left(t\right)\right\}\stackrel{\left(1\right)}{-}{\int }_0^{\infty }\left(f_1\left(t\right)+f_2\left(t\right)\right)e^{st}dt$

$={\int }_0^{\infty }{f}_1\left(t\right)e^{st}\text{d}t+{\int }_0^{\infty }{f}_2\left(t\right)e^{st}\text{d}t$

$\stackrel{\left(1\right)}{=}\mathcal{L}\left\{f_1\left(t\right)\right\}+\mathcal{L}\left\{f_2\left(t\right)\right\}$

$=F_1\left(s\right)+F_2\left(s\right)$

$\mathcal{L}\left\{f_1\left(t\right)+f_2\left(t\right)\right\}$converges when$S$is greater than maximum of${\mathrm{C}}_1$and${\mathrm{C}}_2$

Hence, $\mathcal{L}\left\{f_1\left(t\right)+f_2\left(t\right)\right\}=F_1\left(s\right)+F_2\left(s\right)$is valid if$s>\max \left\{c_1,c_2\right\}$