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There are several functions for which we know that Fourier Transform will exist but still we calculate its Laplace Transform. Can I know the reason why we need to take Laplace transform for which we know its convergence?

Thanks

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There is a large class of functions for which both the Fourier transform and the Laplace transform exist, and for which one can be obtained from the other by setting $s=j\omega$. (Note that even when both exist, the latter need not be the case). So for this class of functions, obtaining the Laplace transform from the Fourier transform (or vice versa) does not require any additional work.

Example: $$\begin{align}x(t)&=e^{-at}u(t),\quad a>0\\ \text{Fourier transform:}\quad X(j\omega)&=\frac{1}{j\omega +a}\\ \text{Laplace transform:}\quad X(s)&=\frac{1}{s +a},\quad |s|> -a\end{align}$$

What the Laplace transform offers is a description in the complex $s$-plane, such as poles and zeros of transfer functions, from which many system properties can be easily deduced. The Fourier transform only shows the behavior on the frequency axis (i.e., the spectrum).

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As I understand (I'm newb in the spectral analysis), using of Laplace Transform is a good idea if you are going to analyze transient signals, because it uses damped exponents as basis.

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  • $\begingroup$ I couldn;t understand what that means? $\endgroup$ – Shinning Eyes Nov 23 '15 at 11:34
  • $\begingroup$ For Fourier the basis functions are sine waves. For Laplace: damped sine waves (en.wikipedia.org/wiki/Damped_sine_wave). With damped sine waves it is easier to describe and analyze transients. $\endgroup$ – sergg Nov 23 '15 at 11:52

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