Laplace transformation and inverse Laplace question

Question

Let $$ F(s)=L\{f(t)\}$$ After a Laplace transform of f(t) the expression is : $$F(s)= \frac{2s-4}{s^2-2s+2} $$ which equals to $$F(s)=2\cdot\left(\frac{s-1}{(s-1)^2+1}-\frac{1}{(s-1)^2+1}\right) $$ let $$x=s-1$$ Therefore $$F(x)= \frac{x}{x^2+1}-\frac{1}{x^2+1}$$ My question is, if I use Inverse Laplace transformation using the table do I get $$ \cos(t)-\sin(t)$$ I ask this because I used Laplace transform from the t domain to the s domain.Now, can I go back to the $t$ domain with the $x$ domain normally ? Thank you

Answer 1

$$F(s)=2\cdot\left(\frac{s-1}{(s-1)^2+1}-\frac{1}{(s-1)^2+1}\right)$$ If you change the variable and substitute $x=s-1$ you have to substitute $s$ on both sides: $$F(x+1)=2 \Bigl(\frac{x}{x^2+1}-\frac{1}{x^2+1}\Bigl)$$ $$\mathcal {L^{-1}}\{F(x+1)\}=2\mathcal {L^{-1}}\left\{ \Bigl(\frac{x}{x^2+1}-\frac{1}{x^2+1}\Bigl)\right \}$$ $$e^{-t}f(t)=2(\cos(t)-\sin(t))$$ Therefore: $$f(t)=2e^{t}(\cos(t)-\sin(t))$$

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So this line is not correct. Because your substitution on LHS is not correct $s=x$. When on RHS you substitute $s=x-1$. $$F(s)=2\cdot\left(\frac{s-1}{(s-1)^2+1}-\frac{1}{(s-1)^2+1}\right)$$ $$\underbrace{F(\color {red}{x})}_{x=s}=\underbrace{ \frac{x}{x^2+1}-\frac{1}{x^2+1}}_{x=s-1}$$ It should be: $$\underbrace{F(\color {red}{x+1})}_{x=s-1}=\underbrace{ \frac{x}{x^2+1}-\frac{1}{x^2+1}}_{x=s-1}$$

Answer 2

I think you're looking for the frequency shift property of the Laplace transform: $$ \mathcal{L}[e^{at} f(t)] = F(s-a). $$ So in your case, you would have $$ F(x) = F(s-1) = \mathcal{L}[e^t f(t)] $$

Answer 3

$$\mathcal{L}^{-1} \{ F(s-\alpha) \}= e^{\alpha t}\mathcal{L}^{-1} \{F(s)\} $$

$$\mathcal{L}^{-1} \{ F(s-1) \}= e^{ t}\mathcal{L}^{-1} \{F(s)\} $$

$$\mathcal{L}^{-1} \left\{ 2\left( \frac{s-1}{(s-1)^2+1}-\frac{1}{(s-1)^2+1}\right) \right\}= 2e^{ t}(\cos t - \sin t). $$

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Where does the "frequency shift" property of the Laplace transform come from?:

$$\mathcal{L} \{ e^{\alpha t} f(t)\} = \int_0^\infty e^{-st}e^{\alpha t} f(t) \, dt = \int_0^\infty e^{-(s-\alpha)t} f(t) \, dt = F(s-\alpha).$$