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I have a doubt about the equivalence between Fourier Transform and Laplace Transform.

It was told me that if I have a function such that:

  • $f(t)=0$ if t<0
  • $f\in L^1(R) \bigcap L^2(R)$

I can define

$F[f(t)]=\int_{0}^{\infty}f(t) e^{-j\omega t}dt$

$L[f(t)]=\int_{0}^{\infty}f(t) e^{-s t}dt$

I can look at the Fourier transform as the Laplace Transform evaluated in $s=j\omega$ IF AND ONLY if the abscisse of convergence is strictly less than zero. (I.e. if the region of convergence includes the imaginary axis.

If the abscisse of congence is $\gamma=0$, then (it was told me), I can have poles on the real axes, and I have to define the Fourier transform with indentations in a proper manner.

In the Papoulis's book there is written "if $\gamma=0, $ the Laplace transform has at least one of the singular points on the imaginary axes.

So, I think that the situation should be like this:

enter image description here

Then, if I extend the frequency in the complex plane, I can consider that -regarding the Fourier transform- the axes are rotated with respect to the axis of the s-plane:

enter image description here

So I should have:

enter image description here

Finally,

enter image description here

I think that these two last steps could explain the word "real" related to the poles at the beginning of the question...

Please, tell me If the reasoning is wrong and where.. many thanks

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  • $\begingroup$ do you know "Wick rotations"? $\endgroup$ – tired Jun 7 '17 at 23:19
  • $\begingroup$ If $f(t) \ge 0$ then $F(s)= \int_0^\infty f(t) e^{-st}dt$ has a singularity at $s = \sigma$ the abscissa of convergence. Otherwise the integral can diverge with (the analytic continuation of) $F(s)$ staying analytic. $\endgroup$ – reuns Jun 8 '17 at 6:25
  • $\begingroup$ @user1952009 but I can't undesrtand why I should have singularities on the real axes... $\endgroup$ – sunrise Jun 8 '17 at 6:36
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If $f \ge 0$ then $$F(s) = \int_0^\infty f(t)e^{-st}dt$$ converges for $\Re(s) > \sigma$ iff $F(s)$ is bounded for $s > \sigma$. Whence if $\sigma_c$ is the abscissa of convergence then $$\lim_{ s \to \sigma_c} |F(s)| = \infty$$ And $F(s)$ is analytic for $\Re(s) > \sigma_c$ and has a singularity at $s= \sigma_c$.

Proof : $$|F(\sigma+i\omega)| \le \int_0^\infty |f(t)e^{-(\sigma+i\omega)t}|dt=\int_0^\infty f(t)e^{-\sigma t}dt = F(\sigma)$$


If $f$ is not non-negative then this is not true anymore.

Let $$f(t) = \sum_{n=1}^\infty (-1)^{n+1} e^n 1_{t \in [\ln n,e^{-n}+\ln n]}$$

Then $$F(s) = \int_0^\infty f(t) e^{-st}dt$$ converges iff $\Re(s) > 0$ but its analytic continuation is entire.

Proof, as $s \to 0^+$ : $$e^n\int_{\ln n}^{e^{-n}+\ln n} e^{-st}dt-e^{n+1}\int_{\ln (n+1)}^{e^{-(n+1)}+\ln (n+1)} e^{-st}dt\sim n^{-s}-(n+1)^{-s}$$

And $\sum_{n=1}^\infty (-1)^{n+1}n^{-s}$ diverges for $\Re(s) = 0$.

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As you already know: $$ \mathscr{L}(j\omega)\{f(t)\}=\mathscr{F}(\omega)\{f(t)\} $$

But the exposed rotations are useless, because the fourier magnitudes are over the page, as shown in this plot.

enter image description here

Edit:

I can look at the Fourier transform as the Laplace Transform evaluated in s=jωs=jω IF AND ONLY if the abscissa of convergence is strictly less than zero. (I.e. if the region of convergence includes the imaginary axis.

The imaginary axis $\sigma=0$ must be in the ROC of the Laplace transform, this is, for causal signals, the Laplace ROC leftmost pole must have negative real part.

If the ROC do not include the imaginary axis, the Fourier integral diverges, do not exist plain and simple.

If the abscisse of congence is γ=0γ=0, then (it was told me), I can have poles on the real axes,

The ROC has anything to do with having or not having poles in the real axis or having or not having the imaginary axis included.

For example:

The causal stable signal $$\frac{1}{s^2 +2sp\xi+p^2}$$ has its poles on $s=-p\xi \pm ip\sqrt{1-\xi^2}$, do not have poles in the real axis, and has the imaginary axis included on its ROC.

The causal unstable signal $$\frac{1}{s^2 -2sp\xi+p^2}$$ has its poles on $s=p\xi \pm ip\sqrt{1-\xi^2}$, do not have poles in the real axis, and do not has the imaginary axis included on its ROC.

The causal critically stable signal $$\frac{1}{s^2 + p^2}$$ has its poles on $s=\pm ip$, do not have poles in the real axis, and do not has the imaginary axis included on its ROC (see below).

and I have to define the Fourier transform with indentations in a proper manner.

This is a theoretical maneuver sometimes cited, for "hacking" a laplace transform having poles in the imaginary axis, drawing a semicircle at the left from these poles, and interrupting the path of the imaginary axis by including those semicircles, hence stabilizing the computation of the Fourier transform.

In the Papoulis's book there is written "if γ=0,γ=0, the Laplace transform has at least one of the singular points on the imaginary axes.

If the ROC limit is the imaginary axis, then it is evident to realize that it must have a pole in this axis. If not, by contradiction, again for causal systems, the ROC can be displaced to the right onto the "most left hand" pole, and start the ROC from there.

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  • $\begingroup$ I'm sorry but if I don't use the rotations, I can't understand why it was told me that I can have singularities on the real axes.... can you help me to understand? $\endgroup$ – sunrise Jun 8 '17 at 6:33

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