I have to find the fourier transform of $f(t)=e^{-a^*t}*u(t)$
For a>0 the signal has an infinite value therefore doesnt have a Fourier transform.For a>0 we have:
$F(w)=\int_0^∞[e^{-a*t}*e^{-j*w*t}]\,dt$ = $\int_0^∞e^{{[-a+jw}]*t}\,dt$=
$[1/-(a+jw)]$*$[e^{-(a+jw)*t}]|_0^∞$
From here,in my book it says $F(w)=1/(a+jw)$=$[a/(a^2+w^2)$-$j*a/(a^2+w^2)]$
But can you please do this integral $[1/-(a+jw)]$*$[e^{-(a+jw)*t}]|_0^∞$ to me step by step?
Also,how did we find the real and the imaginary parts ?
Real part : $[a/(a^2+w^2)]$
Imaginary part:$[j*a/(a^2+w^2)]$
Let $a > 0$ and $g$ be the function defined on $\mathbb{R}$ by :
$$\forall t \in \mathbb{R}, \; g(t)= \begin{cases} e^{-at} & \text{if } t \geq 0 \\[2mm] 0 & \text{otherwise}. \end{cases} $$
$g$ is in $\mathrm{L}^{1}(\mathbb{R})$ and its Fourier transform at some point $\omega \in \mathbb{R}$ is given by :
$$ \begin{align*} \mathcal{F}(g)(\omega) & = {} \int_{\mathbb{R}} g(t) e^{-it\omega} \; dt \\[2mm] & = \int_{0}^{+\infty} e^{-at} e^{-it\omega} \; dt \\[2mm] & = \int_{0}^{+\infty} e^{-(a+i\omega)t} \; dt \\[2mm] & = \lim \limits_{M \to +\infty} \Big[ -\frac{1}{a+i\omega} e^{-(a+i\omega)t} \Big]_{0}^{M} \\[2mm] & = \frac{1}{a+i\omega}. \end{align*} $$
Remember that for any complex number $z$, $\overline{z} z = \vert z \vert^{2}$. So, $\mathcal{F}(g)(\omega)$ writes :
$$ \mathcal{F}(g)(\omega) = \frac{a-i\omega}{\vert a+i\omega \vert^{2}} = \frac{a-i\omega}{a^{2}+\omega^{2}} $$
As a consequence,
$$ \Re \Big( \mathcal{F}(g)(\omega) \Big) = \frac{a}{a^{2}+\omega^{2}} \quad \mathrm{and} \quad \Im \Big( \mathcal{F}(g)(\omega) \Big) = -\frac{\omega}{a^{2}+\omega^{2}}$$
where $\Re(z)$ (resp. $\Im(z)$) is the real (resp. imaginary) part of $z$. I think that answers your question. Now, if $f$ is given by :
$$ f = g \star u $$
where $\star$ is the convolution between $g$ and some integrable function $u$, you will have :
$$ \mathcal{F}(f)(\omega) = \mathcal{F}(g)(\omega) \times \mathcal{F}(u)(\omega) = \frac{\mathcal{F}(u)(\omega)}{a+i\omega} $$