This is the Fourier transform pair I use: \begin{eqnarray} X (\omega) &=& \int_{-\infty}^\infty x(t) e^{-j\omega t} dt,\\ x(t) &=& \frac{1}{2\pi} \int_{-\infty}^\infty X(\omega) e^{j\omega t} d\omega, \end{eqnarray} where, $\mathcal{F}\{x(t)\}=X(\omega)$.
And here are the properties of Fourier transform related to the problem I have:
- Time differentiation: $\mathcal{F}\left\{\dfrac{d x(t)}{dt}\right\} = j\omega X(\omega)$
- Frequency differentiation: $\mathcal{F}\left\{t\,x(t)\right\} = j\dfrac{d X(\omega)}{d\omega}$
Now, I want to calculate $\mathcal{F}\{t\}$.
Using the time differentiation property: \begin{eqnarray} &&x(t) = t\\ \implies &&x'(t) = 1\\ \implies &&\mathcal{F}\{x'(t)\} = \mathcal{F}\{1\}\\ \implies &&j\omega X(\omega) = 2\pi \delta(\omega) \\ \implies &&X(\omega) = \dfrac{2\pi}{j\omega } \delta(\omega) \end{eqnarray}
i.e. $\mathcal{F}\{t\} = \dfrac{2\pi}{j\omega } \delta(\omega)$.
But apparently, this is an incorrect result!
The correct solution is achievable using the frequency differentiation property: \begin{eqnarray} &&\mathcal{F}\{1\} = 2\pi\delta(\omega)\\ \implies &&\mathcal{F}\{t\} = j \dfrac{d}{d\omega}\left( 2\pi\delta(\omega) \right)\\ \implies &&\mathcal{F}\{t\} = 2j\pi\delta'(\omega) \end{eqnarray}
Can someone please explain why the first method is wrong and why the later one is the correct one? Thanks in advance!
The solutions to the distribution equation $x\,u(x) = \delta(x)$ are $u(x) = -\delta'(x) + C\,\delta(x).$ So from $j\omega X(\omega) = 2\pi \delta(\omega)$ you get the solutions $X(\omega) = 2\pi j\,\delta'(\omega) + C\,\delta(\omega).$ Now, since $t$ is an odd function, so must be its Fourier transform, which makes $C=0.$