why there's not a fourier transform to the function $x(t) = -\frac{1}{2}\cdot (e^{-2t}\cdot u(-t) + e^{-4t}\cdot u(t))$

Question

I know that the Laplace transform of a signal is

$X(s)= \frac{1}{s^2+6s+8}$

$ ROC = $ {s|-4 < Re{s} < -2}

therefore $x(t) = -\frac{1}{2}\cdot (e^{-2t}\cdot u(-t) + e^{-4t}\cdot u(t))$

$u(t)$ is the heaviside function.

I need to find its Fourier transform (or to prove why there is not such transform).

by wolfram alpha I see that the Fourier transform doesn't exist for this signal. Someone here can explain why ?

Thank you :)

Answer 1

to check if a fourier transform exists from laplace transform you need to have the line $jw$ inside the ROC (equivalently $0 \in ROC$) which is clearly not the case in your question.

Answer 2

First of all, notice that:

$$\mathcal{L}^{-1}\left[\frac{\alpha}{(s+\beta)^2 + \alpha^2}\right] = e^{-\beta t} \sin(\alpha t)u(t),$$

where $u(t)$ is the Heaviside step function.

Given $X(s)= \frac{1}{s^2+2s+5}$, observe that:

$$s^2 + 2s +5 =s^2 + 2s + 1 + 4 = (s+1)^2 + 2^2.$$

Hence: $$X(s) = \frac{1}{2}\frac{2}{(s+1)^2 + 2^2} \Rightarrow \mathcal{L}^{-1}\left[X(s)\right] = \frac{1}{2}e^{-t}\sin(2t)u(t).$$

Secondly, letting $g(t) = \frac{1}{2}e^{-t}u(t)$, we observe that:

$$\begin{align*} \mathcal{F}\left[g(t)\right]= &\int_{\mathbb{R}}\frac{1}{2}e^{-t}u(t) e^{-i2\pi ft} dt \\ =& \int_{0}^{+\infty}\frac{1}{2}e^{-t} e^{-i2\pi ft} dt \\ =& \frac{1}{2}\int_{0}^{+\infty} e^{-(i2\pi f+1)t} dt \\ =& \frac{1}{2}\left[\frac{e^{-(i2\pi f+1)t}}{-(i2\pi f+1)}\right]_{t=0}^{t=+\infty} \\ =& \frac{1}{-2(i2\pi f+1)}[0-1]\\ =& \frac{1}{i4\pi f+2} = G(f). \end{align*}$$

Finally:

$$\begin{align*} \mathcal{F}\left[g(t) \sin(2t)\right]=& \frac{G\left(f - \frac{2}{2\pi}\right) - G\left(f + \frac{2}{2\pi}\right)}{2 i} = \ldots\end{align*}$$

Answer 3

Given $X(s)= \frac{1}{s^2+6s+8} = \frac{1}{2}\left[\frac{1}{s+2} - \frac{1}{s+4}\right]$, observe that: $$\mathcal{L}^{-1}[X(s)] = \frac{1}{2}\left(e^{-2t} - e^{-4t}\right)u(t) = x(t).$$

Since:

$$\mathcal{F}[e^{-\alpha t}u(t)] = \frac{1}{\alpha + 2\pi i f},$$

then:

$$\mathcal{F}[x(t)] = \frac{1}{2}\left[\frac{1}{2 +2 \pi i f} - \frac{1}{4 +2 \pi i f}\right].$$