8
$\begingroup$

I am trying to understand Fourier Transform and Laplace Transform. What is the difference between $X(j\omega)$ and $X(\omega)$ notation?

what is the meaning of $j\omega$ ? Is it represent frequency? If it is, what is the meaning of imaginary frequency?

Thanks in advance.

Improve this question
$\endgroup$
1
  • $\begingroup$ The Laplace transform covers the entire 2D S plane. The Fourier transform is just the 1D slice of that plane along the jω axis $\endgroup$
    – endolith
    May 7 '15 at 1:51
8
$\begingroup$

Both notations are common and correct. As pointed out by Yuri Nenakhov, the advantage of the argument $j\omega$ is that it coincides with the complex (Laplace transform) variable $s$ when its real-part is zero. Note that in the complex $s$-plane the frequency axis is the imaginary axis. So $j\omega$ has nothing to do with complex frequency (which makes no sense).

So if the Laplace transform $X(s)$ of a signal $x(t)$ exists, and if the imaginary axis is inside its region of convergence, then the Fourier transform is obtained by setting $s=j\omega$.

Note that this does not work in general! In general you can't get the Fourier transform by replacing $s$ with $j\omega$ and vice versa. Two conditions must be satisfied in order for this to lead to a correct result:

  1. Both transforms must exist (in the sense that the corresponding signal $x(t)$ has a Laplace transform and a Fourier transform).
  2. The imaginary axis $s=j\omega$ must be inside the region of convergence of the Laplace transform.

An example where replacing $s$ by $j\omega$ doesn't work, even though both transforms exist, is the step function:

$$\begin{align}&x(t)=u(t)\\\text{Laplace transform: }&X(s)=\frac{1}{s}\\ \text{Fourier transform: }&\hat{X}(j\omega)=\pi\delta(\omega)+\frac{1}{j\omega}\neq X(s)|_{s=j\omega}\end{align}$$

Improve this answer
$\endgroup$
5
  • $\begingroup$ I disagree with your example. Clearly, the Laplace transform of the Heaviside function does not exist at s=0 (and it exists on the imaginary axis also only by analytical continuation), so your own requirements fail. Also note that the Fourier transform and the Laplace transform coincide where both are defined. $\endgroup$
    – Jazzmaniac
    May 7 '15 at 9:24
  • $\begingroup$ @Jazzmaniac: The Laplace transform of the step function exists. It has a pole at $s=0$. This is different from cases where the Laplace transform does not exist (such as for $x(t)=\sin(\omega t)$). If you're not saying that the Fourier transform of $x(t)=u(t)$ can be obtained from $X(s)$ by setting $s=j\omega$ (which is wrong), I'm not sure what it is that you're saying. $\endgroup$
    – Matt L.
    May 7 '15 at 10:02
  • 1
    $\begingroup$ I'm saying that the Laplace transform does not exist at s=0, and strictly speaking it does not even exist at real(s)=0. The improper integral does not converge there. You can fix that by either applying a Cauchy principal value assignment to the integral or by analytically continuing the Laplace transform at real(s)>0. But whatever you do, you can't fix it for s=0. So the Laplace transform does not exist everywhere on the imaginary axis, and if you're really strict, it exists nowhere on the imaginary axis in the proper sense. $\endgroup$
    – Jazzmaniac
    May 7 '15 at 10:43
  • $\begingroup$ That's why the disagreement between the Laplace transform and the Fourier transform at $s=0$ or $\omega=0$ is the least you can expect. It's not much more than a fortunate coincidence following from some nice properties of the analytical continuation that you get the correct result at $s=\omega*j$ for $\omega \neq 0$. So I agree with you example, but not with the premise "...even though both transforms exist...". $\endgroup$
    – Jazzmaniac
    May 7 '15 at 10:46
  • $\begingroup$ @Jazzmaniac: OK, but the idea of the example was to show a simple case where with $X(s)$ the Laplace transform, $X(j\omega)$ does not (for all $\omega$) equal the Fourier transform. And still, both transforms exist for that example. Obviously, the ROC of $X(s)$ is $Re\{s\}>0$. The step function has a Laplace transform and a Fourier transform, that's all I mean to say when I say they both exist. I'll add two more examples later on, where one of the two does not exist, to make my point a bit clearer. $\endgroup$
    – Matt L.
    May 7 '15 at 14:50
4
$\begingroup$

$X(j \omega)$ (frequency response) is a Fourier transform of system's impulse response. It's actually a function of frequency ($\omega$) but usually is written as $X(j \omega)$ because replacing $j \omega$ in the formula with $s$ will give you system's Laplace transform $X(s)$ without any additional conversions. (This works in the opposite direction as well: if you have a Laplace ransform, you can get frequency response by replacing $s$ with $j \omega$.)

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ Note that replacing $s$ with $j\omega$ and vice versa is generally not a valid way to get the Fourier transform from the Laplace transform or the other way around. See my answer. $\endgroup$
    – Matt L.
    May 7 '15 at 7:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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