# If I use Inverse Laplace transformation using the table do I get cos(t)−sin(t). can I go back to the t domain with the x domain normally ?

Let
$F\left(s\right)=L\left\{f\left(t\right)\right\}$
After a Laplace transform of f(t) the expression is :
$F\left(s\right)=\frac{2s-4}{{s}^{2}-2s+2}$
which equals to
$F\left(s\right)=2\cdot \left(\frac{s-1}{\left(s-1{\right)}^{2}+1}-\frac{1}{\left(s-1{\right)}^{2}+1}\right)$
let
$x=s-1$
Therefore
$F\left(x\right)=\frac{x}{{x}^{2}+1}-\frac{1}{{x}^{2}+1}$
My question is, if I use Inverse Laplace transformation using the table do I get$\mathrm{cos}\left(t\right)-\mathrm{sin}\left(t\right)$. Can I go back to the t domain with the x domain normally ?
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Phoenix Morse
${\mathcal{L}}^{-1}\left\{F\left(s-\alpha \right)\right\}={e}^{\alpha t}{\mathcal{L}}^{-1}\left\{F\left(s\right)\right\}$
${\mathcal{L}}^{-1}\left\{F\left(s-1\right)\right\}={e}^{t}{\mathcal{L}}^{-1}\left\{F\left(s\right)\right\}$
${\mathcal{L}}^{-1}\left\{2\left(\frac{s-1}{\left(s-1{\right)}^{2}+1}-\frac{1}{\left(s-1{\right)}^{2}+1}\right)\right\}=2{e}^{t}\left(\mathrm{cos}t-\mathrm{sin}t\right).$
Where does the "frequency shift" property of the Laplace transform come from?:
$\mathcal{L}\left\{{e}^{\alpha t}f\left(t\right)\right\}={\int }_{0}^{\mathrm{\infty }}{e}^{-st}{e}^{\alpha t}f\left(t\right)\phantom{\rule{thinmathspace}{0ex}}dt={\int }_{0}^{\mathrm{\infty }}{e}^{-\left(s-\alpha \right)t}f\left(t\right)\phantom{\rule{thinmathspace}{0ex}}dt=F\left(s-\alpha \right).$
###### Did you like this example?
batejavizb
$F\left(s\right)=2\cdot \left(\frac{s-1}{\left(s-1{\right)}^{2}+1}-\frac{1}{\left(s-1{\right)}^{2}+1}\right)$
If you change the variable and substitute x=s−1 you have to substitute s on both sides:
$F\left(x+1\right)=2\left(\frac{x}{{x}^{2}+1}-\frac{1}{{x}^{2}+1}\right)$
${\mathcal{L}}^{\mathcal{-}\mathcal{1}}\left\{F\left(x+1\right)\right\}=2{\mathcal{L}}^{\mathcal{-}\mathcal{1}}\left\{\left(\frac{x}{{x}^{2}+1}-\frac{1}{{x}^{2}+1}\right)\right\}$
${e}^{-t}f\left(t\right)=2\left(\mathrm{cos}\left(t\right)-\mathrm{sin}\left(t\right)\right)$
Therefore:
$f\left(t\right)=2{e}^{t}\left(\mathrm{cos}\left(t\right)-\mathrm{sin}\left(t\right)\right)$
So this line is not correct. Because your substitution on LHS is not correct s=x. When on RHS you substitute s=x−1.
$F\left(s\right)=2\cdot \left(\frac{s-1}{\left(s-1{\right)}^{2}+1}-\frac{1}{\left(s-1{\right)}^{2}+1}\right)$
$\underset{x=s}{\underset{⏟}{F\left(x\right)}}=\underset{x=s-1}{\underset{⏟}{\frac{x}{{x}^{2}+1}-\frac{1}{{x}^{2}+1}}}$
It should be:
$\underset{x=s-1}{\underset{⏟}{F\left(x+1\right)}}=\underset{x=s-1}{\underset{⏟}{\frac{x}{{x}^{2}+1}-\frac{1}{{x}^{2}+1}}}$