Riya Andrews

2022-09-03

Using laplace trasnform find the convulation of (f*g)(t).

$let\phantom{\rule{thickmathspace}{0ex}}f(t)=\mathrm{sin}(3t)$ and $g(t)={e}^{-2t}$

Using laplace trasnform find the convulation of (f*g)(t).

convulation theorem: $h(t)=(f\ast g)(t)={\int}_{0}^{t}f(u)g(t-u)du$

$L(sin(3t)=\frac{3}{{s}^{2}+9}\phantom{\rule{0ex}{0ex}}L({e}^{-2t})=\frac{1}{s+2}$

then how we processed for this problem

$let\phantom{\rule{thickmathspace}{0ex}}f(t)=\mathrm{sin}(3t)$ and $g(t)={e}^{-2t}$

Using laplace trasnform find the convulation of (f*g)(t).

convulation theorem: $h(t)=(f\ast g)(t)={\int}_{0}^{t}f(u)g(t-u)du$

$L(sin(3t)=\frac{3}{{s}^{2}+9}\phantom{\rule{0ex}{0ex}}L({e}^{-2t})=\frac{1}{s+2}$

then how we processed for this problem

barquegese2

Beginner2022-09-04Added 11 answers

Let $h(t)=(f\ast g)(t)$. We have

$$\mathcal{L}\{f\ast g\}=\mathcal{L}\{f\}\cdot \mathcal{L}\{g\}$$

that is

$$H(s)=F(s)G(s)=\frac{3}{{s}^{2}+9}\cdot \frac{1}{s+2}=-\frac{9}{13}\cdot \frac{s}{{s}^{2}+9}+\frac{6}{13}\cdot \frac{3}{{s}^{2}+9}+\frac{9}{13}\cdot \frac{1}{s+2}$$

and then

$$h(t)={\mathcal{L}}^{-1}\{F(s)G(s)\}=[-\frac{9}{13}\mathrm{cos}(3t)+\frac{6}{13}\mathrm{sin}(3t)+\frac{9}{13}{\mathrm{e}}^{-2t}]u(t)$$

$$\mathcal{L}\{f\ast g\}=\mathcal{L}\{f\}\cdot \mathcal{L}\{g\}$$

that is

$$H(s)=F(s)G(s)=\frac{3}{{s}^{2}+9}\cdot \frac{1}{s+2}=-\frac{9}{13}\cdot \frac{s}{{s}^{2}+9}+\frac{6}{13}\cdot \frac{3}{{s}^{2}+9}+\frac{9}{13}\cdot \frac{1}{s+2}$$

and then

$$h(t)={\mathcal{L}}^{-1}\{F(s)G(s)\}=[-\frac{9}{13}\mathrm{cos}(3t)+\frac{6}{13}\mathrm{sin}(3t)+\frac{9}{13}{\mathrm{e}}^{-2t}]u(t)$$

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My steps:

$F(s)=\frac{2{s}^{2}+(a-6b)s+{a}^{2}-4ab}{(s+a)(s-a)(s-2b)}$

$=\frac{A}{s+a}+\frac{B}{s-a}+\frac{C}{s-2b}+K$

$K=0$

$A=F(s)\ast (s+a)$