10 How to obtain the phase deviation and the frequency deviation for angle modulated signal?

From the general form for angle modulated signal (see above note)

\[ x\left ( t\right ) =\cos \left ( \omega _{c}t+\theta \left ( t\right ) \right ) \]

The phase deviation is \(\theta \left ( t\right ) \). And the maximum phase deviation is simply the maximum of \(\theta \left ( t\right ) \)

Now, to find the frequency deviation, we need a little bit more work. Start with

\[ f_{i}=f_{c}+\Delta f \]

Where \(f_{i}\) is the instantaneous frequency in Hz. But

\begin{align*} f_{i} & =\frac {1}{2\pi }\frac {d}{dt}\left ( \omega _{c}t+\theta \left ( t\right ) \right ) \\ & =f_{c}+\overset {\Delta f}{\overbrace {\frac {1}{2\pi }\frac {d}{dt}\theta \left ( t\right ) }}\end{align*}