# Compare Fourier and Laplace transform

I would like to clarify main difference between Fourier and Laplace transforms and also understand if exponential factor is main difference between this two method. So Fourier transform is following $$F(j\omega)=\int\limits_{-\infty}^\infty f(t)e^{-j\omega t}\mathrm dt$$

and Laplace transform is following one

$$F(s)=\int\limits_{-\infty}^\infty f(t)e^{-st}\mathrm dt$$ where $s=\alpha+j\omega$.

Let us this notation, I can't print symbols exactly, but if we put into equation of Laplace, we will get that because of

$e^{-a-j\omega}=e^{-a}*e^{-j\omega}$.

We get that in integral first function $f(t)$ is multiplied by factor $e^{-at}$ if we put notation of $s$ into Laplace integral and also multiply it by $t$ ,which of course would be some another real function for example $M(t)$ and again it would be back to Fourier transform of this $M(t)$ function . So let us make it more detailed.in fourier transform we have $e^{-j\omega t}$,in laplace we have $e^{-st}$ where again $s=\alpha+j\omega$.

If we put this into Laplace, we get

$f(t)e^{-\alpha t-j\omega t}$

which we can write as

$(f(t)e^{-\alpha t})e^{-j\omega t}$,

but first one is real right? And again we get real transform of function, or we can assign $(f(t)e^{-\alpha t})=M(t)$.

Please help me to clarify main difference between this two transform.

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sorry i could not enlarge symbols,if you could i will be happy –  dato datuashvili Apr 14 '13 at 12:01
thanks @UnkleRhaukus for update –  dato datuashvili Apr 14 '13 at 12:17
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## 1 Answer

Laplace is generalized Fourier transform. It is used to perform the transform analysis of unstable systems. Simply stating, Laplace has more convergence compared to Fourier.

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