I know that $$X_L(s) \Big|_{s=j\omega}=X_F(\omega)$$ if $x(t)$ is one sided and absolutely integrable and hence the 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?


2 Answers 2


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


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 $s=j\omega$:


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:


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

  • $\begingroup$ 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, 2016 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, 2016 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, 2016 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, 2016 at 13:46
  • 1
    $\begingroup$ @jomegaA: If you evaluate the Laplace transform for $s=j\omega$, then yes, you use an imaginary argument, and the function generally returns a complex value. $\endgroup$
    – Matt L.
    Feb 10, 2020 at 12:12

Just as an effort to increase the post's didatics to the future visitors of this question (just as me): I have noticed you commented you were taught the imaginary axis of the Laplace plane is the Fourier plane. This is a very common misleading intuition, that I myself once had.

This intuition is somewhat valid if you are talking about the 3D plot that sometimes is made in which we plot either the magnitude of s (abs(s)) or its phase angle in terms of the values of the arguments of s (the real and imaginary part). These two plots completely define the complex variable s.

So, in the context of this plot, then yes, over the imaginary axis of such plot you will see the absolute (or phase angle) value of the Fourier transform of the function you are working with.

Moreover, do not let this plot mislead you: it only shows half of the information about the variable (either the value of the real argument or the value of the imaginary argument). Generally speaking, all these points are potentially made of real and imaginary parts, even those over the imaginary axis. You need two 3D plots, of absolute and phase angle, to fully get the information about a given instance of s.

A good video on these intuitions is this one.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.