KesseTher12

2022-10-05

How to find the inverse laplace transform of this function:

$$Y(s)=\frac{1}{\tau s+1}\times \frac{\omega}{{s}^{2}+{\omega}^{2}}$$

for $\omega $, $\tau $ constant

$$Y(s)=\frac{1}{\tau s+1}\times \frac{\omega}{{s}^{2}+{\omega}^{2}}$$

for $\omega $, $\tau $ constant

Caiden Brewer

Beginner2022-10-06Added 5 answers

you can apply convolution theorem

and seperate laplace transform of them are

$$\frac{1}{\tau s+1}\times \frac{\omega}{{s}^{2}+{\omega}^{2}}$$

laplace for 1st is

$$g(t)=\frac{1}{\tau}\mathrm{exp}\frac{-t}{\tau}$$

and for

$$\frac{\omega}{{s}^{2}+{\omega}^{2}}$$

is

$$f(t)=\mathrm{sin}\omega t$$

hence apply convolution theorem

$${\int}_{0}^{t}g(v)f(t-v)\mathrm{d}v$$

solve the integral you get the result

and seperate laplace transform of them are

$$\frac{1}{\tau s+1}\times \frac{\omega}{{s}^{2}+{\omega}^{2}}$$

laplace for 1st is

$$g(t)=\frac{1}{\tau}\mathrm{exp}\frac{-t}{\tau}$$

and for

$$\frac{\omega}{{s}^{2}+{\omega}^{2}}$$

is

$$f(t)=\mathrm{sin}\omega t$$

hence apply convolution theorem

$${\int}_{0}^{t}g(v)f(t-v)\mathrm{d}v$$

solve the integral you get the result

sengihantq

Beginner2022-10-07Added 3 answers

$$y(x)=\frac{{e}^{-x/\tau}\omega \tau -\omega \tau \mathrm{cos}(\omega x)+\mathrm{sin}(\omega x)}{1+{\omega}^{2}{\tau}^{2}}$$

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inverse laplace transform - with symbolic variables:

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

My steps:

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

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

$K=0$

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