Problem with Inverse Fourier transform

Question

What's the easiest way to find inverse Fourier transform of the signal $$ \mathscr{F} (j\omega) = -2-j\omega $$ When going by definition of inverse Fourier transform, the integrals get very tricky, so I was wondering is there another way? Maybe stationary-phase metod?

Answer 1

Here are some known fourier transforms

$$ \mathscr{F}\{ \delta(t) \} = 1 $$

$$ \mathscr{F} \left \{ \frac{d^n}{dt^n}f(t) \right \} = (j\omega)^n\mathscr{F}(\omega) $$

Fourier transform is linear. So you get

$$ \mathscr{F} \left \{ -2\delta(t) \right \} = -2 \cdot \mathscr{F} \left \{ \delta(t) \right \} = -2 \cdot 1 \ = -2 $$

Also the first derivative of a delta fourier transform is:

$$ \mathscr{F} \left \{ -\frac{d}{dt}\delta(t) \right \} = -(j\omega) \cdot \mathscr{F} \left \{ \delta(t) \right \} = -(j\omega)\cdot 1 = -j\omega $$

Now the inverse is: $$ f(t)=-2 \delta(t) - \frac{d}{dt}\delta(t) $$