Phase of Dirac delta

Question

How can I get the phase of the following Fourier transform:

$$X(f)=\frac 1 2 \delta\left(f-\frac B 2\right)+\frac 1 2 \delta\left(f+\frac B 2\right)+\frac 1 {2j} \delta\left(f-\frac {3B} 2\right)-\frac 1 {2j} \delta\left(f+\frac {3B} 2\right)$$

Maybe have I phase=$0$ at $f=\pm B/2$, phase=$-\pi /2$ at $f=3B/2$ and phase=$\pi /2$ at $f=-3B/2?$

Thank you in advance.

Answer 1

I found in figure 3.27 at page 99 of the book "L. Verrazzani, G. Corsini - Teoria dei segnali" that the phase of the frequency components in which there are Dirac delta are equal to the phase of the coefficients that multiply the Dirac delta.

Answer 2

This may be useful , you can express them in terms of sin and cosine , using duality property $$1 \Leftrightarrow 2\pi \delta(\omega) \implies e^{j\omega_0 t} \Leftrightarrow 2\pi\delta(\omega-\omega_0) $$ Now if we use this , and break sin and cosine into exponential basis then we'll have $$\cos (\omega_0 t) \Leftrightarrow \pi \{\delta(\omega-\omega_0)+\delta(\omega+\omega_0)\}$$ $$\sin (\omega_0 t) \Leftrightarrow \frac{\pi}{2j} \{\delta(\omega-\omega_0)-\delta(\omega+\omega_0)\}$$ You can also write these in terms of $f$ then there wont a $\pi$ for both cases.