# Polarity inversion in frequency domain (polar coordinates)

I don't fully grasp how polarity inversion is performed in the polar coordinates of frequency domain. The frequency components of the signal do not change, so the amplitude part is going to stay whatever it is. That means the phase must change, but how? If I multiply the phase by -1 the time domain gets reversed. Phase shifting on the other hand only generates a delay...

The only thing I can think off is to reverse the amplitudes, but as far as I am aware the amplitudes are always positive in polar format. Am I wrong on this?

• You either multiply the complex amplitudes by -1 or you shift the phase by $\pi$. Both is fully equivalent to multiplying your time domain signal with -1. – Jazzmaniac Jun 14 '14 at 10:12
• Do you mean inverted when you write reversed? – gauteh Jul 25 '17 at 9:57

This is really basic, but here's how it works. If $X(\omega)$ is the Fourier transform of $x(t)$, then we obviously have

$$\mathcal{F}\{-x(t)\}=-\mathcal{F}\{x(t)\}=-X(\omega)$$

Furthermore,

$$X(\omega)=|X(\omega)|e^{j\phi(\omega)}$$

where $\phi(\omega)$ is the phase. From this we have

$$-X(\omega)=-|X(\omega)|e^{j\phi(\omega)}=|X(\omega)|e^{j(\phi(\omega)\pm\pi)}$$

because $e^{\pm j\pi}=-1$. So Jazzmaniac's comment that the phase needs to be shifed by $\pm \pi$ (or any odd multiple of $\pi$ for that matter) is of course correct.