The phase components of a signal that they are got from the Fourier Transform, are simply, the phase offset of each sinusiod. Which is the $\phi$ in equation $1$. However, in case of Hilbert transform, the instantaneous phase is the argument of the sinusoid function. Which is $x(t)$ in equation $2$. I know that we can get the instantaneous frequency by taking the derivative of the $x(t)$, but how to get the phase offset from $x(t)$? In other words, how to write equation $2$ as equation $(3)$

I understand that the equation $(2)$ should be the complex combinations of $sin$ and $cos$ but I ignored that for the simplicity sake.

$$\begin{align} \sin(2\pi ft +\phi ) \tag 1\\ \sin( x(t) ) \tag2\\ \sin(2\pi x'(t) + \phi) &&\text{or}&& \sin(2\pi x'(t) + \phi(t)) \tag3 \end{align}$$



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