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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).

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

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  • $\begingroup$ Thank you Matt, somehow my mind never went that far ahead. Also, I have been out of touch with by Signal Processing since quite a bit and need to brush up on these stability criteria. $\endgroup$ – Mr. Johnny Doe Jul 14 '18 at 11:47

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