fourier transform - what is the difference between $X(jomega)$ and $X(omega)$ notation

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.

Answer

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:

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