# Bilateral Laplace transform and existence of Fourier transform

I was reading from Athanosios Papoulis' "The Fourier integral and its applications." and they referenced the bilateral Laplace transform and Fourier Transform as:

$$F(p)=\int_{-\infty}^{\infty}e^{-pt}f(t)dt$$ $$F(\omega)=\int_{-\infty}^{\infty}e^{-j\omega t}f(t)dt$$

and stability indicates that the real part of $p$ must lie between $a$ an $b$.(For stability I'm guessing?)

As per the textbook, if we take the values of $p$ to be purely real and ignore the imaginary axis, then $F(\omega)$ doesn't even exist.

logically, I can't see how that could happen. I was wondering if anyone could explain how that is so. As well as why we limit the real part of $p$.(guessing it's similar to the unit circle in Z transform).

• sorry about the formatting of integral, Not sure how to express it as formula on the site – Mr. Johnny Doe Jul 11 '18 at 17:23
• Take a look at how I formatted the formulas, so you can do it yourself the next time. – Matt L. Jul 11 '18 at 18:58

The bilateral Laplace transform converges in a vertical strip $a<\text{Re}\{p\}<b$, called the region of convergence (ROC). Compare this to the bilateral $\mathcal{Z}$-transform which converges in an annulus centered at the origin of the complex plane: $r_1<|z|<r_2$. For causal signals we have $b=\infty$ and $r_2=\infty$.
If the vertical strip $a<\text{Re}\{p\}<b$ does not include the imaginary axis, i.e., if $0<a<b$ or $a<b<0$, the bilateral Laplace transform does not converge for $p=j\omega$ because the imaginary axis is not inside the ROC. Consequently, the Fourier transform does not exist because the corresponding integral does not converge. For the $\mathcal{Z}$-transform the analogous case would be that the ROC does not include the unit circle, in which case the discrete time Fourier transform does not exist.