# Inverse Fourier Transform of $\frac{1}{(1+iw)(2+iw)}$

I am having problems finding the inverse Fourier transform of:

$$\frac{1}{(1+iw)(2+iw)}$$

I think I could use convolution property but I'm stuck. I know that:

$$f(t) = e^{-at} H(t) \implies f(w) = \frac{1}{(a+iw)}$$

Can somebody help me?

• Use partial fraction expansion – Mark Viola Jul 5 '18 at 23:15
• Thanks so much! – emee Jul 5 '18 at 23:32
• You're welcome. My pleasure – Mark Viola Jul 6 '18 at 2:45

We write

$$\frac{1}{(1+j\omega)(2+j\omega)}$$

As

$$\frac{A}{(1+j\omega)}+\frac{B}{(2+j\omega)}$$

Then we have

$$A+B=0$$ $$2A+B=1$$

$$A=1$$ $$B=-1$$

Then

$$\mathcal{F}^{-1}(\frac{1}{(1+j\omega)(2+j\omega)})$$

$$=(e^{-t}-e^{-2t})u(t)$$

Where $u(t)$ is the Heaviside function

Note: there may be a constant of $\sqrt{2\pi}$ depending on conventions being used