2021-12-16

Hello,

i'm here because i didn't found anything general on this topic, i need to know what is the expression of the laplace transform a composed function like that : $L\left\{f\left(x\right)\circ g\left(x\right)\right\}=L\left\{f\left(g\left(x\right)\right)\right\}=??$

my complete equation to transform is : $\phantom{\rule{0ex}{0ex}}x\left(t\right)=4×\left[\stackrel{¨}{a}\left(t\right)cos\left(a\left(t\right)\right)-\stackrel{˙}{a}\left(t{\right)}^{2}sin\left(a\left(t\right)\right)\right]+2500\phantom{\rule{0ex}{0ex}}y\left(t\right)=4×\left[-\stackrel{¨}{a}\left(t\right)sin\left(a\left(t\right)\right)-\stackrel{˙}{a}\left(t{\right)}^{2}cos\left(a\left(t\right)\right)\right]+\frac{3.711}{4}×{t}^{2}+2700$

the goal is to isolate the $a\left(t\right)$ function as a function of $x\left(t\right)$ and $y\left(x\right)$

Robin Luiz

nick1337

To find the Laplace transform of a composite function like $L\left\{f\left(x\right)\cdot g\left(x\right)\right\}$, we can use the property that $L\left\{f\left(g\left(x\right)\right)\right\}=\frac{G\left(s\right)}{g\text{'}\left(s\right)}$, where G(s) is the Laplace transform of f(g(x)).

Now, let's apply this property to the function x(t) given in the prompt: $x\left(t\right)=4\cdot \left[-a\text{'}\text{'}\left(t\right)\mathrm{sin}\left(a\left(t\right)\right)-a\text{'}{\left(t\right)}^{2}\mathrm{cos}\left(a\left(t\right)\right)\right]+\frac{3.777}{4}\cdot {t}^{2}+2700$

First, we need to take the Laplace transform of each term separately. The Laplace transform of $4\cdot \left[-a\text{'}\text{'}\left(t\right)\mathrm{sin}\left(a\left(t\right)\right)\right]$ can be found using the property $L\left\{f\text{'}\text{'}\left(t\right)\right\}={s}^{2}F\left(s\right)-0-f\text{'}\left(0\right)$, and the Laplace transform of $a\text{'}{\left(t\right)}^{2}\mathrm{cos}\left(a\left(t\right)\right)$ can be found using the property $L\left\{f{\text{'}}^{2}\left(t\right)\right\}=-F\text{'}\text{'}\left(s\right)$.

So we have: $L\left\{4\cdot \left[-a\text{'}\text{'}\left(t\right)\mathrm{sin}\left(a\left(t\right)\right)\right]\right\}=4\left[{s}^{2}L\left\{\mathrm{sin}\left(a\left(t\right)\right)\right\}+s\cdot a\text{'}\left(0\right)\cdot L\left\{\mathrm{cos}\left(a\left(t\right)\right)\right\}+a\text{'}\text{'}\left(0\right)\cdot L\left\{\mathrm{sin}\left(a\left(t\right)\right)\right\}\right]L\left\{a\text{'}{\left(t\right)}^{2}\mathrm{cos}\left(a\left(t\right)\right)\right\}$

$=-L\left\{a\text{'}\text{'}\left(t\right)\mathrm{cos}\left(a\left(t\right)\right)\right\}$

Next, we need to take the Laplace transform of the remaining term, $\frac{3.777}{4}{t}^{2}+2700$. The Laplace transform of ${t}^{n}$ is $\frac{n!}{{s}^{n+1}}$, so we have: $L\left\{\frac{3.777}{4}{t}^{2}+2700\right\}=\frac{3.777}{4}\cdot \frac{2!}{{s}^{3}}+\frac{2700}{s}$

Putting it all together, we have:

$L\left\{x\left(t\right)\right\}=L\left\{4\cdot \left[-a\text{'}\text{'}\left(t\right)\mathrm{sin}\left(a\left(t\right)\right)\right]+a\text{'}{\left(t\right)}^{2}\mathrm{cos}\left(a\left(t\right)\right)\right]+L\left\{\frac{3.777}{4}{t}^{2}+2700\right\}$

$=4\left[{s}^{2}L\left\{\mathrm{sin}\left(a\left(t\right)\right)\right\}+sa\text{'}\left(0\right)\cdot L\left\{\mathrm{cos}\left(a\left(t\right)\right)\right\}+a\text{'}\text{'}\left(0\right)\cdot L\left\{\mathrm{sin}\left(a\left(t\right)\right)\right\}\right]-L\left\{a\text{'}\text{'}\left(t\right)\mathrm{cos}\left(a\left(t\right)\right)\right\}+\frac{3.777}{4}\cdot \frac{2!}{{s}^{3}}+\frac{2700}{s}$

Note that we have used the initial conditions a'(0) and a''(0) to simplify some terms.

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