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I know that $\ X_L(s=j\omega)=X_F(\omega)$ if $\ x(t)$ is one sided and absolutely integrable and hence, imaginary axis of the Laplace transform is the Fourier transform. But Fourier transform also has imaginary and real parts. So how could this be right?

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The Laplace transform evaluated at $s=j\omega$ is equal to the Fourier transform if its region of convergence (ROC) contains the imaginary axis. This is also true for the bilateral (two-sided) Laplace transform, so the function need not be one-sided.

As for real and imaginary parts, since $s$ is a complex variable, both the Laplace and the Fourier transform generally have real and imaginary parts. Take as a simple example the function $x(t)=e^{-at}u(t)$, with $a>0$, where $u(t)$ is the unit step function. The Laplace transform is

$$X_L(s)=\frac{1}{s+a}\tag{1}$$

Since $a>0$, the ROC of $X_L(s)$ contains the imaginary axis, and the Fourier transform of $x(t)$ is simply obtained by evaluating $X_L(s)$ on the imaginary axis $s=j\omega$:

$$X_F(\omega)=X_L(j\omega)=\frac{1}{j\omega+a}\tag{2}$$

Since $s=\sigma+j\omega$ is generally complex, not only the Fourier transform but also the Laplace transform $(1)$ has a real and an imaginary part:

$$X_L(\sigma+j\omega)=\frac{1}{\sigma+j\omega+a}=\frac{\sigma+a}{(\sigma+a)^2+\omega^2}-j\frac{\omega}{(\sigma+a)^2+\omega^2}\tag{3}$$

Only when evaluated on the real axis $s=\sigma$ ($\omega=0$) does the imaginary part vanish.

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  • $\begingroup$ Thanx for the answer. I think my confusion was because I was taught that the imaginary axis of the Laplace plane is the Fourier plane. But since the Fourier plane has both imaginary and real parts(and the imaginary axis of the Laplace transform has only one dimension) it didn't make sense to me. May I conclude from your answer that it's just a substitution that works when satisfies proper conditions and has nothing to do with the relation between those planes? Because it's doesn't make any sense that the imaginary axis of the Laplace plane has both imaginary and real axis. $\endgroup$ – Vlad Jan 6 '16 at 12:26
  • $\begingroup$ @Vlad: There is no "Fourier plane", it's just the imaginary axis of the complex $s$ plane. $\endgroup$ – Matt L. Jan 6 '16 at 12:45
  • $\begingroup$ But again, if the Fourier transform it's just an imaginary axis in complex s plane, why does it have real and imaginary parts? Shouldn't it just be solely imaginary? $\endgroup$ – Vlad Jan 6 '16 at 13:17
  • $\begingroup$ @Vlad: No, the argument of the Fourier transform is purely imaginary (if you take $j\omega$ as its argument). The expression for the Fourier transform, i.e. the function, generally has non-zero real and imaginary parts, just like in the example (Eq. (2)) I gave in my answer. $\endgroup$ – Matt L. Jan 6 '16 at 13:46
  • $\begingroup$ It was confusing me for a few months. Can't believe I've been missing such trivial detail. Matt, thank you very much. $\endgroup$ – Vlad Jan 6 '16 at 14:17

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