Inverse Fourier transform of $\frac{\frac{2}{1+2i\pi f}-\frac{2}{4+2i\pi f}}{\frac{1}{1+2i\pi f}+\frac{1}{3+2i\pi f}}$ I wanted to calculated the inverse fourier transform of the transfer function :
\begin{align}
H(f) &amp;= \frac{\frac{2}{1+2i\pi f}-\frac{2}{4+2i\pi f}}{\frac{1}{1+2i\pi f}+\frac{1}{3+2i\pi f}}\\
&amp;\text{
I did some simplification : 
}\\
&amp;= \frac{3}{4}\left(\frac{1}{2+i\pi f}+\frac{1}{1+i\pi f}\right)
\end{align}
and after applying $\mathcal{F}^{-1}\{H(f)\}$ I obtained
$$ h(t)=\frac{3}{2}\left(e^{-4t}+e^{-2t}\right)u(t)$$ $$(this\ formula\ is\ corrected)$$
I used the fact that $$\mathcal{F}\left\{e^{-at} u(t)\right\}= \frac{1}{a+2i\pi f}$$
where $ u(t)$ is the unit step function.
Can I get some help to know if my process is correct?
This looks correct; obviously you need to simplify to obtain a transfer function with polynomials as numerator and denominator. Partial fraction expansion then leads to a form that can be directly transformed to the time domain. The only thing that might be wrong is the minus sign between the two exponentials.