9 How to derive the Phase and Frequency modulation signals?

For any bandpass signal, we can write it as

\[ x\left ( t\right ) =\operatorname {Re}\left ( \tilde {x}\left ( t\right ) e^{j\omega _{c}t}\right ) \]

Where \(\tilde {x}\left ( t\right ) \) is the complex envelope of \(x\left ( t\right ) \). For PM and FM, the baseband modulated signal, \(\tilde {x}\left ( t\right ) \) has the form \(A_{c}e^{j\theta \left ( t\right ) }\) Hence the above becomes

\begin{align*} x\left ( t\right ) & =\operatorname {Re}\left ( A_{c}e^{j\theta \left ( t\right ) }e^{j\omega _{c}t}\right ) \\ & =A_{c}\left ( \cos \omega _{c}t\cos \theta \left ( t\right ) -\sin \omega _{c}t\sin \theta \left ( t\right ) \right ) \end{align*}

But \(\cos \left ( A+B\right ) =\cos A\cos B-\sin A\sin B\), hence the above becomes

\begin{equation} x\left ( t\right ) =\cos \left ( \omega _{c}t+\theta \left ( t\right ) \right ) \tag {1}\end{equation}

The above is the general form for PM and FM. Now, for PM, \(\theta \left ( t\right ) =k_{p}m\left ( t\right ) \) and for FM, \(\theta \left ( t\right ) =k_{f}\int _{0}^{t}m\left ( t_{1}\right ) dt_{1}\). Hence, substituting in (1) we obtain

\[ x_{FM}\left ( t\right ) =\cos \left ( \omega _{c}t+k_{f}\int _{0}^{t}m\left ( t_{1}\right ) dt_{1}\right ) \]

and

\[ x_{PM}\left ( t\right ) =\cos \left ( \omega _{c}t+k_{p}m\left ( t\right ) \right ) \]