Phase spectrum of a Fourier transform
A real-valued negative number has a phase of $\pm\pi$, because $e^{\pm j\pi}=-1$. So for all frequencies for which the real-valued spectrum is negative, the corresponding phase angle equals $\pm \pi$:

$$S(j\omega)=\begin{cases}|S(j\omega)|,&amp;&amp;S(j\omega)\ge 0\\
|S(j\omega)|e^{\pm j\pi},&amp;&amp;S(j\omega)&lt;0\end{cases}$$